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Title: CBSE Chemistry Class 12 Examination Paper upto 2015
Description: This PDF contains a full set of CBSE class 12 question paper from 2007 till 2015 for practice purpose.
Description: This PDF contains a full set of CBSE class 12 question paper from 2007 till 2015 for practice purpose.
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CHEMISTRY
Examination Papers
2008–2014
CONTENT
n
CBSE Examination Paper–2008 (Delhi)
3
n
CBSE Examination Paper–2008 (All India)
23
n
CBSE Examination Paper–2009 (Delhi)
43
n
CBSE Examination Paper–2009 (All India)
69
n
CBSE Examination Paper–2009 (Foreign)
92
n
CBSE Examination Paper–2010 (Delhi)
116
n
CBSE Examination Paper–2010 (All India)
134
n
CBSE Examination Paper–2010 (Foreign)
152
n
CBSE Examination Paper–2011 (Delhi)
171
n
CBSE Examination Paper–2011 (All India)
191
n
CBSE Examination Paper–2011 (Foreign)
209
n
CBSE Examination Paper–2012 (Delhi)
229
n
CBSE Examination Paper–2012 (All India)
249
n
CBSE Examination Paper–2012 (Foreign)
267
n
CBSE Examination Paper–2013 (Delhi)
285
n
CBSE Examination Paper–2013 (All India)
300
n
CBSE Examination Paper–2013 (Foreign)
315
n
CBSE Examination Paper–2014 (Delhi)
331
n
CBSE Examination Paper–2014 (All India)
349
n
CBSE Examination Paper–2014 (Foreign)
367
CBSE EXAMINATION PAPERS
DELHI—2008
Time allowed: 3 hours
Maximum marks: 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Delhi) SET–I
1
...
3
...
5
...
What causes Brownian movement in a colloidal solution?
In which one of the two structures, NO+ and NO- , the bond angle has a higher value?
2
2
Write the IUPAC name of the following compound:
H3 C ¾ CH ¾ CH2 ¾ CH ¾ CH ¾ CH2 OH
½
½
½
CH3
OH CH3
Arrange the following compounds in an increasing order of their acid strengths:
(CH3 ) 2 CHCOOH, CH3 CH2 CH(Br)COOH, CH3 CH (Br) CH2 COOH
7
...
8
...
9
...
OR
Which compound in each of the following pairs will react faster in SN2 reaction with —OH? Why?
(i) CH3 Br or CH3 I
(ii) (CH3 ) 3 CCl or CH3 Cl
10
...
(b) Complete the following chemical equation:
peroxide
CH3 CH2 CH = CH2 + HBr ¾¾¾¾® ¼¼
11
...
12
...
Calculate its t 1/ 2
value
...
What is meant by the ‘rate constant, k’ of a reaction? If the concentration is expressed in mol L-1
units and time in seconds, what would be the units for k (i) for a zero order reaction and (ii) for a
first order reaction?
14
...
List the reactions of glucose which cannot be explained by its open chain structure
...
Assign a reason for each of the following statements:
(i) Ammonia is a stronger base than phosphine
...
17
...
What are biodegradable and non-biodegradable detergents? Give one example of each class
...
What is a semiconductor? Describe the two main types of semiconductors and explain mechanisms
for their conduction
...
Calculate the temperature at which a solution containing 54 g of glucose, (C 6 H12 O6 ), in 250 g of
water will freeze
...
86 K mol -1 kg)
21
...
Which one of these two
types of sols is easily coagulated and why?
22
...
Write chemical equations for the following processes:
(i) Chlorine reacts with a hot concentrated solution of sodium hydroxide
...
OR
Complete the following chemical equations:
(i) Ca 3 P2 ( s) + H2 O (l) ¾¾® ¼
(ii) Cu 2+ ( aq) + NH3 ( aq) ¾¾® ¼
(excess)
(iii) F2 ( g) + H2 O (l) ¾¾® ¼
24
...
(b) Explain as to how the two complexes of nickel, [Ni (CN) 4 ] 2- and Ni (CO) 4 , have different
structures but do not differ in their magnetic behaviour
...
Name the reagents which are used in the following conversions:
(i) A primary alcohol to an aldehyde
(ii) Butan-2-one to butan-2-ol
(iii) Phenol to 2, 4, 6-tribromophenol
26
...
(ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric
hydroxide
...
Examination Papers | 5
27
...
Conductivity of 0
...
896 ´ 10 -5 S cm -1
...
If L°m for acetic acid is 390
...
A steady current of 1
...
45 g of silver were deposited at the cathode of cell B
...
4, Cu = 63
...
Assign reasons for the following:
(i) The enthalpies of atomisation of transition elements are high
...
(iii) From element to element the actinoid contraction is greater than the lanthanoid contraction
...
(v) Scandium ( Z = 21) does not exhibit variable oxidation states and yet it is regarded as a
transition element
...
Indicate relative stability of oxidation states in each case
...
30
...
¾¾¾¾¾® 2
Zn ¾ H 2 O
(ii)
...
KOH, heat
(b) Describe the following reactions:
(i) Cannizaro reaction
(ii) Cross aldol condensation
OR
(a) How would you account for the following:
(i) Aldehydes are more reactive than ketones towards nucelophiles
...
(iii) The aldehydes and ketones undergo a number of addition reactions
...
6 | Xam idea Chemistry–XII
CBSE (Delhi) SET–II
Questions Uncommon to Set-I
1
...
What is primary cell? Give an example
...
Write the IUPAC name of the following compound:
CH3 COCH2 COCH3
9
...
Taking suitable examples explain the meaning
of positive and negative deviations from Raoult’s law
...
Describe how the molecular mass of a substance can be
determined by a method based on measurement of osmotic pressure?
10
...
20 M solution of KCl at 298 K is 0
...
Calculate its molar
conductivity
...
Formulate the galvanic cell in which the following reaction takes place:
Zn(s) + 2Ag + (aq) ¾¾® Zn 2+ (aq) + 2Ag(s)
State:
(i) Which one of its electrodes is negatively charged?
(ii) The reaction taking place at each of its electrode
...
14
...
(i) Why is it that haloalkanes are more reactive than haloarenes towards nucleophiles
...
(a) Derive the general form of the expression for the half-life of a first order reaction
...
What are the rates of
production of N2 and H2 if k = 2
...
(b) The half-life for decay of radioactive 14 C is 5730 years
...
Calculate the age of the artefact
...
(a) How will you bring about the following conversions?
(i) Ethanol to acetone
(ii) Benzene to acetophenone
(iii) Benzoic acid to benzaldehyde
(b) Describe the following giving a suitable example in each case:
(i) Decarboxylation
(ii) Cannizaro’s reaction
Examination Papers | 7
OR
(a) An organic compound contains 69
...
63% hydrogen and the rest is oxygen
...
It does not reduce Tollens’ reagent but forms an
addition compound with sodium hydrogen sulphite and gives positive iodoform test
...
Deduce the possible structure of the
organic compound
...
(ii) Ethanoic acid is a weaker acid than benzoic acid
...
What type of substances exhibit antiferromagnetism?
2
...
3
...
The resistance of a conductivity cell containing 0
...
What is
the cell constant if the conductivity of 0
...
146 ´ 10 -3 S cm -1 ?
19
...
(ii) Schottky defects lower the density of related solids
...
OR
Explain the following properties giving suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
21
...
Explain what is observed when
(i) an electrolyte, KCl, is added to a hydrated ferric oxide sol
...
(iii) a beam of strong light is passed through a colloidal solution
...
2
...
4
...
6
...
This is due to the unequal bombardment of colloidal particles by the molecules of dispersion
medium
...
2
2
2, 5-Dimethyl hexane-1, 3-diol
...
Benzene
diazonium chloride
8
...
+ KCl + N2
Iodobenzene
Phenol, 0
...
In haloarenes C—X bond acquires a partial double bond character due to resonance
...
X
|
+X
+X
X
|
+ X
OR
(i) CH3–I reacts faster than CH3–Br as iodine is a better leaving group because of its larger size
...
10
...
Peroxide
(b)
CH3 - CH2 - CH = CH2 + H - Br ¾¾¾® CH3 - CH2 - CH2 - CH2 - Br
But-1-ene
1-bromo butane
11
...
Examination Papers | 9
Mathematically, m µ P
or
m = KH P
where K H is the Henry’s law constant
...
(ii) To minimize the painful effects accompanying the decompression of deep sea divers, oxygen
diluted with less soluble helium gas is used as breathing gas
...
For a first order reaction
[R ]0
2
...
303
10 2
...
303
K=
log
=
log 1
...
1548
40
7
40
40
K = 8
...
693
0
...
91 ´ 10 -3
t 1/ 2 = 77
...
13
...
rate = K [ A ] n
Q
æ mol
...
ø
(i) For a zero order reaction n = 0
\ Unit of K = mol lit -1 sec -1
\
1- n
s -1
For a first order reaction n = 1
\ Unit of K = sec - 1
14
...
O
½
½
- H2O
H2 N ¾ CH2 ¾ C ¾ OH + H NH ¾ CH ¾ COOH ¾¾¾®
½
CH3
(iii)
O
½
½
H2 NCH2 ¾ C ¾ NH ¾ CH ¾ COOH
½
CH3
Peptide linkage
(Gly -Ala)
10 | Xam idea Chemistry–XII
(ii) Denaturation: When a protein in its native form is subjected to physical change like change in
temperature or chemical change like change in pH, the hydrogen bonds are disturbed
...
During
denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact, e
...
,
coagulation of egg white on boiling
...
The following reactions of glucose cannot be explained by its open chain structure
...
(ii) The pentacetate of glucose does not react with hydroxylamine indicating the absence of free
—CHO group
...
These glucosides do not react with hydrogen cyanide or with hydroxylamine
...
(i) As the atomic size of nitrogen is smaller than phosphorus, therefore electron density on
nitrogen atom is higher than that on phosphorus atom
...
Thus, ammonia is
stronger base than phosphine
...
17
...
F
F
b
...
= 4
5e pairs
S
l
...
= 1
F
F
See-saw shape
(ii) sp 3 d 2 hybridisation
F
b
...
= 4
F
Xe
6e pairs
l
...
= 2
F
F
Square planar
18
...
g
...
CH3 ¾ (CH2 ) 11 ¾
¾ SO- Na +
3
Examination Papers | 11
Non-biodegradable detergents: Detergents having branched hydrocarbon chains are not easily
degraded by micro-organisms and hence are called non-biodegradable detergents, e
...
, sodium-4(1, 3, 5, 7-tetramethyl octyl) benzenesulphonate
...
19
...
Semiconductors are of two types
(i) n-type of semiconductors
(ii) p-type of semiconductors
n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like
P or As, the dopant atom forms four covalent bonds like a Si or Ge atom but the fifth electron, not
used in bonding, becomes delocalised and contribute its share towards electrical conduction
...
p-type semiconductor: When silicon or germanium is doped with group 13 element like B or Al,
the dopant has only with three, valence electrons
...
Here, this hole moves through the crystal like a positive
charge giving rise to electrical conductivity
...
Silicon atom
Mobile electron
Positive hole
(no electron)
As
Perfect crystal
n-type
20
...
86 k mol - 1 kg
K f ´ WB ´ 1000
Applying the formula, D Tf =
M B ´ WA
1
...
23
180 ´ 250
Tf = T ° f - DTf = 0 - ( 2
...
23° C
B
p-type
12 | Xam idea Chemistry–XII
21
...
g
...
Lyophobic sols: In this type of sols the particles of dispersed phase have little or no affinity for
the dispersion medium, e
...
, gold sol, Fe (OH) 3 sol, As 2 O3 sol
...
Lyophobic sols easily coagulate on the addition of small amount of electrolyte because these are
not stable
...
is due to charge as well as solvation of
colloidal particles
...
(i) The principle of froth floatation process is that sulphide ore particle are preferentially wetted by
pine oil, whereas the gangue particles are wetted by water
...
(iii) The principle of refining by liquation is that the impurities whose melting points are higher than
the metal are left behind on melting the impure metal
...
23
...
)
D
4H3 PO3 ¾¾® 3H3 PO4 + PH3
(ii)
PtF6 + Xe ¾¾® Xe + [PtF6 ] -
(iii)
OR
(i)
Ca 3 P2 (s) + 6H2 O(l) ¾¾® 2PH3 + 3Ca (OH) 2
(ii) Cu 2+ (aq) + 4NH3 (aq) r [Cu(NH3 ) 4 ] 2+ (aq)
(blue)
(iii)
(excess)
(deep blue)
2F2 (g) + 2H2 O (l) ¾¾® 4H (aq) + 4F - (aq) + O2 (g)
+
24
...
A ligand should have lone pair of electrons in their valence orbital which can be donated
to central metal atom/ion
...
Structure: Square planar
...
Examination Papers | 13
Orbital of
Ni (O)
4p
4s
3d
sp 3 hybridised
orbitals of Ni
sp 3 hybrid
3d
Ni(CO) 4
(Low spin complex)
Four pairs of electrons from
four CO molecules
3d
Structure: Tetrahedral
Magnetic behaviour: Diamagnetic due to the absence of unpaired electrons
...
(i) Pyridinium chlorochromate (C5 H5 N HCrO3 Cl - ) or Cu/573 K
(ii) LiAlH4 / ether
(iii) Br2 / H2 O
26
...
As a result, the electron density on the nitrogen decreases
...
Therefore
aniline is a weaker base than methyl amine and hence its pK b value is higher than that of
methyl amine
...
FeCl 3 + 3OH– ¾ ® Fe (OH) 3 ¯ + 3Cl ¾
Ferric hydroxide
(Brown ppt)
(iii) Aniline being a Lewis base, reacts with lewis acid AlCl 3 to form a salt
...
27
...
c = 0
...
896 ´ 10 -5 S cm -1 , l¥ = 390
...
896 ´ 10 -5 ´ 1000
lm =
= 32
...
002
lc
32
...
084
¥
l m 390
...
4%
CH3 ¾ COOH r CH3 ¾ COO- + H+
0
0
c
c(1 - a )
ca
ca
[ CH3 ¾ COO- ] [ H+ ] ca
...
0241(0
...
86 ´ 10 -5
(1 - 0
...
45 g of Ag will be deposited by =
´ 1
...
6 C
Q 1295
...
7 s
...
50
Cu 2+ + 2e - ¾¾® Cu
2 ´ 96500 C deposit Cu = 63
...
5
\ 1295
...
6 = 0
...
3 g
65
...
6 C deposit Zn =
´ 1295
...
438 g
2 ´ 96500
29
...
(ii) The catalytic activity of transition metals is attributed to the following reasons:
(a) Because of their variable oxidation states transition metals form unstable intermediate
compounds and provide a new path with lower activation energy for the reaction
...
(iii) This is due to poorer shielding by 5f electrons in actinoids than that by 4f electron in the
lanthanoids
...
(v) This is because scandium has partially filled d orbitals in the ground state (3d 1 4s 2 )
...
S
...
S
...
4FeCr2 O4 + 8Na 2 CO3 + 7O2 ¾® 8Na 2 CrO4 + 2Fe 2 O3 + 8CO2
Chromite ore
Sod
...
2MnO2 + 4KOH + O2 ¾®
2K 2 MnO4 + 2H2 O
Pyrolusite ore
Potassium maganate
30
...
COO- K +
(iii)
(b) (i) Cannizzaro reaction: Aldehydes which do not have an a-hydrogen, undergo self
oxidation and reduction (disproportionation) reaction on treatment with concentrated
alkali
...
conc
...
O
O
¾ CHO +
½
½
OH –
¾ C ¾ CH3 ¾¾¾®
½
½
¾ CH = CH ¾ C ¾
293 K
Benzaldehyde
Acetophenone
Benzal acetophenone
If both of them contain a-hydrogen atoms
...
16 | Xam idea Chemistry–XII
OR
(a) (i) This is due to steric and electronic reasons
...
Electronically two alkyl groups reduce the
positivity of the carbonyl carbon more effectively in ketones than in aldehydes
...
(iii) Due to greater electronegativity of oxygen than carbon the C atom of the C = O bond
acquires a partial positive charge in aldehydes and ketones and hence readily undergo
nucleophilic addition reactions
...
CH3 ¾ CHO + 3NaOI ¾® HCOO- Na + + CHI 3
+ 2NaOH
Iodoform
(yellow ppt)
CHO
½
NaOI
¾¾¾¾® No reaction
(ii) Propanone give orange-red ppt with 2, 4-DNP reagent and yellow ppt of iodoform with
sodium hypoiodite, whereas 1-propanol does not give these tests
...
NaOI
CH3 ¾ CH2 ¾ CH2 ¾ OH ¾ ¾¾® No yellow ppt
...
2
...
9
...
g
...
Pentane-2, 4-dione
...
Mathematically
PA µ x A
PB µ x B
o
o
PA = PA x A
PB = PB x B
Positive deviation from Raoult’s law: In this type of deviation the partial pressure of each
o
component of solution is greater than that calculated from Raoult’s law, i
...
, PA > PA x A &
8 (corner atoms) ´
o
PB > PB x B
...
Negative deviation from Raoult’s: In this type of deviation the partial pressure of each component
o
o
of solution is less than that expected from Raoult’s law, i
...
, PA < PA x A & PB < PB x B
...
Examination Papers | 17
OR
Osmotic pressure ( p) is defined as the extra pressure that must be applied to the solution side in
order to prevent the flow of solvent molecules into it through a semipermeable membrane
...
If WB grams of the solute whose molecular mass M B is present in the solution then
W RT
p= B
M B RT
W RT
MB = B
pV
Thus, knowing WB , T , p and V molecular mass of the solute, M B can be calculated
...
k = 0
...
2 mol L-1
Substituting the values
k ´ 1000 0
...
2
L m = 1248 cm 2 mol -1
11
...
)
Ag + (conc) Ag
(i) Zn electrode is negatively charged
...
* Zn 2+ ions in anodic half cell
...
*
Ions of salt bridge, i
...
, K + and Cl -
...
(i)
H
(ii)
C 6 H5 I
Iodobenzene
H
H
+ Br2 ¾¾¾®
C=C
H
CCl 4
+ KCl + N2
H
H
H
C¾ C
½ ½ H
Br Br
1, 2-Dibromoethane
15
...
As a result
the bond cleavage in haloarenes is difficult than haloalkanes and therefore, they are less reactive
towards nucleophilic substitution reaction
...
28
...
Therefore equation (i) can be written as
ln [ R ] 0 = - K ´ 0 + C
ln [ R ] 0 = C
Substituting the value of C in equation (i)
ln [ R ] = - Kt + ln [ R ] 0
Kt = ln [ R ] 0 - ln [ R ]
ln [ R ] 0
Kt =
[R ]
[R ]0
[R ]0
2
...
303 log
Þ
t=
log
[R ]
K
[R ]
[R ]0
when
t = t 1/ 2
[R ] =
2
[R ]0
2
...
303
2
...
3010
K
K
0
...
5 ´ 10 -4
1 d [ NH3 ] d [ N2 ] 1 d [ H2 ]
Rate = =
=
2
dt
dt
3 dt
d [ N2 ]
-4
-1 -1
= 2
...
5 ´ 10 -4 mol L-1 S -1
dt
dt
d [ H2 ]
= 7
...
693 0
...
209 ´ 10 -4 year -1
t 1/ 2
5730
[R ]0
2
...
303
100
t=
log
=
log
K
[ R ] 1
...
303 ´ 10 4
2
...
2
1
...
303 ´ 0
...
303
t=
´ 10 4 (1 - 3 ´ 0
...
2
1
...
7 years
t=
K 2 Cr2 O7 /H 2 SO4
(O)
29
...
AlCl 3
½
½
+ CH3 ¾ C ¾ Cl ¾¾¾¾¾¾¾¾®
(ii)
Acetyl chloride
Acetophenone
O
½
½
C ¾ Cl
½
PCl5
O
½
½
C¾ H
½
+ H2
¾¾¾®
(iii)
+ HCl
Friedel craft acylation
Benzene
O
½
½
C ¾ OH
½
O
½
½
C ¾ CH3
½
¾¾¾¾®
Pd/BaSO4
Benzoic acid
Benzoyl chloride
Benzaldehyde
20 | Xam idea Chemistry–XII
(b) (i) Decarboxylation: Carboxylic acids lose CO2 to form hydrocarbons when their sodium salts are
heated with soda lime (NaOH and CaO in the ratio 3: 1)
...
In this reaction, one molecule of the aldehyde is
reduced to alcohol while the other is oxidised to carboxylic acid
...
NaOH
2HCHO
¾¾¾¾®
Formaldehyde
CH3 ¾ OH +
Methyl alcohol
–+
HCOO Na
Sodium formate
OR
(a) % C = 69
...
63%
%O = 100 - (69
...
63) = 18
...
77 11
...
6
C : H: O =
:
=
= 5
...
63 :1
...
molecular mass
n=
Empirical formula mass
Molecular mass = 86
Empirical formula mass = 5 ´ 12 + 10 ´ 1 + 1 ´ 16 = 86
86
n=
=1
81
\ Molecular formula = n ( Empirical formula)
= C5 H10 O
The (forms addition compound with NaHSO3 ) given organic compound is methyl ketone as it
gives positive iodoform test and also does not reduce Tollen’s reagent
...
- +
+ CH3 ¾ CH2 ¾ CH2 ¾ CO O Na + 3NaI + 3H2 O
O
½ :
½
K 2 Cr2 O7 /H 2 SO4
CH3 ¾ C ¾ CH2 ¾ CH2 ¾ CH3 ¾¾¾¾¾® CH3 ¾ COOH + CH3 ¾ CH2 ¾ COOH
:
(O)
Ethanoic acid
Propanoic acid
(b) (i) This is because dichloroethanoic acid is a stronger acid than monochloroethanoic acid
...
CBSE (Delhi) SET–III
1
...
k ´ 1000
c
where, L m = Molar conductivity
k = Conductivity
c = Molar concentration
3
...
k = ´ cell constant
R
k = 0
...
146 ´ 10 -3 =
´ cell constant
1500
cell constant = 0
...
219 cm -1
19
...
(ii) As the number of ions decreases as a result of Schottky defect, the mass decreases whereas the
volume remains the same
...
Creation of hole causes p-type semiconductor and creation of electron causes n-type semiconductor
...
g
...
] Ferromagnetism arises
due to spontaneous alignment of magnetic moments in the same direction
...
lm =
Alignment of magnetic moments in ferromagnetic substance
...
It is due to the presence of one or more unpaired electrons, e
...
,O2 , Fe 3+ , Cr 3+ ,
etc
...
g
...
21
...
(ii) Vapour phase refining
...
It is the
decomposed to give pure metal
...
(b) the volatile compound should be easily decomposable, so that the recovery is easy
...
A strip of the same metal in pure
form is used as cathode
...
When electric current is passed, impure metal forms metal ions which are
discharged at cathode forming pure metal
...
(i) The positively charged colloidal particles of Fe(OH) 3 get coagulated by the oppositely charged
Cl - ions provided by KCl
...
(iii) Scattering of light by the colloidal particles takes place and the path of light becomes visible
(Tyndall effect)
...
Question nos
...
Question nos
...
Question nos
...
Question nos
...
Use log tables if necessary, use of calculators is not allowed
...
2
...
4
...
6
...
8
...
10
...
12
...
Define the ‘forbidden zone’ of an insulator
...
Mention all the oxidation states exhibited by chlorine in its compounds
...
Define the term ‘polymerisation’
...
How is it that measurement of osmotic pressures is more widely used for
determining molar masses of macromolecules than the rise in boiling point or fall in freezing point
of their solutions?
OR
Derive an equation to express that relative lowering of vapour pressure for a solution is equal to
the mole fraction of the solute in it when the solvent alone is volatile
...
20 M solution of KCl at 298 K is 0
...
Calculate its molar
conductivity in this solution
...
(ii) Phosphorus (P4 ) is much more reactive than nitrogen (N2 )
...
How would you justify this situation?
(At
...
Cr = 24, Mn = 25, Fe = 26, Co = 27)
What happens when bromine reacts with CH3 ¾ C º CH? How would you justify this reaction?
14
...
Alcohols react both as nucleophiles as well as electrophiles
...
16
...
Write the structures of the monomers of the following polymers :
(i) PVC
(ii) Polypropene
18
...
19
...
If its density is 8
...
(Atomic mass of Nb = 93 u; NA = 6
...
Calculate the amount of KCl which must be added to 1 kg of water so that its freezing point is
depressed by 2 K
...
86 K kg mol -1 , Atomic mass K = 39, Cl = 35
...
In the button cells widely used in watches and other devices the following reaction takes place :
Zn(s) + Ag 2 O + H2 O (l) ¾® Zn 2+ (aq) + 2 Ag(s) + 2OH- (aq)
Determine D r G° for the reaction
...
76 V and E° Ag + Ag = + 0
...
Explain what is observed when
(i) a beam of light is passed through a colloidal solution
...
(iii) an electric current is passed through a colloidal sol
...
Write the chemical reactions which take place in the following operations :
(i) Electrolytic reduction of Al 2 O3
...
(iii) Mond’s process for refining nickel
...
Compare actinoids and lanthanoids with reference to their :
(i) electronic configurations of atoms
(ii) oxidation states of elements
(iii) general chemical reactivity of elements
...
Write the IUPAC name and describe the magnetic behaviour (diamagnetic or paramagnetic) of the
following coordination entities:
(i) [Cr (H2 O) 2 (C 2 O4 ) 2 ] -
Examination Papers | 25
(ii) [Co (NH3 ) 5 Cl] 2+
(iii) [NiCl 4 ] 2(At
...
: Cr = 24, Co = 27, Ni = 28)
26
...
(ii) Aniline does not undergo Friedel–Crafts reaction
...
27
...
(a) A reaction is of first order in A and of second order in B
...
How will its initial rate be affected if the concentration of both A and B are together
doubled?
(b) The rate constant k of a reaction increases four fold when the temperature changes from 300 K
to 320 K
...
(R = 8
...
(b) The half-life for radioactive 14 C is 5730 years
...
Calculate the age of the artefact
...
(a) Assign reasons for the following :
(i) Bi (V) is a stronger oxidising agent than Sb (V)
...
(b) Draw the structures of the following molecules :
(i) H2 S 2 O7
(ii) BrF3
(iii) XeF2
OR
(a) Complete the following chemical reaction equations :
(i) Ca 3 P2 + H2 O ¾®
(ii) XeF4 + H2 O ¾®
(b) How would you account for the following observations :
(i) NH3 is a stronger base than PH3
...
(iii) Hydrogen fluoride has a higher boiling point than hydrogen chloride
...
(a) Illustrate the following reactions giving one example for each :
(i) Cannizzaro reaction
(ii) Decarboxylation
(b) Complete the following reaction equations by giving the indicated missing substances :
H 2 NCONHNH 2
(i) CH3 CHO ¾¾¾¾¾¾®
...
¾¾¾¾¾¾® 2
(ii) Zn - H 2 O
O
26 | Xam idea Chemistry–XII
...
Which crystal defect lowers the density of a solid?
6
...
8
...
Define the following terms giving an example for each:
(i) The order of a reaction
(ii) The molecularity of a reaction
18
...
The rate constant for a first order reaction is 60 s -1
...
Describe the principle involved in the following metallurgical operations:
(i) Zone refining
(ii) Electrolytic refining
(iii) Froth-floatation process of concentrating sulphide ores
27
...
28
...
(ii) what are the carrier of the current in the cell
...
(b) Write the Nernst equation and determine the e
...
f
...
001 M) || Cu 2+ (0
...
375 V, E° Cu 2 + /Cu = + 0
...
How do they vary
when the concentration of electrolyte in the solution increases?
(b) Three conductivity cells A, B and C containing solutions of zinc sulphate, silver nitrate and
copper sulphate respectively are connected in series
...
5 amperes is passed
through them until 1
...
How long did the
current flow? What mass of copper and what mass of zinc got deposited in their respective
cells?
(Atomic mass : Zn = 65
...
5 u)
CBSE (All India) SET–III
Questions Uncommon to Set-I and Set–II
...
Name an element with which silicon may be doped to give a p-type semiconductor
...
What is meant by a pseudo first order reaction? Give an example of a pseudo first order reaction
and write the rate equation for the same
...
Assign a reason for each of the following:
(i) The third ionization energy of Mn (Z = 25) is higher than that of either Cr ( Z = 24) or Fe
( Z = 26)
...
18
...
21
...
How much time will it take to reduce the
initial concentration of reactant to 1/10 th of its value?
24
...
2
...
4
...
6
...
The difference of energy between conduction band and valence band is called forbidden zone and
for insulator its value is averaging between 3–6 eV
...
(ii) by adding a suitable solvent
...
Lower members aldehydes are able to form intermolecular hydrogen bonds with water molecules
...
Due to the presence of a lone pair of electrons on nitrogen atom
...
g
...
The process of formation of polymers from respective monomers is called polymerisation
...
9
...
Osmotic pressure ( p) may be defined as the extra pressure that must be applied to the solution to
prevent the flow of solvent molecules into it through a semipermeable membrane
...
R
...
V
where V is the volume of solution in litre containing n B moles of solute of molecular mass M B
...
The osmotic pressure method has the advantage over rise in boiling point or fall in freezing point
for determining molar masses of macromolecules because
(i) Osmotic pressure is measured at the room temperature and the molarity of solution is used
instead of molality
...
OR
For a solution of volatile liquids Raoult’s law, is given as
P = PA + PB
If solute (component B) is non-volatile then
o
P = PA = PA x A
o
P = PA (1 - x B )
o
PA
o
P=
- PA x B
o
o
PA x B = PA - P
(Q x A + x B = 1)
Examination Papers | 29
o
PA - P
= xB
o
PA
Thus, relative lowering of vapour pressure is equal to the mole fraction of non-volatile solute
...
K = 0
...
(i) It is
(a)
(b)
(ii) As
M = 0
...
0248 ´ 1000
lm =
=
M
0
...
Higher H—F bond dissociation enthalpy than H—Cl
...
4 kJ mol -1 ) in N2
...
On the basis of electrochemical series the standard electrode potential shows the following order
E° Mn 2 + /Mn < E° Cr 2 + /Cr < E° Fe2 + /Fe
< E° Co 2 + /Co
Therefore, Co 2+ gets easily reduced to metallic cobalt while it is difficult to reduce Mn 2+
...
When bromine reacts with propyne, the reddish brown colour of bromine is discharged as long as
propyne is present in excess
...
CH3
½
¾ CH2 ¾ Cl
14
...
Alcohols as nucleophile:
The bond between O—H is broken when an alcohol reacts as a nucleophile
...
R ¾ CH2 ¾ O ¾ H + HCl ¾¾® R ¾ CH2 ¾ Cl + H2 O
Alcohol
Mechanism
··
+
··
··
R ¾ CH2 ¾ O ¾ H + H+ ¾¾® R ¾ CH2 ¾ O H2
-
16
...
Polymer
Monomer
Structure of the monomer
(i)
PVC
Vinyl chloride
CH2 = CH ¾ Cl
(ii)
Polypropene
Propene
CH3 ¾ CH = CH2
18
...
g
...
+
¾ SO- Na
3
CH3 ¾ (CH2 ) 11 ¾
Non-biodegradable detergents: Detergents having branched hydrocarbon chains are not easily
degraded by the micro-organism and hence are called non-biodegradable detergents, e
...
,
sodium-4-(1, 3, 5, 7-tetramethyl octyl) benzenesulphonate
...
19
...
55 g / cm 3 , M = 93 g / mol
For bcc, z = 2, a = ?
N A = 6
...
55 =
3
a ´ 6
...
55 ´ 6
...
55 ´ 3
...
55 ´ 3
...
55 ´ 3
...
55 - log 3
...
9685 - 0
...
4786)
3
1
log x = (1
...
5193
3
x = Antilog ( 0
...
306
a = 3
...
732 ´ 3
...
306 ´ 10 -8 =
4
4
-8
r = 1
...
15 ´ 10 - 10 cm
r = 143
...
Thus silicon or germanium doped with P or As is called n-type semiconductor, n
indicative of negative, since it is the electron that conducts electricity
...
An electron vacancy or a hole is created
at the place of the missing fourth electron
...
Thus Si or Ge doped with B or Al is
called p type of semiconductor, (P stands for positive hole) since it is the positive hole that is
responsible for conduction
...
They impart characteristic
colour to the compound and increase electrical conductivity
...
g
...
] Ferromagnetism arises due
to spontaneous alignment of magnetic moments in the same direction
...
20
...
86 K kg mol -1 , WB = ?
M B = 74
...
86 ´ WB ´ 1000
2=
74
...
5
\
WB =
= 40
...
86
21
...
34 - ( - 0
...
10 V
Examination Papers | 33
also
n=2
DG ° = - nFE° cell
DG° = - 2 ´ 96500 ´ 1
...
123 ´ 105 J
22
...
(ii) The positively charged colloidal particles of Fe(OH) 3 get coagulated by the oppositely charged
Cl - ions provided by NaCl
...
23
...
Characteristics
Lanthanoids
1-14
0 -1
Actinoids
2
1-14
6d 0 -17 s2
(i) Electronic configuration
[Xe] 4f
(ii) Oxidation states
Besides + 3 O
...
lanthanoids Besides +3 O
...
actinoids show
show +2 and +3 O
...
only in a higher O
...
of +4, +5, +6, +7
few cases
...
5d
[Rn] 5f
6s
(iii) General chemical reactivity of These are less reactive metals
elements
Lesser
tendency
complex formation
...
highly
towards Greater tendency
complex formation
...
Compounds are more basic
...
(i) [Cr(H2 O) 2 (C 2 O4 ) 2 ] Diaquadioxalatochromate (III) ion
O
...
of Cr = x + 0
...
2 = - 1, x = + 3
3
0
Electronic configuration of Cr 3+ = 3d 3 4s 0 = t 2 g e g
Unpaired electrons ( n) = 3, Paramagnetic
...
S
...
5 + ( - 1)
...
[NiCl 4 ] 2Tetrachloronickelate (II) ion
O
...
of Ni = x + ( -1) 4 = - 2, x = + 2
0
Electronic configuration of Ni 2+ = 3d 8 4s 0 = t 8 g e g
2
Unpaired electrons ( n) = 2, Paramagnetic
...
(i) In aniline, due to resonance, the lone pair of electrons on the nitrogen atom are delocalized over
the benzene ring
...
On the other hand,
in methyl amine +ve I effect of CH3 increases the electrondensity on the nitrogen atom
...
(ii) Aniline being a Lewis base reacts with lewis acid AlCl 3 to form a salt
...
(iii) Ethyl amine is freely soluble in water because it forms hydrogen bonds with water molecules
...
H ¾ N
...
H ¾ N
...
½
½
½
½
C 2 H5
H
C 2 H5
H
On the other hand in aniline due to large, hydrocarbon part, the extent of hydrogen bonding
decreases considerably and hence aniline is slightly soluble
...
(i) Primary structure of proteins: The sequence in which various amino acids are arranged in a
protein is called its primary structure
...
(ii) Denaturation: When a protein in its native form is subjected to physical change like change in
temperature or chemical change like change in pH, the hydrogen bonds are disturbed
...
During
denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact, e
...
,
coagulation of egg white on boiling
...
Examination Papers | 35
O
½
½
- H2O
H2 N ¾ CH2 ¾ C ¾ OH + H NH ¾ CH ¾ COOH ¾¾¾®
½
CH3
O
½
½
H2 NCH2 ¾ C ¾ NH ¾ CH ¾ COOH
½
CH3
Peptide linkage
(Gly -Ala)
28
...
e
...
K2
(b) K 2 = 4K1 i
...
,
=4
K1
T1 = 300 K
T2 = 320 K
K2
E a æ T2 - T1 ö
log
=
ç
÷
K1 2
...
303 ´ 8
...
147 è 300 ´ 320 ø
2 ´ 0
...
147 ´ 300 ´ 320
Ea =
= 55327 J
20
E a = 55
...
693 0
...
209 ´ 10 -4 year -1
t 1/ 2
5730
[R ]0
2
...
303
100
t=
log
=
log
-4
K
[ R ] 1
...
303 ´ 10 4
2
...
2
1
...
303 ´ 0
...
303
t=
´ 10 4 (1 - 3 ´ 0
...
2
1
...
7 years
29
...
(ii) Xe has least ionization enthalpy among noble gases and hence it readily forms chemical
compounds particularly with O2 and F2
...
p
...
p
...
p
...
p
...
Consequently, the tendency of N in NH3 to
denote its lone pair of electrons is much higher than that of P in PH3
...
(ii) In vapour state sulphur partly exists as S 2 molecule which has two unpaired electrons in
the antibonding p * orbital like O2 and hence exhibits paramagnetic behaviour
...
Due to large size and low electronegativity of chlorine no hydrogen
bonding is present in HCl, only Vander Waal forces are present
...
30
...
KOH
2 H ¾ C ¾ H ¾¾¾¾® CH3 ¾ OH + H ¾ C ¾ O K
Formaldehyde
Methyl alcohol
Potassium formate
Examination Papers | 37
(ii) Decarboxylation
R ¾ COOH + NaOH ¾¾®
R ¾ COONa ¾¾¾¾®
D
R ¾ COONa + H2 O
Sodium carboxylate
NaOH /CaO
R¾ H
Hydrocarbon
+ Na 2 CO3
O
CH3
(b) (i)
O
CH3
½
½
C = O + H2 N NH ¾ C ¾ NH2 ¾ ®
¾
Semi carbazide
H
½
½
C = NNH ¾ C ¾ NH2 + H2 O
Ethanal semicarbazone
H
Ethanal
O3
¾¾¾®
(ii)
2
O
Zn /H 2 O
(i) BH 3 /THF
CH2 ¾¾¾¾¾¾®
(iii)
—CHO
(ii) H 2 O2 / OH
(iii) PCC
OR
(a) (i) Propanone on treatment with I 2 / NaOH (NaOI) undergoes iodoform reaction to give yellow
ppt of iodoform but propanal does not
...
of Iodoform
Propanal
Tollen’s test: Propanal being an aldehyde reduces Tollen’s reagent to silver mirror but
propanone being a ketone does not
...
6 C 6 H5 OH + FeCl 3 ¾¾® [Fe(OC 6 H5 ) 6 ] -3 + 3H+ + 3HCl
Violet complex
NaHCO 3 test: Benzoic acid being a stronger acid than phenol decomposes NaHCO3 to evolve
CO2 but phenol does not
...
H 2 SO4
CH3 ¾ CH ¾ CH3 ¾¾¾¾¾® CH3 ¾ CH = CH2
443 K
Propene
38 | Xam Idea Chemistry–XII
COOH
CHO
O
COCl
C
+
KMnO4 /OH
(ii)
PCl5
anhyd
...
)
Ethanol
Acetaldehyde
3-Hydroxybutanal
CBSE (All India) SET–II
Schottky defect
...
g
...
8
...
It is used
as an antiseptic
...
(i) Order of reaction may be defined as the sum of powers of the concentration of the reactants in
the rate law expression
...
(ii) Molecularity of a reaction may be defined as the number of reacting species (atoms, ions or
molecules) taking part in an elementary reaction which must collide simultaneously in order to
bring about a chemical reaction
...
Simultaneous collision between two HI molecules
...
Hard water contains calcium and magnesium ions
...
Hence
soap cannot be used with hard water
...
6
...
[R ]0
21
...
303
t=
log
K
[R ]
Examination Papers | 39
Substituting the values
[R ]0
2
...
303
2
...
4 log 2
60
60
2
...
6932
t=
´ 0
...
62 ´ 10 -2 s
...
(i) Zone refining is based on the principle that the impurities are more soluble in the liquid state
than in the solid state of metal
...
A strip of the same metal in
pure form is used as cathode
...
When electric current is passed, impure metal forms metal ions which are
discharged at cathode forming pure metal
...
CHO
COOH
½
(CHOH) 4
27
...
(a)
Zn
Zn
2+
(conc
...
(ii) Current carriers of cell are
· electrons in external wire
...
n -Hexane
40 | Xam idea Chemistry–XII
· Ag + ions in cathodic half cell
...
e
...
(iii) At anode
At cathode
(b)
Zn ¾¾® Zn 2+ + 2e 2Ag + + e - ¾® 2Ag
Mg ¾¾® Mg 2+ + 2e Cu 2+ + 2e - ¾¾® Cu
Mg + Cu 2+ ¾¾® Cu + Mg 2+
n=2
According to Nernst equation,
E cell = E° cell -
[ Cu ] [ Mg 2+ ]
0
...
059
log
2
[ Cu 2+ ]
0
...
34 + 2
...
0295 log10
E cell = 2
...
685 V
OR
(a) The conductivity of a solution at any given concentration is the conductance of one unit volume
of solution kept between two platinum electrodes with unit area of cross section at a distance of
unit length
...
Molar conductivity ( L m ) of a solution at a given concentration is the conductance of the
volume V of solution containing one mole of electrolyte kept between two electrodes with area
of cross section A and distance of unit length
...
This is because the total volume,
V of solution containing one mole of electrolyte also increases
...
(b) Ag + + e - ¾® Ag
108 g of Ag are deposited by 96500 C
96500
\ 1
...
45 C
108
= 1295
...
6
t= =
= 863
...
I
1
...
34 - ( -2
...
5 g
63
...
6 C deposit Cu =
´ 1295
...
426 g
2 ´ 96500
Zn 2+ + 2e - ¾¾® Zn
2 ´ 96500 C deposit Zn = 65
...
3
\ 1295
...
6 = 0
...
Boron or Aluminium
...
A reaction which is of higher order but follows the kinetics of first order under special conditions
is called a pseudo first order reaction
...
H+
CH3 ¾ COOC 2 H5 + H2 O ¾¾® CH3 ¾ COOH + C 2 H5 ¾ OH
Here, the rate law is given by expression
Rate = K [ CH3 ¾ COOC 2 H5 ]
The concentration of H2 O is so large that it hardly undergoes any change during the reaction,
therefore, it does not appear in the rate law
...
(i) This is because Mn 2+ is more stable as it has exactly half filled configuration 3d 5 4s 0
...
This is because, although second ionization
enthalpy of copper is large but for Cu 2+ (aq) is much more negative than that of Cu + (aq) and
therefore, it more than compensates for the second ionisation enthalpy of copper
...
2Cu + (aq) ¾® Cu 2+ (aq) + Cu(s)
18
...
Examples:
Aspartame: Aspartame is 100 times as sweet as cane sugar
...
Saccharin: It is about 550 times as sweet as cane sugar
...
21
...
303
t=
log
K
[R ]
[ R ] 0 2
...
303
t=
log
=
log 10 = 0
...
038 s
24
...
Therefore all elements except Sc and Zn, of the
first transition series show a number of oxidation states as shown in table
...
The reason of this variation in atomic radii has been attributed to the increase in nuclear charge
in the beginning of the series
...
When the increased nuclear charge
and the increased screening effect balance each other in the middle of transition series, the
atomic radii becomes almost constant (Mn to Fe)
...
As a result there is an
expansion of the electron cloud; consequently, the atomic size increases
...
The magnetic moment (m) of the elements of the first transition series can be calculated with the
unpaired electrons ( n) by the spin-only formula
m = n ( n + 2) B
...
Ion
Configuration
Unpaired electrons
Magnetic moment (m )
calculated
Mn 2+
3d5 4s0
5
5 (5 + 2) = 5
...
M
...
73 B
...
Zn 2+
3d10 4s0
0
0 ( 0 + 3) = 0
CBSE EXAMINATION PAPERS
DELHI–2009
Time allowed : 3 hours
Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Delhi) SET–I
1
...
Define the term ‘Tyndall effect’
...
Why is the froth floatation method selected for the concentration of sulphide ores?
1
4
...
Give the IUPAC name of the following compound:
CH3—C C—CH2OH
1
CH3 Br
6
...
1
7
...
Give an example of elastomers
...
A reaction is of second order with respect to a reactant
...
Explain the role of
2
(i) Cryolite in the electrolytic reduction of alumina
...
11
...
Complete the following chemical reaction equations:
(i) P4( s) + NaOH (aq ) + H2 O( l ) ¾ ®
¾
(ii) I - ( aq ) + H2 O( l ) + O3( g ) ¾ ®
¾
2
44 | Xam idea Chemistry–XII
13
...
What is the effect of change in temperature
of a solution on its molality and molarity?
2
14
...
Complete the following reaction equations:
2
—OH + SOCl2
(i)
CH2OH
(ii)
+ HCl
OH
16
...
Name two water soluble vitamins, their sources and the diseases caused due to their deficiency in diet
...
Draw the structures of the monomers of the following polymers:
(i) Teflon
2
(ii) Polythene
OR
What is the repeating unit in the condensation polymer obtained by combining HO2 CCH2 CH2 CO2 H
(succinic acid) and H2 NCH2 CH2 NH2 (ethylene diamine)
...
Iron has a body-centred cubic unit cell with a cell edge of 286
...
The density of iron is 7
...
Use this information to calculate Avogadro’s number (At
...
3
20
...
0 mL of solution
...
3 mm Hg at 25oC, what is the molar mass of the protein?
3
(R = 0
...
)
21
...
0051 min–1
...
10M concentration of the
reactant, what concentration of reactant will remain in solution after 3 hours?
3
22
...
3
(i) An aerosol
(ii) A hydrosol
(iii) An emulsion
Examination Papers | 45
23
...
(ii) Sulphur has a greater tendency for catenation than oxygen
...
OR
Explain the following situations:
(i) In the structure of HNO3 molecule, the N—O bond (121 pm) is shorter than N—OH bond
(140 pm)
...
(iii) XeF2 has a straight linear structure and not a bent angular structure
...
For the complex [Fe(en)2Cl2]Cl, (en = ethylene diamine), identify
3
(i) the oxidation number of iron
...
(iii) the magnetic behaviour of the complex
...
(v) whether there is an optical isomer also, and
(vi) name of the complex
...
no
...
Explain the mechanism of the following reactions:
3
(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed
by hydrolysis
...
(iii) Acid catalysed hydration of an alkene forming an alcohol
...
Giving an example for each, describe the following reactions:
3
(i) Hofmann’s bromamide reaction
(ii) Gatterman reaction
(iii) A coupling reaction
27
...
(a) Define molar conductivity of a substance and describe how for weak and strong electrolytes,
molar conductivity changes with concentration of solute
...
001 M) | Ag and Cu2+ (0
...
46 V)
46 | Xam idea Chemistry–XII
OR
(a) State the relationship amongst cell constant of a cell, resistance of the solution in the cell and
conductivity of the solution
...
001 M) and Ni | Ni2+ (0
...
25 V, EoAl/Al3+ = – 1
...
(a) Complete the following chemical reaction equations:
(i)
MnO– aq )
4(
+
C 2 O2– aq )
4(
+ H
+
( aq )
5
¾ ®
¾
2–
(ii) Cr2 O7( aq ) + F 2 + ( aq ) + H+ ( aq ) ¾ ®
¾
(b) Explain the following observations about the transition/inner transition elements:
(i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29)
...
(iii) The members in the actinoid series exhibit a large number of oxidation states than the
corresponding members in the lanthanoid series
...
(ii) The greatest number of oxidation states are exhibited by the members in the middle of a
transition series
...
30
...
77% carbon, 11
...
The
molecular mass of the compound is 86
...
On
vigorous oxidation it gives ethanoic and propanoic acids
...
Examination Papers | 47
OR
(a) How are the following obtained?
(i) Benzoic acid from ethyl benzene
...
(b) Complete each synthesis by giving the missing material, reagent or products:
(i)
H2
Pd-BaSO4
C6H5COCl
...
(ii)
CH3
Anhydrous AlCl3
O
C
+
...
CBSE (Delhi) SET–II
Questions different from Set-I
1
...
Which is a stronger oxidising agent Bi(V) or Sb(V)?
21
...
How are associated colloids different from these two types of colloids?
24
...
(ii) The majority of known noble gas compounds are those of Xenon
...
27
...
28
...
(ii) With the same d4 d-orbital configuration Cr2+ ion is reducing while Mn3+ ion is oxidising
...
48 | Xam idea Chemistry–XII
OR
(a) Complete the following chemical reaction equations:
(i) CrO72–(aq) + H2S(g) + H+(aq) ¾ ®
¾
(ii) MnO2(s) + KOH
(aq)
+ O(2) ¾ ®
¾
(b) Explain the following observations:
(i) Transition metals form compounds which are usually coloured
...
(iii) The actinoids exhibit a greater range of oxidation states than the lanthanoids
...
(a) What type of a cell is the lead storage battery? Write the anode and the cathode reactions and
the overall reaction occurring in a lead storage battery while operating
...
001 M) and Ni | Ni2+ (0
...
Write the equation for the reaction that occurs when the cell generates an electric current and
determine the cell potential
...
25V, E oAl 3+
Al
= –1
...
OR
(a) Express the relationship amongst cell constant, resistance of the solution in the cell and
conductivity of the solution
...
(b) Calculate the equilibrium constant for the reaction
...
40 V, E oFe 2 +
Fe 2 + (aq ) + Cd( s)
Fe
= – 0
...
CBSE (Delhi) SET–III
Questions different from Set-I and Set–II
1
...
What does the part ‘6, 6’ mean in the name nylon-6,6?
19
...
0711m aqueous solution of Na 2 SO4
...
320°C, what would be the value of van't Hoff factor? (K f for water is
1
...
24
...
What are the following substances? Give one example of each type
...
(ii) Nonionic detergents
(iii) Antiseptics
(a) Complete the following chemical reaction equations:
2(i) Cr2O 7 (aq) + I - (aq) + H + (aq) ¾ ®
¾
(ii) MnO - (aq) + Fe2+(aq) + H + (aq) ¾ ®
¾
4
(b) Explain the following observations:
(i) In general the atomic radii of transition elements decrease with atomic number in a given
series
...
34 V)
...
(iii) The Eo value for Mn3+ | Mn2+ couple is much more positive than for Cr3+ | Cr2+ or Fe3+ | Fe2+
couple
...
(ii) Although Co2+ ion appears to be stable, it is easily oxidised to Co3+ ion in the presence of
a strong ligand
...
29
...
How is it related to conductivity of the related solution?
(b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution
of unknown concentration
...
0 M
solution of Zn(NO3)2
...
48 V is measured for this cell
...
(E o
Zn 2 + /Zn
= - 0 × 76V,
Eo
Ag 2 + /Ag
= + 0 × 80V)
OR
(a) Corrosion is essentially an electrochemical phenomenon
...
(b) Calculate the equilibrium constant for the equilibrium reaction
...
Frenkel defect
...
The scattering of light by colloidal particles is known as Tyndall effect
...
As only sulphide ore particles are wetted by oil while gangue particles are wet by water
...
Because Bi (V) is more stable than Sb (V) due to inert pair effect
...
2-Bromo-3-methyl but-2-en-1-ol
...
3-Oxopentanal: CH3 — CH2 — C — CH2 — C — H
7
...
8
...
Let the rate law be, r1 = k [A]2
(i) If [A] is doubled than rate r2 = k(2A)2 = 4k [A]2 = 4r1, i
...
, rate becomes 4 times
...
e
...
(i) Role of cryolite
· It lowers the melting point of the mixture
...
(ii) When nickel is heated with carbon monoxide it forms a volatile complex nickel tetracarbonyl
which on further heating at higher temperature decomposes to give pure nickel
...
p
...
(i) No
...
p
...
b
...
= 3
(ii) No
...
p
...
12
...
Molality is the number of moles of solute per thousand grams of solvent whereas molarity is the
number of moles of solute dissolved in one litre of solution
...
14
...
(ii)
: As iodine is a better leaving group because of its large size, it will
be released at a faster rate in the presence of incoming nucleophile
I
15
...
(i)
HO
O
||
The — C— NH — bond formed between two amino acid molecules with loss of water in a
polypeptide is called peptide linkage
...
17
...
Name of Vitamins
Sources
Deficiency diseases
(i)
Vitamin B12
Meat, fish, egg and curd
Pernicious anaemia
(ii)
Vitamin C
Citrus fruits and amla
Scurvy
Polymer
Teflon
Monomer
Tetrafluoroethene
Polyethene
Ethene
Structure
F F
|
|
F - C = C- F
H H
|
|
H - C = C- H
18
...
d = 7
...
65pm = 286
...
87g cm–3 =
NA
=
=
=
20
...
65 ´ 10 -3 cm) 3 ´ NA
2 ´ 56 g mol -1
(2
...
87 g cm -3
2 ´ 56 g mol -1
23
...
87 g cm -3
112 mol -1
185
...
604 × 1024 mol–1
= 6
...
0821 L atm K–1 mol–1
T
= 25°C = (25 + 273) K = 298K
V
= 10 mL = 10 × 10–3L
13
...
3 mm Hg =
atm
760
p
MB =
MB =
MB =
100 ´ 10 -3 g ´ 0
...
3
atm ´ 10 ´ 10 -3 L
760
100 ´ 10 -3 ´ 0
...
3 ´ 10 ´ 10 -3
13980
...
For a first order reaction
t
Here
=
[R] o
2
...
0051min–1, [R]o = 0
...
303
0
...
0051 min
Log
log
0
...
1
[R]
0
...
3986
= Anti log (0
...
503
[R] =
01
...
03995 M
2
...
04M
22
...
This is because the lone pair of electrons on N atom in NH3 is
directed and not diffused as it is in PH3 due to larger size of phosphorus and hence more
available for donation
...
(iii) Bond dissociation energy of F2 is less than Cl2 this is due to relatively large electron – electron
repulsion among the lone pairs in F2 molecule where they are much closer to each other than in
case of Cl2
...
Dispersed Phase
Aerosol
–
O
O
As a result of resonance, N–O bond length is average of single bond and double bond whereas
N–OH bond has purely single bond character
...
(ii) S atom in SF4 is not sterically protected as it is surrounded by only four F atoms, so attack of
H2O molecules can take place easily and hence hydrolysis takes place easily
...
Therefore does not allow H2O molecules to attack
S atoms
...
54 | Xam idea Chemistry–XII
(iii) In XeF2, Xe is sp3d hybridised having 2 bond pair and 3 lone pair of electrons
...
24
...
\
x + 2 ´ 0 + 2(–1) + 1( -1) = 0
Þ
x =3
4s
3d
4p
(ii) Orbitals of Fe (III)
Six d2sp 3 hybridised orbitals
3d
d2 sp3 hybridised orbitals of Fe (III)
Thus, hybridisation: d2 sp3
Shape of the complex: Octahedral
(iii) Paramagnetic due to presence of three unpaired electrons
(iv) Two, cis and trans isomers
(v) Yes, cis isomer will also show optical isomerism
(vi) Dichlorido bis (ethane-1, 2 diamine) iron (III) chloride or Dichloro bis (ethylenediamine)
iron (III) chloride
...
(i) Step I: Nucleophilic addition of Grignard reagent to Carbonyl group
...
H H H
H H
+
Slow
H— C — C — O — H
H
+
H
H
Ethyl oxonium ion
+
H— C — C
H2O
H
Ethyl carbocation
Step III: Formation of ethylene by elimination of a proton
H H
H
H
+
—
C—C
H— C — C
H
H
H H
+
+ H
Ethylene
Ethyl carbocation
To drive the equilibrium to the right, ethylene is removed as it is formed
...
+
H2O
+
H
H3O +
H
—
C—C
+
H
+
H—O—H
+
Step II: Nucleophilic attack of water on carbocation
...
H
+
H2O
OH
—C—C—
+
H3O +
(i) Hoffman’s bromamide reaction: When a primary acidamide is heated with bromine in an
aqueous or ethanolic solution of sodium hydroxide, it gives a primary amine with one carbon
atom less
...
O
R — C — NH2 + Br2 + 4NaOH
1° acid amide
R — NH2 + Na2CO3 + 2NaBr + 2H2O
1° amine
56 | Xam idea Chemistry–XII
(ii) Gattermann reaction: Chlorine or bromine can be introduced in benzene ring by treating the
diazonium salt solution with corresponding halogen acid in the presence of copper powder
...
The mechanism is basically that of electrophilic
aromatic substitution where the diazonium ion is electrophile
...
+
–
NCl + H—
— NH 2
H
(pH 4-5)
N
Aniline
–
NCl + H —
N—
–
— NH 2 + Cl + H 2O
p-Aminoazobenzene (Yellow dye)
— OH
–
OH
(pH 9-10)
N
Phenol
N—
–
— OH + Cl + H 2O
p-Hydroxyazobenzene (Orange dye)
(i) Cationic Detergents : Cationic detergents are quarternary ammonium salts of amines with
acetates, chlorides or bromides as anions
...
Hence, these are called cationic detergents
...
é
CH3
ê
½
ê
êCH3 (CH2 ) 15 ¾ N ¾ CH3
½
ê
CH3
ê
ë
ù
ú
ú
ú
ú
ú
û
+
Br -
Cetyltrimethyl ammonium bromide
Cationic detergents have germicidal properties and are expensive, therefore, these are of
limited use
...
Preservatives prevent the rancidity of food and inhibit growth or kill the microganisms
...
(iii) Analgesics: Drugs which reduce or abolish pain without causing impairment of consciousness,
mental confusion, incoordination or paralysis or some other disturbances of nervous system
...
(a) Non-narcotic (non-addictive) analgesics: Aspirin, paracetamol etc
...
g
...
(a) Molar Conductivity (Lm): It may be
defined as the conductance of a solution
containing 1 mole of electrolyte such that
the entire solution is placed is between two
electrodes one centimeter apart
...
CH3COOH
200
KCl
Molar conductivity increases with decrease
in concentration or increase in dilution as
number of ions as well as mobility of ions
increased with dilution
...
2
1/2
0
...
Therefore L m increases a
little as shown in graph by a straight line
...
Cu( s) ¾ ® Cu 2 + ( aq) + 2 e
¾
(b) At anode:
At cathode:
2Ag + ( aq) + 2e ¾ ® 2Ag( s)
¾
Cu( s) + 2Ag + ( aq) ¾ ® Cu 2 + ( aq) + 2Ag( s)
¾
[Cu 2 + ]
0 × 0591
log
n
[Ag + ] 2
Here
E cell = E o cell
Here
E o = 0
...
1M
E cell = 0 × 46 -
0 × 0591
0 ×1
log
2
(10 -3 ) 2
E cell = 0 × 46 -
0 × 0591
0
...
312 V
OR
(a) k =
1 æl ö
´ç ÷
R èAø
Where
k = Conductivity
l
= Cell Constant
A
R = Resistance
k ´ 1000
Lm =
M
Where
L m = Molar conductivity
k = Conductivity
M = Molarity of Solution
(b) At anode:
At cathode:
Al( s) ¾ ®Al 3 + ( aq) + 3e ] ´ 2
¾
Ni 2 + ( aq) + 2e ¾ ® Ni ( s)] ´ 3
¾
2Al ( s) + 3Ni 2 + ( aq) ¾ ® 2Al 3 + ( aq) + Ni( s)
¾
E cell = E o cell
[Al 3 + ] 2
0
...
001M = 1× 10–3M, [Ni2+] = 0
...
25V - (-1
...
41V
cell
Eo
cell
( )
10 -3
0
...
41 log
6
(0
...
41 -
0
...
125
0
...
0591
log 10 –6 ´ 8 =1
...
0591
0
...
41 log (– 6 log 10 + 3 log2) = 1
...
3010 )
6
6
0
...
3012
= 1
...
097) = 1
...
41 -
= 1
...
0502
E cell = 1
...
4602V
(
)
Examination Papers | 59
29
...
(ii) The frequent metal-metal bonding in compounds of heavy transition elements is due to their
high enthalpy of atomization
...
OR
(a) (i)
In neutral or faintly alkaline solutions
MnO- + 2H2 O + 3e - ¾ ¾ ® MnO2 + 4OH– ] ´ 8
¾
4
-
S 2 O2 - + 10 O H ¾ ¾ ® 2 SO2 - + 5 H2 O + 8e - ] ´ 3
¾
3
4
-
8MnO- + 3S 2 O2 - + H2 O ¾ ¾ ® 8MnO2 + 6SO2 - + 2 O H
¾
4
3
4
(ii)
In acidic solutions
2
Cr2 O7 - + 14H+ + 6e - ¾ ® 2Cr 3 + + 7H2 O
¾
H2 S ¾ ® S + 2H+ + 2e - ] ´ 3
¾
2
Cr2 O7 - + 3H2 S + 8H+ ¾ ® 2Cr 3 + + 3S + 7H2 O
¾
(b) Fe( s) + Cd 2 + ( aq) l
log k c = n
Fe 2 + ( aq) + Cd( s)
o
E cell
0
...
44) = 0
...
04 0
...
059
0
...
3536
Þ
kc = Antilog 1
...
57
60 | Xam idea Chemistry–XII
30
...
O
||
Conc
...
—
C—O
+
4[H]
Zn/conc
...
of moles
Simplest molar
ratio
C
69
...
63
1
11 × 63
= 11 × 63
1
11 × 63
= 10
1 ×16
O
(100 – 81
...
60
16
18 × 60
= 1 ×16
16
1 ×16
=1
1 ×16
Empirical formula of the compound A = C5 H10 O
Molecular formula of the compound A = n (Empirical formula)
Molecular mass of compound A
n=
Empirical formula mass of compound A
Molecular mass of compound A = 86
Empirical formula mass of compound A = 5 ´ 12 + 1´ 10 + 1´ 16
= 60 + 10 + 16
= 86
86
n =
=1
86
Molecular formula of the compound A = 1(C5 H10 O)
l
l
l
= C5 H10 O
As the compound A forms addition compound with NaHSO3 therefore it must be either an
aldehyde or ketone
...
As on oxidation the compound A gives a mixture of ethanoic acid and propanoic acid, therefore
compound A is
Examination Papers | 61
O
||
CH3 — C— CH2 — CH2 — CH3
Pentan -2-one
The chemical reactions are:
O
OH
||
|
CH3 — C— CH2 — CH2 — CH3 + NaHSO3 ¾ ¾ ® CH3 — C — CH2 — CH2 — CH3
¾
Pentan -2-one
|
+
SO- N a
3
Sodium hydrogen sulphite addition product
O
||
CH3 — C— CH2 — CH2 — CH3 + 3I 2 + 4NaOH
Pentan - 2 - one
- +
Iodoform
¾ ¾¾¾ ®
¾
reaction
CHI 3 ¯
+ CH3 — CH2 — CH2 — CO O Na + 3NaI + 3H2 O
(Iodoform yellow ppt)
O
||
K 2 Cr2 O7
CH3 — C— CH2 — CH2 — CH3 ¾ ¾¾¾ ® CH3 — COOH + CH3 — CH2 — COOH
¾
H 2 SO4
Pentan - 2 - one
Ethanoic acid
OR
–+
CH2—CH3
COOK
COOH
KMnO4 – KOH
(a) (i)
Propanoic acid
H3 O
+
D
Benzoic acid
–
–
O
– O — C — CH3
O — C — CH
O
CrO3
(ii)
H3 O
O
+
–
–
D
–
CH3 – C
CHO
3
–
–
HC
–
CH3
O
CH3 – C –
–
–
Benzylidene
diacetate
Benzaldehyde
O
OR
CH3
CHO
CH(OCrOHCl2)2
+ CrO2Cl2
CS2
H3 O
Chromium complex
+
Benzaldehyde
62 | Xam idea Chemistry–XII
–
–
O
C — Cl
CHO
H2
(b) (i)
+
Pd – BaSO4
Benzoyl chloride
HCl
Benzaldehyde
–
–
O
–
–
O
+ CH3 — C — Cl
(ii)
anhyd
...
AlCl3
Benzoyl chloride
+
HCl
Benzophenone
CBSE (Delhi) SET–II
1
...
Multimolecular colloids: In this type of colloids, colloidal particles are aggregates of atoms or
molecules each having size less than 1nm, e
...
, sulphur sol, gold sol
...
g
...
Associated colloids: There are certain substances which at low concentrations behave as normal
electrolyte, but at higher concentrations exhibit colloidal behaviour due to the formation of
aggregates
...
g
...
(i) Fluorine does not exhibit any positive oxidation state as it is the most electro-negative element
and does not have d orbitals in its valance shell
...
(ii) This is due to low ionization enthalpy of xenon
...
4 kJ mol -1 )
...
Difference between antiseptics and disinfectants
...
Antiseptics are generally applied to living tissues such as wounds, cuts, ulcers and diseased skin
surfaces
...
Examination Papers | 63
Disinfectants are chemical substances which kill micro-organisms or stop their growth but are
harmful to human tissues
...
l
Chlorine in the concentration of 0
...
4 ppm in aqueous solution and SO2 in very low
concentration are disinfectants
...
(a) (i)
MnO- + 8H+ + 5e - ¾ ¾¾® Mn 2 + + 4H2 O
4
Fe 2 + ¾ ¾¾® Fe 3 + + e - ] ´ 5
MnO- + 5Fe 2 + + 8H+ ¾ ¾¾® Mn 2 + + 5Fe 3 + + 4H2 O
4
(ii)
2
Cr2 O7 - + 14H+ + 6e - ¾ ¾¾® 2Cr 3 + + 7H2 O
2I - ¾ ¾¾® I 2 + 2e - ] ´ 3
2
Cr2 O7 - + 6I - + 14H+ ¾ ¾¾® 2Cr 3 + + 3I 2 + 7H2 O
(b)
(i)
Transition elements form many interstitial compounds as they are capable of entrapping
small atoms like H, C or N in the interstitial sites in their crystal lattice
...
On the other hand, the change from Mn 3+ to Mn 2+ results in the
half-filled d 5 configuration which has extra stability therefore Mn 3+ is oxidising
...
OR
(a)
(i)
2
Cr2 O7 -
+
+ 14H + 6e
-
¾ ¾¾® 2Cr 3 + + 7H2 O
H2 S ¾ ¾¾® S + 2H+ + 2e - ] ´ 3
2
Cr2 O7 - + 3H2 S + 8H+ ¾ ¾¾® 2Cr 3 + + 3S + 7H2 O
(ii) 2MnO2 + 4KOH + O2 ¾ ¾¾® 2K 2 MnO4 + 2H2 O
(b)
(i)
This is due to d – d transition as the energy of excitation of d orbital electrons from lower
energy to higher energy level lies in the visible region
...
(iii) This is due to comparable energies of 5f , 6d and 7s orbitals of actinoids
...
(a) The lead storage battery is a secondary cell
The cell reactions when the battery is in use are given below
At anode
:
Pb( s) + SO2 - ( aq) ¾ ¾ ® PbSO4 ( s) + 2e ¾
4
At cathode
:
PbO2 ( s) + SO2 - ( aq) + 4H+ ( aq) + 2e - ¾ ¾ ® PbSO4 ( s) + 2H2 O(l)
¾
4
Overall cell reaction : Pb( s) + PbO2 ( s) + 2H2 SO4 ( aq) ¾ ¾ ® 2PbSO4 ( s) + 2H2 O(l)
¾
64 | Xam idea Chemistry–XII
Al( s) ¾ ®Al 3 + ( aq) + 3e ] ´ 2
¾
(b) At anode:
Ni 2 + ( aq) + 2e ¾ ® Ni ( s)] ´ 3
¾
At cathode:
2Al ( s) + 3Ni 2 + ( aq) ¾ ® 2Al 3 + ( aq) + Ni( s)
¾
E cell = E o cell
[Al 3 + ] 2
0
...
001M = 1× 10–3M, [Ni2+] = 0
...
25V - (-1
...
41V
cell
Eo
cell
( )
10 -3
0
...
41 log
6
(0
...
41 -
0
...
125
0
...
0591
log 10 –6 ´ 8 =1
...
0591
0
...
41 log (– 6 log 10 + 3 log2) = 1
...
3010 )
6
6
0
...
3012
= 1
...
097) = 1
...
41 -
)
= 1
...
0502 = 1
...
46 V
OR
1 æl ö
k=
´ç ÷
R èAø
(a)
Where
k = Conductivity
l
= Cell Constant
A
R = Resistance
k ´ 1000
Lm =
M
Where
L m = Molar conductivity
k = Conductivity
M = Molarity of Solution
( aq) l
E ° cell
log k c = n
0
...
40 – (– 0
...
04 V
2 ´ 0 × 04 0 × 08
log kc =
=
0 × 059
0 × 059
log kc = 1
...
3536
kc = 22
...
Interstitial defect increases the density of a solid
...
It means the two monomers combine to make nylon 6,6, contain six carbon atoms each
...
DTf = K f
...
86 K kg mol -1 × 0
...
132 K
Observed value of DTf
0
...
i = 2
...
S
...
Central
metal ion
Configuration
of metal ion
Hybridisation
of metal ion
Geometry
of the
complex
Number of
unpaired
electrons
Magnetic
behaviour
(i)
[ CoF4 ]2-
Co 2+
d7
sp 3
Tetrahedral
3
Paramagnetic
(ii)
[Cr(H2O)2
(C2O4 )2 ]
Cr 3+
3d 3
d 2sp 3
Octahedral
3
Paramagnetic
(iii)
27
...
g
...
(ii) Non-ionic Detergents: Non-ionic detergents do not contain any ion in their constitution
...
(iii) Antiseptics: These are the chemical substances which prevent the growth of micro-organisms
and may even kill them but are not harmful to living tissues
...
Dettol, soframicine are antiseptics
...
(a) (i)
2I - ¾ ¾¾® I 2 + 2e - ] ´ 3
2
Cr2 O7 - + 6 I - + 14H+ ¾ ¾¾® 2Cr 3 + + 3I 2 + 7H2 O
(ii)
MnO- + 8H+ + 5e - ¾ ¾¾® Mn 2 + + 4H2 O
4
Fe 2 + ¾ ¾¾® Fe 3 + + e - ] ´ 5
MnO- + 5Fe 2 + + 8H+ ¾ ¾¾® Mn 2 + + 5Fe 3 + + 4H2 O
4
(b)
(i) The atomic radii of transition metals decreases with atomic number in a series as the
nuclear charge increases due to poor shielding effect of d orbitals
...
(iii) This is due to much large third ionisation energy of Mn as Mn 2+ is very stable on
account of stable d 5 configuration
...
Cause of lanthanoid contraction: As we move along the lanthanoid series, for every
additional proton in the nucleus, the corresponding electron goes into 4f -subshell, there is poor
shielding of one electron by another in this subshell due to the shapes of these f-orbitals
...
Thus the net result is decrease in size with increase in atomic number
...
(ii) The basic strength of hydroxides decreases from La(OH) 3 to Lu(OH) 3
...
(ii) Because strong ligand cause spin pairing giving rise to diamagnetic octahedral complex
which are very stable and have very large crystalfield stabilization energy
...
(iii) This is due to stability of Mn 2+ as it has half filled d 5 configuration
...
(a) Molar conductivity ( L m ): It may be defined as the conductivity of one molar electrolytic
solution placed between two electrodes one centimeter apart and have enough area of cross
section to hold entire volume
...
48 = 1
...
The H+ ions are produced by
either H2 O or H2 CO3 (formed by dissolution of CO2 in moisture)
2H+ ( aq) + 2e - ¾ ¾ ® 2H
¾
1
2H + O2 ( g) ¾ ¾¾® H2 O
2
Net reaction at cathodic area
1
2H+ ( aq) + O2 + 2e - ¾¾® H2 O
E° H + /O /H O = 1
...
67V
¾
2
The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as
rust in the form of hydrated ferric oxide (Fe 2 O3
...
(b) At anode:
At cathode:
Al( s) ¾ ®Al 3 + ( aq) + 3e ] ´ 2
¾
Ni 2 + ( aq) + 2e ¾ ® Ni ( s)] ´ 3
¾
2Al ( s) + 3Ni 2 + ( aq) ¾ ® 2Al 3 + ( aq) + Ni( s)
¾
68 | Xam idea Chemistry–XII
E cell = E o cell
[Al 3 + ] 2
0
...
001M = 1× 10–3M, [Ni2+] = 0
...
25V - (-1
...
41V
cell
Eo
cell
( )
10 -3
0
...
41 log
6
(0
...
41 -
0
...
125
0
...
0591
log 10 –6 ´ 8 =1
...
0591
0
...
41 log (– 6 log 10 + 3 log2) = 1
...
3010 )
6
6
0
...
3012
= 1
...
097) = 1
...
41 -
(
= 1
...
0502 = 1
...
46 V
)
(
)
CBSE EXAMINATION PAPERS
ALL INDIA–2009
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (All India) SET–I
1
...
What is the ‘coagulation’ process?
1
3
...
Why is red phosphorus less reactive than white phosphorus?
1
5
...
Write the structural formula of 1-phenylpentan-1-one
...
Arrange the following compounds in an increasing order of basic strengths in their aqueous
solutions:
1
NH3, CH3NH2, (CH3)2NH, (CH3)3N
8
...
What type of cell is a lead storage battery? Write the anode and the cathode reactions and the overall
cell reaction occurring in the use of a lead storage battery
...
51V
4
Sn 2 + ( aq) ® Sn 4 + ( aq) + 2e - , E° = + 0
...
10
...
Describe the underlying principle of each of the following metal refining methods:
2
(i) Electrolytic refining of metals
(ii) Vapour phase refining of metals
...
Complete the following chemical reaction equations:
2
(i) XeF2 + H2O ¾ ®
¾
(ii) PH3 + HgCl2 ¾ ®
¾
13
...
Which one in the following pairs undergoes S N 1 substitution reaction faster and why?
Cl
2
Cl
(i)
or
Cl
Cl
or
(ii)
15
...
Name the four bases present in DNA
...
Name two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet
...
Differentiate between molecular structures and behaviours of thermoplastic and thermosetting
polymers
...
2
19
...
0051 min -1
...
10 M concentration of the
reactant, what concentration of the reactant will be left after 3 hours?
3
20
...
Each side of the unit cell has a length of 409 pm
...
)
3
21
...
The copper ion concentration in it is 0
...
The concentration of silver
ion is not known
...
422 V
...
3
(Given: E o
Ag + /Ag
= +0
...
34 V)
Examination Papers | 71
22
...
(ii) A beam of light is passed through a colloidal solution
...
23
...
OR
Compare the following complexes with respect to structural shapes of units, magnetic behaviour and
hybrid orbitals involved in units:
[Co(NH3 ) 6 ] 3 + , [Cr(NH3 ) 6 ] 3 + , Ni(CO) 4
(At No
...
Explain the following observations:
3
(i) The boiling point of ethanol is higher than that of methoxymethane
...
(iii) o-and p-nitrophenols are more acidic than phenol
...
How would you account for the following:
3
(i) Many of the transition elements and their compounds can act as good catalysts
...
(iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids
...
Complete the following reaction equations:
O
||
(i) R - C- NH2 ¾LiAlH 4 ®
¾¾
¾
3
H2O
(ii) C6H5N2Cl + H3PO2+H2O ¾ ®
¾
(iii) C6H5NH2 + Br2 (aq) ¾ ®
¾
27
...
(a) Define the following terms:
(i) Mole fraction
(ii) Van’t Hoff factor
5
72 | Xam idea Chemistry–XII
(b) 100 mg of a protein is dissolved in enough water to make 100 mL of a solution
...
3 mm Hg at 25° C, what is the molar mass of protein?
(R = 0
...
)
OR
What is meant by:
(i) Colligative properties
(ii) Molality of a solution
...
78
...
42 × 10–7 M/mm Hg]
5
29
...
(ii) Sulphur has a greater tendency for catenation than oxygen
...
5
OR
(a) Draw the structures of the following:
(i) H2S2O7
(ii) HClO3
(b) Give an explanation for each of the following observations:
(i) In the structure of HNO3, the N – O bond
(121 pm) is shorter than the N – OH bond (140 pm)
(ii) All the P – Cl bonds in PCl5 are not equivalent
...
30
...
(b) An organic compound A has the molecular formula C8H16O2
...
Oxidation of C with chromic
acid also produced B
...
Write equations for the
reactions involved
...
Which point defect of its crystals decrease the density of a solid?
1
8
...
For a decomposition reaction the values of rate constant k at two different temperatures are given
below:
3
k1 = 2
...
39 × 10–7 L mol–1 s–1 at 700 K
Calculate the value of activation energy for this reaction
...
314 J K–1 mol–1)
27
...
(a) Give chemical tests to distinguish between compounds in the following pairs of substances:
(i) Ethanal and Propanal
(ii) Benzoic acid and Ethyl benzoate
(b) An organic compound contains 69
...
63% hydrogen and rest oxygen
...
It does not reduce Tollen’s reagent but forms an addition
compound with sodium hydrogensulphite and gives positive iodoform test
...
Derive the structure of the compound ‘A’
...
(a) Draw the structure of the following:
(i) H3PO2
(ii) BrF3
(b) How would you account for the following observations:
(i) Phosphorus has a greater tendency for catenation than nitrogen
...
(iii) No chemical compound of helium is known
...
(ii) Nitrogen does not form pentahalides
...
CBSE (All India) SET–III
[Questions different from Set–I and Set–II]
1
...
Describe the role of the following:
1
2
(i) NaCN in the extraction of silver from a silver ore
...
14
...
Differentiate between condensation and addition polymerisations
...
2
21
...
0010 M) and Ni | Ni
3
(0
...
Write the equation for the cell reaction that occurs when the cell generates an electric current and
determine the cell potential
...
25V, E o
Al 3+ /Al
= -1
...
Explain the following:
3
(i) Low spin octahedral complexes of nickel are not known
...
(iii) CO is a stronger ligand than NH3 for many metals
...
Nos,: Ni = 28; Co = 27]
27
...
3
(i) Cationic detergents
(ii) Enzymes
(iii) Sweetening agents
30
...
(ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine
...
OR
(a) Draw the structures of the following:
(i) PCl5(s)
(ii) SO23
(b) Explain the following observations:
(i) Ammonia has a higher boiling point than phosphine
...
(iii) Bi(V) is a stronger oxidising agent than Sb(V)
...
Metallic substances conduct electricity through electrons while ionic substances conduct electricity in
molten state or in solution through ions
...
The process of settling of colloidal particles is called coagulation
...
It is a thermal process of extracting a metal from its ore
...
This is due to polymeric structure of red phosphorus or angular strain in P4 molecule of white
phosphorus where the angle is only 60°
...
H2C = CH – CH – CH2 – CH2 – CH3
|
OH
6
...
O
||
CH3—CH2—CH2—CH2—C—
7
...
8
...
9
...
At anode: Pb( s) + SO2 - ( aq) ¾ ® PbSO4 ( s) + 2e ¾
4
At cathode: PbO2 ( s) + SO2 - ( aq) + 4H+ ( aq) + 2e - ¾ ® PbSO4 ( s) + 2H2 O(l)
¾
4
Overall reactions: Pb( s) + PbO2 ( s) + 2H2 SO4 ( aq) ¾ ® 2PbSO4 ( s) + 2H2 O(l)
¾
OR
At cathode:
MnO- ( aq) + 8H+
4
At anode:
Sn 2 + ¾ ® Sn 4 + ( aq) + 2e - ] ´ 5
¾
Overall reaction:
2MnO- ( aq) + 5Sn 2 + ( aq) + 16H+ ( aq) ¾ ® 2Mn 2+ ( aq) + 5Sn 4+ ( aq) + 8H2 O(l)
¾
4
Eo
Sn 4 + /Sn 2 +
= -E o
Sn 2 + /Sn 4 +
+ 5e
-
¾ ® Mn 2+ ( aq) + 4H2 O(l)] ´ 2
¾
E° = + 1
...
15 V
= -0
...
51 – (– 0
...
66V
As E o is +ve therefore the reaction will take place in forward direction, i
...
, favours the formation of
cell
products
...
(i) Elementary step: Each step of a complex reaction is called an elementary step
...
11
...
A strip of
same metal in pure form is used as cathode
...
When electric current is passed, the metal from the
anode goes into solution as ions due to oxidation while pure metal gets deposited at the
cathode due to reduction of metal ions
...
At anode: M ¾ ® M n + + ne ¾
At cathode: M n + + ne - ¾ ® M
¾
(ii) Vapour phase refining of metals: In this method, crude metal is freed from impurities by first
converting it into volatile compound and collected elsewhere
...
For example refining of nickel by Mond process
...
(i) 2XeF2 + 2H2O ¾ ® 2Xe + 4HF + O2
¾
(ii) 2PH3 +
13
...
(i)
Cl
3° halide reacts faster than 2° halide because of the greater stability of tertiary carbocation
...
CH3
15
...
The four bases present in DNA are adenine (A), guanine (G), cytosine(C) and thymine (T)
...
17
...
Vitamin D
Exposure to sunlight, fish and egg
...
Vitamin E
Vegetable oils like wheat germ oil, Sterility and muscular atrophy
sunflower oil, etc
...
18
...
On heating undergo extensive cross linking in
moulds and become infusible
...
Some common examples are bakelite,
urea-formaldehyde resins, terylene, etc
...
For a first order reaction
[R] o
2
...
0051min–1, [R]o = 0
...
303
0
...
0051 min
Log
0
...
1
= 0
...
1
= Anti log (0
...
503
[R]
[R] =
01
...
03995 M
2
...
04M
Examination Papers | 79
20
...
58 pm
21
...
1 M
Given, E cell = 4
...
80 V
Eo
= 0
...
80 – 0
...
46 V
cell
E cell = E o cell
[Cu 2 + ]
0
...
422 = 0
...
0591
0
...
422 – 0
...
0295 log
– 0
...
0295 log
10 -1
[ Ag + ] 2
10 -1
[ Ag + ] 2
0
...
0295
1
...
2881 = – 1 – 2 log [Ag+]
2 log [Ag +] = 2
...
144
[Ag+] = Antilog 2
...
178 × 10–2 M
- Eo
Cu 2 + /Cu
80 | Xam idea Chemistry–XII
22
...
(ii) The path of light becomes visible due to scattering of light by colloidal particles (Tyndall
effect)
...
23
...
Because of these
interactions the degeneracy of d orbitals of the metal ion is lost and these split into two sets of
orbitals having different energies
...
g
...
g
(ii) Linkage isomerism: The isomers which have same molecular formula but differ in the linkage
of ligand atom to the central metal atom are called linkage isomers, e
...
,
[Co(NH3 ) 5 NO2 ]Cl 2 and [Co(NH3)5ONO]Cl2
Pentaamminenitrito–N–Cobalt (III) chloride, pentaamminenitrito–O–Cobalt(III) chloride
(iii) Ambidentate ligand: A ligand which can bind to the central metal atom through any of the
two donor atoms present in it is called ambidentate ligand, e
...
, NO2 can bind to metal either
O
through nitrogen, i
...
, as nitrito-N (
N
O
) or through oxygen atom, i
...
, as nitrito –O
(¬ O—N=O)
OR
Complex/Ion
Central
metal
ion/atom
Configuration of
metal ion
Hybridisation of
metal ion
involved
[Co(NH3)6]3+
Co3+
d64s0
d2sp3
octahedral
0
Diamagnetic
3+
3+
3
0
d2sp3
octahedral
3
Paramagnetic
3
2
Tetrahedral
0
Diamagnetic
[Cr(NH3)6]
[Ni(CO)4]
24
...
Cr
Ni
3d 4s
3d 4s
3
sp
Geometry
of the
complex
Number
of
unpaired
electrons
Magnetic
behaviour
(i) Ethanol undergoes intermolecular hydrogen bonding due to the presence of a hydrogen
attached to oxygen atom
...
(ii) Phenol is stronger acid than ethanol because the phenoxide ion left after the release of proton is
stabilized by resonance but ethoxide ion is not
...
(iii) Due to –I effect or –R effect of –NO2 group, the resulting phenolate ion is more stable than
phenoxide ion
...
(i) The catalytic activity of transition metals is attributed to the following reasons:
(a) Because of their variable oxidation states transition metals form unstable intermediate
compounds and provide a new path with lower activation energy for the reaction
...
(b) In some cases, the transition metal provides a suitable large surface area with free
vacancies on which reactants are adsorbed
...
(iii) This is due to comparable energies of 5f, 6d and 7s orbital in actinoids
...
(i) Non-ionic detergents: These are the esters of high molecular mass alcohols with fatty acids
...
g
...
(ii) Food preservatives: These are the substances which prevent spoilage of food due to microbial
growth, e
...
, sodium benzoate, potassium metabisulphite, salts of sorbic acid and propanoic
acid, etc
...
g
...
2 to
0
...
28
...
(ii) Van’t Hoff factor (i): It may be defined as the ratio of normal molar mass to the observed
molar mass, i
...
, Van’t Hoff factor
Normal molar mass
i=
Observed molar mass
OR
Observed colligative property
i=
Calculated colligative property
(b)
p =
WB ´ R ´ T
M B ´V
82 | Xam idea Chemistry–XII
MB=
WB ´ R ´ T
p´ V
Here
WB = 100 mg = 100 × 10–3 g
R
= 0
...
3
= 13
...
0821 L atm K -1 mol -1 ´ 298K
13
...
0821 ´ 298 ´ 760 g mol -1
13
...
286 g mol–1
OR
(i) Colligative properties: Those properties which depend on the number of solute particles
irrespective of their nature relative to the total number of particles present in the solution are
called colligative properties of solutions
...
78 atm = 0
...
8 mm Hg
KH = 8
...
42 × 10–7 M/mmHg × 592
...
376 × 10–7
X N 2 = 4
...
99 ´ 10 -4 = 0
...
77 × 10–2 = 2
...
(a) ( i)
O
S
O
O
S
O
OH
HO
O
Peroxodisulphuric acid
(H2S2O8)
H
O
(ii)
Cl
O
O
O
Perchloric acid
(HClO4)
(b) (i) This is because the lone pair of electrons on N atom in NH3 is directed and not delocalised as it
in PH3 due to larger size of P
...
(iii) Due to small size of fluorine atom there are strong interelectric repulsions in the relatively
small size 2p orbitals of fluorine and thus, the incoming electron does not experience much
attraction
...
(ii) PCl5 has trigonal bipyramidal structure in which the three equatorial P—Cl bonds are
equivalent, while the two axial bonds are longer than equational bonds
...
(iii) This is because I—Cl bond has lower bond dissociation enthalpy than Cl—Cl bond
...
(a)
(i) Cannizzaro Reaction:
CH3
CH3
CH3
|
|
|
– +
Conc
...
-
CH3 — CH2 — CHO + 2[Ag(NH3 ) 2 ] + + 3 O H
Tollens reagent
-
¾ ® CH3 — CH2 — CO O+ 2Ag + 4NH3 + 2H2 O
¾
Propanoate ion
Tollens reagent
CH3 – COCH3 ¾ ¾ ¾ ¾ ¾ ® No silver mirror
¾
Propanone
(ii) Benzophenone and Acetophenone:
Iodoform test: Acetophenone being a methyl ketone on treatment with NaOI (I2/NaOH) gives
yellow ppt of iodoform but benzophenone does not
...
)
(iii) Phenol and Benzoic acid:
FeCl3 test: Phenol gives a violet colouration with neutral FeCl3 solution while benzoic acid
gives buff coloured ppt ferric benzoate
...
H SO
CH3 — CH2 — CH2 — CH2 — OH ¾ ¾ ¾2¾ 4 ® CH3 — CH2 — CH = CH2 + H2 O
¾
Butan –1– ol
CBSE (All India) SET–II
1
...
8
...
19
...
87 g cm–3
a = 286
...
65 × 10–10 cm = 2
...
86 | Xam idea Chemistry–XII
NA =
NA =
z´ M
a3 ´ d
, we get
2 ´ 56 g mol –1
( 286
...
87 g cm –3
= 6
...
log
k2
E a æ T2 - T1 ö
ç
÷
=
k1 2
...
303 ´ R ´ T1 ´ T2
k
log 2
T2 – T1
k1
Ea =
2
...
314J mol -1 K -1 ´ 650 K ´ 700 K
2
...
Ea =
19147 ´ 650 ´ 700
...
9 – log 2
...
7(13783 – 0
...
E a = 174237
...
0459 J mol–1 = 182235
...
24 kJ mol–1
...
(i)
+ CH3—C—Cl
COCH3
(ii) C 2 H5 NH2 + C 6 H5 SO2 Cl ¾ ® C 6 H5 SO2 NHC 2 H5
¾
(iii) C 2 H5 — NH2 + HNO2 ¾ ® C 2 H5 — OH + H2 O + N2
¾
28
...
– +
C 6 H5 — COOH+ NaHCO3 ¾ ® C 6 H5 — CO O N a + CO2 + H2 O
¾
Benzoic acid
Sodium benzoate
C 6 H5 — COOC 2 H5 + NaHCO3 ¾ ® No evolution of CO2 gas
¾
Ethyl benzoate
Examination Papers | 87
(b)
Element
Percentage
Atomic mass
No
...
77
12
H
11
...
4) = 18
...
As it does not reduce Tollens reagent and give positive Iodoform test therefore it must be a
methyl ketone
...
(ii) Acid strength: (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH
< CH3CH2CH(Br)COOH
O
OH
||
|
LiAlH 4
Conc
...
(a) (i)
(b)
(ii)
P
H
CH—CH3
O
Br
F
OH
H
H3PO2
F
BrF3: Bent “T”
(i) Due to greater bond strength of P—P single than N—N single bond
...
(iii) Due to very high ionization enthalpy of helium
...
(ii) Due to non-availability of d orbitals in valence shell nitrogen does not form pentahalide
...
CBSE (All India) SET–III
1
1
1
...
(i) Dilute NaCN forms a soluble complex with Ag or Ag2S while the impurities remain unaffected
which are filtered off
...
l
l
14
...
It increase the electrical conductivity of the mixture
...
(ii) Activation energy of a reaction: It may be defined as the extra amount of energy over and
above the average energy of reactants which must be supplied to them to undergo a chemical
reaction
...
Addition Polymerisation
Condensation Polymerisation
(i)
Monomers are unsaturated molecules
(i)
Monomers have di or polyfunctional groups
...
(iii)
Formed by adding monomers to a growing (iii)
chain without loss of any molecules
...
Examples: Polyethene, Polystyrene, Teflon
etc
...
90 | Xam idea Chemistry–XII
21
...
0591
log
n
[ Ni 2 + ] 3
Here n = 6, [Al3+] = 0
...
5M
E o = E o Ni 2 +
cell
/ Ni
-Eo
Al
3+
/ Al
= - 0
...
66V )
E o = 1
...
0591
= 1
...
5) 3
2
= 1
...
0591
10 -6
log
6
0
...
0591
0
...
41 –
log10 –6 + log 2 3
6
6
0
...
0591
= 1
...
41 –
(– 6 + 3 ´ 0
...
0591
0
...
41 ( -5
...
41 +
6
6
(
= 1
...
41 + 0
...
4602V
E cell = 1
...
(i) Ni in its atomic ionic state can not afford two vacant 3d orbitals hence d 2 sp 3 hybridisation is
not possible
...
g
...
Thus dp - Pp
bonding is possible
...
OR
Complex
ion
Central
metal ion
Configuration
of metal ion
Hybridisation
of metal ion
involved
Geometry of Number of
complex ion unpaired
electrons
Magnetic
behaviour
[Ni(CN)4]2
Ni2+
d8
dsp2
Square planar
0
Diamagnetic
[Ni(Cl)4]2–
Ni2+
d8
sp3
Tetrahedral
2
Paramagnetic
[CoF6]3–
Co3+
d6
sp3d2
Octahedral
4
Paramagnetic
Examination Papers | 91
27
...
g
...
(ii) Enzymes: Enzymes are globular proteins with high molecular mass ranging from 15,000 to
1,000,000 g mol–1, and form colloidal solution in water
...
(b) Sweetening agents: These are non-nutritive substances which are used as substitute of sugar
in food and beverages, e
...
, ortho-sulphobenzimide (saccharin), sucrolose, aspartame, etc
...
(a) (i)
O
(ii)
Xe
F
S
O
F
O
S
OH
O
OH
Pyrosulphuric acid
(H2S2O7)
Square planar
(XeF4)
(b)
O
(i) This is because P–P single bond is stronger than the N–N single bond
...
(iii) There is H-bonding in HF molecules due to high electronegativity and small size of
fluorine atom but there is no H-bonding in HCl
...
(ii) This is due to small size and high ionization enthalpy of helium
...
CBSE EXAMINATION PAPERS
FOREIGN–2009
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Foreign) SET–I
1
...
What is an emulsion?
1
3
...
Give the IUPAC name of the following compound:
H3C
CH3
1
H
H
CH3
Br
5
...
1
6
...
1
7
...
Write the name of an antacid which is often used as a medicine
...
Differentiate between molality and molarity of a solution
...
Describe the role of the following:
2
(i) NaCN in the extraction of silver
(ii) CO in the purification of nickel
11
...
With the help of a diagram explain the
reactions occurring during the corrosion of iron kept in open atmosphere
...
2
Examination Papers | 93
12
...
State reasons for the following observations:
2
(i) The enthalpies of atomisation of transition elements are quite high
...
14
...
Explain the mechanism of each of the following processes:
2
(i) Acid catalysed dehydration of an alcohol
(ii) Hydration of ethene to yield ethanol
...
Name two water soluble vitamins, state their sources and the diseases caused due to their deficiency in
diet
...
What are the following substances?
(i) Invert sugar
2
(ii) Polypeptides
18
...
(ii) Synthetic detergents are preferred to soaps in washing machines
...
19
...
Each side of the unit cell has a length of 409 pm
...
)
3
20
...
40V;E o
Fe2 + /Fe
= -0
...
Calculate the freezing point depression for 0
...
If this solution actually freezes at – 0
...
86°C mol–1)
3
22
...
(ii) a beam of light is passed through a colloidal solution
...
3
94 | Xam idea Chemistry–XII
23
...
3+
ion is a reducing agent while Mn
ion is an
(ii) Cu+ ion is not stable in aqueous solutions
...
24
...
Draw molecular structure of
these three isomers and indicate which one of them is chiral
...
Complete the equations for the following reactions:
3
H
(i)
+ HBr
H
H
CH3
+ HI
(ii)
—OH + SOCl2
(iii)
26
...
Differentiate between the modes of formation of an addition polymer and a condensation polymer
...
3
28
...
¾
Pd– BaSO4
Examination Papers | 95
O
C
(ii)
+
...
...
77% carbon, 11
...
The
molecular mass of the compound is 86
...
On
vigorous oxidation it gives a mixture of ethanoic and propanoic acids
...
29
...
)
(b) Explain the following observations:
(i) +3 oxidation state becomes more and more stable from As to Bi in the group
...
(iii) Fluorine does not exhibit any positive oxidation state
...
(ii) Phosphorus has a greater tendency for catenation than nitrogen
...
96 | Xam idea Chemistry–XII
30
...
How is the rate of reaction affected if
the concentration of this reactant is
(i) doubled,
(ii) reduced to half?
(b) A first order reaction has a rate constant of 0
...
If we begin with 0
...
(ii) Elementary step in a reaction
(b) For a decomposition reaction, the values of rate constant k at two different temperature are
given below:
k1 = 2
...
39 × 10–7 L mol–1 s–1 at 700 K
Calculate the value of activation energy (Ea) for this reaction
...
314 J K–1 mol–1)
CBSE (Foreign) SET–II
[Questions different from Set–I]
2
...
Why is it that sulphide ores are concentrated by the ‘froth floatation process’?
1
8
...
Complete the following chemical equations:
(i)
2
C 2 O7 - ( aq) + C 2 O2 - ( aq) + H+ ( aq)
4
2
¾®
¾
(ii) MnO- ( aq) + Fe 2 + ( aq) + H+ ( aq) ¾ ®
¾
4
17
...
Iron has a body-centred cubic unit cell with a cell edge of 286
...
The density of iron is 7
...
Use this information to calculate Avogadro’s number
...
mass of Fe = 56 g mol–1)
3
22
...
3
24
...
No
...
Decomposition of phosphine (PH3) at 120°C proceeds according to the equation:
4PH3(g) ¾ ® P4(g) + 6H2(g)
¾
It is found that this reaction follows the following rate equation:
Examination Papers | 97
Rate = k[PH3]
The half-life of PH3 is 37
...
(i) How much time will be required for 3/4 of PH3 to decompose?
(ii) What fraction of the original amount of PH3 will remain undecomposed after 1 minute?
5
OR
Hydrogen peroxide, H2O2 (aq) decomposes to H2O(l) and O2(g) in a reaction that is of first order in
H2O2 and has a rate constant, k = 1
...
(i) How long will it take 15% of a sample of H2O2 to decompose?
(ii) How long will it take 85% of a sample of H2O2 to decompose?
CBSE (Foreign) SET–III
[Questions different from Set–I and Set–II]
1
...
Write the structure of pent-2-enal
...
State the principles on which the following operations are based:
(i) Zone refining
(ii) Vapour phase refining
...
Draw the structures of the following molecules:
(i) XeF4
(ii) H2S2O7
2
16
...
State their sources and the diseases caused due to their deficiency in
diet
...
State what the following are and how they differ from each other:
(i) a nucleotide, and
(ii) a nucleoside
...
Explain the following giving an example in each case:
(i) Linkage isomerism
(ii) An outer orbital complex
(iii) A bidentate ligand
3
27
...
(ii) Teflon
(iii) Neoprene
(a) Complete the following reaction equations:
(i) SO2 + MnO - + H2O ¾ ®
¾
4
(ii) HgCl2 + PH3 ¾ ®
¾
(b) Explain the following observations:
(i) Sulphur has a greater tendency for catenation than oxygen
...
3
98 | Xam idea Chemistry–XII
(iii) The +5 oxidation state becomes less stable down the group in group 15 of the periodic
table
...
(ii) iron dissolves in HCl to form FeCl2 and not FeCl3
...
Emulsion is a colloidal solution in which both the dispersed phase and dispersion medium are liquids
e
...
milk, cod liver oil, etc
...
8 (Corner atoms) ×
3
...
CH3Br
4
...
O
O
||
||
5
...
Aluminium occurs as Al2O3
...
NH3 < R–NH2 < R3N < R2NH
8
...
9
...
Molality is independent of temperature whereas molarity changes with change in temperature as
volume changes with temperature
...
(i) Dilute NaCN forms a soluble complex with Ag or Ag2S while the impurities remain unaffected
which are filtered off
...
330– 350K
Ni + 4CO ¾ ¾ ¾ ¾ ®
¾
Impure
Ni(CO) 4
Nickel tetracarbonyl
450 - 470K
Ni(CO) 4 ¾ ¾ ¾ ¾ ® Ni + 4CO
¾
Pure
11
...
The H+ ions are produced by
either H2 O or H2 CO3 (formed by dissolution of CO2 in moisture)
2H+ ( aq) + 2e - ¾ ¾ ® 2H
¾
100 | Xam idea Chemistry–XII
1
O ( g) ¾ ¾¾® H2 O
2 2
Net reaction at cathodic area
1
2H+ ( aq) + O2 + 2e - ¾¾® H2 O
E° H + /O /H O = 1
...
67V
¾
2
2H +
The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in
the form of hydrated ferric oxide (Fe 2 O3
...
OR
Molar Conductivity (Lm): It may be defined as the conductance of a solution containing 1 mole of
electrolyte such that the entire solution is placed is between two electrodes one centimeter apart
...
Therefore L m increases a
little as shown in graph by a straight line
...
400
CH3COOH
200
KCl
0
0
...
4
l/2
c / (mol / L)
For weak electrolyte the number of ions as well
as mobility of ions increases on dilution as shown by curve in the figure
...
p
...
(i) No
...
p
...
O
S
(ii)
O
OH
F
O
O
S
O
OH
Pyrosulphuric acid
(H2S2O7)
F
Examination Papers | 101
13
...
(ii) This is because in transition elements incoming electron goes into d-orbitals of inner shell
whereas in main group elements, the incoming electron goes to outermost shell
...
(i) Williamson’s Synthesis:
CH3
CH3
CH3—C—ONa+CH3—Br
CH3—C—O—CH3+NaBr
CH3
CH3
Methyl tert-butyl (ether)
Sodium tert-butoxide
H
H
H
H— C — C — O — H + H
H
H
+
Fast
+
H
H— C — C — O — H
H
H
Ethyl alcohol
H
Ethyl oxonium ion
(ii) Reimer-Tiemann reaction:
– +
– +
ONa
OH
CHCl3+NaOH(aq)
ONa
CHCl2
OH
CHO
Phenol
15
...
H H H
H H
+
H— C — C — O — H
H
Slow
H
Ethyl oxonium ion
+
+
H— C — C
H
H2O
H
Ethyl carbocation
Step II: Formation of protonated alcohols
Step III: Formation of ethylene by elimination of a proton
H H
H
H
+
—
C—C
H— C — C
H
H
H H
Ethyl carbocation
+
+ H
Ethylene
To drive the equilibrium to the right, ethylene is removed as it is formed
...
+
H2O + H
H3O +
H
—
C—C
+
H
+
+
Step II: Nucleophilic attack of water on carbocation
...
B group vitamin and vitamin C are soluble in water
...
Sources
Vitamin C
Citrus fruits and amla
Scurvy
(i) Invert Sugar: Sucrose is dextrorotatory, on hydrolysis in the presence of HCl or enzyme
invertase, it produces a mixture of D–C(+)–glucose and D–(–)–fructose which is laevorotatory
called invert sugar
...
18
...
(ii) This is because detergents can be used in hard water as well as in acidic solutions, as sulphuric
acid and their calcium and magnesium salts are soluble in water but the fatty acids and their
calcium and magnesium salts are insoluble
...
For fcc unit cell
a
r=
2 2
Given a = 409 pm
\
r=
409
2 2
=
409 2
4
r = 144
...
Fe( s) + Cd 2 + ( aq) l
Fe 2 + ( aq) + Cd( s)
E ° cell
log k c = n
0
...
40 – (– 0
...
04V
2 ´ 0 × 04 0 × 08
log kc =
=
0 × 059
0 × 059
log kc = 1
...
3536
kc = 22
...
DTf = [ 273 ×15 - ( - 0 × 320 + 273 ×15)]K = 0 × 320 K
DTf = K f
...
86 K kg mol -1 × 0
...
132 K
Observed value of DTf
0 × 320K
i =
=
Calculated value of DTf 0 ×132K
i = 2 × 42
+ 2nH2 O
104 | Xam idea Chemistry–XII
22
...
(ii) The path of light becomes visible due to scattering of light by colloidal particles (Tyndall
effect)
...
23
...
On the other hand, the change from Mn 3+ to Mn 2+ results in the half-filled d 5
configuration which has extra stability therefore Mn 3+ is oxidising
...
24
...
3+
3+
3+
NH3
NH3
en
H 2N
H 3N
H 2O
en
CO
en
CO
H 2O
H 2O
CO
H 3N
OH2
H 2O
25
...
NaNO2/HCl
273–278K
(i)
Aniline
+
N2Cl
–
N2BF4
HBF4
NO2
NaNO2
Cu, D
Nitrobenzene
Examination Papers | 105
O
O
||
||
(ii) CH3 — CH2 — NH2 + CH3 — C— Cl ¾ ® CH3 — C— NH — CH2 — CH3 + HCl
¾
Ethanamine
Ethanoyl chloride
N – Ethyl ethanamine
(iii) CH3 — CH2 — Cl ¾alc
...
CH3 — NH2 + CHCl 3 + 3KOH ¾ ® CH3 — CH2 — NC + 3KCl + 3H2 O
¾
(alc)
Methyl amine
(1°amine)
Ethyl isocyanide
(Offensive smell)
CHCl 3 /KOH
(CH3 ) 2 NH ¾ ¾ ¾ ¾¾® No reaction
...
OH
NH2
NaNO2/HCl
0–5°C
+
–
b-Naphthol
N = NCl dil NaOH
N = N—
Aniline
Orange dye
(iii) Add Br2(aq), aniline forms white ppt while ethyl amine does not form such ppt
...
Addition Polymerisation
Condensation Polymerisation
(i)
Monomers are unsaturated molecules
(i)
Monomers have di or polyfunctional groups
...
(iii)
Formed by adding monomers to a growing (iii) Monomers combine together with the loss of
chain without loss of any molecules
...
Examples: Polyethene, Polystyrene, Teflon
etc
...
106 | Xam idea Chemistry–XII
(a)
(i)
–+
CH2—CH3
COOK
COOH
KMnO4 – KOH
H3 O
+
D
Benzoic acid
O
–
–
(ii)
HC
– O — C — CH3
O — C — CH
O
CrO3
H3 O
O
+
–
–
D
–
CH3 – C
CHO
3
–
–
CH3
–
O
CH3 – C –
–
–
Benzylideue
diacetate
Benzaldehyde
O
OR
CH3
CHO
CH(OCrOHCl2)2
CS2
+ CrO2Cl2
H3 O
+
Chromium complex
Benzaldehyde
O
–
–
C — Cl
H2
(i)
+
Pd – BaSO4
Benzoyl chloride
HCl
Benzaldehyde
–
–
O
O
+ Ar/R — C — Cl
(ii)
anhyd
...
C
anhyd
...
of moles
Simplest molar
ratio
C
69
...
63
1
11 × 63
= 11 × 63
1
11 × 63
= 10
1 ×16
O
(100 – 81
...
60
16
18 × 60
= 1 ×16
16
1 ×16
=1
1 ×16
Empirical formula of the compound A = C5 H10 O
Molecular formula of the compound A = n (Empirical formula)
Molecular mass of compound A
n=
Empirical formula mass of compound A
Molecular mass of compound A = 86
Empirical formula mass of compound A = 5 ´ 12 + 1´ 10 + 1´ 16
= 60 + 10 + 16
= 86
108 | Xam idea Chemistry–XII
n =
86
=1
86
Molecular formula of the compound A = 1(C5 H10 O)
= C5 H10 O
As the compound A forms addition compound with NaHSO3 therefore it must be either an
aldehyde or ketone
...
As on oxidation the compound A gives a mixture of ethanoic acid and propanoic acid,
therefore compound A is
O
||
CH3 — C— CH2 — CH2 — CH3
Pentan -2-one
The chemical reactions are:
O
OH
||
|
CH3 — C— CH2 — CH2 — CH3 + NaHSO3 ¾ ¾® CH3 — C — CH2 — CH2 — CH3
Pentan -2-one
|
+
SO- N a
3
Sodium hydrogen sulphite addition product
O
||
CH3 — C— CH2 — CH2 — CH3 + 3I 2 + 4NaOH
Pentan - 2 - one
- +
Iodoform
¾ ¾ ¾¾®
reaction
CHI 3 ¯
+ CH3 — CH2 — CH2 — CO O Na + 3NaI + 3H2 O
(Iodoform yellow ppt)
O
||
K 2 Cr2 O7
CH3 — C— CH2 — CH2 — CH3 ¾ ¾ ¾¾® CH3 — COOH + CH3 — CH2 — COOH
Pentan - 2 - one
29
...
)
(b)
(i) This is due to inert pair effect
...
(iii) This is because fluorine is the most electronegative element and does not have d orbitals
in its valence shell
...
(ii) This is because P–P single bond is stronger than N–N single bond
...
30
...
e
...
2
A
1
1
1
(ii) If [A] is reduced to half then rate, r 3 = k é ù = k [ A ] 2 = r1 , i
...
, rate becomes
ê2ú
4
4
4
ë û
times
(b) For a first order reaction
[R] o
2
...
0051min–1, [R]o = 0
...
303
0
...
0051min
Log
0
...
1
= 0
...
1
= Anti log (0
...
503
[R]
[R]=
01
...
03995 M
2
...
04M
OR
(a)
(i) Elementary step: Each step of a complex reaction is called an elementary step
...
(b)
log
k2
E a æ T2 - T1 ö
=
ç
÷
k1 2
...
303 ´ R ´ T1 ´ T2
k
log 2
T2 – T1
k1
Ea =
2
...
314J mol -1 K -1 ´ 650 K ´ 700 K
2
...
Ea =
19147 ´ 650 ´ 700
...
9 – log 2
...
7(13783 – 0
...
E a = 174237
...
0459 J mol–1 = 182235
...
24 kJ mol–1
...
Reverse osmosis: If the pressure applied on the solution is greater than the osmotic pressure then the
solvent molecules start to move from solution into solvent through semipermeable membrane
...
6
...
8
...
g
...
2
13
...
(i) Cationic Detergents: Cationic detergents are quaternary ammonium salts of amines with
acetates, chlorides or bromides as anions
...
Hence, these are called cationic detergents
...
é
CH3
ê
½
êCH (CH ) ¾ N ¾ CH
2 15
3
ê 3
½
ê
CH3
ê
ë
ù
ú
ú
ú
ú
ú
û
+
Br -
Cetyltrimethyl ammonium bromide
Cationic detergents have germicidal properties and are expensive, therefore, these are of
limited use
...
Preservatives prevent the rancidity of food and inhibit growth or kill the microorganisms
...
19
...
87 g cm–3, For bcc, Z = 2, M = 56 g mol–1
a = 286
...
65 × 10–10cm, NA = ?
Z´ M
d=
a 3 ´ NA
7
...
65 ´ 10 -3 cm) 3 ´ NA
2 ´ 56 g mol -1
(2
...
87 g cm -3
2 ´ 56 g mol -1
23
...
87 g cm -3
112 mol -1
185
...
604 × 1024 mol–1
= 6
...
22
...
(ii) Chemisorption is highly specific in nature whereas physisorption is not specific
...
into
unimolecular layer whereas physisorption results into
(iv) Chemisorption is irreversible in nature whereas physisorption is reversible in nature
...
24
...
112 | Xam idea Chemistry–XII
(ii) Orbitals of Fe2+ ion
3d
d2sp3
hybridised orbitals of
4p
4s
Fe2+
d 2sp 3hybrid
3d
[Fe(CN)6]4–
Shape: Octahedral
Magnetic behaviour: Diamagnetic
3d
Six pairs of electrons
from six CN–ions
(iii) Orbitals of Ni2+ ions
3d
sp3
hybridised orbitals of
4p
4s
Ni2+
3d
sp 3hybrid
[Ni(Cl)4]2–
3d
Shape: Tetrahedral
Magnetic behaviour: Paramagnetic
30
...
95
0
...
693
t1/2 =
= 1
...
693 -1
2
...
9
k
[A ]
k=
[ A ] 0 2
...
183 ´ 10 s
...
303 ´ 100 ´ 2 ´ 0
...
64
=
=
s
183
...
t 3/ 4 =
t 3/ 4
2
...
76s
0
...
693 -1
(ii) k =
=
s = 183 ´ 10 –2 s -1
...
9
t = 1 minute = 60 s
[A ]0
2
...
303
183 ´ 10
...
=
= 0
...
303
Four pairs of electrons
from four Cl–ions
Examination Papers | 113
log
[A ]
= -0
...
5232
[A ]0
[A ]
= Anti log 1
...
3336
[A ]0
[A ]
= 0
...
303
(i) t =
log
k
[A ]
Given: k = 1
...
303
-3
-1
log
[ A ] 0 100
=
[A ]
85
100 2303
=
[ 2 log10 - log 85] min
85 1
...
06 ´ 10 min
2303
2303 ´ 0
...
9294] =
= 153
...
06
1
...
4 min
...
06 × 10–3 mm–1,
t=
2
...
06
1
...
8239
=
[ 2 ´ 1 -11761] =
...
06
1
...
CBSE (Foreign) SET–III
1
...
O
||
2
...
(i) Zone refining based on the principle that impurities are more soluble in the melt than in the
solid state of the metal
...
It is then decomposed to give
pure metal
...
114 | Xam idea Chemistry–XII
330 - 350K
Ni + 4CO ¾ ¾ ¾ ¾ ® Ni(CO) 4
¾
impure
Tetracarbonyl nickel (volatile)
450 - 470K
Ni(CO) 4 ¾ ¾ ¾ ¾ ® Ni + 4CO
¾
Pure
b
...
= 4
12
...
of electron pairs around central atom (Xe) =
F
6
F
l
...
= 2
Xe
The shape would be square planar
...
Fat soluble vitamins
Sources
Deficiency diseases
Vitamin A
Xerophthalmia, Night blindness
...
Rickets and osteomalacia
...
Vitamin K
18
...
increased
blood
(i) Nucleotides: The monomeric unit of nucleic acid is called nucleotide
...
(ii) Nucleoside: A nucleoside is the condensation product of purine or pyrimidine base with
pentose sugar
...
Base
O—P—O—H2C
4¢H
H
3¢
OH
H1¢
2¢
H
OH
Nucleotide
(i) Linkage isomerism: The isomers which have same molecular formula but differ in the linkage
of ligand atom to the central metal atom are called linkage isomers, e
...
,
[Co(NH3 ) 5 NO2 ]Cl 2 and [Co(NH3)5ONO]Cl2
Pentaamminenitrito–N–Cobalt (III) chloride pentaamminenitrito–O–Cobalt (III) chloride
Examination Papers | 115
(ii) Outer orbital complex: When ns, np and nd orbitals are involved in hybridisation, outer
orbital complex is formed, e
...
, [CoF6 ] 2- in which cobalt is sp 3 d 2 hybridised
...
g
...
4
27
...
(a) (i) SO2 + 2H2 O
MnO- + 8H+ + 5 e
4
¾ ® SO2 ¾
4
CH2 = C — CH = CH2
|
Cl
+ 4H+ + 2e - ] ´ 5
¾ ® Mn 2 + + 4H2 O] ´ 2
¾
2MnO- + 5SO2 + 2H2 O ¾ ® 2Mn 2 + + 5SO2 - + 4H+
¾
4
4
(ii) 3HgCl 2 + 2PH3 ¾ ® Hg 3 P2 + 6HCl
¾
(b) (i) Sulphur has a greater tendency for catenation than oxygen because S-S bond is stronger than
O – O bond due to less interelectronic repulsions
...
l
high hydration enthalpy of F–
...
OR
(a) ( i) P4 + 3NaOH + 3H2O ¾ ® PH3 + 3NaH2PO2
¾
(ii) 3Cu + 8HNO3 (dilute) ¾ ® 3Cu(NO3)2 + 2NO + 4H2O
¾
(b) (i) Because of small size and high electro negativity of oxygen, molecules of water are highly
associated through H-bonding resulting in its liquid state
...
(iii) Because of its very low solubility in blood it prevents ‘bends’
...
(ii) Question nos
...
(iii) Question nos
...
(iv) Question nos
...
(v) Question nos
...
(vi) Use log tables if necessary, use of calculators is not allowed
...
Write a feature which will distinguish a metallic solid from an ionic solid
...
Define 'order of a reaction'
...
What is an emulsion?
4
...
Give an example of linkage isomerism
...
A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl
...
Draw the structural formula of 1-phenylpropan-1-one molecule
...
Give the IUPAC name of H2N – CH2 – CH2 – CH = CH2
...
Non-ideal solutions exhibit either positive or negative deviations from Raoult's law
...
10
...
How is the rate of this
reaction affected when (i) the concentration of B alone is increased to three times (ii) the
concentrations of A as well as B are doubled?
11
...
0030 mol L–1 s–1
...
10 M to 0
...
Draw the structures of white phosphorus and red phosphorus
...
Explain the following observations:
(i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29)
in the first series of transition elements
...
Examination Papers 117
14
...
Illustrate the following reactions giving a chemical equation for each:
(i) Kolbe's reaction,
(ii) Williamson synthesis
...
How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol,
(ii) Methyl magnesium bromide to 2-methylpropan-2-ol
...
Explain the following terms:
(i) Invert sugar
(ii) Polypeptides
OR
Name the products of hydrolysis of sucrose
...
What are essential and non-essential amino acids in human food? Give one example of each type
...
The well known mineral fluorite is chemically calcium fluoride
...
The F– ions
fill all the tetrahedral holes in the face centred cubic lattice of Ca2+ ions
...
46 × 10–8 cm in length
...
18 g cm–3
...
08 g mol–1)
20
...
25 g of oil of winter green (methyl salicylate) in 99
...
31°C
...
(B
...
of pure Benzene =
80
...
53°C kg mol–1)
21
...
How are associated colloids different from these two types of colloids?
22
...
(ii) Zinc oxide into metallic zinc
...
OR
Describe the role of
(i) NaCN in the extraction of gold from gold ore
...
23
...
118 Xam idea Chemistry—XII
(ii) The E° value for the Mn 3 + / Mn 2 + couple is much more positive than that for Cr 3 + / Cr 2 +
couple or Fe 3 + / Fe 2 + couple
...
24
...
(ii) Which compound in the following couples will react faster in S N 2 displacement and why?
(a)
1-Bromopentane or 2-bromopentane
(b)
1-Bromo-2-methylbutane or 2-bromo-2-methylbutane
...
In the following cases rearrange the compounds as directed:
(i) In an increasing order of basic strength:
C 6 H5 NH2 , C 6 H5 N(CH3 ) 2 , (C 2 H5 ) 2 NH and CH3 NH2
(ii) In a decreasing order of basic strength:
Aniline, p-nitroaniline and p-toluidine
(iii) In an increasing order of pKb values:
C 2 H5 NH2 , C 6 H5 NHCH3 , ( C 2 H5 ) 2 NH and C 6 H5 NH2
26
...
27
...
(a) State Kohlrausch law of independent migration of ions
...
(b) Calculate L°m for acetic acid
...
(b) A copper-silver cell is set up
...
10 M
...
The cell potential when measured was 0
...
Determine the
concentration of silver ions in the cell
...
80V, E° Cu 2 + /Cu = 0
...
(a) Complete the following chemical equations:
(i) NaOH(aq) + Cl 2(g) ¾ ®
¾
(Hot and conc
...
(ii) NF3 is an exothermic compound but NCl3 is endothermic compound
...
OR
(a) Complete the following chemical reaction equations::
(i) P4 + SO2 Cl 2 ¾ ®
¾
(ii) XeF4 + H2 O ¾ ®
¾
(b) Explain the following observations giving appropriate reasons:
(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table
...
(iii) Halogens are strong oxidizing agents
...
(a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone
...
Oxidation of (C) with chromic
acid also produced (B)
...
Write the equations for the
reactions involved
...
CBSE (Delhi) SET–II
Questions Uncommon to Set-I
1
...
What is the oxidation number of phosphorus in H3 PO2 molecule?
5
...
12
...
Describe the shape and magnetic behaviour of following complexes:
(i) [Co(NH3)6] 3+
(ii) [Ni(CN4)] 2- , (At
...
Co = 27, Ni = 28)
15
...
16
...
22
...
This solution has a boiling point of 100
...
What mass of glycerol was
dissolved to make this solution? K b for water = 0
...
24
...
Write the name and structure of the monomer of each of the following polymers:
(i) Neoprene
(ii) Buna-S
(iii) Teflon
CBSE (Delhi) SET–III
Questions Uncommon to Set-I and Set–II
...
Which point defect in crystals of a solid decreases the density of the solid?
2
...
3
...
4
...
5
...
13
...
(ii) Zinc is not regarded as a transition element
...
State clearly what are known as nucleosides and nucleotides
...
The density of copper metal is 8
...
If the radius of copper atom is 127
...
mass of
Cu=63
...
022 × 1023 mol -1 )
Examination Papers 121
20
...
(i) Aerosol
(ii) Emulsion
(iii) Hydrosol
26
...
Give one example of each
...
Explain the following terms with one suitable example in each case
...
Metals are malleable and ductile whereas ionic solids are hard and brittle
...
The order of a reaction can be defined as the sum of the powers of the concentration terms as
expressed in rate law
...
An emulsion is a colloidal dispersion of one liquid in another liquid
...
4
...
l
262 K
2NO2
Paramagnetic
N2 O4
Colourless and diamagnetic
5
...
–
–
6
...
CH3 — CH2 — C —
8
...
9
...
These deviation are caused when solute-solvent molecular interactions ( A - B) are either weak or
stronger than solvent-solvent ( A - A) or solute-solute ( B - B) molecular interactions
...
For example a mixture of ethanol and acetone
...
For example a mixture of chloroform and acetone
...
According to the given reaction:
Rate = k [A] [B]2
(i) When the concentration of reactant ‘B’ is increased three times the rate of reaction becomes 9
times
...
e
...
i
...
, Rate = k [2A]1 [2B]2 = 8k[A] [B]2
11
...
0030 mol L–1s–1,
[R]0 = 0
...
075 M
We know that
[R] = – kt + [R]0
0
...
0030 t + 0
...
33 second
12
...
or
Red phosphorus is less reactive due to its polymeric structure
...
(ii) Red phosphorus
(i) The density of elements from titanium to copper increase in the first series of transition
elements
...
(ii) Many transition metals and their components show catalytic properties
...
14
...
(i) Kolbe’s reaction
OH
ONa
OH
½
½
½
NaOH
400 K
4-7 atm
...
ethoxide
+
ICH3
Methyl iodide
¾ ¾¾® C 2 H5 OCH3 + NaI
Methyl ethyl ether
(i)
16
...
(i) Invert Sugar: The dextrorotatory sucrose when hydrolysed by boiling with mineral acid
produces an equal number of molecules of dextrorotatory fructose
...
C12 H22 OH
+
H2O
H+
¾ ¾ ¾®
Sucrose
C6H12O6
+
C6H12O6
Glucose
Fructose
[a]D = 52
...
5°
[a]D = – 93°
Invert sugar
(ii) Polypeptides: Polypeptides are polymers formed by condensation of more than ten amino
acids
...
Polypeptides are amphoteric
because of the presence of terminal-ammonium and carboxylate ions as well as the ionized
side chains of amino acid residues
OR
Sucrose gives glucose and fructose on hydrolysis
...
124 Xam idea Chemistry—XII
18
...
For example: valine, leucine
...
For example: Glycine, Alanine
...
Given:
\
d = 3
...
46 × 10–8 cm
M = 78
...
08
NA =
(5
...
NA =
313 × 32 ´ 10 24
517 × 61
NA = 6
...
Given DTB = 80
...
10 = 0
...
21 K
WB = 1
...
53°C kg mol–1
MB = ?
WA = 99g
K ´ WB
MB = b
´ 1000
DTb ´ WA
=
\
2 × 53 ´ 1× 25 ´ 1000 3162
...
79
MB = 152 g/mol
21
...
g
...
Macromolecular colloids: In this type of colloids, colloidal particles are themselves large molecules
of colloidal dimensions, e
...
, starch, proteins, polyethene, etc
...
Such colloids are known as associated colloids, e
...
, soaps and detergents
...
(i) Pig iron into steel: Pig iron is converted into steel by heating in a converter
...
Oxygen reacts with impurities and
raised the temperature to 2173K
...
Oxides of silicon and Mg form slag
...
(ii) Metallic zinc can be obtained from zinc oxide
...
Examination Papers 125
ZnO
+
C
D
¾ ®
¾
Zn
+ CO
The oxide is then made into brickettes with coke and clay and heated by producer gas in
vertical retorts at 1673 K, zinc, boiling point is 1183 K, distills off and is collected by rapid
chilling
...
Ti (impure)
+
2 I2
¾ ¾ ® TiI4
¾
D
¾ ¾ ® Ti (pure)
¾
+
I2
OR
(i) Role of NaCN in the extraction of gold is to do the leaching of gold ore in the presence of air
from which the gold is obtained later by replacement
...
FeO
+
SiO2
¾¾®
FeSiO3 (Slag)
(iii) Iodine is heated with impure Zr to form volatile compound which on further heating
decomposes to give pure zirconium
...
D
¾¾ ®
¾
ZrI4
D
¾¾ ®
¾
Zr (pure) +
2I2
(i) This is due to lanthanide contraction
...
(ii) The E° value for the Mn3+/Mn2+ couple is much positive than Cr3+/Cr2+ couple or Fe3+/Fe2+
couple because Mn3+ ion receiving an electron gets d-subshell half-filled which is highly
stable
...
(iii) This is because fluorine and oxygen are highly electronegative elements and have small size
...
(i) DDT: It is used as insecticide to control flies, mosquitoes, etc
...
(ii) 1-Bromopentane, as it is a primary alkyl halide
...
25
...
(i) Polythene, PVC
...
(iii) Buna-S, Buna-N
...
Analgesics: They are the drugs used for relieving pain
...
(b) Non-narcotics: Non-narcotic drugs are effective in relieving skeletal pain, reducing fever,
preventing heart attack
...
(a) Kohlrausch Law: “The molar conductance of an electrolyte at infinite dilution is equal to the
sum of the molar conductances of the two ions, i
...
, the cation and the anion
...
(1)
Lo
m
...
(3)
Subtracting (2) from the sum of (1) and (3), we get:
Lo ( CH3 COO- ) + Lo ( Na + ) + Lo ( H+ ) + Lo ( Cl - ) - Lo (( Na + ) + Lo Cl - )
m
m
m
m
m
m
= Lo (CH3 COO- ) + Lo (H+ ) = Lo (CH3 COOH)
m
m
m
\ Lo ( CH3 COOH) = 91
...
0 S cm2 mol–1
OR
(a) When lead battery operates, the following cell reactions occur:
Anode half cell reaction:
Pb (s)
+
SO2- (aq)
4
¾ ®
¾
PbSO4 (s)
+
2e–
Cathode half cell reaction:
PbO2 (s) + 4 H+ (aq) + SO2- + 2e– ¾ ®
¾
4
PbSO4 (s) + 2H2O (l)
Net reaction, Pb(s) + PbO2 + 2H2SO4 (aq) ¾ ® 2PbSO4 (s) + 2H2O (l)
¾
Examination Papers 127
EoAg+/Ag = 0
...
34V
[Cu2+] = 0
...
80 – 0
...
46V
From Nernst Equation:
Ecell = Eocell -
[ Cu 2 + ]
0
...
422 = 0
...
[Ag + ] 2
[ 010]
...
591
log
2
[Ag + ] 2
= 1
...
0051
[Ag+] = 7
...
(i) 6 NaOH
+
3Cl2
(ii) XeF6 +
3H2O
¾¾®
or
XeO3
XeF6 +
H2O
¾¾®
XeOF4 + 2HF
¾¾®
5NaCl + NaClO3 + 3H2O
+ 6HF
(i) Due to smaller size of oxygen the electron cloud is distributed over a small region of space,
making electron density high which repels the incoming electrons
...
(iii) ClF3 molecule has a T-shaped structure
...
OR
(a) Complete the following chemical reaction equations:
(i) P4 + 8SO2Cl2
¾ ®
¾
4PCl3 + 4SO2 + 2S2Cl2
(ii) 6XeF4 + 12H2O ¾ ® 4Xe + 2XeO3 + 24HF + 3O2
¾
(b) (i)
The stability of +5 oxidation state decreases down the group in group 15 of the periodic
table
...
This is because of inert pair effect
...
(iii) Halogens are strong oxidising agents because they have high electron affinities, so, they
pick up electrons from other substances
...
(a) Mechanism of nucleophilic addition reactions:
A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a
direction approximately perpendicular to the plane of sp2 hybridized orbitals of carbonyl
carbon
...
The intermediate captures a proton from the medium to give the
neutral product
...
O
O
||
||
dil
...
H SO
2
4
¾¾¾¾ ®
¾
CH3 CH2 CH2 COOH + CH3 CH2 CH2 CH2 OH
Butanoic acid
(B)
chromic acid
(ii) CH3 CH2 CH2 CH2 OH
¾¾¾¾ ®
¾
oxidation
Butanol
(C)
Butanol
(C)
CH3 CH2 CH2 COOH
Butanoic acid
(B)
(iii) CH3 — CH2 CH2 CH2 OH ¾dehydration ®
¾¾¾
¾
CH3 — CH2 — CH
Butanol
CH2
But-1-ene
OR
(a) Chemical tests to distinguish between ethanal and propanal
Iodoform Test: Ethanol gives iodoform test positive while propanal does not give this test
...
Benzoic acid does not give violet colour
with FeCl3
...
C
...
NaOH
D
CH3 CHO ¾ ¾ ¾ ¾® CH3 — CH— CH2 — CHO ¾ ® CH3 — CH
¾
-H2O
aldol
(iii) Propanone to propane
CH3
C
O + 4[H]
CH3
Zn–Hg/HCl(conc
...
Frankel defects do not change the density of the solids
...
Oxidation number of phosphorous in H3PO2 is +1
...
Example of coordination isomerism
...
(i) Structure of BrF3
F
F
Br
F
Bent T-shaped
(ii) Structure of XeF4 molecule
F
F
Xe
F
F
Square planar
CH — CHO
But - 2-en - al
+ H2O
(i) \
Co 3+
l
14
...
(ii) [Ni(CN)4] 2-
l
l
(atom number 28)
l
Ni
l
The electronic configuration of the nickel atom and Ni 2+ ion are depicted below:
3d2
4s2
4p0
Since coordination number in the complex is 4
...
l
l
l
\ Ni 2+ (after rearrangement)
l
However, experiments show that the complex is diamagnetic
...
15
...
) NaOH ê
¾¾¾¾¾® ê
340 K
ê
ê
ê
ê
ë
ONa
½
ù
ú
CHCl 2ú
ú NaOH
ú ¾¾®
ú
ú
ú
ú
û
ONa
½
OH
CHO
½
H 3O
CHO
+
¾®
Salicyladehyde
(2-Hydroxy benzaldehyde)
(ii) Friedel - Craft Reactions
Acylation:
O
||
C— CH3
O
||
Anhyd
...
AlCl
3
+ CH3 Cl ¾ ¾ ¾ ¾¾®
Methyl
chloride
Benzene
16
...
H 2 SO4
CH2 ¾ ¾ ¾ ¾ ® CH3 — CH — CH3
¾
Propene
(ii)
+ HCl
Propane-2-ol
C 2 H5 MgCl
Ethyl magnesium chloride
( i ) HCHO
¾ ¾ ¾ ¾® CH3 — CH2 — CH2 OH
(ii) H +
Propan -1-ol
MB = 92 g mol -1
22
...
42 – 100 = 0
...
42 K
WB = ?
\
Kb = 0
...
42 =
\
0
...
42 ´ 92 ´ 500
WB =
0
...
73 g
512
+
24
...
)
1° amine
¾¾¾®
R ¾ NC
Alkyl isocyanide
+ 3KCl + 3H2 O
132 Xam idea Chemistry—XII
25
...
Cl
|
CH2
CH — C CH2
chloroprene
(ii) The monomer of Buna-S is Buta-1,3-diene and styrene
...
F
|
C— F
Tetrafluoroethene
CBSE (Delhi) SET–III
1
...
2
...
3
...
4
...
8°
pm
O
7
...
Example of ionisation isomerism is –
[Pt(NH3)4Cl2]Br2
13
...
When visible (white) light falls on a compound, it absorbs certain
radiations of white light and transmit the remaining ones
...
(ii) Because the atoms or simple ions of zinc never have partially filled d orbitals
...
(i) Nucleoside: A nucleoside is the condensation product of purine or pyrimidine base with
pentose sugar
...
When a nucleoside is
linked to phosphoric acid at 5’ position of sugar moiety, we get nucleotide
...
Given,
Radius (r) = 127
...
723 ´ 10 -23 cm 3
\
Atomic mass of Cu = 63
...
022 × 1023 mol -1
Density = 8
...
95 =
Z ´ 63
...
023 ´ 10 23 ´ 4
...
95 ´ 6
...
723
Z=
Þ Z = 4
...
54
Thus copper has fcc structure
...
Aerosol
Emulsion
Hydrosol
1
...
A colloidal system in which 1
...
liquids
...
Milk, cod liver oil
...
26
...
Thermoplastics
Thermosetting plastics
(i)
These polymer are linear or slightly (i)
branched chain molecules
These polymers are cross linked or heavily
branched molecules
(ii)
Soften on heating and harden on cooling (ii)
and can be remoulded
...
(iii) Some common examples are polyethene, (iii) Some common examples are bakelite,
PVC, polystyrene, etc
...
27
...
For example: Cetyltrimethyl-ammonium chlorides
...
For example,
invertase, zymase, pepsin, trypsin, etc
...
For example, norethindrone, novestrol, etc
...
(ii) Question nos
...
(iii) Question nos
...
(iv) Question nos
...
(v) Question nos
...
(vi) Use log tables if necessary, use of calculators is not allowed
...
What type of interactions hold the molecules together in a polar molecular solid?
2
...
Fluorine does not exhibit any positive oxidation state
...
Give the IUPAC name of the following compound:
CH3
H3C
H
H
CH3 Br
5
...
What is Tollen’s reagent? Write one usefulness of this reagent
...
What is meant by ‘reducing sugars’?
8
...
Define the terms, ‘osmosis’ and ‘osmotic pressure’
...
10
...
How is the conductivity of a solution related to its molar conductivity?
11
...
93 V, Ag + / Ag = 0
...
34 V, Mg 2+ / Mg = -2
...
74V,
Fe 2+ / Fe = - 0
...
Arrange these metals in an increasing order of their reducing power
...
51 V
E° = + 0
...
12
...
What is meant by coagulation of a colloidal solution? Name any method by which coagulation of
lyophobic sols can be carried out
...
Complete the following chemical reaction equations:
(i) I2 + HNO3 ¾ ®
¾
(ii) HgCl2 + PH3 ¾ ®
¾
(conc
...
Draw the structural formulae of the following compounds:
(i) H4P2O5
(ii) XeF4
16
...
Identify A and B in each of the following processes:
(i) CH3CH2Cl ¾NaCN ® A ¾Reduction ® B
¾¾
¾
¾¾¾
¾
Ni/H 2
NaNO2 / HCl
C H NH
6 5
2
(ii) C6H5NH2 ¾ ¾¾¾¾® A ¾ ¾¾¾ ® B
¾
OH –
18
...
The density of copper metal is 8
...
If the radius of copper atom be 127
...
54 g mol–1 and NA = 6
...
What mass of NaCl (molar mass = 58
...
5°C? The freezing point depression constant, Kf , for water is 1
...
Assume van’t Hoff factor for NaCl is 1
...
21
...
Explain the following cases giving appropriate reasons:
(i) Nickel does not form low spin octahedral complexes
...
(iii) Co2+ is easily oxidised to Co3+ in the presence of a strong ligand
...
How would you differentiate between SN1 and SN2 mechanisms of substitution reactions? Give one
example of each
...
How would you convert the following:
(i) Phenol to benzoquinone
(ii) Propanone to 2-methylpropan-2-ol
(iii) Propene to propan-2-ol
25
...
(ii) XeF2 is a linear molecule without a bend
...
26
...
How does this happen? What are essential and
non-essential amino acids? Name one of each type
...
Explain the following terms with one example in each case:
(i) Food preservatives
(ii) Enzymes
(iii) Detergents
28
...
9 s at 120°C
...
Calculate the energy of activation of reaction, assuming that it does not change with
temperature
...
314 J K–1 mol–1)
29
...
2
4
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high
...
OR
(a) Complete the following chemical equations:
(i) MnO– ( aq) + S 2 O2– ( aq) + H2 O(l) ¾ ®
¾
4
3
2–
(ii) Cr2 O7 ( aq) + Fe 2+ ( aq) + H+ ( aq) ¾ ®
¾
(b) Explain the following observations:
(i) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions
...
(iii) Cu+ ion is not known in aqueous solutions
...
(a) Illustrate the following name reactions giving a chemical equation in each case:
(i) Clemmensen reaction
(ii) Cannizzaro’s reaction
(b) Describe how the following conversions can be brought about:
(i) Cyclohexanol to cyclohexan-1-one
(ii) Ethylbenzene to benzoic acid
(iii) Bromobenzene to benzoic acid
OR
(a) Illustrate the following name reactions:
(i) Hell–Volhard–Zelinsky reaction
(ii) Wolff–Kishner reduction reaction
(b) How are the following conversions carried out:
(i) Ethylcyanide to ethanoic acid
(ii) Butan-1-ol to butanoic acid
(iii) Methylbenzene to benzoic acid
Write chemical equations for the involved reactions
...
What type of semiconductor is obtained when silicon is doped with arsenic?
2
...
Why?
6
...
What is meant by ‘copolymerisation’?
13
...
Complete the following chemical reaction equations:
(i)
NaOH + Cl2 ¾ ®
¾
(cold and dilute)
(ii) XeF6
+
H2O ¾ ®
¾
(excess)
17
...
Draw the structures of the monomers of the following polymers:
(i) Bakelite
(ii) Nylon-6
19
...
Each side of the unit cell has a length of 409 pm
...
What mass of ethylene glycol (molar mass = 62
...
50 kg of water to lower
the freezing point of water from 0°C to – 10
...
86 K kg mol–1)
27
...
Write a distinguishing feature of metallic solids?
3
...
8
...
Silver crystallises in fcc lattice
...
07 × 10–8 cm and the density of
the crystal is 10
...
(NA = 6
...
15 g of an unknown molecular substance was dissolved in 450 g of water
...
34°C
...
86 K kg mol–1)
...
How would you account for the following:
(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur
...
(iii) Fluorine never acts as the central atom in polyatomic interhalogen compounds
...
Write the name, the state of hybridization, the shape and the magnetic behaviour of the following
complexes:
[CoCl4]2–, [Ni(CN)4]2–, [Cr(H2O)2(C2O4)2]–
(Atomic number: Co = 27, Ni = 28, Cr = 24)
26
...
What is meant by the denaturation of a
protein?
27
...
The interactions hold the molecules together in a polar molecular solid is dipole-dipole attractions
...
When concentration of an electrolyte approaches zero, the molar conductivity is known as limiting
molar conductivity (Lo )
...
Fluorine does not exhibit any positive oxidation state because it is the most electronegative element in
the periodic table
...
4–Bromo–3–methyl–pent–2–ene
5
...
Tollen’s reagent – A solution of AgNO3 dissolved in NH4OH is known as Tollen’s reagent
...
For example:
RCHO + 2Ag (NH3)2 OH ¾ ® RCOONH4 + 2Ag¯ + H2O + 3NH3
...
Reducing Sugars: The carbohydrates which reduces Fehling’s Reagent and Tollen’s Reagent are
referred to as reducing sugar
...
8
...
9
...
140 Xam idea Chemistry—XII
Osmotic Pressure: The hydrostatic pressure exerted by the column of the solution which is just
sufficient to prevent the osmosis is called osmotic pressure
...
Hence, results are obtained in a very short time
...
(iii) This is ideal method for proteins to find their molar mass since it happens at room temperature
and no coagulation occurs
...
Conductivity (K) =
× Cell constant
Resistance ( R)
Molar Conductivity ( L m ) =
Conductivity ( k) ´ 1000
Molarity ( M )
OR
Lm =
K
=
K ´1000 L m ´ M
=
M
1000
138
...
5 mol L–1
3
1000 cm L
–1
= 0
...
Increasing order of reducing power:
Ag < Cu < Fe < Cr < Mg < K
OR
(i) MnO- + 8H+ + Se ¾ ® Mn2+ + 4H2O, DE° = + 1
...
15V
¾
Multiplying equation (i) by 2 and (ii) by 5 we get:
Sn2+ + 2MnO- + 16H+ ¾ ® 5 Sn4+ + 8H2O + 2Mn2+
¾
4
E° = E° - E°
cell
R
L
= 1
...
15 = 1
...
E°
cell
is positive, therefore it will favour product formation
...
This phenomenon is called Tyndall effect
...
Zeolites are good shape-selective catalysts because of their honey comb-like structures
...
ZSM -5
x CH3 — OH ¾ ¾¾ ® (CH2 ) x + xH2 O
¾
Gasoline
Examination Papers 141
13
...
If the charge is removed, they come together and are precipitated
...
The coagulation of a lyophobic sols can be done by mixing of two opposite charged sols or by
addition of electrolyte
...
(i) I2 + 10HNO3 ¾ ® 2HIO3 + 10NO2 + 4H2O
¾
(ii) 3HgCl2 + 2PH3 ¾ ® Hg3P2 + 6HCl
¾
15
...
(i) To distinguish Ethylamine and aniline
...
HCl at 273 - 278K
followed by NaOH and b-naphthol
No reaction
C6H5NH2
Yellow orange dye is obtained
...
Aniline
No evolution
of N2 gas
Benzyl amine
N2 gas is evolved
CH2NH2
CH2OH
(i) HONO
(ii) H2O
17
...
OH –
p-Amino-azobenzene
(B)
(i) PVC — structure of monomer: CH2 = CH – Cl (vinylchloride)
(ii) Teflon — structure of monomer:
F
F
C
C
F
F
(Tetrafluoroethylene)
19
...
8 pm
For fcc a = 2 2 r
a 3 = ( 2 2r ) 3 = 4
...
54 g mol -1
NA = 6
...
95 g cm -3
Z´M
d=
NA ´ a3
Þ
\
Z ´ 63
...
95 =
6
...
723 ´ 10 -23
8
...
022 ´ 4
...
00
63
...
20
...
5 g mol–1
WA = 65g,
WB = ?
DTf = 7
...
5 K,
Kf = 1
...
87
We know that,
DTf = i Kf ×
WB 1000
´
M B WA
WB 1000
´
58
...
5 ´ 58
...
87 ´ 1
...
75
WB =
3478
...
199 g
...
5 = 1
...
86 ×
\
NH2
Examination Papers 143
21
...
Argentite is leached with dilute aqueous
solution of NaCN in the presence of air
Ag2S + 4NaCN ¾ ® 2 Na [Ag (CN)2] + Na2S
...
The crude metal is heated with iodine in an evacuated vessel to form volatile compound
...
The pure metal deposit
on the filament
...
OR
(i) Froth floatation method: This method is used to remove gangue from sulphide ores
...
A rotating paddle agitates the mixture and draws air in it
...
(ii) Electro refining of metals: In this method, the impure metal is made to act as anode
...
They are put in a suitable electrolyte containing
soluble salt of the same metal
...
(iii) Zone refining: This method is based on the principle that the impurities are more soluble in
the melt than in the solid state of the metal
...
The molten zone moves along with the heater
...
22
...
(ii) The p-complexes are known for the transition metals only because they have ‘sandwich’
structure in which the metal ion lies between two planar C5H5 rings
...
So that all M–C
bonds are identical for their stability
...
144 Xam idea Chemistry—XII
SN1 Reactions: They are substitution nucleophilic unimolecular reactions in which the rate of
the reaction depends on the concentration of only one reactant
...
For example:
23
...
For example:
OH -
R — Cl ¾ ¾ ® R — OH + Cl ¾
Rate µ [R – Cl][OH– ]
The product formation takes place by formation of transition state
...
OH
O
K2Cr2O7/H2SO4
O
Phenol
p-Benzoquinone
(ii)
CH 3
C
OH
OMgCl
O
CH 3
CH 3MgCl
CH 3
Propanone
C
CH 3
H 2O/H
+
CH 3
C
CH 3
CH 3
CH 3
2-Methyl propan-2-ol
OH
(iii)
CH3— CH
Propene
CH2
dil H2SO 4
CH 3—CH—CH 3
Propan-2-ol
(i) This is because of (a) bond dissociation enthalpy of F2 is lower than Cl2 and (b) fluorine forms
stronger bond with nitrogen due to comparable size
...
(ii) As xenon is sp3 hybridised with three lone pair of electrons
...
26
...
O
l
Amino acids which contain two —C—O—H group and one —NH2 group are called acidic amino
acid, e
...
, aspartic acid
...
Amino acids which contain two —NH2 group and one —C—O—H group are called basic amino
acids, e
...
, lysine
...
g
...
Amino acids that cannot be synthesized by the body and must be supplied in the diet are called
essential amino acids, e
...
, lysine, valine, leucine, etc
...
g
...
(i) Food preservatives: Chemical substance which are used to prevent food spoilage due to
microbial growth are called preservative
...
(ii) Enzymes: Enzymes are the essential biological catalysts which catalyse specific biological
reactions at a very high rate under mild conditions of temperature and pH
...
(iii) Detergents: They are very similar to the salts of fatty acids found in soap but they are
manufactured from materials other than animal fats
...
28
...
For a hypothetical reaction R ® P
– D[R]
+D [P]
Rate =
Rate =
Dt
Dt
(ii) Activation energy: The amount of energy which the reactants must absorb to pass over the
energy barrier between reactants and products is known as the activation energy
...
303
(b) (i) We know that k =
log
t
[A ]t
0
...
693
k=
\
t1/ 2
\
\
\
0
...
9
k = 0
...
303
k=
log
t
[A]
2
...
0183
14
k=
=
2 ´ 2
...
303
log 2 2 =
log 2
0
...
0183
146 Xam idea Chemistry—XII
2 ´ 2
...
3010
0
...
8 s
...
303
t=
log
k
[A]
[A] o
2
...
0183
[A]
[A] o
60 ´ 0
...
303
[A]
t=
Taking antilog,
0
...
997 =
[A]
[A]
1
=
= 0
...
333%
[A] o 2
...
3%
Antilog (0
...
(ii) Molecularity: It may be defined as the number of reacting species (molecules, atoms or
ions) taking part in an elementary reaction, which must collide simultaneously in order to
bring about a chemical reaction
...
134 JK–1 mol–1
Ea = ?
é TT ù
k
We know that, Ea = 2
...
303 × 8
...
15 [4800] × 0
...
32 kJ mol–1
...
(a)
(i) Cr2O72– + 3H2S + 8H+ ¾ ® 2Cr3+ + 3S + 7H2O
¾
(ii) 2Cu2+ + 4I
–
¾ ® Cu2I2 + I2
¾
Examination Papers 147
(b)
(i) The oxidising power of oxonions are in the order VO2+ < Cr2 O72– < MnO4–, this is due
to increase in the oxidation state of the metal ion
...
(iii) Cr2+ stronger reducing agent than Fe2+ as its configuration changes from d4 to d3, a more
stable half filled t2g configuration
...
¾
(b)
(i) La3+ and Lu3+ ions do not show any colour in solution because they do not contain any
unpaired electrons
...
(iii) Cu+ ion is not known in aqueous solutions because Cu2+ ions are more stable due to more
negative DHhyd of Cu2+ than Cu+, which compensates for the second ionisation enthalpy
of Cu
...
(a)
(i) Clemmensen reaction
O
CH 3—C—CH 3
Zn–Hg/HCl (conc
...
)
CrO3
or PCC
(i)
Cyclohexanol
Cyclohexanone
C 2H 5
COOH
KMnO4/OH
–
(ii)
Ethyl benzene
Benzoic acid
Br
CuCN
Py,473K
(iii)
Bromobenzene
COOH
CN
H2O/H +
Benzoic acid
+ CH3OH
Methanol
148 Xam idea Chemistry—XII
OR
(a) (i)
Hell-Volhard-Zelinsky reaction
(i) X 2 /red P
R—CH2—COOH ¾ ¾¾¾¾® R—CH—COOH
(ii) H 2 O
Carboxylic acid
(ii)
( X == Cl, Br)
|
X
a -Halocarboxylic acid
Wolf-Kishner reduction
(i) NH - NH
2
2
CH3 COCH 3 ¾ ¾¾¾¾¾¾¾® CH3 CH2 CH3
¾
(ii) KOH, Ethylene glycol, D
Propanone
(b) (i)
H+
Propane
NaOH /CaO
Cl /hv
2
C 2 H5 CN ¾ ¾ ® C 2 H5 COOH ¾ ¾¾¾¾® C 2 H6 ¾ ¾¾ ® C 2 H5 Cl
¾
¾
D
Ethyl cyanide
H O/H +
alc
...
n-type semiconductor
...
Due to presence of triple bond whose dissociation energy is very high
...
They are the simplest carbohydrates which do not undergo further hydrolysis, for example: glucose
...
(i) Peptization: The process of converting a precipitate into colloidal sol by shaking it with
dispersion medium in the presence of a suitable electrolyte
...
15
...
(i) To distinguish CH3 NH2 and CH3– NH – CH3
Reaction with HONO
CH3 NH2 gives methyl alcohol and N2 gas
...
(CH3 ) 2 NH + HONO ¾ ®
¾
(CH3 ) 2 N – N = O + H2 O
N-Nitroso
dimethyl amine
(yellow oil)
(iii) Aniline and N-methyl aniline
Test: Carbyl amine test:
Aniline on warming with chloroform and KOH gives offensive smell of carbylamine while
N-methyl aniline does not
...
(i) Bakelite
Monomers: Phenol and formaldehyde
Structure:
OH
O
and
Phenol
(ii) Nylon–6
Monomers: Caprolactum
H
½
N
H2 C
½
H2 C
C=O
½
CH2
H2 C ¾ CH2
Caprolactum
19
...
414 2
...
62 pm
–1
20
...
0 g mol , WA = 5
...
86 K kg mol–1, D Tf = 10 K
H—C—H
Formaldehyde
150 Xam idea Chemistry—XII
We know that,
D Tf = Kf ×
WB 1000
´
M B WA
WB 1000
´
62 5500
62 ´ 5500 ´ 10
WB =
1000 ´ 186
...
3 g
\
1860
27
...
Analgesics are classified into two categories:
10 = 1
...
(b) Non-narcotics: Non-narcotic drugs are effective in relieving skeletal pain, reducing fever,
preventing heart attack
...
Metallic solids have high melting and boiling points
...
Molality of a solution does not change on changing the temperature of the solution while molarity
does change
...
C12 H22 O11 + H2 O ¾H¾® C 6 H12 O6 + C 6 H12 O6
¾
(Glucose)
(Fructose)
19
...
07 × 10–8 cm, d = 10
...
02 × 1023
We know that,
d=
\
20
...
5 ´ 6
...
07 ´ 10 -8 ) 3
4
M = 106
...
34)]K = 0
...
86 K kg mol–1
W ´ 1000
MB = Kf × B
DTf ´ WA
= 1
...
3 g mol–1
0
...
(i) This is due to smaller size of oxygen the electron cloud is distributed over a small region of
space, making electron density high which repels the incoming electrons
...
(iii) Fluorine never acts as the central atom in polyatomic inter-halogen compounds since it is the
most electronegative element of the group
...
(i) Tetrachloridocobaltate (II) ion
sp3 hybridisation
Tetrahedral
Paramagnetic
(ii) Tetracyanonickelate (II) ion
dsp2 hybridisation
Square plannar
Diamagnetic
...
26
...
Consist of linear thread-like molecules 1
...
like structure
...
Insoluble in water
...
Soluble in water
...
Keratin in hair, fibroin in silk, etc
...
Albumin in eggs, insulin, etc
...
Due to this, globules unfold and helix get uncoiled and protein loses its biological
activity
...
During denaturation 2° and 3° structures are destroyed but
1° structures remain intact, e
...
, coagulation of egg white on boiling, curdling of milk, etc
...
(i) Antibiotics
The chemical substances produced from some micro-organism and are used to inhibit the
growth of other microorganism or even kill them are called antibiotics
...
(ii) Antiseptics
Antiseptics are the chemicals which prevent or destroy the growth of the harmful microorganism
...
(iii) Analgesics
These are the substances which are used to get relief from pain
...
CBSE EXAMINATION PAPERS
FOREIGN–2010
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Foreign) SET–I
1
...
Identify the order of reaction from the following unit for its rate constant:
L mol–1s–1
3
...
4
...
What is an ambidentate ligand? Give an example
...
Draw the structure of the following compound:
4-Bromo-3-methylpent-2-ene
7
...
What happens when glucose is treated with bromine water?
9
...
How is the overall
reaction represented?
10
...
11
...
0030 mol L–1 s–1
...
10 M to 0
...
Draw the structures of the following molecules:
(i) BrF3
13
...
Describe the state of hybridization, the shape and the magnetic behaviour of the following complexes:
(i) [Cr(H2O)2(C2O4)2]–
(ii) [Co(NH3)2(en)2]3+, (en = ethane-1,2-diamine)
(At
...
15
...
16
...
17
...
(ii) Why do primary amines have higher boiling points than the tertiary amines?
Complete the following reaction equations:
(i) C6H5NH2 + CHCl 3 + KOH (alc) ¾ ®
¾
(ii) C6H5N2 Cl + H3PO2 + H2O ¾ ®
¾
19
...
Iron has a body-centred cubic unit cell with a cell edge of 286
...
The density of iron is
7
...
Use this information to calculate Avogadro’s number
...
mass of iron = 56 g mol–1)
One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution
of unknown concentration
...
10 M solution
of Zn (NO3) 2
...
48 V is measured for this cell
...
°
(Given E ° 2 + /Zn = – 0
...
80 V)
Zn
21
...
What is the difference between multimolecular and macromolecular colloids? Give one example of
each type
...
(ii) There is a close similarity in physical and chemical properties of the 4 d and 5 d series of the
transition elements, much more than expected on the basis of usual family relationship
...
(iii) The members in the actinoid series exhibit larger number of oxidation states than the
corresponding members in the lanthanoid series
...
154 Xam idea Chemistry—XII
(ii) Among the following compounds, which one is more easily hydrolysed and why?
CH3CHCICH2CH3
or
CH3CH2CH2CH2Cl
2
(iii) Which of these will react faster in SN displacement and why?
1-bromopentane
24
...
25
...
OR
Mention three such facts/reactions about glucose which cannot be explained by its open end structure
...
Identify the four groups into which the polymers are classified on the basis of the magnitude of
intermolecular forces present in them
...
How do antiseptics differ from disinfectants? Name a substance that can be used as an antiseptic as
well as a disinfectant
...
(a) Non-ideal solutions exhibit either positive or negative deviations from Raoult's law
...
5 g mol–1) must be dissolved in 65g of water to lower
the freezing point by 7
...
86 K kg
mol–1
...
87
...
What is the advantage of using osmotic
pressure as compared to other colligative properties for the determination of molar masses of
solutes in solution?
(b) A solution prepared from 1
...
0 g of benzene
has a boiling point of 80
...
Determine the molar mass of this compound
...
10°C and Kb for benzene = 2
...
29
...
(ii) The stability of +3 state increases down the group in group 15 of the periodic table
...
Examination Papers 155
OR
(a) Complete the following chemical reaction equations:
(i) AgCl(s) + NH3(aq) ¾ ®
¾
(ii) P4(s) + NaOH(aq) + H2O(l) ¾ ®
¾
(b) Explain the following observations:
(i) H2S is less acidic than H2Te
(ii) Fluorine is a stronger oxidising agent than chlorine
(iii) Noble gases are the least reactive elements
...
(a) How will you prepare the following compounds starting with benzene?
(i) Benzaldehyde
(ii) Acetophenone
...
OR
(a) Explain the mechanism of nucleophilic attack on a carbonyl group of an aldehyde or a ketone
...
CBSE (Foreign) SET–II
1
...
Questions uncommon to Set I
State a feature to distinguish a metallic solid from an ionic solid
...
14
...
How are the following conversions carried out?
(i) Propene to Propan-2-ol
(ii) Ethyl magnesium chloride to Propan-1-ol
19
...
If the edge length of the unit cell is 4
...
5g cm–3, calculate the atomic mass of silver
...
02 × 1023 atoms mol–1)
20
...
The copper ion concentration is 0
...
The concentration of silver
ions is not known
...
422 V
...
(Given
E ° Ag + / Ag = + 0
...
34 V )
156 Xam idea Chemistry—XII
21
...
(ii) an electric current is passed through a colloidal solution
...
26
...
Is there any
structural difference between the two?
27
...
CBSE (Foreign) SET–III
Questions uncommon to Set I and Set II
2
...
3
...
Give an example of coordination isomerism
...
What happens when glucose reacts with nitric acid?
12
...
Illustrate the following name reactions giving a chemical reaction equation for each:
(i) Kolbe’s reaction of phenol
(ii) Friedel-Crafts’ acetylation of anisole
...
The well-known mineral fluorite is calcium fluoride
...
The F– ions fill all the tetrahedral holes in the fcc lattice of Ca2+ ions
...
46 × 10–8 cm in length
...
18g cm–3
...
(Molar mass of CaF2 = 78
...
A voltaic cell is set up at 25°C with the following half-cells, Al3+ (0
...
50 M)
...
(Given E° Ni 2 + / Ni = -0
...
66V)
21
...
Illustrate your answer
with suitable examples
...
Explain the following:
(i) The transition elements have great tendency for complex formation
...
(iii) Lanthanum and Lutetium do not show colouration in solutions
...
No
...
One atom
...
Second order
3
...
HI is stronger acid then HCl because dissociation energy of H–I bond is less than that of H–Cl
...
An ambidentate ligand is one which can link through either of the two donor atoms to the central
metal ion
...
2
6
...
4-Methyl-pent-3-en-2-one
8
...
Br /H O
2
2
HOCH2 — (CHOH) 4 — CHO ¾ ¾ ¾¾® HOCH2 — (CHOH) 4 — COOH
Gluconic acid
9
...
(i) Some reactions occur by a series of elementary steps and such simple steps are known as
elementary reactions in a process
...
Rate =
11
...
10 M
or +
D[P ]
Dt
[R] = 0
...
0030 mol L–1 s–1, t = ?
[R ] – [R]
We know that
k= o
t
[R o ] - [R] 010 - 0
...
or
t =
=
= 8
...
k
0
...
(i) Structure of BrF3 molecule
F
Br
F
F
Bent T–shaped
(ii) Structure of XeOF4 molecule
...
(i) Preparation of potassium dichromate from sodium chromate
(a)
2Na 2 CrO4 + H2 SO4 ¾ ® Na 2 Cr2 O7 + Na 2 SO4 + H2 O
¾
(b)
Na 2 CrO7 + 2KCl ¾ ® K 2 Cr2 O7 + 2NaCl
¾
(ii) Preparation of KMnO4 from K2MnO4
2K2MnO4 + Cl2 ¾ ® 2KMnO4 + 2KCl
¾
OR
14
...
According to C
...
T at first there is an increase in the energy of d-orbitals relative to that of the
free ion just as would be the case in spherical field
...
The degenerate set of d-orbitals get split
into two sets: the lower energy t2g set and eg set
...
The
actual configuration adopted is decided by the relative values of D o and P
...
(a) If D o < P then fourth electron goes to e g
(b) When D o > P, then fourth electron pairs the t2g orbital
...
Five degenerate
d-orbitals in free
metal ion
(ii) Spectrochemical Series: It is a series in which ligands can be arranged in the order of
increasing field strength or in order of increasing magnitude of splitting they produce
...
(i) Mechanism of hydration of ethene to ethanol
Step-I: Protonation of alkene
H
H
H
H
+
O
+ H
C
H
H
C
H
+
C
H
C
H
H + H 2O
H
Step-II: Nucleophilic attack of water as carbocation
H
H
C
H
H
H
H + HO
2
H
H
C
C
H
+
C
H
Step-III: Deprotonation of alcohol
H
H
H
C
H
+
H +
H
C
H
O
HO
2
H
OH
C
H
C
H
H
H
+
O
H
+
H + H3 O
(ii) Mechanism of dehydration of ethanol to ethene
...
Formation of ethene
H
H
H H
|
|
|
|
¾ ¾ —
¾
H — C— C Å ¬¾® C — C + H+
|
|
|
|
H
H
H H
ethene
16
...
) NaOH ê
¾¾¾¾¾® ê
340 K
ê
ê
ê
ê
ë
ù
ONa
ú
½
CHCl 2
ú
ú NaOH
ú ¾¾®
ú
ú
ú
ú
û
(ii) Williamson synthesis
C2H5ONa + ICH3 ¾ ®
¾
Sodium ethoxide
17
...
H
H
+
+I
+
N
N
R
+ H
R
H
H
H
H
H
H
+
+ H
N
+
H
H
N
H
H
(ii) The boiling points of primary amine are higher than the tertiary amines because strong
intermolecular hydrogen bonding takes place between the molecules of primary amine
...
(i) C 6 H5 NH2 + CHCl 3 + 3KOH(alco) ¾ ®
¾
Aniline
(ii)
19
...
874 g cm–3
a = 286
...
874 =
NA =
\
NA = ?
ZM
2 ´ 56
3
(286
...
874 ´ ( 286
...
5 ´ 10 -23
=
112 ´ 10 23
18
...
054 × 1023
20
...
10 M) | | Ag+ (conc
...
80 (– 0
...
563 V
We know that
E o = (E o - E o ) R
L
cell
1
...
563 –
log
0
...
10
[Ag + ] 2
21
...
10]
0
...
083
= 2
...
02955
= antilog of 2
...
7
[Ag+]2 =
\
[Zn 2 + ]
0
...
= 1
...
7
[Ag+] = 1
...
Where in size
of the macromolecules may fall within the colloidal range
...
The colloids are quite stable and in many
respect they resemble true solutions
...
162 Xam idea Chemistry—XII
(ii) Multimolecular colloids: A large number of atoms or smaller molecules (diameter < 1 nm) of
a substance on dissolution aggregate together to form species having size in the colloidal
range
...
Example: A sulphur sol contains
thousands of S8 sulphur molecules, a platinum or gold sol may have particles of various sizes
having many atoms
...
Such colloids are associated colloids and
aggregated particles as micelles
...
22
...
(ii) There is a close similarity in physical and chemical properties of the 4d and 5d series of the
transition elements much more than expected on the basis of usual family relationship
...
Due to equality in size of Zr and Hf, Nb and Ta, Mo and W, etc
...
(iii) The members in the actinoid series exhibit larger number of oxidation states than the
corresponding members in the lanthanide series due to the fact that the 5f, 6d and 7s levels are
of comparable energies
...
(i)
CH3 —C* CH2 CH3
—
|
OH
(Chiral)
H
|
CH3 — C — CH3
|
OH
(Non -chiral)
(ii) CH3CHClCH2CH3, in more easily hydrolysed as it forms secondary carbocation which is more
stable than primary carbocation
...
24
...
A blast of oxygen
diluted with carbon dioxide is blown through the converter
...
Oxides of silicon and Mg form slag
...
(ii) Bauxite into pure alumina
...
xH2 O( s) + 2NaHCO3 ( aq)
¾
K
(c) Al 2 O3
...
In this method, the impure metal is made
to act as anode
...
They are put in a
suitable electrolyte bath containing soluble salt of the same metal
...
(i) Invert Sugar: The dextrorotatory sucrose when hydrolysed by boiling with mineral acid
produces an equal number of molecules of dextrorotatory fructose
...
C12 H22 OH
+
H2O
H+
¾¾¾ ®
¾
Sucrose
[a]D = 66
...
5°
Fructose
[a]D = – 93°
Invert sugar
O
||
(ii) Peptide linkage: The amide linkage (— C — NH —) formed between two a–amino acid
molecules with the loss of a water molecule in a polypeptide is called a peptide linkage
...
Due to this, globules
unfold and helix get uncoiled and protein loses its biological activity
...
During denaturation 2° and 3° structures are destroyed but 1°
structures remain intact, e
...
, coagulation of egg white on boiling, curdling of milk, etc
...
(i) Despite having the aldehyde group, glucose does not give 2, 4– DNP test, Schiffs test and it
does not form the hydrogensulphite addition product with NaHSO3
...
(iii) D–Glucose on treatment of methyl alcohol in the presence of dry HCl gas gives two isomeric
nonomethyl derivatives known as a-D-glucoside and methyl b– D–glucoside
...
26
...
Polymers are classified into four groups based on the magnitude of intermolecular forces
...
Bakelite— It is a thermosetting polymer
...
Antiseptics: Antiseptics are the chemicals which prevent the growth of the harmful microorganism
...
For example dettol
...
1% solution of phenol is a disinfectant whereas 0
...
28
...
164 Xam idea Chemistry—XII
Cause of positive deviation: This type of deviation is observed by solution in which the
forces of attraction between A–A molecules and between B–B molecules is greater then the
forces of attraction between A–B molecules
...
Causes of negative deviation: This type of deviation is shown by solutions in which the
forces of attraction between A–A and B–B molecules is less than the forces of attraction
between A and B molecules
...
5 g mol
WA = 65g,
WB = ?
DTf = 7
...
5 K,
Kf = 1
...
87
We know that,
WB 1000
´
M B WA
WB 1000
7
...
87 × 1
...
5 65
7
...
5 ´ 65
WB =
1
...
86 ´ 1000
28518
...
2
WB = 8
...
DTf = iKf ×
\
OR
(a) Osmosis: The phenomenon in which solvent molecules flow through a semipermeable
membrane from a solution of low concentration to a solution of higher concentration is termed
as osmosis
...
Advantage of using osmotic pressure as compared to other collegative properties:
(i) The equilibrium is established very quickly
...
(ii) The concentration of the solution does not change during determination of osmotic
pressure
...
Examination Papers 165
(b) Given
DTB = 80
...
10 = 0
...
21 K
WB = 1
...
53°C kg mol–1
MB = ?, WA = 99g
K ´ WB
MB = b
´ 1000
DTb ´ WA
=
MB = 152 g/mol
\
29
...
5
=
0 × 21´ 99
20
...
(ii) The stability of +3 state increases down the group in group 15 due to inert pair effect
...
OR
(a) (i) AgCl + 2NH3 ¾ ® [Ag (NH3)2]+ Cl–
¾
(ii) P4 + 3NaOH + 3H2O ¾ ® PH3 + 3NaH2PO2
¾
(b) (i) This is because bond dissociation enthalpy of H–Te bond is less than H–S as the size of
Te is larger than S
...
(iii) Noble gases are the least reactive elements due to fully filled outermost shells and zero
electron gain enthalpy
...
CHO
CH3Cl, Anhy
...
AlCl
3
(ii)
Benzene
Acetophenone
(b) (i) Ethanol and propanal
Iodoform test: Ethanol will give iodoform test positive while propanal will not
...
)
166 Xam idea Chemistry—XII
(ii) Benzaldehyde and Acetophenone
Iodoform test: Treat both the compounds with I2 and NaOH
...
of iodoform
will be obtained in acetophenone
...
A violet colouration will be obtained in case of phenol
...
The hybridization of carbon changes from sp2 to sp3 in this process and a tetrahedral
intermediate is produced
...
Nu
Nu
d+
C
O
d–
+
Nu
step-1
Aldehyde or ketone
(Planar)
(b)
l
slow
C
l
fast
–
O
H
Tetrahedral
intermediate
+
C
step-2
O
H
Addition product
OH
O
|
||
SO4
(i) CH3 — C— CH3 ¾LiAlH 4 ® CH3 — CH — CH3 ¾conc
...
AlCl3
Benzaldehyde
Toluene
Benzophenone
Examination Papers 167
CBSE (Foreign) SET–II
1
...
2
...
9
...
8°
pm
O
7
...
(i) Propan-2-ol can be prepared from propene by hydration as shown below:
CH3 — CH = CH2 + conc
...
chloride
H
Addition product
Methanal
H+, H O
¾ ¾ ¾ 2 ® CH3 CH2 CH2 OH + Mg(OH)Cl
¾
Hydrolysis
3
19
...
5 g cm
d ´ a ´ NA
=
Z
10
...
767 ´ 10
-3
-24
´ (4
...
02 ´ 10 23 mol -1 )
4
3
cm ´ 6
...
09 g mol
20
...
80V,
Eo Cu 2 + /Cu = + 0
...
10 M
Given
[Ag+] = ?
Eocell = EoR – EoL = 0
...
34V = 0
...
0591
log
2
[Ag + ] 2
0
...
46 log
010
...
0
...
2881
[Ag+]2 = 0
...
[Ag+] = 7
...
(ii) On passing direct current, colloidal particles move towards the oppositely charged electrode
where they lose their charge and get coagulated
...
26
...
g
...
+
CH3 ¾ (CH2 ) 11 ¾
¾ SO- Na
3
Non-biodegradable detergents: Detergents having branched hydrocarbon chains are not easily
degraded by micro-organisms and hence are called non-biodegradable detergents, e
...
,
sodium-4-(1, 3, 5, 7-tetramethyl octyl) benzenesulphonate
...
27
...
Second order
...
Those colloids which can be separated back into dispersed phase and dispersion medium
...
[Co(NH2)6] [Cr(CN)6] and [Cr(NH2)6] [Co(CN)6]
8
...
(i)
(ii)
Xe
Cl
O O O
HClO4 (Perchloric acid)
F
XeF2 (Linear shape)
16
...
+ CO2
Sodium
phenoxide
COONa
Ethanoyl
chloride
Given:
Salicylic acid
(2-Hydroxybenzoic acid)
Major product
OCH3
Anhyd
...
18 g cm–3, Z = 4,
a = 5
...
08 g mol NA = ?
Z´M
d=
a3 ´ NA
\
NA =
Z´M
a3 ´ P
Substituting the given values, we get
4 ´ 78
...
46 ´ 10 -8 ) -3 ´ 318
...
035 × 1023 mol -1
20
...
OH
OH
[Al 3 + ] 2
0
...
001M = 1× 10–3M, [Ni2+] = 0
...
25 V - (-1
...
41V
(10 -3 ) 2
0
...
0591
10 -6
log
= 1
...
125
( 0
...
0591
0
...
41 log (10 –6 ´ 8) =1
...
0591
0
...
41 ( – 6 log 10 + 3 log 2) =1
...
3010 )
6
6
0
...
3012
= 1
...
097) = 1
...
41 + 0
...
4602V
Ecell = 1
...
41 -
21
...
(iii) It is highly specific in nature
...
(i) The transition elements have great tendency for complex formation due to presence of vacant
d-orbitals of suitable energy, small size of cations and higher nuclear charge
...
(iii) Lanthanum and Lutetium do not show colouration in solutions because both the element
exhibit +3 oxidation state in their compound thus their cations do not possess any unpaired
electrons in them
...
(ii) Question numbers 1 to 8 are very short answer questions and carry 1 mark each
...
(iv) Question numbers 19 to 27 are also short answer questions and carry 3 marks each
...
(vi) Use log tables, if necessary
...
CBSE (Delhi) SET–I
1
...
’ What does this statement mean?
2
...
Define ‘electrophoresis’
...
Draw the structure of XeF2 molecule
...
Write the IUPAC name of the following compound:
(CH3)3 CCH2Br
6
...
7
...
What are biodegradable polymers?
9
...
Explain the
reactions occurring during the corrosion of iron in the atmosphere
...
Determine the values of equilibrium constant (KC) and DG° for the following reaction:
Ni(s) + 2Ag+(aq) ¾ ® Ni2+(aq) + 2Ag(s), E° = 1
...
Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction
...
State reasons for each of the following:
(i) The N–O bond in NO– is shorter than the N–O bond in NO–
...
OR
State reasons for each of the following:
(i) All the P—Cl bonds in PCl5 molecule are not equivalent
...
172 Xam idea Chemistry—XII
13
...
(ii) Actinoids exhibit greater range of oxidation states than lanthanoids
...
Explain the following giving one example for each:
(i) Reimer–Tiemann reaction
...
15
...
What is essentially the difference between a-form of glucose and b-form of glucose? Explain
...
Describe what you understand by primary structure and secondary structure of proteins
...
Mention two important uses of each of the following:
(i) Bakelite
(ii) Nylon-6
19
...
Each side of this unit cell has a length of 400 pm
...
(Assume the atoms just touch each other on the diagonal across
the face of the unit cell
...
)
20
...
¾
This first order reaction was allowed to proceed at 40°C and the data below were collected:
[N2O5] (M)
Time (min)
0
...
00
0
...
0
0
...
0
0
...
0
0
...
0
(a) Calculate the rate constant
...
(b) What will be the concentration of N2O5 after 100 minutes?
(c) Calculate the initial rate of reaction
...
Explain how the phenomenon of adsorption finds application in each of the following processes:
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth floatation process
OR
Define each of the following terms:
(i) Micelles
(ii) Peptization
(iii) Desorption
Examination Papers 173
22
...
(ii) Electrolytic refining of a metal
...
23
...
Write the name, stereochemistry and magnetic behaviour of the following:
(At
...
Mn = 25, Co = 27, Ni = 28)
(i) K4[Mn(CN)6]
(ii) [Co(NH3)5Cl]Cl2
(iii) K2[Ni(CN)4]
25
...
(iii) Of the two bromoderivatives, C6H5CH(CH3)Br and C6H5CH(C6H5)Br, which one is more
reactive in SN1 substitution reaction and why?
26
...
(b) How would you convert
(i) Aniline to nitrobenzene
(ii) Aniline to iodobenzene?
27
...
(a) Differentiate between molarity and molality for a solution
...
50 g of MgBr2 in 200 g of
water
...
86 K kg mol–1)
OR
(a) Define the terms osmosis and osmotic pressure
...
(b) Calculate the boiling point of a solution prepared by adding 15
...
00 g of
water
...
512 kg mol–1),
(Molar mass of NaCl = 58
...
(a) Give chemical tests to distinguish between
(i) Propanal and propanone,
(ii) Benzaldehyde and acetophenone
...
(a) Explain the following:
(i) NF3 is an exothermic compound whereas NCl3 is not
...
(b) Complete the following chemical equations:
(i) C + H2 SO4 (conc
...
(ii) Tendency to form pentahalides decreases down the group in group 15 of the periodic
table
...
Which stoichiometric defect in crystals increases the density of a solid?
3
...
Draw the structure of XeF4 molecule
...
Explain what is meant by (i) a peptide linkage, (ii) a glycosidic linkage
...
Name the bases present in RNA
...
Explain the role of each of the following in the extraction of metals from their ores:
(i) CO in the extraction of nickel
...
(iii) Silica in the extraction of copper
...
For the complex [Fe(en)2 Cl2] Cl, identify the following:
(i) Oxidation number of iron
...
(iii) Magnetic behaviour of the complex
...
(v) Whether there may be optical isomer also
...
27
...
(ii) Enzymes
(iii) Analgesics
28
...
(ii) Raoult’s law in its general form in reference to solutions
...
95 mg of a gene fragment in 35
...
335 torr at 25°C
...
OR
(a) Differentiate between molarity and molality in a solution
...
21g of it dissolved in 24
...
04°C
...
7°C and the boiling point elevation constant, Kb for chloroform is 3
...
176 Xam idea Chemistry—XII
CBSE (Delhi) SET-III
Questions Uncommon to Set-I and II
2
...
4
...
8
...
What type of a battery is lead storage battery? Write the anode and the cathode reactions and the
overall reactions occurring in a lead storage battery
...
Two half-reactions of an electrochemical cell are given below:
MnO– ( aq) + 8H+ (aq) + 5e – ¾ ® Mn 2+ ( aq) + 4H2 O(l ) , E o = + 1
...
15 V
¾
Construct the redox equation from the standard potential of the cell and predict if the reaction is
reactant favoured or product favoured
...
Assign reasons for each of the following:
(i) Transition metals generally form coloured compounds
...
18
...
19
...
35 g cm–3 and the metal crystallizes with fcc unit cell
...
(At
...
02 × 1023 mol –1)
26
...
Answer the following questions :
(i) Why do soaps not work in hard water ?
(ii) What are the main constituents of dettol?
(iii) How do antiseptics differ from disinfectants?
Examination Papers 177
Solutions
CBSE (Delhi) SET - I
1
...
, are different in different directions
...
2
...
The movement of colloidal particles under an applied electric potential is called electrophoresis
...
Total no
...
of bond pairs = 2
No
...
CH3
5
...
CH3
O
|
||
CH3 — C H — CH2 — C — H
7
...
Polymers which get disintegrated by themselves in biological systems during a certain period of time
by enzymatic hydrolysis or to some extent by oxidation are called biodegradable polymers
...
e
...
9
...
In this
cell pure iron acts as anode and impure iron surface acts as cathode
...
The reactions are given below
...
44 V
¾
Fe
/ Fe
1
At cathode:
2H + O2 + 2e – ¾ ® H2 O; E ° H + / O / H O = 1
...
67V
¾
2
The Fe2+ ions are further oxidised by atmospheric oxygen to Fe3+ ions, which comes out in the form of
hydrated ferric oxide (rust)
...
Ni(s) + 2Ag+(aq) ¾ ® Ni2+ (aq) + 2Ag(s); E° = 1
...
059
2
log Kc =
´ 1
...
5932
0
...
5932 = 3
...
92 × 1039
o
DG o = – nFE cell
DG o = – 2 × 96500 × 1
...
65 kJ
11
...
Consider a general reaction
aA + bB ¾ ® c C + dD
¾
The rate expression for this reaction is
Rate = k [A]m [B]n
Where the proportionality constant k is called rate constant
...
(i) This is because the N—O bond in NO– is an average of a single bond and a double bond
2
whereas the N—O bond in NO– is an average of two single bonds and a double bond
...
Further, F does not have d-orbitals to accept the electrons denoted by H2O
molecules
...
OR
(i) In PCl5 the two axial bonds are longer than the three equatorial bonds
...
(ii) The property of catenation depends upon bond strength of the element
...
13
...
2 Cu+ (aq) ¾ ® 2 Cu2+(aq) + Cu(s)
¾
The higher stability of Cu2+ in aqueous solution may be attributed to its greater negative
D hydH° than that of Cu+
...
(ii) Actinoids exhibit greater range of oxidation states than lanthanoids
...
14
...
Examination Papers 179
OH
½
é
ê
CHCl 3 + NaOH
ê
¾¾¾¾¾¾® ê
340K
ê
ê
ê
ë
Phenol
ù
ONa
ú
½
ú NaOH
— CHCl 2 ú ¾¾®
ú
ú
ú
û
ONa
½
— CHO H O+
3
¾®
OH
½
— CHO
Salicyaldehyde
–
–
–
–
(ii) Friedel–Crafts acylation of anisole: Anisole on treatment with acylchloride in the presence
of anhydrous AlCl3 undergo electrophilic substitution in the ring at ortho and para positions
...
AlCl3
+ CH3 — C — Cl
+
Ethanoyl chloride
o-Methoxy
acetophenone
C — CH3
–
–
Anisole
O
p-Methoxy acetophenone
(major product)
OH
½
OH
½
H 2 SO4 (conc
...
(i)
OH
½
— SO3 H
¾¾¾¾¾®
HNO3 (conc
...
KOH ® CH3 — C — CH2
¾
¾¾
¾
2-Methyl propanol
16
...
Such a pair of stereo-isomers which differ in the configuration only around C1 are called
anomers
...
Primary Structure: The specific sequence in which the various a-amino acids present in a protein
are linked to one another is called its primary structure
...
Secondary Structure: The conformation which the polypeptide chain assumes as a result of
hydrogen bonding is known as secondary structure
...
In a-helix structure, the polypeptide chain forms all the possible hydrogen bonds by twisting into a
right handed screw (helix) with the —NH groups of each amino acid residue hydrogen bonded to the
C= groups of an adjacent turn of the helix
...
18
...
(ii) Nylon-6 is used for making tyre cords, ropes and fabrics
...
a = 400 pm
For fcc, r =
a
2 2
400
=
400
2
r=
Þ
2 2
2 2
2
r = 100 × 1
...
4 pm
´
=
400 2
= 100 2
4
\
20
...
289 mol L–1
Also, [A]0 = 0
...
303
k=
log
t
[A]
2
...
400
k=
log
\
20
0
...
303
4
...
89
2
...
00 – log 2
...
303
k=
[0
...
4609]
Þ
20
2
...
20
k = 2
...
00706 = 0
...
6259 × 10–2 min–1
Þ
Examination Papers 181
(b) t =
[A ]0
2
...
400 mol–1, t = 100 min, k = 1
...
303
0
...
626 ´ 10
100 ´ 1
...
4
= log
[A ]
2
...
7060 = log
Antilog (0
...
4
[A ]
5
...
4
[A ]
0
...
4
= 0
...
082
(c) Initial rate, i
...
, rate of reaction when t = 0
[A] =
Þ
When,
Also,
\
t = 0
...
400 mol L–1
k = 1
...
626 × 10–2 min–1 × 0
...
504 × 10–3 mol L–1 min–1
...
(i) Production of Vacuum: Adsorption can be successfully applied to create conditions of high
vacuum
...
The remaining traces of air inspite
of low pressure are adsorbed by the charcoal almost completely
...
Manufacture of ammonia using iron as a catalyst, manufacture of
H2SO4 by contact process using V2O5 catalyst and use of finely divided nickel in the
hydrogenation of vegetable oils are the excellent examples
...
As a result, the concentration of the reactants increases on
the surface of the catalyst and hence the rate of reaction increases
...
It
is then mixed with pine oil (a frother)
...
Now, a stream of air is blown through the mixture from below when froth is formed
at the water surface
...
The foam is
separated out and is collected and in the course, the ore particles also settle down
...
The aggregated particles thus formed are called micelles
...
Surface active
agents such as soap and synthetic detergents belong to this class
...
During peptization, the precipitate absorbs one of the ions of the electrolyte on its
surface
...
(iii) Desorption: The process of removing an adsorbed substance from a surface on which it is
adsorbed is called desorption
...
(i) Vapour phase refining of a metal: In this method, the metal is converted into its volatile
compound and collected elsewhere
...
For example,
refining of nickel by Mond process
...
A
strip of same metal in pure form is used as cathode
...
When electric current is passed the metal from the anode
goes into solution as ions due to oxidation while pure metal gets deposited at the cathode due to
reduction of metal ions
...
–
At anode:
M ¾ ® Mn+ + n e
¾
At cathode:
Mn+ + n e ¾ ® M
¾
–
(iii) Recovery of silver after silver ore was leached with NaCN: During leaching Ag is
oxidised to Ag+ which then combines with CN– ions to form soluble complex, [Ag(CN)2]–
...
2[Ag(CN)2]–(aq) + Zn(s) ¾ ® 2Ag(s) + [Zn(CN)4]2–(aq)
¾
23
...
Name of the Complex
Potassium hexacyano
manganate (II)
Pentaammine Chloridocobalt
(III) Chloride
Potassium tetra
cyanonikelate (II)
25
...
(ii) An equimolar mixture of a pair of enantiomers is called racemic mixture
...
A racemic mixture is optically inactive due to external compensation
...
Therefore, C6H5CH (C6H5)Br is more
reactive than C6H5(CH3) Br
...
H
H
(a) R — N
+
+
+
H
R—N —H
H
H
Due to electron releasing nature, the alkyl group (R) pushes electrons towards nitrogen in alkyl
amine and thus makes the unshared electron pair more available for sharing with the proton of
the acid
...
+
NH2
+
–
N2Cl
HBF4
NaNO2/HCl
(b)
(i)
–
N2BF4
NO2
NaNO2
Cu, D
273-278 K
Aniline
Nitro benzene
+
NH2
NaNO2/HCl
(ii)
I
+KI
273-278 K
Aniline
27
...
g
...
These can be used both in
soft and hard water as they give foam even in hard water
...
Sodium benzoate, sodium metabisulphite are some common preservatives
...
(iii) Antacids: These are the chemical substances which neutralize the excess acid and raise the pH
to an appropriate level in the stomach
...
28
...
Molarity decreases with increase in
temperature as volume increases with increase in temperature
...
i ´ K f ´ WB ´ 1000
(b) DTf =
M B ´ WA
i = 3, Kf = 1
...
5 g
MB = 184 g mol–1, WA = 200 g
Substituting these values in above equation, we get
3 ´ 1
...
5 g ´ 1000 g kg -1
DTf =
184 g mol -1 ´ 200 g
DTf = 1
...
15 K – 1
...
56 K
OR
(a) Osmosis: The spontaneous movement of the solvent molecules either from the pure solvent to
the solution or from a less concentrated solution to a more concentrated solution through a
semi-permeable membrane
...
Osmotic pressure is a colligative property as it depends on the number of moles of solute
particles and not on their identity
...
512 K kg mol–1, WB= 15g
MB = 58
...
512 K kg mol -1 ´ 15 g ´ 1000 g Kg -1
58
...
05 K
Therefore, boiling point of aqueous solution
= 373
...
05 K
= 374
...
(a) (i) Propanal and propanone
Tollen’s reagent test: Propanal being an aldehyde reduces Tollen’s reagent to silver mirror
but propanone being a ketone does not
...
of iodoform but benzaldehyde does not
...
)
OH
(b)
dil
...
+
2 HCHO
Formaldehyde
¾Conc
...
formate
CH3 - OH
Methyl alcohol
(ii) Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbons when their
sodium salts are heated with sodalime
...
benzoate
COOH
COCl
SOCl2
(ii)
COOH
heat
COCl
Phthalic acid
(iii) C6H5CONH2
Phthaloyl chloride
H O+
heat
3¾
¾¾ ®
Benzamide
C6H5COOH
+
NH3
Benzoic acid
30
...
(ii) Because of low F—F bond dissociation enthalpy
...
)
¾ ® CO2 + 2SO2 + 2H2O
¾
(ii) P4 + 3NaOH + 3H2O ¾ ® PH3
+ 3NaH2PO2
¾
Phosphine
(iii) Cl 2 +
3F2
(excess)
573 K
¾¾ ®
¾
2ClF3
OR
(a)
(i) As bond dissociation enthalpy of H—Cl bond is lower than H—S which is lower than
P—H
...
(ii) This is due to inert pair effect
...
(b) (i) P4 + 10SO2Cl2
¾ ® 4PCl5 + 10SO2
¾
(ii) 2Xe F2 + 2H2O ¾ ® 2Xe + 4HF + O2
¾
(iii) 2HNO3 (conc
...
) ¾ ® 2HIO3 + 10NO2 + 4H2O
¾
CBSE (Delhi) SET - II
1
...
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant
and product molecules is called shape-selective catalysis
...
Total number of electron pairs in the valence shell of central
F
F
8+4
Xe atom =
=6
2
Xe
No
...
of lone pairs = 2
F
F
Therefore, the shape of molecule would be square planar
...
(i) Peptide linkage: The amide linkage ( — C — NH —) formed between two a–amino acid
molecules with the loss of a water molecule in a polypeptide is called a peptide linkage
...
10
...
Uracil is not
present in DNA
...
(i) CO in the extraction of nickel: Impure nickel is heated in a stream of carbon monoxide when
volatile nickel tetracarbonyl is formed and the impurities are left behind in the solid state
...
K
K
Ni
+ 4CO ¾330 - 350¾® Ni(CO) 4
¾¾¾
¾450 - 470¾® Ni + CO
¾¾¾
Impure nickel
Pure nickel
Nickel tetracarbonyl
(ii) Zinc in the extraction of silver: Silver present in the ore is leached with dilute solution of
NaCN in the presence of air or oxygen to form a soluble complex
...
2[Ag(CN)2]– (aq) + Zn(s) ¾ ® 2Ag(s) +
[Zn(CN)4]2–(aq)
¾
(iii) Silica in the extraction of copper: During smelting and bessemerisation the impurity ferrous
sulphide oxidised to ferrous oxide which is then reacted with silica (flux) to form slag ferrous silicate
...
(i) [Fe (en)2 Cl2]Cl
x + 0 × 2 + (–1) × 2 + (–1) × 1 = 0
x =3
Oxidation number of iron = 3
(ii) d2sp3 hybridization, Octahedral
(iii) Paramagnetic
(iv) Two
(v) Yes, cis-[Fe(en)2Cl2]
(vi) Bis–(ethane-1, 2-diamine)–dichlorido iron (III) chloride
27
...
For example, saccharin
...
188 Xam idea Chemistry—XII
(ii) Enzymes: Enzymes are catalysts of biological origin which accelerate various cellular reactions
without themselves undergoing any apparent change during the course of action
...
Almost all the enzymes are globular proteins
...
Analgesics
relieves pain by acting on central nervous system or on peripheral pain mechanism, without
significantly affecting consciousness
...
28
...
W ´ R ´T
(b)
MB = B
p´ V
WB = 8
...
95 × 10–3g, R = 0
...
335
T = (25 + 273 ) K = 298 K , p =
atm
760
V = 35 × 10–3 L
Substituting these values in the above equation
8
...
0821 L atm K -1 mol –1 ´ 298 K
We get,
MB =
0
...
29 g mol–1
OR
Here,
(a) Molarity (M): It is defined as number of moles of solute dissolved in one litre of solution
...
Molarity changes with change in temperature as
volume changes with change in temperature
...
Molality is independent of temperature
...
63K kg mol–1,
WB = 6
...
04°C – 61
...
34°C, WA = 24
...
63 K kg mol -1 ´ 6
...
34 K ´ 24
...
15 g mol–1
CBSE (Delhi) SET - III
2
...
Total number of electron pairs in the valence shell of central
7 +3
Br atom =
=5
2
Number of bond pairs = 3
Number of lone pairs = 2
Therefore, the shape of molecule would be that of slightly bent-T
...
In nylon-6, 6 both the monomers hexamethylene diamine and adipic acid contain six carbon atoms
each
...
Secondary cell
...
At anode:
At cathode:
Cell reaction :
Sn2+ (aq)
MnO4– (aq) + 8H+ (aq) + 5e–
¾ ® 2 PbSO4(s) + 2H2O(l)
¾
¾ ® Sn4+ (aq) + 2e–] × 5
¾
¾ ® Mn2+ (aq) + 4H2O (l)] × 2
¾
2MnO4–(aq) + 5Sn2+ (aq) + 16 H+ (aq) ¾ ® 2Mn2+ (aq) + 5Sn4+(aq) + 8H2O(l)
¾
o
o
o
E Cell = E cathode – E anode = 1
...
15 V
= 1
...
13
...
When visible (white) light falls on a transition metal compounds,
they absorb certain radiation of visible light and transmit the remaining ones
...
(ii) As manganese has maximum number of unpaired electrons (5) in 3d subshell in addition to 2
electrons in the 4s subshell
...
On the basis of magnitude of intermolecular forces polymers are classified into following four
sub-groups:
(i) Elastomers
(ii) Fibers
(iii) Thermoplastic polymers
(iv) Thermosetting polymers
190 Xam idea Chemistry—XII
19
...
(i)
For fcc unit cell, Z = 4
M = 207 g mol–1 NA = 6
...
35 g cm–3
Substituting these values in equation (i), we get
4 ´ 207 g mol -1
a3 =
11
...
02 ´ 10 23 mol -1
1/ 3
4 ´ 207 ´ 10
a3 =
11
...
02 ´ 10
24
cm 3
æ
ö
8280
Þ a =ç
ç 11
...
02 ÷
÷
è
ø
´ 10 -8 cm
1/ 3
26
...
35 ´ 6
...
35 – log 6
...
9180 - 1
...
7796]
3
3
1
log x = [2
...
6945
3
x = Antilog (0
...
949
a = 4
...
9 pm
a
For fcc,
r=
2 2
494
...
414
494
...
9 2
r=
\
pm =
pm =
pm
4
4
2 2
r = 174
...
(i) Hard water contains calcium and magnesium salts
...
(ii) Chloroxylenol and a-terpineol in a suitable solvent
...
Examples are dettol and savlon
...
Furacin and
soframycin are well-known antiseptic creams
...
These are generally used to kill microorganisms present in the toilets, drains,
floors, etc
...
2
to 0
...
CBSE EXAMINATION PAPERS
ALL INDIA–2011
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
(iii) Question numbers 9 to 18 are short answer questions and carry 2 marks each
...
(v) Question numbers 28 to 30 are long answer questions and carry 5 marks each
...
Use of calculators is not allowed
...
Define ‘order of a reaction’
...
What is meant by ‘shape selective catalysis’?
3
...
4
...
Write the IUPAC name of the following compound:
CH2 = CHCH2Br
6
...
7
...
Rearrange the following in an increasing order of their basic strengths:
C6H5NH2, C6H5N(CH3)2, (C6H5)2 NH and CH3NH2
9
...
10
...
11
...
(ii) Henry’s law about partial pressure of a gas in a mixture
...
What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction
if the units of its rate constant are:
(i) L–1 mol s–1
(ii) L mol–1 s–1
...
The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2
...
Calculate how long will it take for three-fourths of initial quantity of HCO2H to
decompose
...
25 = – 0
...
Describe the principle controlling each of the following processes:
(i) Vapour phase refining of titanium metal
...
15
...
(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation
occurs in the middle of the series
...
Complete the following chemical equations:
(i) MnO4–(aq) + S2O32–(aq) + H2O(l) ¾ ®
¾
2–
2+
+
(ii) Cr2O7 (aq) + Fe (aq)
+ H (aq) ¾ ®
¾
OR
State reasons for the following:
(i) Cu(I) ion is not stable in an aqueous solution
...
17
...
18
...
Of the four bases, name those which are
common to both DNA and RNA
...
A solution prepared by dissolving 8
...
0 mL of water has an osmotic
pressure of 0
...
Assuming that the gene fragment is a non-electrolyte, calculate its
molar mass
...
Classify colloids where the dispersion medium is water
...
OR
Explain what is observed when
(a) an electric current is passed through a sol
...
(c) an electrolyte (say NaCl) is added to ferric hydroxide sol
...
How would you account for the following:
(i) H2S is more acidic than H2O
...
2
3
(iii) Both O2 and F2 stabilise high oxidation states but the ability of oxygen to stablise the higher
oxidation state exceeds that of fluorine
...
Explain the following terms giving a suitable example in each case:
(i) Ambident ligand
(ii) Denticity of a ligand
(iii) Crystal field splitting in an octahedral field
...
Rearrange the compounds of each of the following sets in order of reactivity towards SN2
displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2methylbutane,
3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2, 2-dimethylpropane,
1-Bromo-2-methylbutane
24
...
State reasons for the following:
(i) pKb value for aniline is more than that for methylamine
...
(iii) Primary amines have higher boiling points than tertiary amines
...
Draw the structures of the monomers of the following polymers:
(i) Polythene
(ii) PVC
(iii) Teflon
27
...
(i) Food preservatives
(ii) Synthetic detergents
(iii) Antacids
28
...
(b) Calculate the potential for half-cell containing
0
...
20Mcr3+ (aq) and 1
...
33V
...
0 M
...
00 A for 3 hours?
(b) A voltaic cell is set up at 25° C with the following half-cells Al3+ (0
...
50 M)
...
o
o
(Given: E Ni 2+ / Ni= –0
...
66 V)
29
...
(a) Illustrate the following name reactions:
(i) Cannizzaro’s reaction
(ii) Clemmensen reduction
(b) How would you obtain the following:
(i) But-2-enal from ethanal
(ii) Butanoic acid from butanol
(iii) Benzoic acid from ethylbenzene
OR
(a) Give chemical tests to distinguish between the following:
(i) Benzoic acid and ethyl benzoate
(ii) Benzaldehyde and acetophenone
...
What are lyophobic colloids? Give one example for them
...
Why is it that only sulphide ores are concentrated by froth floatation process?
Examination Papers 195
5
...
Draw the structure of 2, 6-Dimethylphenol
...
Define the following terms in relation to crystalline solids:
(i) Unit cell
(ii) Coordination number
Give one example in each case
...
A reaction is of second order with respect to a reactant
...
Describe the principle controlling each of the following processes:
(i) Zone refining of metals
...
15
...
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that
for the 3d series
...
What mass of NaCl must be dissolved in 65
...
50°C? The freezing point depression constant (Kf) for water is 1
...
Assume van’t Hoff factor
for NaCl is 1
...
(Molar mass of NaCl = 58
...
22
...
Define ‘activation energy’ of a reaction
...
What is meant by ‘reverse osmosis’?
3
...
Differentiate between molality and molality values for a solution
...
Describe the principle controlling each of the following processes:
(i) Preparation of cast iron from pig iron
...
196 Xam idea Chemistry—XII
15
...
(ii) The chemistry of actinoids is not as smooth as that of lanthanoids
...
Write such reactions and facts about glucose which cannot be explained by open chain structure
...
How would you account for the following:
(i) NF3 is an exothermic compounds but NCl3 is not
...
22
...
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Dacron
(iii) Neoprene
Solutions
CBSE (All India) SET–I
1
...
2
...
3
...
4
...
5
...
CH3 — CH— CH2 — C — CH3
7
...
¾ 2¾ 4 ® CH2 = CH2 + H2O
¾ H SO
¾
Ethanol
443K
Ethene
8
...
We can determine the atomic mass of an unknown metal by using the formula of density of its unit
cell
...
of atoms per unit cell) ´ M (atomic mass)
d (density) =
a 3 (cell edge) ´ NA (Avogadro number)
By knowing d, Z, a and NA, we can calculate M, the atomic mass of metal
...
Packing efficiency =
´ 100
Volume of cubic unit cell
4
Z ´ pr 3
3
=
´ 100
a3
For a simple cubic lattice, a = 2r and Z = 1
4
1´ pr 3
p
\ Packing efficiency = 3 3 ´ 100 = ´ 100
6
( 2r )
= 52
...
4%
11
...
(ii) Henry’s law: It states that the partial pressure of a gas in vapour phase (P) is proportional to its
mole fraction (x) in the solution
...
An experimentally determined expression which relates the rate of reaction with the concentration of
reactants is called rate law while the rate of reaction when concentration of each reactant is unity in a
rate law expression is called rate constant
...
e
...
e
...
For a first order reaction
[A] o
2
...
4 × 10–3 s–1, [A] = [A] o - [A] o =
,t=?
4
4
Substituting these values in the equation, we get
[A]o
2
...
4 ´ 10 s
Here,
4
t=
2
...
4 ´ 10 -3 s -1
2
...
6021 s
2
...
7 s = 578 s
198 Xam idea Chemistry—XII
14
...
It is
then decomposed to give pure metal
...
15
...
(ii) It is due to greater number of unpaired electrons in (n–1)d and ns orbitals at the middle of the
series
...
(i) 8MnO4–(aq) + 3S2O32–(aq) + H2O (l) ¾ ® 8MnO2(s) + 6SO42–(aq) + 2OH–(aq)
¾
(ii) Cr2 O2– ( aq) + 14H+ ( aq) + 6Fe 2+ ( aq) ¾ ® 2Cr 3+ ( aq) + 6Fe 3+ ( aq) + 7H2 O(l)
¾
7
OR
(i) This is because Cu(I) ion is unstable in aqueous solution and undergo disproportionation
...
O
||
17
...
(ii) The six membered cyclic structure of glucose is called pyranose structure ( a - or b -), in
analogy with heterocylic compound pyran
...
Structural difference between DNA and RNA
DNA
RNA
1
...
1
...
2
...
2
...
The common bases present in both DNA and RNA are adenine (A), guanine (G) and cytosine (C)
...
MB =
WB ´ R ´ T
p´ V
...
95 mg = 8
...
0821 L atm mol–1 K–1
0
...
335 torr =
atm
760
V = 35 mL = 35 × 10–3 L
Substituting these values in the equation (i), we get
8
...
0821 L atm mol -1 K -1 ´ 298K ´ 760
MB =
= 14193
...
335 atm ´ 35 ´ 10 -3 L
20
...
Nature: Reversible
Examples: Starch, gum, etc
...
Nature: Irreversible
Examples: Metal hydroxide like Fe(OH)3 and metal sulphide like As2S3
...
(ii) Scattering of light by the colloidal particles takes place and the path of light becomes visible
(Tyndall effect)
...
21
...
(ii) This is because two bonds share a double bond in the resonance hybrid structure of NO– while
2
three bonds share a double bond in the resonance hybrid structure of NO–
...
5 while
NO–
3
This is because
has bond order 1
...
(iii) This is due to tendency of oxygen to form multiple bonds with metal atom
...
(i) Ambidentate ligand: A ligand which can coordinate to central metal atom through two different
atoms is called ambidentate ligand
...
(ii) Denticity: The number of coordinating groups present in ligand is called the denticity of ligand
...
··
··
H2 N— CH2 — CH2 — N H2
Ethane–1, 2–diamine
200 Xam idea Chemistry—XII
(iii) The spliting of the degenerated d-orbitals into three orbitals of lower energy, t2g set and two
orbitals of higher energy eg, set due to the presence of ligand in a octahedral crystal field is
known as crystal spliting in a octahedral field
...
(i) 1-Bromopentane > 2-Bromopentane > 2-Bromo-2methyl butane
...
OH
24
...
(i) In aniline, the lone pair of electrons on N-atom are delocalised over benzene ring due to
resonance
...
In contrast, in
–
methylamine, +I-effect of CH3 group increases electron density on the nitrogen atom
...
(ii) Ethylamine is soluble in water due to formation of inter-molecular hydrogen bonds with water
molecules
...
e
...
(iii) Due to the presence of two H-atoms on N-atom, primary amines undergo extensive
intermolecular hydrogen bonding whereas tertiary amines have no H-atoms on the nitrogen
atom, do not undergo H-bonding
...
26
...
(i) Food Preservatives: These are the chemical substances which are added to the food materials
to prevent their spoilage due to microbial growth and to retain their nutritive value for long
time
...
Preservatives prevent rancidity and kill or inhibit the growth of microorganism
...
g
...
These can be
used both in soft and hard water as they give foam even in hard water
...
Sodium hydrogen carbonate or a mixture of aluminium
and magnesium hydroxide are some common antacids
...
(a) The lead storage battery is a secondary cell
...
0591
log
n
[Cr2 O2– ] [H+ ]14
7
Here, E o = 1
...
2 M
Cell
[Cr2 O2- ] = 0
...
20) 2
0
...
33 V–
log
6
( 01) (10 -4 ) 14
...
0591
log (4 × 10–55)
6
0
...
33 V–
[log 4 + log 1055]
6
0
...
33 V –
[2 log 2 + 55 log 10]
6
0
...
33 V –
[2 × 0
...
33 V – 0
...
782 V
= 1
...
119 × M g
m = 0
...
of moles of mercury produced =
0
...
119 mol
At anode:
Al (s)
¾®
¾
Al 3+ ( aq) + 3e – ] ´ 2
At cathode:
Ni 2+ + 2 e –
¾®
¾
Ni (s)
Cell reaction:
(b)
2Al(s) + 3Ni2+(aq)
¾®
¾
2Al3+(aq) + 3Ni(s)
]×3
Applying Nerst equation to the above cell reaction
E cell = E o –
cell
[Al 3+ ] 2
0
...
25 V – (–1
...
41V, n = 6
cell
Ni
\
[Al3+] = 1×10–3M, éNi ê
ë
E cell = 1
...
5 M
ú
û
(10 -3 ) 2
0
...
0591
= 1
...
5)
0
...
0591
= 1
...
0591
= 1
...
3010 – 6]
6
= 1
...
050 V
= 1
...
41 V –
E cell
F
O
O
O
P
O
29
...
) ¾ ® 5NaCl + NaClO3 + 3H2O
¾
(Hot)
(ii) 2Fe3+ + SO2 + 2H2O ¾ ® 2Fe2+ + SO42– + 4H+
¾
O
P
(b) (i)
H
OH
OH
Two, as the structure of H3PO4 has two P—OH bonds
...
(iii) Noble gases being mono atomic gases have no interatomic forces except weak dispersion
forces, therefore they have low boiling points
...
+
2 HCHO
Formaldehyde
¾Conc
...
formate
CH3 - OH
Methyl alcohol
(ii) Clemmensen reduction: The carbonyl group of aldehydes and ketones is reduced to CH2
group on treatment with zinc-amalgam and concentrated hydrochloric acid
...
)
OH
(b)
(i) CH3CHO
dil
...
of iodoform but benzaldehyde does not
...
)
COOH
(b)
COCl
SOCl2
(i)
+SO2+HCl
heat
COOH
COCl
(ii) C6H5 —CHO ¾H 2NCONHNH 2 ® C6H5CH = NNHCONH2 + H2O
¾¾¾¾
¾
(i) B2H6,H2O2/OH
CH2
(iii)
CHO
(ii) PCC
CBSE (All India) SET– II
2
...
g
...
3
...
5
...
9
...
(ii) Coordination number: The number of nearest neighbours of any constituent particle in a
packing is called its coordination number
...
12
...
(i)
If the concentration of the reactant reduced to half, then
Rate,
éRù
r ¢ = kê ú
ë2û
2
...
e
...
4
4
The unit of rate constant is mol–1 L s–1
...
(i) Zone refining is based on the principle that the impurities are more soluble in the melt than in
the solid state of the metal
...
A
strip of same metal in pure form is used as cathode
...
When electric current is passed the metal from the anode
goes into solution as ions due to oxidation while pure metal gets deposited at the cathode due to
reduction of metal ions
...
–
At anode:
M ¾ ® Mn+ + n e
¾
At cathode:
Mn+ + n e ¾ ® M
¾
–
(i) The catalytic activity of transition metals and their compounds is attributed to the following
reasons:
Because of their variable oxidation states transition metals form unstable intermediate
compounds and provide a new path with lower activation for the reaction
...
(ii) In general in the same group of d block elements, the 4d and 5d transition element has larger
size than that of 3d elements
...
i ´ K f ´ WB ´ 1000
DTf =
19
...
Here,
i = 1
...
0 g, DTf = 7
...
86 K kg mol–1, MB= 58
...
5 K =
WB =
1
...
86 K kg mol -1 ´ WB ´ 1000 g kg -1
58
...
0 g
7
...
5 ´ 65
...
199 g
187 ´ 186 ´ 1000
...
WB = 8
...
(i)
en
en
en
Co
en
en
d-Tris-(ethane1,2-diamine)
cobalt (III)
l-Tris-(ethane1,2-diamine)
cobalt (III)
NH3
Cl
(ii)
Cl
NH3
Pt
Cl
Pt
NH3
Cl
H3N
cis-Diamminedichloro
platinum (II)
trans-Diamminedichloro
platinum (II)
+
NH3
(iii)
en
Co
H3N
Cl
H3N
Fe
H3N
+
Cl
NH3
Fe
Cl
NH 3
cis-Tetraamminedichloro
iron (III)
NH3
H3N
Cl
trans-Tetraamminedichloro
iron (III)
CBSE (All India) SET– III
1
...
Activation energy = Threshold energy – Average energy of the reactants
2
...
This phenomenon is called reverse osmosis
...
If either the ore or the gangue is capable of being attracted by magnetic field, then such separation are
carried out by magnetic separation
...
Molarity is the number of moles of solute dissolved in one litre of solution whereas molality is the
number of moles of solute per kilogram of the solvent
...
Molality is independent of temperature
because mass does not depend on temperature
...
(i) Pig iron is melt with scrap iron and coke using hot air blast
...
(ii) Bauxite is soluble in concentrated NaOH solution whereas impurities are not
...
(i) This is due to presence of unpaired electrons in the (n–1) d orbitals of transition elements
...
18
...
(i) Despite having the aldehyde group, glucose does not give 2, 4-DNP test, Shiff’s test and it does
not form the hydrogen sulphite addition product with NaHSO3
...
21
...
(ii) The acidic strength of compounds increases because of increase in polarity of E—H bond from
P—H to H—Cl, which is due to increase in electronegativity of E
...
Further, F does not have d-orbitals to accept the electrons denoted by H2O
molecules
...
22
...
—
—
(i) CH2—CH—CH—CH2
and
Buta-1, 3-diene
Styrene
CH2
208 Xam idea Chemistry—XII
COOH
(ii) HO—CH2—CH2—OH
and
Ethylene glycol
COOH
Terephthalic acid
(iii) CH2 — C — CH — CH2
|
Cl
2-chloro-1, 3-butadiene
CBSE EXAMINATION PAPERS
FOREIGN–2011
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Foreign) SET–I
1
...
State Henry’s law about partial pressure of a gas in a mixture
...
What do you understand by ‘denticity of a ligand’?
4
...
Give the IUPAC name of the following compound:
CH3 — C — C — CH2 OH
|
|
CH3 Br
6
...
7
...
What is meant by a ‘broad spectrum antibiotic’?
9
...
Explain how molarity value of a solution
can be converted into its molality
...
A 0
...
93°C
...
86°C kg mol–1
...
Determine the values of equilibrium constant (Kc) and DG° for the following reaction:
Ni ( s) + 2Ag + ( aq) ¾ ® Ni 2+ ( aq) + 2Ag( s), E° = 1
...
Define the following terms giving an example of each:
(i) Emulsion
(ii) Hydrosol
13
...
How would you account for the following:
(i) The following order of increase in strength of acids:
PH3 < H2S < HCI
210 Xam idea Chemistry—XII
(ii) The oxidising power of oxoacids of chlorine follows the order:
HClO4 < HClO3 < HClO2 < HClO
15
...
no
...
Explain what is meant by the following:
(i) Peptide linkage
(ii) Pyranose structure of glucose
17
...
OR
Mention three such properties of glucose which cannot be explained by its open chain structure
...
State the reason in each of the following cases:
(i) Soaps do not work well in hard water
...
19
...
Radius of the atom in the metal is 125 pm
...
A voltaic cell is set up at 25°C with the following half-cells, Al3+ (0
...
50 M)
...
o
(Given : E o 2+ / Ni = -0
...
66 V)
Ni
21
...
(ii) Electrolytic refining of a metal
...
22
...
)
(ii) XeF4 + O2F2 ¾143¾®
¾K
(iii) Br2 + F2 ¾ ®
¾
(excess)
23
...
(b) Explain the following giving appropriate reasons:
(i) Sulphur vapour exhibits paramagnetic behaviour
...
OR
Draw the structures of the following molecules:
(i) NF3
(ii) H2S2O8
(iii) H3PO3
Examination Papers 211
24
...
Illustrate the following reactions giving a chemical equation in each case:
(i) Gabriel phthalimide synthesis
(ii) A coupling reaction
(iii) Hoffmann’s bromamide reaction
26
...
Mention two important uses for each of the following polymers:
(i) Bakelite
(ii) Nylon 6,6
(iii) PVC
28
...
(b) Nitrogen pentoxide decomposes according to the equation
2N2 O5 ( g) ¾ ® 4NO2 ( g) + O2 ( g)
¾
This first order reaction was allowed to proceed at 40°C and the data given below were
collected:
[N2O5](M)
Time (min)
0
...
00
0
...
00
0
...
00
0
...
00
0
...
00
(i) Calculate the rate constant for the reaction
...
(ii) Calculate the initial rate of reaction
...
350 M?
OR
(a) Define
(i) Order of a reaction
...
00510 min–1
...
10 M concentration of the reactant, how much of the reactant will remain
after 3
...
(a) Complete the following reactions in an aqueous medium:
(i) MnO– + C 2 O2- + H+ ¾ ®
¾
4
4
(ii) Cr2 O2- + H2 S + H+ ¾ ®
¾
7
(b) How would you account for the following:
(i) Metal-metal bonding is more extensive in the 4d and 5d series of transition elements than
the 3d series
...
(iii) Co (II) is easily oxidised in the presence of strong ligands
...
(ii) With same (d4) configuration Cr (II) is reducing whereas Mn (III) is oxidising
...
30
...
“Crystalline solids are anisotropic in nature
...
State Henry’s law about partial pressure of a gas in a mixture
...
Why is CO a stronger ligand than Cl–?
8
...
12
...
16
...
17
...
Explain the following terms giving an example of each:
(i) Antacids
(ii) Sweetening agents
21
...
(ii) Refining of zirconium by van Arkel method
...
22
...
(a) What does the designation ‘6,6’ in nylon 6, 6 polymer mean?
(b) Which polymer is obtained when free radical polymerisation of chloroprene occurs? Write the
structure of the polymer thus obtained
...
How many atoms are there in one unit cell of a body centred cubic crystal?
2
...
3
...
Which will react faster in SN1 displacement, 1-bromobutane or 2-bromobutane, and why?
12
...
Explain giving a reason for each of the following situation:
(i) In aqueous medium, HCl is a stronger acid than HF
...
15
...
Silver crystallises with face-centred cubic unit cell
...
What is the radius of silver atom? Assume the atoms just touch each other on the diagonal across the
face of the unit cell
...
Complete the following chemical equations:
(i) C + H2SO4 (conc
...
How would you obtain
(i) Benzoquinone from phenol?
(ii) Propan-2-ol from propene?
(iii) 2- Methylpropan-2-ol from methyl magnesium bromide?
27
...
Solutions
CBSE (Foreign) SET–I
1
...
2
...
3
...
For
example, bidentate ligand ethane-1, 2-diamine has two donor nitrogen atoms which can link to central
metal atom
...
1–Bromopentane, as it is a primary alkyl halide
...
2–Bromo –3– methyl-but –2–en–1–ol
...
CH3 — CH2 — C— CH2 — C — H
Examination Papers 215
H
|
7
...
Hence alkyl amine is more basic than ammonia
...
The antibiotic which is effective against a wide range of microorganisms is known as broad spectrum
antibiotic
...
9
...
If MB is the molar mass of solute, d is the density of solution then molarity (M) value of a solution can
be converted into its molality (m) by using the following for formula
...
DTf = K f ´ m
Kf = 1
...
561 mol kg–1
Substituting these values in the above equation,
We get,
DTf = 1
...
561 mol kg–1 = 1
...
04 °C, (DTf) Observed = 2
...
93 o C
1
...
82
...
Ni(s) + 2Ag+(aq) ¾ ® Ni2+ (aq) + 2Ag(s); E° = 1
...
059
2
log Kc =
´ 1
...
5932
0
...
5932 = 3
...
92 × 1039
o
DG o = – nFE cell
DG o = – 2 × 96500 × 1
...
65 kJ
12
...
g
...
)
(ii) Hydrosol: A sol in which solid is the dispersed phase and water in dispersion medium is called
hydrosol
...
(i) Production of Vacuum: Adsorption can be successfully applied to create conditions of high
vacuum
...
The remaining traces of air inspite
of low pressure are adsorbed by the charcoal almost completely
...
Manufacture of ammonia using iron as a catalyst, manufacture of
H2SO4 by contact process using V2O5 catalyst and use of finely divided nickel in the
hydrogenation of vegetable oils are the excellent examples
...
As a result, the concentration of the reactants increases on
the surface of the catalyst and hence the rate of reaction increases
...
(i) Larger the difference in electronegativity greater will be polarity and hence greater will be
acidic character
...
15
...
(i) Peptide linkage: The amide (— C— NH—) linkage between two a-amino acids formed with
the loss of a water molecule is called a peptide linkage
...
Examination Papers 217
CH2OH
H
H
OH
H
O
H
OH
HO
H
OH
a – D – (+) – Glucopyranose
17
...
(i) Despite having the aldehyde group, glucose does give 2, 4-DNP test, Shiff’s test and it does not
form the hydrogen sulphite addition product with NaHSO3
...
(i) Hard water contains calcium and magnesium salts
...
(ii) As synthetic detergents can be used in hard water as well as acidic solutions
...
(i) For fcc (or ccp), a = 2 2r = 2 × 1
...
54 × 10–8 cm
Volume of one unit cell = a3 = (3
...
436 × 10–23cm3 = 4
...
=
1 cm 3
4
...
At anode:
At cathode:
Cell reaction:
-23
cm
3
= 2
...
0591
log
n
[Ni 2+ ] 3
o
Here, E o = E o 2+ / Ni - EAl 3+ /Al = –0
...
66V) = 1
...
5 M
ú
û
218 Xam idea Chemistry—XII
E cell = 1
...
0591
0
...
41 V–
log
log (8 ´ 10 -6 )
6
6
( 0
...
0591
(log 23 + log 10–6)
6
0
...
41 V –
[ 3 × log 2 + (– 67) log 10]
6
0
...
41 V –
[3 × 0
...
41 V + 0
...
46 V
21
...
Silver is
then recovered from this complex by displacement method using more electropositive zinc
metal
...
A
strip of same metal in pure form is used as cathode
...
When electric current is passed the metal from the anode
goes into solution as ions due to oxidation while pure metal gets deposited at the cathode due to
reduction of metal ions
...
= 1
...
It is then decomposed to give pure metal
...
Ni + 4CO ¾330-¾¾® Ni(CO)4
¾ 350 K
K
Ni(CO)4
¾450 – 470¾® Ni + 4CO
¾¾¾
22
...
(ii) XeF4 + O2F2 ¾143¾® XeF6 + O2
¾K
(iii) Br2 + 5F2
¾ ¾ ® 2 BrF5
¾
(excess)
23
...
Pressure: 200 bar
Temperature: 723–773 K
Catalyst: Finely divided iron and molybdenum as promoter
...
Examination Papers 219
(ii) White phosphorus is more reactive than red phosphorus due to its discrete tetrahedral
structure and angular strain
...
OR
(i) Total no
...
of bond pairs = 3
F
F
No
...
F
O
O
S
(ii)
O
S
P
O
O
O
O
OH
OH
Peroxodisulphuric acid
(H2S2O8)
24
...
(i) Gabriel phthalimide synthesis: This reaction is used for the preparation of aliphatic primary
amines
...
Potassium phthalimide on treatment with alkyl halide gives N-alkyl phthalimide,
which on hydrolysis with dilute hydrochloric acid gives a primary amine as the product
...
)
NH ¾¾¾¾¾®
NK
¾C
¾C
½
½
½
½
O
O
+ R¾ X
220 Xam idea Chemistry—XII
¾ COOH
¾ COOH
O
½
½
¾C
H 2 O/ H +
+ R ¾ NH2 ¬¾¾¾¾¾
N¾ R
1 ° amine
¾C
½
½
O
Phthalic acid
(ii) Coupling reactions:
Diazonium salts react with aromatic amines in weakly acidic medium and phenols in weakly
alkaline medium to form coloured compounds called azo dyes by coupling at p-position of
amines or phenol
...
+
-
OH
¾ N º N Cl + H ¾
¾ OH ¾¾¾¾¾®
(pH 9-10)
Phenol
¾ OH + Cl - + H2 O
¾ N = N¾
p -Hydroxy -azobenzene
(orange dye)
+
H+
-
¾ N º N Cl + H ¾
¾ NH2 ¾¾¾¾¾®
(pH 4 -5)
Aniline
¾ NH2 + Cl - + H2 O
¾ N = N¾
p -Amino azobenzene
(yellow dye)
(iii) Hoffman’s bromamide reaction:
When a primary acid amide is heated with an aqueous or ethanolic solution of NaOH or KOH
and bromine (i
...
, NaOBr or KOBr), it gives a primary amine with one carbon atom less
...
Aniline
O
Na2Cr2O7/H2SO4
(i)
Phenol
O
Benzoquinone
Examination Papers 221
H+
(ii) CH3 — CH = CH2 + H2 O ¾Markovnikov's ® CH3 — C H— CH3
¾¾¾
¾
|
addition
Propene
OH
Propan -2-ol
O
||
(iii) CH3 — C— CH3 +
Propanone
é
OMgBr ù
OH
ê
ú
|
|
2O
CH3 MgBr ¾ ® êCH3 — C— CH3 ú ¾H ¾® CH3 — C— CH3
¾
¾
ê
ú
|
|
ê
ú
Methyl magnesium
CH3
CH3
bromide
ê
ú
ë
û
2-methyl propan -2-ol
27
...
It is used widely in making electrical goods such as switches, plugs, handles of various utensils
...
It is used for making carpets and fabrics in textile industry
...
It is used for insulating electric wires
...
(a) An experimentally determined expression which relates the rate of reaction with the
concentration of reactants is called rate law while the rate of reaction when concentration of
each reactant is unity in a rate law expression is called rate constant
...
e
...
e
...
289 mol L
Also,
[A]0 = 0
...
303
k=
log
t
[A]
2
...
400
k=
log
\
20
0
...
303
4
...
89
2
...
00 – log 2
...
303
k=
[0
...
4609]
Þ
20
2
...
20
k = 2
...
00706 = 0
...
6259 × 10–2 min–1
222 Xam idea Chemistry—XII
(ii) Initial rate, i
...
, rate of reaction when t = 0
When,
t = 0
...
400 mol L–1
Also,
k = 1
...
626 × 10–2 min–1 × 0
...
504 × 10–3 mol L–1 min–1
...
303
(iii) t =
log
k
[A]
Here,
k = 1
...
400 M, [A] = 0
...
303
1
...
400
2
...
350 1
...
583 [1
...
5441]
= 141
...
0580
= 141
...
0580 = 8
...
OR
(a) (i) Order of a reaction may be defined as the sum of the powers of the concentration terms of
the reactants in the rate law expression
...
[A] o
2
...
10 × 10–3 min–1, t = 3 × 60 × 60 min= 10800 min
[A]o = 0
...
303
0
...
1 ´ 10 min
...
1 10800 ´ 51 ´ 10 -3
=
= 23
...
303
0
...
9166 = 8
...
[A] =
= 1
...
255 ´ 10
+ 8H+ + 5e
¾ ® Mn2+ + 4H2O] × 2
¾
2–
C2O4
¾ ® 2CO2 + 2e ] × 5
¾
log
29
...
In some cases, the transition metal provides a suitable large surface area with free
valencies on which reactants are adsorbed
...
Thus, the valence electrons are less lightly held and
hence can form metal-metal bond more frequently
...
That is why, Mn3+ undergoes disproportionation reaction
...
e
...
In the
presence of strong ligands, two unpaired electrons in 3d subshell pair up and third
unpaired electron shift to higher energy subshell from where it can be easily lost and
hence oxidised to Co(III)
...
(a)
(i) In aqueous solution Cu+ undergoes disproportionation to form a more stable Cu2+ ion
...
It compensates the second ionization enthalpy of Cu involved in
the formation of Cu2+ ions
...
(iii) The catalytic activity of transition metals and their compounds is attributed to the
following reasons:
Because of their variable oxidation states transition metals form unstable intermediate
compounds and provide a new path with lower activation for the reaction
...
(i) Propanal and propanone
Tollen’s reagent test: Propanal being an aldehyde reduces Tollen’s reagent to silver
mirror but propanone being a ketone does not
...
of iodoform but benzaldehyde does not
...
)
OH
(b)
dil
...
+
¾Conc
...
formate
(ii) Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbons when their
sodium salts are heated with sodalime
...
benzoate
COOH
COCl
SOCl2
(ii)
COOH
heat
COCl
Phthalic acid
(iii) C6H5CONH2
Phthaloyl chloride
H O+
heat
3¾
¾¾ ®
C6H5COOH
+
NH3
CBSE (Foreign) SET–II
1
...
, are different in different directions
...
Henry’s law : It states that “the partial pressure of the gas in vapour phase ( p) is proportional to the
mole fraction of the gas ( x ) in the solution” and is expressed as
p = KH x
where, KH is the Henry’s law constant
...
Higher the value of Henry’s law constant KH, the lower is the solubility of the gas
in the liquid
...
Because CO has p bonds
...
Chemical substances which prevent the growth of microorganisms or kill them but are not harmful to
living tissues are called antiseptics
...
12
...
It arises because of van der waals’ forces
...
It is reversible
...
It is not specific in nature
...
surface under high pressure
...
Primary structure refers to the sequence in which various amino acids are arranged in a protein while
secondary structure refers to the shape in which a long polypeptide chain can exist as a result of
O
||
regular folding of the backbone of the polypeptide chain due to hydrogen bonding between — C—
and —NH—groups of the peptide bond
...
(i) Geometrical isomers of [Cr(NH3)4Cl2]+
+
Cl
H3N
Cl
H3N
Cr
H3N
+
Cl
NH3
Cr
NH 3
NH 3
cis-Tetraamminedichlorido
chromium (III) ion
(ii) Optical isomers of [Co(en)3]3+
NH3
H3N
Cl
trans-Tetraamminedichlorido
chromium (III) ion
226 Xam idea Chemistry—XII
en
en
en
Co
Co
en
en
en
dextro-Tris-(ethane-1,2,-diamine)
cobalt (III) ion
laevo-Tris-(ethane-1,2-diamine)
cobalt (III) ion
18
...
Sodium hydrogen carbonate or a mixture of aluminium
and magnesium hydroxide are some common antacids
...
g
...
These can be used both in
soft and hard water as they give foam even in hard water
...
(i) Leaching of bauxite ore to prepare pure alumina:
Al2O3(s) + 2NaOH(aq) + 3H2O(l) ¾ ® 2Na[Al (OH)4] (aq)
¾
2Na[Al(OH)4](aq) + CO2(g) ¾ ® Al2O3
...
xH2O(s) ¾ ¾ ® Al2O3(s) + xH2O(g)
¾
(ii) Refining of zirconium by van Arkel method
...
2[Au (CN)2]– (aq) + Zn (s) ¾ ®2 Au (s)+[Zn (CN4+)]2– (aq)
¾
22
...
(a) In nylon-6, 6 both the monomers hexamethylene diamine and adipic acid contain six carbon
atoms each
...
CH2 = C - CH = CH2 ¾O2 peroxide ® — 2 - C = CH - CH2 ] n
¾ ¾¾
[CH
—
or O2
|
|
Cl
Cl
Chloroprene
Neoprene
CBSE (Foreign) SET–III
1
...
Raoult’s law: It states that for any solution the partial pressure of each volatile component in the
solution is directly proportional to its mole fraction
...
Potassium fluorido chromate (III)
...
2-bromobutane as secondary carbocation formed is more stable as compared to the primary
carbocation in case of 1-bromobutane
...
(i) Aerosols: These are the colloidal systems in which dispersion medium is gas and dispsed,
phase is either solid or liquid e
...
, smoke, insecticide sprays
...
14
...
(ii) White phosphorus is more reactive than red phosphorus due to its discrete tetrahedral structure
and angular strain
...
15
...
Magnetic behaviour: Diamagnetic
...
Magnetic behaviour: Diamagnetic
...
As the atoms just touch each other on the diagonal across the face of unit cell, therefore
b2 = a2 + a2 = 2a2
b = 2a
...
(ii)
r
From (i) and (ii), we get
2
4r = 2a Þ r =
a
a
4
1
...
58 pm
4
22
...
) ¾ ® CO2(g) + 2H2O (l) + 2SO2(g)
¾
(ii) P4 + 3NaOH + 3H2O ¾ ® PH3 + 3NaH2 PO2
¾
b
a
228 Xam idea Chemistry—XII
(iii) Cl2 + 3F2 ¾573K ®2ClF3
¾
¾
(excess)
OH
26
...
Polymers which are formed by the repeated addition of monomers molecules possessing double or
triple bonds are called the addition polymers
...
g
...
n CH2 — CH2 ¾ ® — 2 — CH2 —
¾
(CH
)n
Ethene
Polyethene
(ii) Copolymers: The polymers made by addition polymerisation from two different monomers are
known as copolymers
...
g
...
CH
nCH2
CH—CH
CH2
(CH2—CH2—CH—CH2—CH—CH2)n
CH2 +
Buta-1,3-diene
Styrene
Buna-S
CBSE EXAMINATION PAPERS
DELHI–2012
Time allowed : 3 hours
Maximum marks: 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Delhi) SET–I
1
...
What is the role of graphite in the electrometallurgy of aluminium?
3
...
Give the IUPAC name of the following compound:
CH2—C—CH2Br
|
CH3
5
...
6
...
7
...
Define the term ‘homopolymerisation’ giving an example
...
A 1
...
The
solution has the boiling point of 100
...
Determine the van’t Hoff factor for trichloroacetic acid
...
512 K kg mol–1)
OR
Define the following terms:
(i) Mole fraction
(ii) Isotonic solutions
(iii) Van’t Hoff factor
(iv) Ideal solution
10
...
230 Xam idea Chemistry—XII
11
...
Give an example of each
group with the chemical equation involved
...
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which
coagulation of lyophobic sols can be carried out
...
Describe the principle involved in each of the following processes
...
(ii) Column chromatography for purification of rare elements
...
Explain the following giving an appropriate reason in each case
...
(ii) Structures of Xenon fluorides cannot be explained by Valence Bond approach
...
Complete the following chemical equations:
(i) Cr2O72– + H+ + I– ¾ ®
¾
–
–
+
(ii) MnO4 + NO2 + H ¾ ®
¾
16
...
Write any two reactions of glucose which cannot be explained by the open chain structure of glucose
molecules
...
Draw the structure of the monomer for each of the following polymers:
(i) Nylon-6
(ii) Polypropene
19
...
If the edge of the unit cell is 316
...
65 pm
...
874 g cm–3
...
(At
...
845 u)
20
...
(Kf for water = 1
...
For the reaction
2NO(g) + Cl2(g) ¾ ® 2NOCl(g),
¾
the following data were collected
...
Initial [NO] (M)
Initial [Cl2] (M)
Initial rate of disappearance
of Cl2 (M/min)
1
0
...
15
0
...
15
0
...
20
3
0
...
15
2
...
25
0
...
(b) Calculate the value of rate constant and specify its units
...
How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds
...
(iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in
their compounds, +4 or even +6 being typical
...
Give the formula of each of the following coordination entities:
(i) Co3+ ion is bound to one Cl–, one NH3 molecule and two bidentate ethylene diamine (en)
molecules
...
Write the name and magnetic behaviour of each of the above coordination entities
...
nos
...
Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in electrophilic
aromatic substitution reactions
...
Draw the structure and name the product formed if the following alcohols are oxidized
...
(i) CH3CH2CH2CH2OH
(ii) 2-butenol
(iii) 2-methyl-1-propanol
26
...
What are the following substances? Give one example of each one of them:
(i) Tranquillisers
(ii) Food preservatives
(iii) Synthetic detergents
28
...
(b) In the button cell, widely used in watches, the following reaction takes place
Zn(s) + Ag2O(s) + H2O(l) ¾ ® Zn2+(aq) + 2Ag(s) + 2OH–(aq)
¾
Determine Eo and DGo for the reaction
...
80 V, E o 2+ / Zn = – 0
...
(b) The resistance of a conductivity cell containing 0
...
What
is the cell constant if the conductivity of 0
...
146 × 10–3 S cm–1?
232 Xam idea Chemistry—XII
29
...
)
(ii) XeF4 + O2F2 ¾ ®
¾
(b) Draw the structures of the following molecules:
(i) H3PO2
(ii) H2S2O7
(iii) XeOF4
30
...
Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre
...
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol of hydrated ferric oxide
...
Examination Papers 233
13
...
(ii) The role of carbon monoxide in the refining of crude nickel
...
Write the main structural difference between DNA and RNA
...
A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water
...
42 °C while pure water boils at 100 °C
...
512 K kg mol–1)
22
...
(ii) CO is a stronger complexing reagent than NH3
...
23
...
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in
the lanthanoid series
...
30
...
However, only one such group is involved
in the formation of semicarbazones
...
(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces
Tollen’s reagent and undergoes Cannizzaro’s reaction
...
Identify the compound
...
CrO2Cl 2 ®
¾¾¾
+
(iii)
2
...
Draw the structure of hex-1-en-3-ol compound
...
15
...
The resulting solution was
found to freeze at – 0
...
What is the molar mass of this material? (Kf for water = 1
...
Explain the following observations giving an appropriate reason for each
...
(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals
(i
...
, 3rd series)
...
23
...
nos
...
Explain the following terms giving one example of each type:
(i) Antacids
(ii) Disinfectants
(iii) Enzymes
30
...
(ii) All the bonds in PCl5 molecule are not equivalent
...
OR
(a) Complete the following chemical equations:
(i) XeF4 + SbF5 ¾ ®
¾
(ii) Cl2 + F2(excess) ¾ ®
¾
(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus
...
(iii) The bond angles (O—N—O) are not of the same value in NO2– and NO2+
...
The process of addition of an appropriate amount of suitable impurity to an intrinsic semiconductor to
enhance its conductivity is called doping
...
Graphite is used as electrodes in the electrometallurgy of aluminium
...
PCl4–, as phosphorus has 10 electrons which can’t be accommodated in sp3 hybrid orbitals
...
3-Bromo-2-methylpropene
CH3
|
5
...
Butanone < Propanone < Propanal < Ethanal
7
...
Homopolymerisation is a polymerisation reaction in which monomers of one kind are allowed to
polymerise and form a homopolymer
...
DTb = 100
...
18°C or 0
...
512 K kg mol–1; m = 1 mol kg–1
DTb = i Kb m
(i)
(ii)
(iii)
(iv)
0
...
512 K kg mol–1 × 1 mol kg–1
0
...
35
0
...
The solutions of the same osmotic pressure at a given temperature are called isotonic solutions
...
A solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution
...
The sum of powers of the concentration of the reactants in the rate law expression is called order of
reaction
...
The two groups into which phenomenon of catalysis can be divided are:
(i) Homogeneous catalysis: When the reactants and the catalyst are in the same phase, the
catalysis is said to be homogeneous catalysis
...
2SO2(g) + O2(g) ¾NO(g) ® 2SO3(g)
¾¾
(ii) Heterogeneous catalysis: When the reactants are in a different phase than the catalyst, the
catalysis is said to be heterogeneous
...
N2(g) + 3H2(g) ¾Fe(s ) ® 2NH3(g)
¾
¾
12
...
Three methods by which coagulation of lyophobic sols
can be carried out are:
(i) Electrophoresis: During electrophoresis the colloidal particles move towards oppositely
charged electrodes, get discharged and coagulated
...
This reduces the charge on the particles which ultimately
settle down in the form of a precipitate
...
This causes
neutralisation leading to their coagulation
...
(i) In Mond process, nickel is converted into its volatile complex, teracarbonyl nickel and gets
collected elsewhere
...
Ni + 4CO ¾330–¾ ¾® Ni(CO)4; Ni(CO)4 ¾450–¾ ¾® Ni + 4CO
¾ 350 K
¾ 470 K
Impure
nickel
Tetracarbonyl
nickel (volatile)
Pure
nickel
(ii) Column chromatography is based on the principle that different components of a mixture are
differently adsorbed on an adsorbent
...
14
...
(ii) This is because the energy required for the promotion of electrons in xenon is very high
...
(i)
Cr2O72– + 14H+ + 6e–
2I–
¾®
¾
¾®
¾
2Cr3+ + 7H2O
I2 + 2e–]×3
Cr2O72– + 6I– + 14H+
¾®
¾
2Cr3+ + 3I2 + 7H2O
Examination Papers 237
MnO4– + 8H+ + 5e–
NO2– + H2O
(ii)
¾ ® Mn2+ + 4H2O]×2
¾
¾ ® NO3– + 2H+ + 2e–]×5
¾
2MnO4– + 5NO2– + 6H+ ¾ ®
¾
2Mn2+ + 5NO3– + 3H2O
16
...
(ii) A number of reactions that occur in the bodies of animals and plants to maintain the life process
are catalysed by enzymes
...
Almost all the enzymes
are globular proteins
...
For example,
activation energy for acid hydrolysis of sucrose is 6
...
15 kJ mol–1 when hydrolyzed in the presence of enzyme sucrose
...
(i) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free
—CHO group
...
+
C6H12O6 + CH3—OH ¾H ® (C6H11O5)OCH3 + H2O
¾
D-glucose
18
...
a = 316
...
5 pm = 137
...
65 pm = 286
...
845 g mol–1; d = 7
...
845 g mol –1
( 286
...
874 g cm –3
NA = 6
...
i = 2, WA = 1 kg = 1000 g; DTf = 2K; Kf = 1
...
5 g mol–1
Substituting these values in the expression,
DTf ´ M B ´ WA
WB =
i ´ K f ´1000
WB =
2 K ´ 74
...
05 g
2 ´ 1
...
Suppose order w
...
t
...
r
...
Cl2 is n
...
60 = k(0
...
15)n
...
(ii)
m
n
...
20 = k(0
...
30)
2
...
30) (0
...
40 k ( 0
...
=
m
0
...
...
20 k ( 0
...
30) n
=
0
...
15) m ( 0
...
60 mol L–1 min–1 = k(0
...
15 mol L–1)
\
(c)
k = 177
...
77 mol–2 L2 min–1 × (0
...
25 mol L–1)
= 2
...
(i) Interstitial compounds are well known for many of the transition elements because the
transition elements are capable of entrapping small sized atoms such as H, C and N in the
interstitial sites in their crystal lattices
...
TiC, Fe3H, Mn4N, etc
...
(iii) This is due to very small energy gaps between 5f, 6d and 7s subshells in actinoids
...
(i) [Co(NH3)Cl(en)2]2+ : Amminechloridobis (ethane-1, 2-diamine)-cobalt(III) ion
Magnetic behaviour : Diamagnetic
(ii) [Ni(H2O)2(OX)2]2– : Diaquadioxalatonickelate (II) ion
Magnetic behaviour : Paramagnetic
24
...
Through negative inductive effect, chlorine destabilizes the intermediate carbocation
formed during the electrophilic substitution
...
The resonance effect tends to oppose the inductive effect for the attack at
ortho and para positions and hence makes the deactivation less for ortho and para attack
...
25
...
NO2
2-Methylpropanoic acid
NH2
–
N2Cl
NaNO2/HCl
Sn/HCl
(i)
+
CN
COOH
H2O/H
CuCN/KCN
+
273-278 K
Nitrobenzene
Benzoic acid
CH2Cl
CH2CN
KCN(alc)
(ii)
CH2—CH2—NH2
Na(Hg)/C2H5OH
Benzyl chloride
+
NH2
Aniline
–
N2Cl
NaNO2/HCl
273-278 K
(iii)
2-Phenylethanamine
CN
CuCN/KCN
COOH
+
H2O/H
CH2OH
LiAlH4
Benzyl alcohol
240 Xam idea Chemistry—XII
Alternatively,
+
NH2
–
N2Cl
CH3
CHCl3/
H3PO2/H2O
NaNO2/HCl
273-278 K
anhyd
...
28
...
These relieve anxiety, stress, irritability or excitement by
inducing a sense of well-being, e
...
, iproniazid, chlordiazopoxide, equanil, luminal, etc
...
Preservatives prevent the rancidity of food and inhibit growth or kill the microorganisms
...
(iii) Synthetic detergents are cleansing agents, which have all the properties of soaps but actually do
not contain any soap
...
These are mainly classified into three categories:
l Anionic detergents, e
...
, sodium dodecylbenzene sulphonate
l Cationic detergents, e
...
, cetyltrimethyl ammonium bromide
l Non-ionic detergents, e
...
, polyethylene glycol stearate
(a) Lead storage battery is a secondary battery
...
80 V – (– 0
...
56 V
DGo = – nF E o
cell
\
DGo = – 2 × 96500 C mol–1 × 1
...
08 kJ mol–1
n=2
Examination Papers 241
OR
Lm/(S cm2 mol–1)
(a) Molar conductivity of a solution at a given concentration is the conductance of the volume V of
the solution containing one mole of electrolyte kept between two electrodes with area of cross
section A and distance of unit length
...
Molar conductivity increases with decrease
400
in concentration or increase in dilution as the
Weak electrolyte (CH3COOH)
number of ions as well as the mobility of
ions increases with increase in dilution
...
Therefore,
Strong electrolyte (KCl)
L m increases a little as shown in figure by a
straight line
...
2
0
...
e
...
(b) R = 1500 W; k = 0
...
146 × 10–3 S cm–1 × 1500 W
= 0
...
(a) (i) P4 + 10SO2Cl2 ¾ ® 4PCl5 + 10SO2
¾
(ii) XeF6 + 3H2O ¾ ® XeO3 + 6HF
¾
(b) (i) There are three bond pairs and one lone pair of electrons around S atoms
in SO32–
...
The angle O—S—O is greater than 90°
...
Therefore according to VSEPR theory, ClF3 should be
bent T-shaped
...
O
–
F
F
ClF3
242 Xam idea Chemistry—XII
F
(iii) There are two bond pairs and three lone pairs of electrons around
Xe atoms in XeF2
...
The angle F—Xe—F is greater than 90°
...
(a)
O
O
H
F
S
Xe
O
F
HO
H2S2O7
F
XeOF4
(i) The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with
zinc-amalgam and concentrated hydrochloric acid
...
This reaction is called Hell–Volhard–Zelinsky reaction
...
O
O
O
–
C—H +
C—CH3
OH
293 K
Benzaldehyde
CH—CH—C—
Benzalacetophenone
(Major product)
(ii) Sodium salt of carboxylic acids on heating with soda lime lose carbon dioxide and form
hydrocarbons
...
– +
R — CO O Na ¾NaOH and CaO ® R — H + Na 2 CO3
¾¾¾¾
D
Sodium carboxylate
(b)
Hydrocarbon
(i) Pentan-2-one when treated with NaOI (I2/NaOH) gives yellow precipitate of iodoform
but pentan-3-one does not give this test
...
)
(ii) Benzaldehyde being an aldehyde gives silver mirror with Tollen’s reagent but
acetophenone being a ketone does not give this test
...
C6H5COOH + NaHCO3 ¾ ® C6H5COONa + CO2 + H2O
¾
Benzoic acid
Sodium
bicarbonate
CBSE (Delhi) SET–II
1
...
11
...
(ii) The path of light is not visible when light is passed through NaCl solution but visible when
light is passed through a sol due to scattering of light by colloidal particles
...
(i) The role of electrolyte is two-fold:
l It covers the melting point of the mixture to about 1250 K
...
(ii) When nickel is heated with carbon monoxide it forms a volatile compound tetracarbonyl nickel,
Ni(CO)4 which on further heating at higher temperature decomposes to give pure nickel
...
Structural difference
DNA
RNA
1
...
1
...
2
...
2
...
3
...
and
thymine
as
3
...
4
...
4
...
Thymine is present in DNA
...
DTb = 100
...
00°C = 0
...
42 K; WA = 500 g
Kb = 0
...
0
...
512 K kg mol –1 ´ 1000 g kg –1
= 37
...
This splitting energy
overcomes the ionization enthalpy
...
This type of p interaction increases the
value of crystal field stabilization energy
...
(iii) Ni in [Ni(CO)4] is sp3 hybridized
...
In [Ni(CN)4]2–, the Ni2+ is dsp2 hybridized
...
4s
3d
4p
Tetrahedral
geometry
Ni in [Ni(CO)4]
CO
sp3
CO CO CO
hybridization
Square planar
geometry
Ni in [Ni(CN)4]2–
CN
CN
sp2
23
...
On the other hand, the change from Mn3+ to Mn2+ results in the extra stable half-filled (d5)
configuration and hence Mn3+ is an oxidizing agent
...
(iii) In aqueous solutions, the transition metal ions which have partially filled d-orbitals undergo d-d
transition by absorbing light from visible region and radiate complementary colour
...
(a) (i)
O
O
O
+
H2N — C — NH — NH2
—
H2N — C — NH — NH2
+
H2N — C — NH — NH2
—
Semicarbazide
Although semicarbazide has two —NH2 group but one which is directly attached to C—O
is involved in the resonance as shown above
...
On the other hand, the
lone pair of electrons on the other —NH2 group is not involved in resonance and hence is
available for nucleophilic attack on the C—O group of aldehydes and ketones
...
As there is no such steric
hindrance in cyclohexanone so nucleophilic attack by the CN– ion occurs readily and
hence cyclohexanone cyanohydrin is obtained in good yield
...
6C6H5OH + FeCl3 ¾ ® [Fe(OC6H5)6]3– + 3H+ + 3HCl
¾
Violet complex
3C6H5COOH + FeCl3 ¾ ® (C6H5COO)3Fe + 3HCl
¾
Benzoic acid
Buff coloured ppt
...
C6H5COCH3 + 3NaOI ¾ ® C6H5COONa + CHI3 + 2NaOH
¾
Acetophenone
Sodium benzoate
Iodoform
(Yellow ppt
...
CH3—CH2—CH2—CH—CH—CH2
—
|
OH
20
...
34 °C) = 0
...
34 K
WB = 15
...
86 K kg mol–1
Substituting these values in the expression,
K f ´ WB ´ 1000
MB =
, we get
DTf ´ WA
MB =
22
...
1
...
0 g ´ 1000 g kg –1
= 182
...
34 K ´ 450 g
(i) Because of large number of unpaired electrons in their atoms, the transition metals have strong
interatomic interactions and hence stronger bonding between atoms, resulting in higher
enthalpies of atomization
...
Hence, the valence electrons are less tightly held and form
metal–metal bond more frequently
...
That is why Mn2+ is more resistance than Fe2+ towards oxidation
...
(i) Antacids: Chemical substances which remove excess acid in the stomach and raise the pH to
appropriate level, e
...
, sodium hydrogencarbonate, a mixture of aluminium and magnesium
hydroxide, ranitidine, etc
...
g
...
2 to
0
...
(iii) Enzymes: Enzymes are globular proteins with high molecular mass ranging from 15,000 to
1,000,000 g mol–1, which form colloidal solution in water
...
30
...
Cl
(ii) In gaseous and liquid phases PCl5 has a trigonal bipyramidal
Cl
structure
...
P 202 pm
This is due to the fact that the axial bond pairs suffer greater
Cl
Cl
repulsion as compared to equatorial bond pairs
...
248 Xam idea Chemistry—XII
OR
(i) XeF4 + SbF5 ¾ ® [XeF3] [SbF6]–
¾
(ii) Cl2 + 3F2(excess) ¾573¾® 2ClF3
¾K
(b) (i) As N N triple bond (941
...
(ii) Due to inert pair effect stability of +5 oxidation decreases down the group 15
...
As lone pair-bond pair repulsion is greater
than bond pair-bond repulsion, thus O—N—O bond angle in NO2– is less than NO2+
...
(ii) Question numbers 1 to 8 are very short answer questions and carry 1 mark each
...
(iv) Question numbers 19 to 27 are also short answer questions and carry 3 marks each
...
(vi) Use log tables, if necessary
...
CBSE (All India) SET–I
1
...
Define ‘peptization’
...
How is copper extracted from a low grade ore of it?
4
...
What happens when bromine attacks CH2—CH—CH2—C CH?
6
...
Write the structure of the product obtained when glucose is oxidised with nitric acid
...
Differentiate between disinfectants and antiseptics
...
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the
solution
...
5 M solution of an electrolyte is found to be 138
...
Calculate the conductivity of this solution
...
A reaction is of second order with respect to a reactant
...
Which methods are usually employed for purifying the following metals
(i) Nickel
(ii) Germanium
Mention the principle behind each of them
...
Explain the following facts giving appropriate reason in each case:
(i) NF3 is an exothermic compound whereas NCl3 is not
...
250 Xam idea Chemistry—XII
13
...
Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol
...
Explain the following behaviours:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses
...
16
...
Complete the following reaction equations:
(i) C6H5N2Cl + H3PO2 + H2O ¾ ®
¾
(ii) C6H5NH2 + Br2(aq) ¾ ®
¾
18
...
19
...
If the radius of copper atom is 127
...
(Atomic mass of Cu = 63
...
02 × 1023 mol–1)
OR
Iron has a body centred cubic unit cell with a cell dimension of 286
...
The density of iron is
7
...
Use this information to calculate Avogadro’s number
...
mass of Fe = 56
...
The electrical resistance of a column of 0
...
55 × 103 ohm
...
21
...
At 1500 K, equilibrium constant K for it is 1
...
Suppose in a case [N2]=
0
...
20 mol L–1 before any reaction occurs
...
22
...
How would you account for the following:
(i) Among lanthanoids, Ln (III) compounds are predominant
...
(ii) The E o 2+ / M for copper is positive (0
...
Copper is the only metal in the first series of transition
M
elements showing this behaviour
...
24
...
Answer the following questions:
(i) What is meant by chirality of a compound? Give an example
...
What is essentially the difference between a-glucose and b-glucose? What is meant by pyranose
structure of glucose?
27
...
Give one example of each
...
(a) Define the following terms:
(i) Mole fraction
(ii) Ideal solution
(b) 15
...
The resulting solution
freezes at – 0
...
What is the molar mass of the material? (Kf for water = 1
...
(ii) Boiling point elevation constant for a solvent
...
This solution has a boiling point of 100
...
What mass of glycerol was dissolved to
make this solution? (Kb for water = 0
...
(a) Draw the molecular structures of the following compounds
(i) N2O5
(ii) XeOF4
(b) Explain the following observations:
( i) Sulphur has a greater tendency for catenation than oxygen
...
(iii) Despite lower value of its electron gain enthalpy with negative sign, fluorine (F2) is a
stronger oxidising agent than Cl2
...
(ii) Oxygen is a gas but sulphur a solid
...
30
...
252 Xam idea Chemistry—XII
(b) An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces
Tollen’s reagent and undergoes Cannizzaro’s reaction
...
Identify the compound
...
Which stoichiometric defect increases the density of a solid?
2
...
What is the role of collectors in Froth Floatation process?
6
...
17
...
Why do soaps not work in hard water?
20
...
001 M) and Ni/Ni2+ (0
...
o
o
E Ni 2+ / Ni = – 0
...
66 V
...
54)
23
...
(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29)
...
27
...
(a) Draw the structures of the following molecules:
(i) H3PO2
(ii) ClF3
(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus
...
(iii) Sulphur has a greater tendency for catenation than oxygen in the same group
...
(ii) Fluorine does not exhibit any positive oxidation state
...
CBSE (All India) SET–III
Questions Uncommon to Set–I and II
1
...
What is the basicity of H3PO2 acid and why?
5
...
Write a reaction which shows that all the carbon atoms in glucose are linked in a straight chain
...
What is the cause of a feeling of depression in human beings? Name a drug which can be useful in
treating this depression
...
Explain the role of each of the following:
(i) NaCN in the extraction of silver
...
18
...
Give one example of each group
...
Write three distinct features of chemisorptions which are not found in physisorptions
...
Explain each of the following observations:
(i) With the same d-orbital configuration (d4), Cr2+ is a reducing agent while Mn3+ is an oxidising
agent
...
(iii) There is hardly any increase in atomic size with increasing atomic numbers in a series of
transition metals
...
Name the following coordination entities and describe their structures:
(i) [Fe(CN)6]4–
(ii) [Cr(NH3)4Cl2]+
2–
(iii) [Ni(CN)4]
(Atomic Numbers Fe = 26, Cr = 24, Ni = 28)
26
...
The conductivity is increased by adding an appropriate amount of impurity which is electron rich or
election deficient as compared to intrinsic semiconductor
...
The process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the
presence of small amount of electrolyte is called peptisation
...
Copper is extracted by hydrometallurgy from low grade copper ores
...
The solution containing Cu2+ ion is treated with scrap iron or H2
...
BiH3, as BiH3 has lower bond dissociation enthalpy than SbH3
...
The reddish brown colour of bromine is discharged
...
Pent-2-enal
7
...
Disinfectants are applied to non-living objects whereas antiseptics are applied to living tissues
...
Conductivity (k) =
× Cell constant
Resistance ( R)
Conductivity ( k) ´ 1000
Molar Conductivity ( L m ) =
Molarity ( M )
OR
k ´1000 L m ´ M
=
Lm =
M
1000
k=
138
...
5 mol L–1
1000 cm 3 L–1
= 0
...
Rate = k[R]2
(i) If [R] is doubled, Rate = k[2R]2 = 4k[R]2 = 4 times
2
(ii) If [R] is reduced to
11
...
In this method nickel is heated in a stream of
carbon monoxide forming a volatile complex, nickel tetracarbonyl (Ni(CO)4), which on further
heating decomposes to give pure nickel
...
This method is based on the principle that the
impurities are more soluble in molten state than in the solid state of the metal
...
(i) This is due to lower bond dissociation enthalpy of fluorine than chlorine and small and strong
bond formed by fluorine with nitrogen
...
The two equatorial S–F bonds are equivalent, while the two axial S–F
bonds are longer than equatorial bonds
...
13
...
I2 + 2e–]×3
C—C
+ H2O
H
¾ ® 2Mn2+ + 5NO3– + 3H2O
¾
+
C—C
H OH
The mechanism of the reaction involves the following three steps:
(i) Protonation of alkene to form carbocation by electrophilic attack of H3O+
...
H
+
+ H2O
—C—C
H
+
—C—C—O—H
(iii) Deprotonation to form an alcohol
...
H OH
+
—C—C— + H3O
(i) This is due to their ability to form hydrogen bonds with water molecules
...
On the other hand, due to +R
256 Xam idea Chemistry—XII
effect of the —OCH3 group, the electron density in the O—H bond in ortho-methoxyphenol
increases, thereby making the release of H+ ion difficult
...
(i) Carbylamine reaction: Aliphatic and aromatic primary amines when heated with chloroform and
ethanolic potassium hydroxide form carbylamines or isocyanides which are foul smelling
substances
...
R—NH2 + CHCl3 + 3KOH (alc
...
) ¾D ® C6H5NC + 3KCl + 3H2O
¾
Aniline
Phenyl
isocyanide
(ii) Hoffmann’s bromamide reaction: When a primary acid amide is heated with bromine in an
aqueous or ethanolic solution of NaOH, a primary amine is obtained
...
RCONH2 + Br2 + 4NaOH ¾ ® R—NH2 + Na2 CO3 + 2NaBr + 2H2O
¾
Acid amide
C6H5CONH2 + Br2 + 4NaOH ¾ ® C6H5NH2 + Na2CO3 + 2NaBr + 2H2O
¾
Benzamide
17
...
Food preservatives are the chemical substances which are added to food materials to prevent their
spoilage and to retain their nutritive value for long periods
...
Two such substances are sodium benzoate and
salts of ascorbic acid
...
For fcc,
a = 2 2r
a = 2 2 × 127
...
4 pm = 361
...
55 g mol–1; a = 3
...
02 × 1023 mol –1
Substituting the values in the expression,
z´ M
d= 3
, we get
a ´ NA
4 ´ 63
...
614 ´ 10 cm) ´ 6
...
87 g cm
a = 286
...
65 × 10–10cm = 2
...
95 g cm–3
Examination Papers 257
NA =
NA =
z´ M
a3 ´ d
, we get
2 ´ 56 g mol –1
( 286
...
87 g cm –3
= 6
...
A = pr 2 = 3
...
785 cm2; l = 50 cm
è2
ø
R ´ A 5
...
785cm 2
=
= 87
...
01148 s cm–1
r 87
...
01148 s cm –1 ´ 1000 cm 3 L–1
=
M
0
...
6 s cm2 mol–1
N2
+ O2 ¾ ® 2NO
¾
0
...
2
0
21
Initial conc
...
in mol L–1
–x
–1
Equilibrium conc
...
8–x
2
KC =
[NO]
[N2 ][O2 ]
+2x
0
...
8 - x ) ( 0
...
2, therefore 0
...
8 and 0
...
2
\
1 × 10–5 =
4x 2
016
...
324 × 10–4 mol L–1,
2x = 2 × 6
...
648 × 10–4 mol L–1
Thus at equilibrium
[N2] = 0
...
2 mol L–1 and [NO] = 1
...
(i) A colloidal system in which dispersion medium is gas and dispersed phase is either solid or
liquid is called aerosol, e
...
, smoke, fog, etc
...
g
...
(iii) There are some substances such as soap which at low concentration behave as normal
electrolytes, but at higher concentration exhibit colloidal behaviour due to the formation of
aggregates
...
23
...
(ii) The main reason for positive E°(0
...
258 Xam idea Chemistry—XII
(iii) This is due to lanthanoid contraction which arises due to filling of 4 f-orbitals which have poor
shielding effect
...
(i) [CO(en)2Cl2]+ : Dichloride bis (ethane-1,2-diamine) cobalt (III) ion
+
Cl
+
Cl
Cl
Cl
en
en
Co
Cl
Co
en
Co
en
Cl
trans
+
en
en
cis
Optical isomers
Geometrical isomers
(ii) [Cr(C2O4)3]3– : Trioxalato chromium (III) ion
3–
3–
OX
OX
OX
Cr
Cr
OX
OX
OX
mirror
dextro
laevo
(iii) [CO(NH3)3Cl3] : Triammine trichloride cobalt (III)
NH 3
NH 3
NH 3
Cl
NH 3
Co
Cl
Co
NH 3
Cl
cis-or-fac-isomer
25
...
Butan-2-ol is an example of chiral molecule
...
butyl chloride is more stable than the 1° carbonium ion CH3—CH2—CH2 derived from
n-propyl chloride
...
butyl chloride gets hydrolyzed more easily than n-propyl
chloride under SN1 conditions
...
26
...
The six membered cyclic structure of glucose is
called pyranose structural (a-or b-), in analogy with pyrane
...
6 CH2OH
5
O
H
6 CH2OH
O
H
H
4
HO
5
H
1
OH
3
H
O
OH
H
4
OH
HO
2
1
OH
3
H
H
2
Pyran
H
OH
H
a-D-(+) Glucopyranose
OH
b-D-(+) Glucopyranose
27
...
(i) These polymers are cross linked or heavily
branched molecules
...
(ii) On heating undergo extensive cross linking in
moulds and become infusible
...
(iii) Some common examples are bakelite,
urea-formaldehyde resins, terylene, etc
...
(a)
(i) Mole fraction of a particular component in a solution may be defined as the ratio of
number of moles of that component to the sum of the moles of all the components present
in the solution
...
Thus for an ideal solution,
°
°
(i) Raoult’s law is obeyed, i
...
, PA = PA xA and PB = PB xB
(ii) DHmix = 0
(iii) DVmix = 0
(b) WB = 15
...
86 K kg mol L–1
D Tf = 0°C – (– 0
...
34°C or 0
...
86 K kg mol –1 ´ 15
...
35 g mol
450 g ´ 0
...
(ii) The boiling point elevation constant may be defined as the elevation in boiling point when
one mole of a non-volatile solute is dissolved in one thousand grams of solvent
...
512 K kg mol–1
DTb = 100
...
42°C or 0
...
( a) (i)
O
92 g mol -1 ´ 0
...
73 g
0
...
(ii) This is because the bond dissociation enthalpy of I–Cl bond is lower than that of I–I bond
...
OR
(a) (i) 3Cu + 8HNO3 (dilute) ¾ ® 3Cu (NO3)2 + 2NO + 4H2O
¾
(ii) XeF4 + O2F2 ¾ ® XeF6 + O2
¾
(b) (i) It is because P–P single bond is stronger than N–N single bond
...
The intermolecular forces in oxygen are weak van der Waals force, due to
which it is a gas at room temperature
...
structure
...
Hence sulphur is a solid at room temperature
...
This is due to absorption of radiation in visible region which
results in the excitation of outer electrons to higher energy level while the remaining light
is transmitted
...
CrO3 - H2 SO4
(a) (i) CH3CH2CH2CH2
...
H2SO4
– K2SO4
–+
COOK
CH3
COOH
4-Methyl
acetophenone
Benzene-1,4dicarboxylic acid
(b) The compound is 2-ethyl benzaldehyde and the reactions involved in the question are given
below:
–
CH2OH
COO
CH2CH3
CH2CH3
+
–
OH
COO
CHO
CH2CH3
Ag +
Silver
mirror
–
2-Ethyl
benzoate
COOH
+–
CH2CH3
[Ag(NH3)2] OH
Tollens’ reagent
2-ethyl benzaldehyde
Benzene-1,2dicarboxylic acid
NO2
H2NNH
NO2
COOH
(O)
NO2
CH—NNH
NO2
CH2CH3
+ H2O
2,4-DNP derivative
OR
(a)
(i) Propanone on treatment with I2/NaOH (NaOI) undergoes iodoform reaction to give yellow
ppt of iodoform but propanol does not
...
–
–
Heat
C6H5CHO + 2[Ag (NH3)2]+ + 3OH ¾ ¾ ® C6H5COO + 2H2O + 2Ag¯ + 4NH3
¾
Benzaldehyde
Tollen’s reagent
Silver mirror
262 Xam idea Chemistry—XII
(b)
(i) Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) 4-Methoxy benzoic acid < Benzoic acid < 3,4-Dinitrobenzoic acid
(iii) (CH3)2CHCOOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH
CBSE (All India) SET–II
1
...
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant
and product molecule is known as shape-selective catalysis
...
Collectors enhance non-wettability of the mineral particles
...
3-Phenyl prop-2-enol
17
...
Since the polar groups
can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed
from the dirty surface
...
(a)
Grease in cloth
(b)
Stearate ions arranging
around the grease droplet
(c)
Grease droplet surrounded
by stearate ions (micelle formed)
Hard water contains calcium and magnesium salts
...
Therefore soaps do
not work in hard water
...
At anode:
At Cathode:
Cell reaction:
Eo
Cell
Al
¾ ® Al3+ + 3e– ]×2
¾
2+
–
Ni + 2e ¾ ® Ni]×3
¾
2+
2Al + 3Ni ¾ ® 2Al3+ + 3Ni
¾
o
o
o
= E Cathode – E o
anode = E Ni 2+ / Ni – EAl 3+ /Al
= – 0
...
66 V) = 1
...
5 M; n = 6
Substituting the values in the Nernst equation,
[Al 3+ ] 2
0
...
41 V –
= 1
...
059
log
6
( 0
...
059
log 8 × 10–6
6
Examination Papers 263
= 1
...
059
(– 0
...
41 V + 0
...
4153 V
23
...
These trapped atoms get bonded to the atoms of
transition elements, for example, TiC, Fe3H and Mn4N, etc
...
(iii) This is due to very small energy gaps between 5f, 6d and 7s subshells in actinoids
...
(i) Elastomers: These are the polymers having the weakest intermolecular forces of attraction
between the polymer chains
...
A few ‘cross
links’ are introduced between the chains, which help the polymer to retract to its original
position after the force is released as in vulcanised rubber
...
g
...
(ii) Condenstaion Polymers: The condensation polymers are formed by the repeated condensation
reaction between different bifunctional or trifunctional monomer units usually with elimination
of small molecules such as water, alcohol hydrogen chloride, etc
...
(iii) Addition polymers: Addition polymers are formed by repeated addition of same or different
monomer molecules
...
g
...
Polythene is an example of addition polymer
...
Polyethene
O
(a)
(i)
P
OH
H
H
H3PO2
(ii) No
...
of bond pairs = 3
No of lone pairs = 2
The shape would be slightly bent T
...
(ii) This is because in H2O hydrogen bond formed is three dimensional whereas in H–F it is
linear
...
OR
H
O
O
(a)
(i)
O
O
N
O
O
O
N
O
N
(N2O5)
(b)
O
O
(ii)
N
O
Cl
O
O
O
Perchloric acid
(HClO4)
(i) Due to smaller size of O as compared to S, the bond dissociation enthalpy of O–H bond is
higher than that of S–H bond
...
Hence H2S is more acidic than H2O
...
(iii) This is because the valence shell orbital of He is completely filled (1s2) and it has high
ionisation enthalpy and more positive electron gain enthalpy
...
Silicon or germanium doped with group 15 elements like P or As are called n type of semiconductors
...
H3PO2 is monobasic as it contains only one ionisable H-atom which is present as OH group
...
H2C—C—CH2—Br
3-Bromo-2-methyl propone
7
...
CHO
|
HI
(CH
...
Human beings suffer from depression when they have low levels of noradrenaline
...
Low levels of noradrenaline lower the
signal-sending activity and make human beings suffer from depression
...
11
...
Dilute NaCN
solution forms a soluble complex with Ag while the impurities remain unaffected which are
then filtered off
...
Sulphide ore of copper contains iron as impurity
which is removed as iron silicate (slag)
...
Difference between antiseptics and disinfectants:
Antiseptics
z Antiseptics are chemical substances which prevent the growth of microorganisms and may even
kill them but are not harmful to living tissues
...
z Dettol, furacine, soframicine are antiseptics
...
z Disinfectants are applied to inanimate objects such as floor, drainage system, instrument, etc
...
2 to 0
...
22
...
(ii) High specificity: Chemisorption is highly specific and it will only occur if there is some
possibility of chemical bonding between adsorbent and adsorbate
...
(iii) Irreversibility: Chemisorption is usually irreversible in nature as it involves compound
formation
...
23
...
On the other hand, the change from Mn2+ to Mn3+ results in the extra stable
half-filled, d5 configuration
...
(ii) This is due to small energy gap between 5f, 6d and 7s subshells in actinoids
...
266 Xam idea Chemistry—XII
24
...
(ii) [Cr(NH3)4Cl2]+ = Tetraamminedichloridochromium (III) ion, Cr3+(3d3)
4s
3d
×× ××
4p
××
×× ×× ××
× × = Electron pair from NH3
molecule or Cl– ion
d 2sp3 hybrid
d2sp3 hybridisation in [Cr(NH3)4Cl2]+ leads to octahedral structure
...
26
...
(ii) Primary structure of proteins: The sequence in which various amino acids are arranged in a
protein is called its primary structure
...
(iii) Denaturation of Proteins: When a protein in its native form is subjected to a change in
temperature or a change in pH, the hydrogen bonds are disturbed
...
This is called denaturation of
protein
...
g
...
CBSE EXAMINATION PAPERS
FOREIGN–2012
Time allowed : 3 hours]
[Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
1 to 8 are very short answer questions and carry 1 mark each
...
9 to 18 are short answer questions and carry 2 marks each
...
19 to 27 are also short answer questions and carry 3 marks each
...
28 to 30 are long answer questions and carry 5 marks each
...
CBSE (Foreign) SET–I
1
...
Why is the adsorption phenomenon always exothermic?
3
...
4
...
Why?
5
...
Draw the molecular structure of the compound, 4-methylpent-3-en-2-one
...
Write the full form of DNA and RNA
...
What is meant by ‘narrow spectrum antibiotics’?
9
...
OR
–1
The density of water of a lake is 1
...
what
is the molarity of Na+ ions in the water of the lake? (Atomic mass of Na = 23
...
Define the following terms:
(i) Order of a reaction
(ii) Activation energy of a reaction
268 Xam idea Chemistry—XII
11
...
Name the method used for concentration of these
two ores
...
Explain the following:
(i) The chemical reactivity of nitrogen is much less than that of phosphorus
...
13
...
What are ambident nucleophiles? Explain giving an example
...
Explain as to why
(i) Alkyl halides, though polar, are immiscible with water
...
16
...
Complete the following reaction equations:
(i) C6H5N2Cl + H3PO2 + H2O ¾ ®
¾
(ii) C6H5NH2 + Br2(aq) ¾ ®
¾
18
...
(ii) The use of the sweetner aspartame is limited to cold foods and drinks
...
Iron has a body centred cubic (bcc) unit cell with a cell dimension of 286
...
The density of iron is
7
...
Use this information to calculate Avogadro’s number
...
845 u)
OR
Silver crystallises in face centred cubic (fcc) unit cell
...
At 25°C the saturated vapour pressure of water is 3
...
75 mm Hg)
...
(Molar mass of urea = 60
...
Consider the reaction:
2A + B ¾ ® C + D
¾
Following results were obtained in experiments designed to study the rate of reaction:
Experiment
No
...
10
0
...
5 × 10–3
2
0
...
20
3
...
20
0
...
0 × 10–3
Examination Papers 269
(a) Write the rate law for the reaction
...
(c) Which of the following possible reaction mechanism is consistent with the rate law found
in (a)?
I
...
B + C ¾ ® E (slow)
¾
A + E ¾ ® F (fast)
¾
A + F ¾ ® D (fast)
¾
22
...
Complete the following chemical equations:
(i) NH4Cl(aq) + NaNO2(aq) ¾ ®
¾
(ii) P4 + NaOH + H2O ¾ ®
¾
(iii)
+ F2(g) ¾673 K ®
Xe( g)
¾
¾
1 Bar
(Xenon in excess)
24
...
(ii) Nickel(II) does not form low spin octahedral complexes
...
25
...
What is glycogen? How is it different from starch? How is starch structurally different from
cellulose?
27
...
(a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions
and the overall reaction occurring in a lead storage battery sending out an electric current
...
001 M) | Ag and Cu2+ (0
...
46 V, log 105 = 5)
cell
270 Xam idea Chemistry—XII
OR
(a) Define the term molar conductivity and explain how molar conductivity changes with solution
concentration for weak and strong electrolytes
...
An electrochemical cell is created when the two
solutions are connected by a salt bridge and the two strips are connected by wires to a
voltmeter
...
100 molar and the initial concentration of AgNO3 is 1
...
o
[E o 2+ / Ni = – 0
...
80 V, log 10–1 = – 1]?
Ni
29
...
(ii) The higher oxidation states are usually exhibited by the members in the middle of a series
of transition elements
...
OR
(a) Calculate the number of unpaired electrons in the following gaseous state ions:
Mn2+, Cr3+, V3+ and Fe2+
Which one of these is the most stable in aqueous solutions?
(At
...
V = 23, Cr = 24, Mn = 25, Fe = 26)
(b) Explain the following observations:
(i) The transition metal ions are usually coloured in aqueous solutions
...
(iii) The highest oxidation state of a transition metal is exhibited in its oxide or fluoride
...
(a) Describe the mechanism of the addition of Grignard reagent to the carbonyl group of a
compound to form an adduct which on hydrolysis yields an alcohol
...
(i) Ethanol to 3-hydroxybutanal
(ii) Benzoic acid to m-nitrobenzyl alcohol
(iii) Benzaldehyde to benzophenone
CBSE (Foreign) SET–II
Questions Uncommon to Set–I
1
...
What is meant by ‘shape selective catalysis’?
4
...
Of the two alcohols; (a) CH2=CH—CH2OH and (b) CH2=CH—CH2—CH2OH, which one will react
more easily with conc
...
Of the two bases named below, which one is present in RNA and which one is present in DNA?
Thymine, Uracil
13
...
What are biodegradable and non-biodegradable detergents? Give one example of each
...
Differentiate among a homogeneous solution, a suspension and a colloidal solution, giving a suitable
example of each
...
Complete the following chemical equations:
(i) HgCl2 + PH3 ¾ ®
¾
(ii) NaOH + Cl2 ¾ ®
¾
(hot & conc
...
Name the following complexes and draw the structures of one possible isomer of each:
(i) [Cr(C2O4)3]3–
(ii) [Pt(NH3)2Cl2]
(iii) [Co(en)2Cl2]+
26
...
Define paramagnetism with an example
...
What is the role of a depressant in Froth Floatation process for the concentration of a sulphide ore?
4
...
Write the IUPAC name of the following compound:
CH3—O—CH2—CH—CH3
|
CH3
6
...
12
...
(ii) Helium does not form any real chemical compound
...
Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride
...
20
...
0711 m aqueous solution of sodium sulphate
...
320 °C, what is the actual value of van’t Hoff factor at this
temperature? (Kf for water = 1
...
Present a classification of colloids where dispersion medium is water
...
23
...
Write three such behaviours of glucose which cannot be explained by an open chain structure of
glucose molecule
...
The energy gap between valance band and conduction is known as forbidden zone
...
As the adsorption progresses, the residual forces at the surface decreases resulting in the decrease of
surface energy which appears as heat
...
2[Ag(CN) 2 ]( aq) + Zn( s) ¾ ® [Zn(CN) 4 ] 2- ( aq) + 2Ag( s)]
¾
4
...
5
...
CH3
O
|
|
—
CH3 —C — CH — C— CH3
7
...
Antibiotics which are mainly effective against gram-positive or Gram-negative bacteria are known as
narrow spectrum antibiotics, e
...
, ampicillin G
...
(a) Van’t Hoff factor:
(i) May be defined as the ratio of normal molecular mass to observed molecular mass or the
ratio of observed colligative property to calculated colligative property
...
(iii) Ebuthoscopic constant may be defined as the elevation in boiling point when one mole of
a non-volatile solute is added to 1000 grams of solvent
...
Number of moles of solute, Na+ ions =
OR
92g
= 4 mol
23g mol -1
1000 g
800 mL
Mass of solution
Volume of solution =
= 0
...
25 g mL
1000 mL L-1
Number of moles Na + ions 4 mol
=
Volume of solution in liter 0
...
Molarity =
10
...
For a general reaction
aA + bB ¾ ® products
¾
274 Xam idea Chemistry—XII
Let rate = K[ A ] m [ B ] n
Order of the reaction = m + n
(ii) The minimum extra energy absorbed by the reactant molecular so that their energy becomes
equal threshold energy is called activation energy
...
Metal
Chief ore
Method of concentration
Copper
Copper pyrite (CuFeS2)
Froth floatation
Aluminium
Bauxite [AlOx (OH)3–2x]
Leaching
where 0 < x < 1
12
...
4 kJmol–1) is much stronger than p–p single bond (213 kJmol–1),
therefore nitrogen is much less reactive than phosphorus
...
13
...
Nucleophiles which can attack through the different nucleophilic centres present in it are called
ambident nucleophiles
...
15
...
(ii) This is because Grignard reagent forms alkanes by reacting with moisture
...
(i) Carbylamine reaction: Aliphatic and aromatic primary amines when heated with chloroform and
ethanolic potassium hydroxide form carbylamines or isocyanides which are foul smelling
substances
...
R—NH2 + CHCl3 + 3KOH (alc
...
) ¾D ® C6H5NC + 3KCl + 3H2O
¾
Aniline
Phenyl
isocyanide
(ii) Hoffmann’s bromamide reaction: When a primary acid amide is heated with bromine in an
aqueous or ethanolic solution of NaOH, a primary amine is obtained
...
Examination Papers 275
RCONH2 + Br2 + 4NaOH ¾ ® R—NH2 + Na2 CO3 + 2NaBr + 2H2O
¾
Acid amide
C6H5CONH2 + Br2 + 4NaOH ¾ ® C6H5NH2 + Na2CO3 + 2NaBr + 2H2O
¾
Benzamide
17
...
2,4,6-Tribromoaniline
(i) In hard water soaps get precipitated as calcium and magnesium soap which being insoluble in
water stick to the clothes as gummy mass
...
19
...
87 g cm–3
a = 286
...
65 × 10–10 cm = 2
...
65 ´ 10 -10 cm) 3 ´ 7
...
042 × 1023 mol–1
OR
a = 2 2r
\
a = 2 2 × 145 pm
= 2 × 1
...
06 pm
...
WB = 5 g, WA = 95 g, M B = 60
...
165 kPa
Substituting the values in the expression;
o
PA - P
o
PA
=
WB ´ M A
, we get
M B ´ WA
-1
5 g ´ 18 g mol
3
...
015
3
...
05 g mol -1 ´ 95g
P = 3
...
015 × 3
...
118 k Pa
276 Xam idea Chemistry—XII
21
...
5 × 10–3 mol L–1 min–1 = K( 01) m ( 01) n
\
...
3
...
0 × 10
-1
min
-1
mol L
–1
m
= K( 0
...
2)
m
–1
min
… (i)
n
= K( 0
...
4)
… (ii)
n
…(iii)
Dividing equation (iii) by (ii), we get
6
...
2) m ( 0
...
0 ´ 10 -3 k( 0
...
2) n
2 = 2n Þ n = 1
Dividing equation (ii) by (i), we get
3
...
5 ´ 10 -3
=
k( 0
...
2) n
k( 01) m ( 01) n
...
2 = 2m2n = 2m
...
23
...
= 1
...
(i) The movement of colloidal particles towards oppositely charged electrodes in an electric field
is called electrophoresis
...
The aggregated particles thus formed are known as micelles or associated colloids
...
(i) NH4 Cl( aq) + NaNO2 ( aq) ¾ ® N2 ( g) + 2H2 O(l) + NaCl( aq)
¾
(ii) P4 + 3NaOH + 3H2 O ¾ ® PH3 + 3NaH2 PO2
¾
(iii) Xe( g) + F2 ( g) ¾673K ® XeF2 ( s)
¾
¾
1bar
(Xenon in excess)
24
...
g
...
(ii) As only one inner d-orbital is available in nickel for bonding in the presence of strong ligand,
e
...
, CO
...
Examination Papers 277
(iii) In both the complexes, Fe is in +2 oxide state with d6 configuration
...
Hence, the transmitted colours are different in dilute solutions
...
(i) CH3—CH2—CH2—O—CH2—CH3—H—Br ¾ ® CH3—CH2—Br—CH3—CH2—CH2—OH
¾
OC2H5
OH
(ii)
+ HBr
C2H5Br
+
(iii) (CH3 ) 3 C — OC 2 H5 + HI ¾ ® (CH3 ) 3 C — I + C 2 H5 — OH
¾
26
...
The carbohydrates are stored in animal body as glycogen
...
Amylose is linear chain polymer of a - D –glucose
...
Strarch is
the main storage polysaccharide of plants
...
27
...
No
...
Tetrafluoro ethene
Chloroprene
CF2 — CF2
CH2 — C — CH — CH2
|
Cl
(a) Lead storage battery is a secondary battery
...
46 V; [Ag+] = 1 × 10–3 M, [Cu2+] = 0
...
059
log
, we get
n
[Ag + ] 2
E cell = 0
...
0
...
46 V – (0
...
46V – 0
...
3125 V
OR
Lm/(S cm2 mol–1)
(a) Molar conductivity of a solution at a given
400
concentration is the conductance of the
volume V of the solution containing one
Weak electrolyte (CH3COOH)
mole of electrolyte kept between two
electrodes with area of cross section A and
distance of unit length
...
0
0
...
4
Molar conductivity increases with decrease
c1/2/(mol/L)1/2
in concentration or increase in dilution as the
number of ions as well as the mobility of ions increases with increase in dilution
...
Therefore, L m increases a
little as shown in figure by a straight line
...
e
...
(b) (i) At Anode:
Ni ¾ ® Ni 2++ 2 e–
¾
+
At Cathode:
Ag + e - ¾ ® Ag] ´ 2
¾
Cell reaction:
Ni + 2Ag + ¾ ® Ni2++ 2Ag
¾
o
E cell = E°cathode – E°anode
o
o
= EAg + /Ag - E Ni 2+ / Ni = 0
...
25 V)
o
E cell = 1
...
059
log
n
[ Ag + ] 2
o
Here, n = 2, E cell = 1
...
1 M, [Ag+] = 1
...
05V –
\
( 0
...
059
log
2
(1) 2
Ecell = 1
...
0295 log 10–1 = 1
...
0295 V
Ecell = 1
...
(a)
(i) Cr2 O2- + 14H+ + 6e - ¾ ® 2Cr 3+ + 7H2 O
¾
7
2I–
¾ ® I2 + 2e–] × 3
¾
Cr2 O2- + 6I - + 14H+ ¾ ® 2Cr 3+ + 3I 2 + 7H2 O
¾
7
(ii) MnO- + 8H+ + 5e - ¾ ® Mn 2+ + 4H2 O] ´ 2
¾
4
SO2- + H2 O
3
¾ ® SO2- + 2H+ + 2e - ] × 5
¾
4
2MnO- + 5SO2- + 6H+ ¾ ® 2Mn 2+ + 5SO2- + 3H2 O
¾
4
3
4
(b)
(i) The catalytic activity of transition metals and their compounds is attributed to the
following reasons:
Due to their tendency to show variable oxidation states transition metals form instable
intermediate compounds and provide a new path for the reaction with lower activation
energy
...
(ii) This is due to presence of maximum number of unpaired electrons in a transition metal
which is present in the middle of a series
...
Hence, the valence electrons are less tightly held and form
metal–metal bond more frequently
...
(b) (i) Ni(CO)4 : Tetracrabonyl nickel(o); Ni = 3d8 4s2
4s
4p
××
3d
×× ×× ××
× × = Electron pair from
ligand CO
sp3 hybrid
Structure = Tetrahedral; Magnetic behaviour : Diamagnetic
280 Xam idea Chemistry—XII
(ii) Cu2+ (aq) is much more stable than Cu(aq)
...
Due to his Cu+(aq) undergo disproportionation as follows:
2Cu + ( aq) ¾ ® Cu 2+ ( aq) + Cu( s)
¾
(iii) This is due to high electronegativity oxygen and fluorine
...
(a) Mechanism:
(i) Nucleophilic addition of Grignard reagent to carbonyl group to form an adduct
...
–
+
C—O—Mg—X
H 2O
C—O + Mg(OH)X
R
R
(b) Name of compound
(i) S–Methyl butanal
Structure
CH3—CH—CH2—CHO
CH3
(ii)
Hexane–1, 6–dioic acid
HOOC—CH2—CH2—CH2—CH2—COOH
CO—CH2—CH3
(iii) P–Nitropropiophenone
NO2
OR
(a)
(i) Cannizzaro reaction: Aldehydes which do not have an–a hydrogen atom, undergo
disproportionation reaction on treatment with concentrated alkali
...
–
CH3 OH + HCO O K +
2HCHO ¾conc
...
R—CH2—COOH
(i) X2/Red P
(ii) H2O
R—CH—COOH (X = Cl, Br)
X
a-Halocarboxylic acid
Examination Papers 281
CH3
|
(b) (i) CH3 CHO ¾dil NaOH ® CH3 — CH — CH2 — CHO
¾¾
¾
Ethanol
(ii)
3 - Hydroxybutanol
COOH
COOH
HNO3 (conc
...
) ; D
CONH2
NO2
+NH3
D
Benzoic acid
NO2
LiAlH4
CH2OH
CH2—NH2
HNO2
NO2
p-Nitro benzyl alcohol
NO2
O
CHO
COOH
K2Cr2O7/H2SO4
(iii)
COCl
SOCl2
(O)
+
C
AlCl3 (anhyd
...
Metallic solid conducts electricity in solid state but ionic solid conducts electricity only in solution or
in molten state
...
The catalytic reaction which depends upon the pore structure of the catalyst and the size of the
reaction and product molecules is called shape selective catalysis
...
H2S, due to low bond dissociation enthalpy of H–S bond
...
CH2 = CH - CH2 OH
8
...
(i) Thymine is present in DNA
...
(i) H2S2O8 (Peroxodisulphuric acid)
(ii) XeF2
O
O
F
S
S
Xe
O
O
OH
O
O
HO
F
282 Xam idea Chemistry—XII
22
...
Particle Size
Colloidal solution
Suspension
Less than 1nm
Between 1 nm to 1000 nm
more than 1000 nm
2
...
Settling of particles
Do not settle
Settle only on configuration
Settle under gravity
4
...
Example
23
...
)
XeF4 + O2F2 ¾143K ® XeF6 + O2
¾
¾
24
...
(i) Sucrose is dextrorotatory, on hydrolysis in the presence of hydrochloric acid or enzyme
invertase, it gives a mixture of D-(+)-glucose and D-(–)-fructose which is laeorotatory called
invert sugar
...
(iii) Enzymes are biocatalyst
...
Enzymes are specific in
nature and efficiency in their action
...
SET–III
1
...
Thus exhibited is called paramagnetism
...
g
...
3
...
4
...
5
...
6
...
(i) This is because fluorine is most electronegative element and does not have d-orbitals in its
valence shell
...
14
...
Since chlorobenzene has lower magnitude
of negative charge on Cl atom and shorter C—Cl bond than cyclohexyl chloride due to
resonance therefore chlorobenzene has lower dipole moment than cyclohexyl chloride
...
20
...
320°C) = 0
...
320 K
m = 0
...
86 K Kg mol–1, DTf = 0
...
320 K = i´186 K kg mol–1 × 0
...
0
...
42
i=
0
...
When the dispersion medium is water colloids are classified as hydrophilic colloids and hydrophobic
colloids
...
g
...
Hydrophobic colloids are solvent (water repelling, irreversible and less stable, e
...
,
metal sulphides sol, metal hydroxide sol, etc
...
(i) Sn + 2PCl5 ¾heating ® SnCl4 + 2PCl3
¾¾
(ii) 2Fe3+ + SO2 + 2H2O ¾ ® 2Fe2+ + SO2- + 4H+
¾
4
(iii) 2XeF2(s) + 2H2O(e) ¾ ® 2Xe + 4HF + O2
¾
26
...
(i) Despite having the aldehyde group, glucose does not give 2, 4– DNP test, Schiffs test and it
does not form the hydrogensulphite addition product with NaHSO3
...
(iii) D–Glucose on treatment of methyl alcohol in the presence of dry HCl gas gives two isomeric
nonomethyl derivatives known as a-D-glucoside and methyl b– D–glucoside
...
A ring structure called pyranose structure (a- or b-) is proposed for the glucose molecule
...
(ii) Question numbers 1 to 8 are very short answer questions and carry 1 mark each
...
(iv) Question numbers 19 to 27 are also short answer questions and carry 3 marks each
...
(vi) Use log tables, if necessary
...
1
...
Name the method used for the refining of nickel metal
...
What is the covalency of nitrogen in N2O5?
4
...
|
Cl
5
...
Write the structure of 3–methyl butanal
...
Arrange the following in increasing order of their basic strength in aqueous solution:
CH3
...
What are three types of RNA molecules which perform different functions?
9
...
At
what temperature will this solution boil?
(Kb for water = 0
...
15 K)
10
...
20 M solution of KCl at 298 K is 0
...
Calculate its molar
conductivity
...
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the
addition of small amounts of electrolytes?
286 Xam idea Chemistry—XII
12
...
(a) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of
silver?
(b) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast
furnace to extract iron from the oxide ore?
14
...
(ii) H3PO3 is heated?
15
...
nos
...
Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction
...
17
...
How will you convert:
(i) Propene to Propan–2–ol?
(ii) Phenol to 2, 4, 6 – trinitrophenol?
19
...
Determine the osmotic pressure of a solution prepared by dissolving 2
...
(R = 0
...
21
...
001 M) || H+ (1M) | H2 (g) (1 bar), Pt (s)
o
(Given E cell = + 0
...
How would you account for the following?
(i) Transition metals exhibit variable oxidation states
...
(iii) Transition metals and their compounds act as catalyst
...
Write the IUPAC names of the following coordination compounds:
(i) [Cr(NH3)3Cl3]
(ii) K3[Fe (CN)6]
(iii) [CoBr2(en)2]+, (en = ethylenediamine)
24
...
Write the names and structures of the monomers of the following polymers:
(i) Buna–S
(ii) Neoprene
(iii) Nylon–6, 6
26
...
Principal immediately instructed the canteen contractor to replace the fast food with the fibre and
vitamins rich food like sprouts, salad, fruits, etc
...
After reading the above passage, answer the following questions:
(a) What values are expressed by Sonali and the Principal of the school?
(b) Give two examples of water-soluble vitamins
...
(a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
28
...
(i) Write the differential rate equation
...
Calculate t1/2 for this reaction
...
428 = 0
...
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
288 Xam idea Chemistry—XII
log k = log A –
Ea æ 1 ö
ç ÷
2
...
When a graph is plotted for log k vs
...
Calculate ‘Ea’ for the reaction
...
314 JK–1 mol–1)
29
...
(ii) PH3 has lower boiling point than NH3
...
(ii) Fluorine does not exhibit positive oxidation state
...
(b) Draw the structures of the following molecules:
(i) XeF2
(ii) H2S2O8
30
...
Give two reasons
...
(i) 2H—C—H ¾ ¾ ¾ KOH ®
¾¾
¾
||
O
Br
(ii) CH3COOH ¾ ¾ 2 P ®
¾
¾
CHO
(iii)
HNO3/H2SO4
273–283 K
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal
(ii) Benzoic acid and Phenol
Examination Papers 289
SET–II
(Questions Uncommon to Set–I)
1
...
Write the IUPAC name of CH3CH—CH — C —CH3
...
What type of bonding helps in stabilising the a-helix structure of proteins?
6
...
Bartlett for carrying out reaction between Xe and PtF6?
7
...
Write the structure of 4–chloropentan–2–one
...
How will you convert the following?
(i) Propan–2–ol to propanone
...
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an
example of each type
...
(a) Which of the following ores can be concentrated by froth floatation method and why?
Fe2O3, ZnS, Al2O3
...
(a) Why does p-dichlorobenzene have a higher m
...
than its o- and m- isomers?
(b) Why is (±) – Butan-2-ol is optically inactive?
23
...
Write the types of isomerism exhibited by the following complexes:
(i) [Co(NH3)5Cl]SO4
(ii) [Co(en)3]3+
(ii) [Co(NH3)6] [Cr(CN)6]
SET–III
(Questions Uncommon to Set-I and II)
1
...
What is the composition of ‘Copper matte’?
5
...
Write the IUPAC name of (CH3)2CH
...
7
...
Write the structure of p-Methylbenzaldehyde molecule
...
What is the difference between multimolecular and macromolecular colloids? Give one example of
each
...
(a) Give an example of zone refining of metals
...
Account for the following:
(i) The C—Cl bond length in chlorobenzene is shorter than that in CH3—Cl
...
18
...
Give the structures of products A, B and C in the following reactions:
LiAlH
HNO
KCN
4
2
(i) CH3 CH2 Br ¾ ¾ ¾ ® A ¾ ¾ ¾ ¾ ® B ¾ ¾ ¾¾® C
¾
¾
0 °C
NH
D
NaOH + Br
CHCl + alc
...
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
(ii) Nylon-6
(iii) Polythene
zzz
Examination Papers 291
Solutions
SET–I
1
...
Mond Process
...
4
4
...
¾
5
...
CH3 ¾ CH¾ CH2 ¾ C HO
½
CH3
7
...
There are three types of RNAs:
(i) Ribosomal RNA (rRNA)
(ii) Messenger RNA (mRNA)
(iii) Transfer RNA (tRNA)
9
...
1 mol
Mass of solvent = 1 kg
Hence, molality of glucose solution = 0
...
52 K kg mol–1 × 0
...
052 K
...
15 K at 1
...
15 +
0
...
202 K
...
L m = k ´ 1000 = 0
...
20 mol L- 1
11
...
Lyophobic sols can be easily coagulated
...
Physisorption
It is not specific in nature
...
It decreases with increase of
temperature
...
Enthalpy of adsorption is low
(20–40 kJ mol -1) in this case
...
High temperature is favourable for
adsorption
...
It is irreversible
...
13
...
(b) CO
...
(i) When PCl5 is heated, the less stable axial bonds break to form PCl 3
...
¾
(ii) Orthophosphorous acid on heating disproportionates to give orthophosphoric acid and
phosphine
...
(a) Cu has the electronic configuration 3d 10 4s 1
...
Hence, it shows +1 oxidation state
...
16
...
(i) Due to
cleave
...
unstable phenyl cation
...
H
½·
··
(i) CH3 ¾ CH2 ¾ O H + HÅ ¾® CH3CH2 ¾ O · ¾ H
··
(ii) CH3—CH2—O
Å
+
CH3—CH2—O
H
H
CH3—CH2—O—CH2—CH3 + H2O
H
H
+
(iii) CH3—CH2—O—CH2—CH3
18
...
)
(ii)
OH
Phenol
HNO3 (conc
...
(a) p-type semiconductor
...
(c) Impurity defect/cation vacancy defect
...
K 2 SO4 dissolved = 0
...
Cell reaction:
3 ´ 0
...
0821 L atm K -1 mol -1 ´ 298 K
174 g mol -1 ´ 2 L
= 5
...
44 -
[Fe 2 + ]
0
...
0591
10 - 3
log
2
(1) 2
0
...
44 + 0
...
52856V
2
22
...
(ii) This is due to filling of 4f orbitals which have poor shielding effect (lanthanoid contraction)
...
= 0
...
(i) Triamminetrichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromidobis-(ethane-1,2-diammine)cobalt(III) ion
24
...
S
...
(i)
Polymers
Buna-S
Monomer names
Monomer structures
1, 3-Butadiene,
CH2—CH—CH—CH2
Styrene
C6H5CH—CH2
Cl
½
—
CH2 — C — CH — CH2
—
(ii)
Neoprene
Chloroprene
(iii)
Nylon-6, 6
Hexamethylene diamine,
NH2—(CH2)6—NH2
Adipic acid
COOH—(CH2)4—COOH
26
...
Values shown by principal: caring, commanding and serious about the welfare of students
...
27
...
(b) Bithional acts as an antiseptic agent and reduces the odour produced by bacterial decomposition
of organic matter on the skin
...
d[ R ]
= k[ A ] 2 [ B ]
28
...
e
...
(ii) When concentration of both A and B is doubled
Rate = k [2A]2[2B] = 8k [A]2[B]
i
...
, Rate of reaction will become 8 times
...
303
k=
log
t
[R ]
[R ]0
100
10
when t = 40 minutes,
=
=
[R ]
100 - 30
7
2
...
303
k=
log
=
log 1
...
303
k=
´ 0
...
91 ´ 10 t1/2 =
3
min - 1
0
...
693
=
k
8
...
78 min
...
303
log
k
[R ]
2
...
303
100
t 90%
log 10
log
k
10
Hence t 99% = 2t 90%
Ea
(b)
Slope = = - 4250 K
2
...
303 × 8
...
375 kJ mol–1
29
...
The F—F bond
dissociation enthalpy is, however, smaller than that of Cl—Cl and even smaller than that of
Br—Br
...
Hence, the bond
dissociation enthalpy increases in the order: I 2 < F2 < Br2 < Cl 2
...
That is why the boiling point of PH3 is lower than NH3
...
(ii) This is because fluorine is the most electronegative element
...
(b) (i) Refer to Q
...
(ii) H2S2O8
O
O
S
S
O
O
O
O
OH
OH
Peroxodisulphuric acid
(H2S2O8)
30
...
(ii) Negative charge is dispersing on two electronegative oxygens in carboxylate ion whereas it
is on one oxygen atom in phenoxide ion
...
NaOH
( Aldol condensation )
4
3
2
1
2CH3 CHO ¾ ¾ ¾ ¾ ¾ ¾ ® CH3 ¾ CHOH ¾ CH2 ¾ CHO
¾
Ethanal
3 - Hydroxybutanal
+
H O / Heat
¾ 3 ¾¾® CH3 ¾ CH — CH ¾ CHO
¾
—
- H 2O
OR
(a)
(i)
2H—C—H
conc
...
(ii) Phenol and benzoic acid: Benzoic acid reacts with NaHCO3 giving CO2 gas with
effervescence, whereas phenol does not Phenol decolourises Br2 water giving white
precipitate, whereas benzoic acid does not
...
Frenkel defect
...
4-Bromo-4-methylpent-2-ene
...
The a-helix structure of proteins is stabilised by intramolecular H-bonding between C—O of one
amino acid residue and the N—H of the fourth amino acid residue in the chain
...
Neil Bartlett observed that PtF6 reacts with O2 to yield an ionic solid, O+ PtF6
2
O2 ( g) + PtF6 ( g) ¾ ¾ ® O+ [PtF6 ] ¾
2
Here, O2 gets oxidised to O+ by PtF6
...
This inspired Bartlett to
carry out the reaction between Xe and PtF6
...
Xe + PtF6
278 K
¾ ¾ ¾¾® Xe + [PtF6 ] -
Hydrolysis
7
...
CH3—CH—CH2—C—CH3
|
Cl
9
...
)
SO3H
HNO3 (conc
...
(i) Oil in water (o/w) type emulsions: In such emulsions oil is the dispersed phase and water is
the dispersion medium, e
...
, milk, vanishing cream
...
g
...
17
...
This is because the sulphide ore particles are preferentially wetted by oil and gangue
particles are preferentially wetted by water
...
FeO + SiO2
¾ ¾ ® FeSiO3
¾
Slag
Flux
18
...
(b) (±)-Butan-2-ol exist in two eventiomeric forms which rotate the plane of polarised light in
opposite directions in equal amounts
...
23
...
(i) [Co(NH3)5Cl]SO4
— Ionisation isomerism
(ii) [Co(en)3]3+
— Optical isomerism
(ii) [Co(NH3)6] [Cr(CN)6] — Coordination isomerism
SET–III
(Questions Uncommon to Set-I and II)
1
...
3
...
5
...
6
...
7
...
H3C
CHO
9
...
g
...
Multimolecular colloids
are generally lyophobic in nature
...
g
...
Macromolecular colloids are generally lyophilic in nature
...
(a) Zone refining is used for production of semiconductors and other metals of very high purity like
germanium, silicon, boron, gallium and indium
...
/Acts as an electrolyte
...
(Any one)
17
...
23(iii) of CBSE Examination Paper (All India) 2013
...
Light
2CHCl3 + O2 ¾ ¾ ¾ ® 2COCl2 + 2HCl
¾
Chloroform
18
...
H
C
OH
CH3
Propan-2-ol
(i) A = CH3CH2CN, B = CH3CH2CH2NH2, C = CH3CH2CH2OH
(ii) A = CH3CONH2, B = CH3NH2, C = CH3NC
27
...
1
...
Name the method used for refining of copper metal
...
Name two poisonous gases which can be prepared from chlorine gas
...
Write the IUPAC name of the following compound:
CH3
|
CH3— C — CH — CH3
|
|
CH3
Cl
5
...
Write the structure of n-methylethanamine
...
What are the products of hydrolysis of sucrose?
(CH
8
...
Account for the following:
(i) Schottky defects lower the density of related solids
...
10
...
Atomic radius of the metal is 125 pm
...
The standard electrode potential (E°) for Daniell cell is + 1
...
Calculate the DG° for the reaction
Zn( s) + Cu 2 + ( aq) ¾ ¾ ® Zn 2 + ( aq) + Cu( s)
¾
(1 F = 96500 C mol–1)
...
(a) For a reaction A + B ® P, the rate law is given by,
r = k[ A ]1 / 2 [ B ] 2
...
5 × 10–14 s–1
...
13
...
(b) How is wrought iron different from steel?
14
...
How are interhalogen compounds formed? What general compositions can be assigned to them?
16
...
443 K
17
...
Define thermoplastic and thermosetting polymers
...
OR
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester
...
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K
...
[R = 8
...
6021]
20
...
Give reasons for the following:
(i) Where R is an alkyl group, R3P—O exists but R3N—O does not
...
22
...
(ii) the hybridisation type
...
(Atomic no
...
Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide
...
(iii) C–X bond length in halobenzene is smaller than C–X bond length in CH3–X
...
Complete the following reactions:
(i) CH3CH2NH2 + CHCl3 + alc
...
+ HCl(aq)
(i) What class of drug is Ranitidine?
(ii) If water contains dissolved Ca2+ ions, out of soaps and synthetic detergents, which will you use
for cleaning clothes?
(iii) Which of the following is an antiseptic?
0
...
26
...
46 V and log 10n = n
...
Shanti, a domestic helper of Mrs
...
Mrs
...
The
doctor prescribed an iron rich diet and multivitamins supplement to her
...
Anuradha supported her
financially to get the medicines
...
After reading the above passage, answer the following questions:
(i) What values are displayed by Mrs
...
(iii) Give an example of a water soluble vitamin
...
(a) State Raoult’s law for a solution containing volatile components
...
00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of
benzene by 0
...
Find the molar mass of the solute
...
12 kg mol–1)
OR
(a) Define the following terms:
(i) Ideal solution
(ii) Azeotrope
(iii) Osmotic pressure
(b) A solution of glucose (C6H12O6) in water is labelled as 10% by weight
...
(a) Give reasons for the following:
(i) Mn3+ is a good oxidising agent
(ii) E o 2+
M
/M
values are not regular for first row transition metals (3d series)
...
(b) Complete the following equations:
(i) 2CrO2 – + 2H+ ¾ ¾ ®
¾
4
heat
(ii) KMnO4 ¾ ¾ ¾®
OR
(a) Why do transition elements show variable oxidation states?
(i) Name the element showing maximum number of oxidation states among the first series of
transition metals from Sc (Z = 21) to Zn (Z = 30)
...
(b) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid
metals
...
(a) How will you convert the following:
(i) Propanone to Propan-2-ol
(ii) Ethanal to 2-hydroxy propanoic acid
(iii) Toluene to benzoic acid
(b) Give simple chemical test to distinguish between:
(i) Pentan-2-one and Pentan-3-one
(ii) Ethanal and Propanal
OR
(a) Write the products of the following reactions:
Zn –
(i) CH3—C—CH3 ¾ ¾ ¾Hg ® ?
¾
¾
conc
...
Write the structure of 2-aminotoluene
...
Which aerosol depletes ozone layer?
4
...
Why?
5
...
Write the name of linkage joining two amino acids
...
Give one example of a condensation polymer
...
(a) Why does presence of excess of lithium makes LiCl crystals pink?
(b) A solid with cubic crystal is made of two elements P and Q
...
What is the formula of the compound?
14
...
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Vapour phase refining
19
...
Write the main products of the following reactions:
H PO + H O
(i) C6H5N+ Cl– ¾ ¾3¾ 2 ¾2¾ ® ?
¾
¾
2
NH2
(ii)
Br2(aq)
?
Br + NaOH
(iii) CH3—C—NH2 ¾ ¾ 2¾ ¾ ¾ ® ?
¾
||
O
27
...
(ii) O3 acts as a powerful oxidising agent
...
Examination Papers 305
SET–III
(Questions Uncommon to Set–I and II)
1
...
What is the basicity of H3PO3 and why?
3
...
What are the products of hydrolysis of lactose?
6
...
Write the structure of prop-2-en-1-amine
...
Draw the structures of the following molecules:
(i) N2 O5
(ii) XeF2
13
...
Name the principal ore of aluminium
...
19
...
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide
...
(iii) The two oxygen-oxygen bond lengths in ozone molecule are identical
...
Write the main products of the following reactions:
HNO
0°C
2
(i) CH3 CH2 NH2 ¾ ¾ ¾¾® ?
O
(ii)
S—Cl + H—N—C2H5
O
(iii)
N—H
H
?
H
O
CH3—C—Cl
Base
?
zzz
306 Xam idea Chemistry—XII
Solutions
SET–I
1
...
2
...
3
...
4
...
5
...
6
...
7
...
8
...
(i) In Schottky defect as the number of ions are missing from their normal lattice sites, the mass
decreases whereas the volume remains the same
...
(ii) This is due to availability of additional unpaired electrons on doping with phosphorous
...
For an fcc unit cell, r =
2 2
a = 2 2r = 2 ´ 1
...
5 pm
...
D r G o = - nFE o
9
...
1 V
D r G o = - 2 ´ 1
...
3 kJ mol–1
1
5
12
...
0
...
693
t1/ 2 =
=
s = 1
...
k
5
...
14
...
(b) Wrought iron is the purest form of iron whereas steel is an alloy of iron
...
Interhalogen compounds are prepared by direct combustion or by the action of halogen on lower
interhalogen compounds
...
16
...
H H
H
½ ½
½+
H — C —C — O — H
··
½ ½
H H
Slow
¾¾¾®
¬¾¾¾
H H
½ ½
··
H — C — C + + H2 O
··
½ ½
H H
Step 3: Formation of ethane by elimination of a proton
H H
½ ½
H
H
¾¾¾®
—
H — C — C + ¬¾¾¾
C—C
H
H
½ ½
Ethene
H H
+
H+
The acid used in step 1 is released in step 3
...
17
...
) NaOH
NaOH
OH
CHO
H 3O +
CHO
340 K
Salicylaldehyde
(2-Hydroxybenzaldehyde)
Phenol
(ii) Williamson synthesis:
+
R¾ X
Alkyl halide
+ Na ¾ O ¾ R¢ ¾¾® R ¾ O ¾ R¢ + NaX
Sodium alkoxide
Ether
18
...
Example, polythene, polystyrene, etc
...
These cannot be reused
...
308 Xam idea Chemistry—XII
OR
Polymers which disintegrate by themselves over a period of time due to environmental degradation by
microorganisms are called biodegradable polymers
...
19
...
éT2 – T1 ù
ú
ê
ë T1 ´ T2 û
log
k2
Ea
=
k1 2
...
303R ´ 8
...
é313 – 293 ù
ê 293 ´ 313 ú
û
ë
E a é 20 ù
19
...
6021 ´ 19
...
2177 J mol–1 or 52
...
Example, sulphur sol, gold sol
...
Example, sols of metal and their sulphides and
hydroxides, As2S3 sol, Fe(OH)3 sol
...
Example, milk, cod liver oil, etc
...
(i) N due to the absence of d-orbitals, cannot form pp - dp multiple bonds
...
So, the compound
R3N—O does not exist
...
Therefore, P forms
R3P—O in which the covalency of P is 5
...
(iii) N2 is less reactive at room temperature because of strong pp - pp overlap resulting into the
triple bond (N ºº N), consequently high bond dissociation enthalpy
...
(i) Tetrachloridonickelate(II) ion
...
(iii) Tetrahedral
...
4
0
(i) t 2 g e g
3
(ii) t 2 g e1
g
Examination Papers 309
23
...
(i) Since I - ion is a better leaving group than Br - ion, hence, CH3 I reacts faster than CH3 Br in
SN2 reaction with OH– ion
...
e
...
The
rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the
mixture optically inactive
...
The sp2 hybridised carbon is more electronegative due to greater s-character and
holds the electron pair of C–X bond more tightly than sp3 hybridised carbon with less
s-character
...
(i) CH3CH2NH2 + CHCl3 + 3KOH(alc
...
NH3Cl
+ HCl(aq)
(i) Antacid/antihistamine
...
2% phenol
...
The given cell notation in the question is incorrect
...
0591
[ Cu 2 + ]
log
2
[Ag + ] 2
o
Given: E cell = 0
...
46 –
0
...
1]
log
2
[10 –3 ] 2
= 0
...
02955 log
[ 0
...
46 – 0
...
46 – 0
...
46 – 0
...
314 V
...
46 –
0
...
0591
[10 –3 ] 2
log
2
[ 0
...
46 – 0
...
1]
= 0
...
0295 ´ 5
= 0
...
27
...
(ii) Vitamin B12
...
28
...
e
...
p = KHx
On comparing it with Raoult’s law it can be seen that partial pressure of the volalite component
or gas is directly proportional to its mole fraction in solution
...
Thus, it becomes a special case of Henry's
o
law in which KH = p A
...
12 K kg mol -1 ´ 1
...
40 K ´ 50 g
OR
(a)
(i) Ideal solution: The solution which follows Raoult’s law at all concentrations and temperatures
...
(iii) Osmotic pressure: The minimum excess pressure that has to be applied on the solution
side when the solution and solvent are separated by a semipermeable membrane to stop
osmosis
...
e
...
1000 ´ 10
1000 ´ wt%
=
m=
= 0
...
(a)
(i) Mn 3 + (3d4) is a good electron acceptor as the resulting species is more stable (3d5)
...
(iii) Due to multiple bond formation ability of oxygen with Mn in Mn2O7
...
(i) Manganese
(ii) Scandium
...
Misch metal
...
(a) (i) CH3 — C— CH3 ¾ ¾ ¾ ¾ ¾ ¾ ¾ ® CH3 — CH — CH3
Propanone
Propan - 2 - ol
CN
|
HCN
(ii) CH3 — C— O ¾ ¾ ¾ ® CH3 — C — OH
¾
|
|
H
H
COOH
|
H O/ H
¾ ¾2¾ ¾ ® CH3 — CH — OH
¾
+
2 - hydroxy propanoic acid
Ethanal
CH3
COOH
+
H /KMnO4
(iii)
Toluene
(b)
Benzoic acid
(i) Pentan-2-one and pentan-3-one: Pentan-2-one responds to iodoform test and gives yellow
coloured precipitate with I2 and NaOH, while pentan-3-one does not
...
OR
(a)
Zn –
(i) CH3—C—CH3 ¾ ¾ ¾Hg ® CH3—CH2—CH3 + H2O
¾
¾
conc
...
OH
(ii) CH3COOH >
, because CH3COO– is a more stable conjugate base
...
NH2
2
...
4
...
Hence, they are soluble in water
...
2-Bromo-4-chloropentane
...
Peptide linkage
...
Nylon-6,6/Dacron/Glyptal
...
(a) Excess of lithium leads to metal excess defect
...
These electrons diffuse into the crystal and form F-centres
...
(b) Number of P atoms per unit cell = 1 (at the body centre) ´ 1 = 1
1
Number of Q atoms per unit cell = 8 (at the corners) ´ = 1
8
Hence, the formula of the compound = PQ
...
(i) XeF6
F
(ii) H2S2O7
O
O
S
F
S
F
Xe
F
F
O
OH
F
O
O
OH
18
...
(ii) Vapour phase refining: In this, metal is converted into its volatile compound and collected
elsewhere
...
19
...
Such colloids are known as associated colloids or micelles, e
...
, soaps
and detergents
...
g
...
(iii) Adsorption: The accumulation of molecular species at the surface rather than in the bulk of a
solid or liquid is termed adsorption, e
...
, water vapours on silica gel, poisonous gases on
charcoal
...
H PO + H O
(i) C6H5N+ Cl– ¾ ¾3¾ 2 ¾2¾ ® C6H6 + N2 + H3PO3 + HCl
¾
¾
2
Examination Papers 313
NH2
NH2
Br2(aq)
(ii)
Br
Br
Br
Br + 4NaOH
(iii) CH3—C—NH2 ¾ ¾ 2¾ ¾ ¾¾® CH3—NH2 + Na2CO3 + 2NaBr + 2H2O
||
O
27
...
Due to small size and high electronegativity oxygen
exists as diatomic (O2 ) molecules
...
(ii) Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidising
agent
...
SET–III
1
...
2
...
3
...
5
...
6
...
8
...
(i) N2 O5
(ii) XeF2
F
O
N
O
O
O
Xe
N
O
F
Linear
13
...
(b) p-type semiconductor
...
Bauxite (Al2O3
...
Leaching is significant as it helps in removing the impurities like SiO2 , Fe 2 O3 , TiO2 etc
...
19
...
g
...
314 Xam idea Chemistry—XII
(ii) Peptisation: The process of conversion of a freshly prepared precipitate into a colloidal
solution by shaking it with dispersion medium in the presence of a suitable electrolyte is called
peptisation
...
(iii) Emulsion: These are the colloidal systems in which both the dispersion medium and the
dispersion phase are liquid, e
...
, milk is an emulsion of fat in water
...
27
...
That
is why it does not form pentahalide
...
(iii) Because of resonance
...
1
...
Measurement of which colligative property is preferred for determination of molar mass of biomolecules?
3
...
4
...
Arrange the following compounds in the increasing order of their acid strengths:
4-nitrophenol, phenol, 2,4,6-trinitrophenol
6
...
7
...
Write the names of monomers of the polymer
— NH - (CH2 ) 6 - NH - C- (CH2 ) 4 - C] n
[
||
||
O
O
9
...
27 ×105 mm Hg
...
10
...
The rate constant for a first order reaction is 60 s–1
...
(a) What happens when a freshly precipitated Fe(OH)3 is shaken with water containing a small
quantity of FeCl3?
(b) Why is a finely divided substance more effective as an adsorbent?
13
...
Give one example of each type of sol
...
How can you separate alumina (Al2O3) from silica present in bauxite ore? Write the chemical
equations for the reactions involved
...
Arrange the following in the order of property indicated for each set:
(a) HF, HCl, HBr, Hl — increasing acid strength
...
16
...
(a) What type of bonding helps in stabilising the a-helix structure of proteins?
(b) What is the structural difference between a nucleoside and a nucleotide?
18
...
An element with molar mass 27 g mol–1 forms a cubic unit cell with edge length 4
...
If its
density is 2
...
For a chemical reaction R ® P, the variation in the concentration ln [R] vs
...
Draw the structures of the following molecules:
(i) N2O5
(ii) H3PO2
(iii) XeF6
22
...
(ii) SnCl4 is more covalent than SnCl2
...
OR
Complete the following equations:
(i) PC13 + H2O ¾ ¾ ®
¾
heat
(iii) NaN3 ¾ ¾ ¾ ®
¾
(ii) XeF2 + PF5 ¾ ¾ ®
¾
Examination Papers 317
23
...
(b) What type of isomerism is shown by the complex [Co(NH3 ) 5 SO4 ]Br?
(c) Why is CO a stronger ligand than NH3 in complexes?
24
...
C2H5
H
C2H5
Y
Y
CH3
C2H5
H
X
(B)
Y
H
Y
Y
CH3
(A) + (B)
CH3
(A)
This can result in giving compound (A) or (B) or both
...
(a) Explain the mechanism of the following reaction:
+
H
2CH3CH2OH ¾ ¾¾® CH3CH2—O—CH2CH3 + H2O
413 K
(b) Name the reagent used in the oxidation of ethanol to ethanoic acid
...
How will you convert the following:
(i) Aniline to chlorobenzene
(ii) Ethanoic acid to methanamine
(iii) Benzene diazonium chloride to phenol
27
...
Naresh works in a multi-national company
...
Mr
...
Mr
...
With regular Yoga
sessions, Mr
...
After reading the above passage, answer the following questions:
(i) Write the values shown by Mr
...
(ii) Which class of drugs is used in sleeping pills?
(iii) Why is it not advisable to take sleeping pills without consultation with the doctor?
28
...
Mention one application of
Kohlrausch’s law
...
What is
the cell constant if conductivity of 10–3 M KCl solution at 25°C is 1
...
36 V and E o
(1 F = 96500 C mol–1)
Cu 2+ Cu
= + 0
...
Assign reasons for the following:
(i) In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of Zn is the lowest
...
(iii) Transition metals show variable oxidation states
...
34 V)
...
OR
Describe the preparation of KMnO4 from pyrolusite ore (MnO2)
...
30
...
(ii) (CH3)3C—CHO does not undergo aldol condensation
...
(b) Give a simple chemical test to distinguish between
(i) Acetophenone and Benzophenone
(ii) Benzaldehyde and Ethanal
OR
Write the structures of products of the following reactions:
O
(i)
C—Cl
+ H2
Pd/BaSO4
?
HCN
(ii) CH3— C— H ¾ ¾ ¾ ®?
¾
||
O
Conc
...
Why is glass considered a supercooled liquid?
2
...
Why?
4
...
Examination Papers 319
6
...
On what principle is chromatography based?
8
...
What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is
formed by mixing ethanol and acetone?
14
...
What is special about the following terms:
(i) Kraft Temperature
(ii) Sorption
22
...
(a) Write the formulae for the following coordination compounds:
(i) Tetraammineaquachloridocobalt (III) chloride
(ii) Potassiumtetracyanonickelate (II)
(b) Write the hybridisation of the complex [NiCl4]2–
...
How are the following conversions carried out:
(i) Aniline to fluorobenzene
(ii) Benzene diazonium chloride to benzene
(iii) Methyl chloride to ethylamine
SET–III
Questions Uncommon to Set-I and II
1
...
Write the IUPAC name of the following compound:
Cl
NO2
NO2
3
...
Why are diazonium salts of aromatic amines more stable than those of aliphatic amines?
320 Xam idea Chemistry—XII
7
...
Name the method used for the refining of Titanium metal
...
Out of two 0
...
Write the structural difference between DNA and RNA
...
Write the dispersed phase and dispersion medium of the following colloids:
(i) Cheese
(ii) Fog
21
...
23
...
How are the following conversions carried out:
(i) Aniline to iodobenzene
(ii) Ethyl nitrile to ethyl amide
(iii) Benzene diazonium chloride to benzonitrile
zzz
Examination Papers 321
Solutions
1
...
2
...
3
...
4
...
5
...
H3 C ¾ C H ¾ CH2 ¾ C ¾ H
7
...
8
...
9
...
27 ´ 105 mm Hg
p = 760 mm Hg
According to Henry’s law, p = k H x CH 4
p
760 mm Hg
\
x CH 4 =
=
= 1
...
27 ´ 10 mm Hg
Mole fraction of methane in benzene; x CH 4 = 1
...
2
(i) The bonds between chloroform molecules and molecules of acetone are dipole–dipole
interactions but on mixing, the chloroform and acetone molecules, they start forming hydrogen
bonds which are stronger bonds resulting in the release of energy
...
(ii) When a non-volatile solute is dissolved in a solvent, the vapour pressure decreases
...
[R ]0
[R ]0
2
...
as [ R ] =
t =
log
k
[R ]
10
[R ]0
2
...
303
2
...
838 ´ 10 - 2 s
\
[R ]0
k
60
60
10
10
...
(a) It is converted into colloidal state
...
Greater
the surface area, greater is the adsorption
...
(i) Lyophobic sols: Particles of dispersed phase have no affinity for dispersion medium, rather
they hate dispersion medium
...
They are irreversible
...
They
are also called extrinsic colloids
...
They are self-stabilized because of strong attractive forces operating between the suspended
particles and the dispersion medium
...
Examples: gums, gelatin,
starch, albumin, etc
...
14
...
Al2O3 is leached out as sodium
aluminate (and SiO2 too as sodium silicate) leaving impurities behind
...
At
this stage, the solution is seeded with freshly prepared samples of hydrated Al2O3 which induces the
precipitation
...
xH2 O + 2NaHCO3 ( aq)
¾
The sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to get
back pure Al2O3
...
xH2 O( s) ¾ ¾ ¾ ¾® Al 2 O3 ( s) + xH2 O( g)
OR
(a) Cu2S and FeS
...
15
...
(b) NH3 < PH3 < AsH3 < SbH3 < BiH3
...
(a) HCN
CHO
|
(CHOH) 4
|
CH2 OH
D-Glucose
(b) Br2-water
CHO
|
(CHOH) 4
|
CH2 OH
D-Glucose
CN
CH
OH
|
HCN
¾ ¾ ¾ ® (CHOH) 4
¾
|
CH2 OH
Glucose cyanohydrin
COOH
|
Br water
¾ ¾ 2 ¾¾® (CHOH) 4
¾
|
CH2 OH
Gluconic acid
17
...
(b) A nucleoside is formed of pyrimidine or purine base connected to C-1 of sugar (ribose or
deoxyribose) by a b-linkage
...
e
...
O
–
O —P—O––H 2C 5'
–
O
O
4'
H
H
3'
OH
Base
H1'
2'
H
OH
18
...
(b) Neoprene < PVC < Nylon-6
...
Density, d = Z ´ M
a3 ´ NA
or
Z =
d ´ a3 ´ NA
M
...
05 ´ 10 -8 cm
d = 2
...
022 ´ 10 23 mol -1
Substituting these values in expression (i), we get
Z =
2
...
05 ´ 10 -8 cm) 3 ´ 6
...
99 = 4
Thus, there are 4 atoms of the element present per unit cell
...
20
...
(i) N2O5
(ii) H3PO2
O
N
O
F
O
O
O
(iii) XeF6
F
N
O
P
H
OH
H
F
Xe
F
F
F
324 Xam idea Chemistry—XII
22
...
(ii) The oxidation states of central atom Sn in SnCl4 and SnCl2 are +4 and +2 respectively
...
(iii) Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidising
agent
...
(a) Dibromobis (ethane-1, 2-diamine) cobalt
(b) Ionisation isomerism
...
24
...
(a) CH3 ¾ CH2 ¾ O H + H
··
CH3—CH2—O
+
H
½·
¾® CH3 — CH2 ¾ O ·¾ H
Å
CH3—CH2—O
H
CH3—CH2—O—CH2—CH3 + H2O
H
H
H
+
CH3—CH2—O—CH2—CH3
CH3—CH2—O—CH2—CH3 + H
H
(b) Tollens’ reagent
...
(i) Aniline to chlorobenzene
+
NH2
–
Cu2Cl2/HCl
NaNO2/HCl
273-278 K
Aniline
Cl
N2Cl
Diazonium
salt
Chlorobenzene
Examination Papers 325
(ii) Ethanoic acid to methanamine
NH
D
Br / KOH
(Hoffmann bromamide
reaction)
3
CH3 COOH ¾ ¾ ¾ ® CH3 CONH2 ¾ ¾ ¾ ¾ 2 ¾ ¾ ¾ ¾ ® CH3 NH2
¾
¾
¾
Ethanoic acid
Ethanamide
Methanamine
(iii) Benzene diazonium chloride to phenol
+
–
N2Cl
OH
boil
+ H2O
+ HCl + N2
Diazonium
chloride
27
...
(ii) Tranquilizers
...
Therefore, a doctor should always be consulted before taking the medicine
...
(a) Kohlrausch’s Law: It states that the limiting molar conductivity of an electrolyte can be
represented as the sum of the individual contributions of the anion and cation of the electrolyte
...
+
Applications of Kohlrausch’s Law:
Calculation of molar conductivities of weak electrolyte at infinite dilution: For example, molar
conductivity of acetic acid (weak acid) at infinite dilution can be obtained from the knowledge of
molar conductivities at infinite dilution of strong electrolytes like HCl, CH3COONa and NaCl as
illustrated below
...
e
...
5 ´ 10 -4 S cm -1 , R = 1500 ohm
1
1
...
5 ´ 10 -4 ´ 1500 = 0
...
0591
log
2
[Cu 2 + ]
E cell = 0
...
36) -
0
...
70 – 0
...
67 V
o
DG ° = – nFE cell
DG ° = – 2 × 96500 × 2
...
23 × 105 J mol–1
= – 5
...
(i) In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc,
while in all other metals of the 3d series, electrons from the d-orbitals are always involved in
the formation of metallic bonds
...
(ii) This is due to filling of 4f orbitals which have poor shielding effect (lanthanoid contraction)
...
(iv) This is because the sum of enthalpies of sublimation and ionisation is not balanced by
hydration enthalpy
...
OR
It is prepared by fusion of pyrolusite, MnO2, with KOH in the presence of an oxidising agent like
KNO3
...
2MnO2 + 4KOH + O2 ¾ ¾ ® K2MnO4 + 2H2O
¾
3MnO42– + 4H+ ¾ ¾ ® 2MnO4– + MnO2 + 2H2O
¾
Commercially, it is prepared by alkaline oxidative fusion of MnO2 followed by the electrolytic
oxidation of manganate (VI)
...
2Mn2+ + 5S2O82– + 8H2O ¾ ¾ ® 2MnO4– + 10SO42– + 16H+
¾
Peroxodisulphate
Permanganate
Examination Papers 327
(i) MnO- + 8H+ + 5Fe 2 +
4
¾ ¾ ® Mn 2 + + 4H2 O + 5Fe 3 +
¾
–
(ii)
2MnO4
+ 16H
+
COO
+5 ½
COO–
¾ ¾ ® 2Mn 2 + + 8H2 O + 10CO2
¾
30
...
Sterically, the presence of two relatively large
substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in
aldehydes having only one such substituent
...
(ii) This is because for aldol condensation to take place, at least one a-hydrogen (i
...
, hydrogen
at carbon adjacent to carbonyl carbon) should be available, which is not present in
(CH3)3C—CHO
...
The hydrogen bonds do not break completely even in the vapour phase
...
COCH3
COONa
I2 /NaOH
+ 3NaOI
+ CHI3 ¯ + 2NaOH
Heat
Yellow ppt
...
–
+
CH3CHO + 3I2 + 3NaOH ¾ ¾ ® HCOO Na + CHI3 ¯ + 3NaI + 3H2O
¾
Iodoform
(Yellow ppt
...
KOH
(iii) 2H— C— H ¾ ¾ ¾ ¾ ¾ ® CH3 OH +
¾
||
Methanol
O
HCOOK
Potassium methanoate
328 Xam idea Chemistry—XII
O
NNHCONH2
H2NCONHNH2
(iv)
–
+
COO Na
(v)
+ H2O
NaOH, CaO
D
+ Na2CO3
SET–II
(Questions Uncommon to Set–I)
1
...
2
...
—
4
...
5-Chloro-4-methyl pent-1-ene
...
Is is based on the principle that different components of a mixture are differently adsorbed on an
adsorbent
...
Ethylene glycol and terephthalic acid
9
...
Minimum boiling azeotropes
...
(a)
(ii)
16
...
(ii) Sorption is the process in which adsorption and absorption take place simultaneously, e
...
,
dyeing of cotton fibres by azo dyes
...
(i) PCl5
(ii) H4P2O7
O
Cl
O
P
Cl
P
P
Cl
Cl
(iii) ClF3
Cl
O
HO
Cl
F
OH
OH
F
OH
F
25
...
27
...
KCN
LiAlH 4
CH3—Cl ¾ ¾ ¾ ¾ ® CH3—CN ¾ ¾ ¾ ¾ ® CH3CH2NH2
¾
¾
SET–III
1
...
2
...
3
...
4
...
OH
7
...
Vapour phase refining
...
0
...
14
...
The sugar present in DNA is 2-deoxy D-(–) ribose
...
The sugar present in RNA is D-(–)-ribose
...
DNA has double stranded a-helix structure
...
RNA has single a-helix structure
...
Dispersed phase
Dispersion medium
Liquid
(ii) Fog
Solid
Liquid
(i) Cheese
Gas
21
...
(b) Vitamin B12 (Cyanocobalamine), the antipernicious anaemia factor, is a complex of cobalt
...
(i) Solid PCl5
In solid state PCl5 exist as [PCl4]+ [PCl6]– in which the cation, [PCl4]+ is tetrahedral and the
anion, [PCl6]– is octahedral
...
Xe
O
O
O
O
O
HO
O
O
(i) Aniline to iodobenzene
+ –
N2Cl
NH2
NaNO2/HCl
273-278 K
I
+KI
Aniline
Iodobenzene
(ii) Ethyl nitrile to ethyl amide
CH3—CN
H2O/H
+
CH3—COOH
NH3,
D
CH3CONH2
(iii) Benzene diazonium chloride to benzonitrile
+
–
N2Cl
CN
KCN,
CuCN
zzz
CBSE
Examination
Paper Delhi-2014
Time allowed : 3 hours
Maximum marks : 70
General Instructions:
(i) All questions are compulsory
...
(iii) Question numbers 9 to 18 are short answer questions and carry 2 marks each
...
(v) Question numbers 28 to 30 are long answer questions and carry 5 marks each
...
Use of calculators is not allowed
...
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion
...
Which reducing agent is employed to get copper from the leached low grade copper ore?
3
...
Write the IUPAC name of compound
...
Which of the following isomers is more volatile:
o-nitrophenol or p-nitrophenol?
6
...
Arrange the following compounds in increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2
8
...
An element with density 11
...
c
...
lattice with edge length of 4 × 10–8 cm
...
(Given: NA = 6
...
Examine the given defective crystal
A+
B–
A+
B–
A+
B–
0
B–
A+
B–
+
–
+
A
B
A
0
A+
B–
A+
B–
A+
B–
Answer the following questions:
(i) What type of stoichiometric defect is shown by the crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect?
11
...
48 K (Kf = 5
...
12
...
13
...
14
...
Complete the following chemical equations:
(i) Ca3P2 +
H2O ¾ ¾ ®
¾
(ii) Cu + H2SO4 (conc
...
Write the IUPAC name of the complex [Cr(NH3)4Cl2]+
...
(i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction?
Br
Br
(a)
(b)
(ii) Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation
18
...
(a) In reference to Freundlich adsorption isotherm write the expression for adsorption of gases on
solids in the form of an equation
...
(c) Based on type of particles of dispersed phase, give one example each of associated colloid and
multimolecular colloid
...
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus
...
Account for the following:
(i) PCl5 is more covalent than PCl3
...
(iii) The two O—O bond lengths in the ozone molecule are equal
...
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a
constant volume:
SO2Cl2 ¾ ¾ ®
¾
SO2(g)
+ Cl2(g)
Time/second–1
Experiment
Total pressure/atm
1
0
0
...
7
Calculate the rate constant
...
6021, log 2 = 0
...
(i) Give two examples of macromolecules that are chosen as drug targets
...
(iii) Why is use of aspartame limited to cold foods and soft drinks?
24
...
(iii) Glucose on reaction with HI gives n-hexane
...
After the ban on plastic bags, students of one school decided to make the people aware of the
harmful effects of plastic bags on environment and Yamuna River
...
All students pledged not to use polythene
bags in future to save Yamuna River
...
(iii) Is polythene a condensation or an addition polymer?
26
...
27
...
)
28
...
1 mol L–1 KCl solution is 100 W
...
02 mol L–1KCl solution is 520 W, calculate the conductivity
and molar conductivity of 0
...
The conductivity of 0
...
29 × 10–2 W–1 cm–1
...
How much charge in terms of Faraday is required for the
reduction of 1 mol of Cu2+ to Cu?
(b) Calculate emf of the following cell at 298 K:
Mg(s) Mg 2 + (0
...
01) Cu(s)
o
[Given E cell = +2
...
(a) How do you prepare:
(i) K 2 MnO4 from MnO2 ?
(ii) Na 2 Cr2 O7 from Na 2 CrO4 ?
(b) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state
...
(iii) Actinoid elements show wide range of oxidation states
...
Why does it show so?
(ii) Which transition metal of 3d series has positive E o (M 2 + / M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidising agent and why?
(iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state
...
(a) Write the products of the following reactions:
(i)
H+
O + H2N — OH ¾ ¾¾®
(ii) 2C 6 H5 CHO + conc
...
(ii) Carboxylic acid is a stronger acid than phenol
...
Give one example each of sol and gel
...
Write the IUPAC name of the compound
...
Some liquids on mixing form ‘azeotropes’
...
Arrange the following in increasing order of basic strength
...
Which component of starch is a branched polymer of a-glucose and insoluble in water?
9
...
What is the effect of temperature on the solubility of a gas in a liquid?
10
...
Write the principle behind the following methods of refining:
(i) Hydraulic washing
(ii) Vapour phase refining
22
...
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V)
...
(iii) Noble gases have very low boiling points
...
(i) Name the sweetening agent used in the preparation of sweets for a diabetic patient
...
(iii) Give two examples of macromolecules that are chosen as drug targets
...
(i) Deficiency of which vitamin causes rickets?
(ii) Give an example for each of fibrous protein and globular protein
...
336 Xam idea Chemistry—XII
SET–III (Questions Uncommon to Set-I and II)
1
...
2
...
CH3—CH—CH2— C — CH3
|
||
OH
O
3
...
Arrange the following in increasing order of basic strength:
C6H5NH2, C6H5NHCH3,C6H5N(CH3)2
6
...
9
...
What is the similarity between
Raoult’s law and Henry’s law?
10
...
Write the principles of the following methods:
(i) Froth floatation method
(ii) Electrolytic refining
20
...
Account for the following:
(i) Sulphur in vapour form exhibits paramagnetic behaviour
...
(iii) H3PO2 is a stronger reducing agent than H3PO3
...
(i) What are disinfectants? Give an example
...
(iii) What are anionic detergents? Give an example
...
(i) Deficiency of which vitamin causes scurvy?
(ii) What type of linkage is responsible for the formation of proteins?
(iii) Write the product formed when glucose is treated with HI
...
Oil in water emulsion: milk, vanishing cream
...
2
...
In both of the given complexes, Co is in +3 oxidation state
...
4
...
o-nitrophenol, is more volatile as intermolecular hydrogen bonding causes association molecules in
p-nitrophenol
...
Two solutions which have the same osmotic pressure at a given temperature are called isotonic solutions
...
The solubility increases in the order in which molecular mass decreases with increase in hydrogen
atoms on nitrogen atom which undergo hydrogen bonding, i
...
,
C 6 H5 NH2 < (C 2 H5 ) 2 NH < C 2 H5 NH2
8
...
9
...
2 g cm–3, a = 4 ´ 10 -8 cm, NA = 6
...
d ´ a3 ´ NA
z
11
...
022 ´ 10 23 mol -1
= 107
...
9 u
4
(i) Schottky defect
(ii) Decreases
(iii) This type of defect is shown by ionic compounds which have high coordination number and
small difference in size of cations and anions e
...
NaCl, KCl etc
...
Here, DTf = 0
...
12 K kg mol–1
...
48 K ´ 256 g mol -1 ´ 75 g
Mass of solute, WB =
= 1
...
12 K kg mol –1 ´ 1000 g kg –1
338 Xam idea Chemistry—XII
12
...
The important characteristics of an ideal solution are
(i) The enthalpy of mixing of pure components to form the solution is zero i
...
, D mix H = 0
(ii) The volume of mixing is zero i
...
, D mix V = 0
13
...
Order
Molecularity
(i) It is the sum of the powers of the concentration
of the reactants in the rate law expression
...
(ii) It is determined experimentally
...
(iii) It can be zero or a fraction
...
(iv) Order is applicable to elementary as well as
complex reactions
...
For complex reactions it has no
meaning
...
(i) Zone refining method is based on the principle that the impurities are more soluble in the melt
than in the solid state of the metal
...
The adsorbed components are removed by using
suitable solvents
...
(i) Ca 3 P2 ( s) + 6H2 O(l) ¾¾® 2PH3 + 3Ca(OH) 2
(ii) Cu + 2H2SO4(conc
...
As the
atomic size increases down the group the E–H (E = F, Cl, Br, I) bond length increases and
hence the bond dissociation enthalpy increases in the reverse order i
...
,
HI < HBr < HCl < HF
(ii) H2O < H2S < H2Se < H2Te
The increase in acidic character from H2 O to H2 Te is due to decrease in bond enthalpy for
dissociation of H - E (E = O, S, Se, Te) bond down the group
...
[Cr(NH3 ) 4 Cl 2 ] + : Tetraamminedichloridochromium (III) ion
...
(i)
, 2-bromobutane is a chiral molecule
...
(ii) (a) SN2 reaction occurs with inversion of configuration
...
18
...
(a) Freundlich adsorption isotherm equation for adsorption of gases on solids:
x
= kp1 / n ( n > 1)
m
x
1
or
log
= log k + log p
m
n
where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p, k and n are
constants which depends on the nature of the gas and adsorbent at a particular temperature
...
e
...
(ii) They are quite stable and are not easily coagulated
...
Multimolecular colloid
Sulphur sol, gold sol
O
20
...
P
P
60°
P
P
340 Xam idea Chemistry—XII
Red phosphorus has polymeric structure in which P4 tetrahedra are linked together through P—P
covalent bond to form chain
...
P—P
P
P—
P
(i) The oxidation state of central atom phosphorus is +5 in PCl5 whereas it is +3 in PCl3
...
(ii) Iron reacts with HCl to form FeCl2 and H2
...
(iii) Ozone is a resonance hybrid of the following two main structures:
O
O
O
O
O
O
As a result of resonance, the two O—O bond lengths in O3 are equal
...
303
22
...
4 atm, t = 100 s, Pt = 0
...
303
0
...
4 - 0
...
303
0
...
303
2
...
6021
100
0
...
386 ´ 10 -2 s -1
23
...
(ii) Antiseptics are the chemical substances which prevent the growth of microorganism or may
even kill them but are not harmful to living human tissues, e
...
, dettol, soframycin, boric acid,
hydrogen peroxide etc
...
24
...
25
...
(ii) The polymers which can be broken into small segments by enzyme catalysed reactions or to
some extent by oxidation over a period of time are called biodegradable polymers
...
g
...
(iii) Polyethene is an addition polymer
...
(a) SN2 Mechanism
(i)
–
H + + Br
H–Br
+
CH3–CH2 O H + H +
CH3–CH2 O H2
Ethyl oxonium ion
(Protonated 1° alcohol)
CH3
–
(ii)
+
d
Br + CH2–OH2
CH3
–
d
+
Br…CH2…OH2
Br–CH2–CH3 + H2O
Ethyl bromide
Transition state
–+
ONa
OH
CHCl2
CHCl3 + NaOH (aq)
(b)
–+
ONa
OH
CHO
NaOH
CHO
H+
Salicylaldehyde
Phenol
Intermediate
27
...
(a)
(i) The molar conductivity when concentration approaches to zero is called limiting molar
conductivity
...
directly into electrical energy
...
(b) For 0
...
29 × 10–2 W–1 cm–1, Resistance, R = 100 W
Cell constant = Conductivity × resistance
= 1
...
29 cm–1
For 0
...
29 cm–1,
Cell constant
Conductivity, k =
Resistance
=
1
...
00248 W–1 cm–1
520 W
Molar conductivity, L m =
=
Conductivity ( k) ´ 1000 cm 3 L-1
Molarity
0
...
02 mol L-1
= 124 W -1 cm 2 mol -1
OR
(a) It states that the amount of chemical reaction which occurs at any electrode during electrolysis
by a current is proportional to the quantity of electricity passed through the electrolyte (solution
or melt)
...
(b) At anode:
Mg ¾® Mg 2 + + 2e –
At cathode: Cu 2 + + 2e –
Mg + Cu
2+
¾® Cu
¾® Mg 2 + + Cu
o
Nernst equation: Ecell = E cell –
[Mg 2 + ]
0
...
71 V, [Mg2+] = 0
...
01 M, n = 2
0
...
Ecell = 2
...
71 - 0
...
01
= 2
...
(a) (i) Pyrolusite is fused with KOH in the presence of atmospheric oxygen to give K 2 MnO4
...
2Na 2 CrO4
Sodium chromate
+ H2 SO4 ¾ ¾ ®
¾
Na 2 Cr2 O7
Sodium dichromate
+ Na 2 SO4 + H2 O
Examination Papers 343
(b)
(i) The electronic configuration of Mn 2 + is [Ar] 3d5 which is half-filled and hence stable
...
On the other hand, Fe 2 + has electronic
configuration [Ar] 3d6
...
Hence Mn2+ is more stable than Fe2+ towards oxidation
...
That is why, the enthalpy of atomisation of
zinc is the lowest in the series
...
OR
(i) Manganese (3d 5 4s 2 ) shows maximum number of oxidation states as its atoms have 5 unpaired
electrons in 3d orbitals
...
(ii) Copper has positive Eo (M2+/M) value, as high energy ( D a H + Di H ) to transform Cu(s) to
Cu2+(aq) is not balanced by hydration enthalpy
...
(iv) Europium, Eu
...
(v) MnO- + 8H+ + 5e - ¾ ¾ ® Mn 2 + + 4H2 O
¾
4
30
...
)
Cl2/P
(iii) CH3COOH
+
– +
C6H5COO Na
Sodium benzoate
Cl – CH2 – COOH + HCl
2-chloro
ethanoic acid
(b)
(i) Benzoic acid being an acid reacts with NaHCO3 solution to give brisk effervescence due
to evolution of CO2 while benzaldehyde does not response to this test
...
CH3CH2COO– + 2Ag + 4NH3 + 2H2O
CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH –
Propanal
Silver
mirror
344 Xam idea Chemistry—XII
CH3COCH3
Tollens reagent
No silver mirror
Propanone
OR
(a)
(i) The methyl group due to its +I effect reduce the magnitude of positive charge on carbonyl
carbon atom
...
Since in
acetaldehyde there is one methyl while in acetone there are two methyl groups attached to
carbonyl group therefore acetaldehyde is more reactive than acetone towards nucleophilic
addition with HCN
...
(b)
(i) Wolff–Kishner reduction
NH NH
2
C — O ¾ ¾ ¾ ¾2 ®
—
¾
- H 2O
Aldehyde
or ketone
CH3
Hydrazone
NH NH
- H 2O
2
C — O ¾ ¾ ¾ ¾2 ®
—
¾
CH3
KOH / Ethylene glycol
D
C — NNH2 ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ®
—
¾
CH3
C — NNH2
—
CH2
Hydrocarbon
+ N2
KOH / Ethylene glycol
¾¾¾¾¾¾¾¾®
D
CH3
Acetone
CH3 ¾ CH2 ¾ CH3 + N2
Propane
(ii) Aldol condensation
O
½
½
CH3 ¾ C ¾ H + H ¾ CH2 ¾ CHO
r
dil
...
KOH
¾¾¾¾¾ ®
¾
Formaldehyde
Methyl alcohol
CHO
2
CH3 OH
Benzaldehyde
Potassium formate
–
+
COO Na
CH2OH
Conc
...
Sol: Paints, cell fluids
Gel: Butter, cheese
3
...
Azeotropes are the binary mixtures of liquids having the same composition in liquid and vapour phase
and boil at a constant temperature e
...
a mixture of 68% nitric acid and 32% water by mass
...
C 6 H5 NH2 < C 6 H5 NHCH3 < C 6 H5 CH2 NH2
7
...
It states that “the partial pressure of the gas in vapour phase (p) is directly proportional to the mole
fraction of gas (x) in the solution”
...
The solubility of a gas in liquid decreases with rise in temperature as dissolution of a gas in a liquid is
an exothermic process
...
(i) A reaction which is not truely of first order but under certain conditions becomes a reaction of
first order is called pseudo first order reaction e
...
acid hydrolysis of ethyl acetate
...
(ii) The half life (t ½ ) of a reaction is the time in which the concentration of reactant is reduced to
one half of its initial concentration [R]o
...
693
For a first order reaction t ½ =
i
...
, independent of [R]o
...
e
...
2k
11
...
(ii) Vapour phase refining: In this, metal is converted into its volatile compound and collected
elsewhere
...
So, the two requirements are:
(a) The metal should form a volatile compound with an available reagent
...
346 Xam idea Chemistry—XII
22
...
Therefore according to VSEPR Theory XeF2 should be linear
...
Therefore according to VSEPR theory BrF3 should be slightly bent “T”
...
(i) Due to inert pair effect +3 oxidation state of Bi is more stable than its +5 oxidation state while
+5 oxidation state of Sb is more stable than its +3 oxidation state
...
(ii) N–N single bond is weaker than P–P single bond due to large interelectronic repulsion between
the lone pairs of electrons present on the N atoms of N–N bond having small bond length
...
Hence, they have very low
boiling points
...
(i) Saccharin, sucrolose
...
A few examples of antibiotics are chloramphenicol, ofloxacin, penicillin
...
27
...
Lyophobic sol — Gold sol, As2S3 sol
Lyophilic sol — Sol of starch, sol of gum
2
...
Hydrogen bonds
5
...
C12 H22 O11 + H2 O ¾ ¾¾®
Sucrose
C 6 H12 O6
D -(+) - Glucose
+ C 6 H12 O6
D -(–) - Fructose
9
...
According to Raoults law, for a volatile
component, A of the solution
o
0
PA µ x A or PA = PA x A , where PA is the vapour pressure of pure component A
...
e
...
Thus the similarity between Raoult’s law and Henry’s law is that both state that the partial vapour
pressure of the volatile component or gas is directly proportional to its mole fraction in the solution
...
(i) Rate constant, k may be defined as the rate of reaction when concentration of each reactant is
taken as unity
...
10 (ii) Set-II CBSE Delhi 2014
...
(i) This method of concentration of ore is based on the principle that the surface of sulphide ores is
preferentially wetted by oils while that of gangue is preferentially wetted by water
...
A strip of same
metal in pure form is used as cathode
...
On passing electric current, metal ions from the electrolyte solution are
deposited at the cathode in the form of pure metal while an equivalent amount of metal
dissolves from the anode and goes into the electrolyte solution as metal ions, i
...
,
At cathode: M n + ( aq) + ne – ¾ ¾ ® M( s)
¾
At anode:
M( s) ¾ ¾ ® M n + ( aq) + ne –
¾
The voltage applied for the electrolysis is such that the impurities of more electropositive
metals remain in the solution as ions where impurities of less basic metals settle down under
the anode as anode mud
...
, are refined by this method
...
(a)
(i) There are four bond pairs and two lone pairs of electrons around
central Xe atom in XeF4
...
(ii) O
O
O
O
N
O
N
O
O
(N2O5)
F
Xe
O
O
N
F
F
O
(XeF4) Square planar
N
F
348 Xam idea Chemistry—XII
22
...
(ii) The oxidation of central atom Sn is +4 in SnCl4 while it is +2 in SnCl2
...
(iii) Acids which contains P—H bonds have reducing character
...
23
...
For example, chlorine in concentrations of
0
...
4 ppm is used for sterilisation of water to make it fit for drinking
...
(iii) Anionic detergents: These are so named because large part of their molecules are anions and it
is the anionic part of the molecule which is involved in the cleansing action
...
For example, sodium lauryl sulphate,
sodium dodecylbenzene sulphonate, etc
...
24
...
CHO
|
HI
(CHOH) 4 ¾ ¾¾® CH3 CH2 CH2 CH2 CH2 CH3
D
|
n - Hexane
CH2 OH
zzz
CBSE
Examination
Paper All India-2014
Time allowed : 3 hours
Maximum marks: 70
General Instructions: Same as CBSE Examination Paper Delhi-2014
...
What is the effect of temperature on chemisorption?
2
...
What is the basicity of H3PO3?
4
...
Which of the following is a natural polymer?
Buna-S, Proteins, PVC
6
...
7
...
Write the structure of p-methylbenzaldehyde
...
An element with density 2
...
c
...
unit cell with edge length 4 × 10–8 cm
...
(Given: NA = 6
...
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl?
OR
How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell
11
...
Why does the conductivity of a solution
decrease with dilution?
350 Xam idea Chemistry—XII
12
...
time (t) plot is given as
R
t
(i) Predict the order of the reaction
...
Explain the principle of the method of electrolytic refining of metals
...
14
...
Draw the structures of the following:
(i) XeF2
(ii) BrF3
16
...
Write the mechanism of the following reaction:
HBr
CH3CH2OH ¾ ¾ ¾ ® CH3CH2Br + H2O
¾
18
...
(a) Calculate D r G o for the reaction
Mg(s) + Cu2+(aq)
¾¾®
¾
Mg2+(aq) + Cu(s)
o
Given: E cell = +2
...
20
...
4
2
100
0
...
(Given: log 4 = 0
...
3010)
21
...
Examination Papers 351
22
...
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur
...
23
...
(ii) What type of isomerism is exhibited by the complex [Co(en)3]3+?
(en = ethane-1,2-diamine)
(iii) Why is [NiCl4]2– paramagnetic but [Ni(CO)4] is diamagnetic?
(At nos
...
(a) Draw the structures of major monohalo products in each of the following reactions:
PCl5
(i)
CH2OH
(ii)
CH2— CH = CH2 + HBr
(b) Which halogen compound in each of the following pairs will react faster in SN2 reaction?
(i) CH3Br or CH3I
(ii) (CH3)3 C—C1 or CH3—C1
25
...
(ii) Aniline does not undergo Friedel–Crafts reaction
...
OR
Give the structures of A, B and C in the following reactions:
Sn +
NaNO2 + HCl
H O
(i) C6H5NO2 ¾ ¾ ¾HCl ® A ¾ ¾ ¾ ¾ ¾ ¾® B ¾ ¾2¾ ® C
¾
¾
¾
273K
H 2O / H +
NH 3
Br
(ii) CH3CN ¾ ¾ ¾ ¾¾® A ¾ ¾ ¾ ® B ¾ ¾ 2 +KOH ® C
¾
¾¾¾
¾
D
26
...
On the occasion of World Health Day, Dr
...
After check-up, he was shocked to see that most of the farmers suffered
from cancer due to regular exposure to pesticides and many were diabetic
...
Dr
...
On the suggestions of NHRC, the government decided to provide medical
care, financial assistance, setting up of super-speciality hospitals for treatment and prevention of
the deadly disease in the affected villages all over India
...
Satpal
(b) NHRC
(ii) What type of analgesics are chiefly used for the relief of pains of terminal cancer?
(iii) Give an example of artificial sweetener that could have been recommended to diabetic
patients
...
(a) Define the following terms:
(i) Molarity
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the
same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–l) in water
...
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason
...
What
would be the molality and molarity of the solution?
(Density of solution = 1
...
(a) Complete the following equations:
(i) Cr2O72– + 2OH– ¾ ¾ ®
¾
(ii) MnO4– + 4H+ + 3e– ¾ ¾ ®
¾
(b) Account for the following:
(i) Zn is not considered as a transition element
...
(iii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple
...
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state
...
: Mn = 25, Cr = 24)
30
...
(ii) Carboxylic acids do not give reactions of carbonyl group
...
Why is adsorption always exothermic?
2
...
3
...
Based on molecular forces what type of polymer is neoprene?
5
...
Write the structure of 4-chloropentan-2-one
...
Write the name of monomers used for getting the following polymers:
(i) Terylene
(ii) Nylon-6, 6
10
...
Complete the following equations:
(i) Ag + PCl5
(ii) CaF2 +
¾¾®
¾
H2SO4
¾¾®
¾
12
...
(i) Write the type of magnetism observed when the magnetic moments are oppositively aligned
and cancel out each other
...
Define the following terms:
(i) Fuel cell
(ii) Limiting molar conductivity (L°m )
19
...
What are the dispersed phase and dispersion medium in milk?
2
...
354 Xam idea Chemistry—XII
3
...
Which of the following is a fibre?
Nylon, Neoprene, PVC
6
...
8
...
9
...
H2SO4
(ii) XeF2 + H2O
¾¾®
¾
¾¾®
¾
10
...
Write the name of monomers used for getting the following polymers:
(i) Teflon
(ii) Buna-N
13
...
(ii) Which stoichiometric defect decreases the density of the crystal?
14
...
Write the principle behind the froth floatation process
...
Define the following terms:
(i) Nucleotide
(ii) Anomers
(iii) Essential amino acids
zzz
Examination Papers 355
Solutions
SET–I
1
...
The initial increase is due to
the fact that heat supplied acts as activation energy
...
2
...
It reduces Ag + to Ag and itself get oxidised to
Zn 2 +
...
As H3PO3 contains two ionizable P–OH bonds
...
O
P
O
OH
OH
H3PO3
4
...
Proteins
6
...
7
...
H+
C12 H22 O11 + H2 O ¾ ¾¾® C 6 H12 O6 + C 6 H12 O6
Sucrose
8
...
For f
...
c
...
8 gcm–3, NA = 6
...
8 g cm -3 ´ 6
...
98 g mol–1
M=
356 Xam idea Chemistry—XII
10
...
(ii) Schottky defect
...
It is much smaller than the size of spheres in the close packing
...
The size of the octahedral
void is smaller than that of the spheres in the close packing but larger than the octahedral void
...
A unit cell is the smallest portion of a crystal lattice, which when repeated over and again in
different directions produces the complete crystal lattice
...
Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the
sum of the individual contributions of cation and anion of the electrolyte
...
Conductivity of a solution decreases with dilution
...
12
...
(i) Order of the reaction is zero
...
A strip of same metal in
pure form is used as cathode
...
On passing electric current, metal ions from the electrolyte solution are deposited at the
cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and
goes into the electrolyte solution as metal ions, i
...
,
At cathode:
M n + ( aq) + ne – ¾ ¾ ® M( s)
¾
At anode:
M( s) ¾ ¾ ® M n + ( aq) + ne –
¾
The voltage applied for the electrolysis is such that the impurities of more electropositive metals
remain in the solution as ions whereas impurities of less basic metals settle down under the anode as
anode mud
...
, are refined
by this method
...
+
H2O
¾¾®
¾
No reaction
(ii) XeF4
15
...
Therefore according to VSEPR theory XeF2 should be
linear
...
Therefore according to VSEPR theory BrF3 should be slightly bent “T”
...
(i) Reimer–Tiemann reaction:
–+
ONa
OH
– +
ONa
CHCl2
CHCl3 + NaOH (aq)
OH
CHO
NaOH
H3O+
CHO
340 K
Salicylaldehyde
(2-hydroxybenzaldehyde)
Phenol
(ii) Williamson synthesis:
+
R¾ X
Alkyl halide
–
+ Na ¾ O ¾ R¢ ¾¾® R ¾ O ¾ R¢ + NaX
Sodium alkoxide
+
–
Ether
CH3 ¾ CH2 ¾ Br + Na ¾ O ¾ CH2 ¾ CH3 ¾® CH3 ¾ CH2 ¾ O ¾ CH2 ¾ CH3 + NaBr
Ethyl bromide
Sodium ethoxide
Diethyl ether
(Symmetrical ether)
17
...
18
...
(a) Mg(s) + Cu2+(aq) ¾ ¾ ® Mg2+(aq) + Cu(s)
¾
o
D r G o = – nFE cell
o
Here, n = 2, F = 96500 C mol–1 and E cell = 2
...
71 V
= – 523030 J mol–1
= – 523
...
Refer to Ans
...
21
...
Emulsion can be classified into two types
...
g
...
(ii) Water in oil (W/O) type emulsion: In this type of emulsions water acts as disperse phase and
oil acts as dispersion medium e
...
, butter, cod liver oil, cold cream
...
(i) Phosphorus due to presence of d orbitals in its valence shell forms pp - dp multiple bonds
...
Therefore, phosphorus form
R3P—O in which its covalency is 5
...
Therefore,
(CH3 ) 3 N—O does not exist
...
Thus when an electron is added to isolated
gaseous O atom, the interelectronic repulsions encountered in the smaller 2p-orbitals of O are
larger than those encountered in the larger 3p-orbitals of S
...
Hence oxygen has less electron
gain enthalpy with negative sign than sulphur
...
Since H3PO2 contains two P–H
bonds while H3PO3 contains only one P–H bond therefore H3PO2 is a stronger reducing agent
than H3PO3
...
(i) Tetraamminedichloridochromium (III) chloride
...
3+
en
en
3+
en
Co
en
Co
en
d-Tris-(ethane1,2-diamine)
cobalt (III)
en
l-Tris-(ethane1,2-diamine)
cobalt (III)
(iii) In complex [NiCl4]2–, Ni is in +2 oxidation state with the electronic configuration 3d8 4s0
...
In [Ni(CO)4], Ni is in zero oxidation state with the electronic configuration
3d8 4s2
...
Since there is no unpaired electron in the complex Ni(CO)4, therefore, it is diamagnetic
...
(a)
PCl5
(i)
CH2OH
(ii)
CH2— CH = CH2 + HBr
CH2Cl + POCl3 + HCl
CH2— CH — CH3
Br
(b)
(i) CH3—I
...
(ii) CH3 Cl
...
25
...
Hence there is no hydrogen bonding in
tertiary amines
...
Examination Papers 359
(ii) Aniline being a Lewis base reacts with AlCl3 a Lewis acid and catalyst used in Friedel crafts
+
-
reaction to form salt C6H5NH2AlCl3
...
It reduces the electron density in the
benzene ring as a result of this aniline does not undergo Friedel-crafts reaction
...
Here the combination of three factors, +ve I effect of CH3 groups, hydrogen
bonding and steric hindrance favour greater stability for ammonium cation of dimethyl amine
than ammonium cation of trimethyl amine
...
OR
+
–
(i)
NH2
OH
N NCl
A = C6H5NH2
Aniline
O
||
(ii) A = CH3 — C— OH ,
Ethanoic acid
+
, B = C6H5N2Cl
–
Benzene
diazoniumchloride
, C = C6H5OH
Phenol
O
||
B = CH3 — C— NH2 , C = CH3 — NH2
Ethanamide
Methanamine
O
||
26
...
(ii) The specific sequence in which various a-amino acids present in a protein are linked to one
another is called its primary structure
...
(iii) Denaturation of Proteins: When a protein in its native form is subjected to a change, such as
change in temperature or change in pH, the hydrogen bonds are disturbed
...
This is called
denaturation of protein
...
g
...
27
...
Satpal are concern for the health of others, dedicated towards
work, kind, compassionate
...
(ii) Narcotic analgesics such as morphine, heroin, codeine are used for the relief of pains of
terminal cancer
...
28
...
Moles of solute
Molarity =
Volume of solution (in litre)
(ii) Molal elevation constant may be defined as the elevation in boiling point when one mole
of solute is dissolved in 1000 grams of the solvent
...
In pure ethanol hydrogen bond exist between the molecules
...
Weakening of molecular
interactions leads to increase in vapour pressure resulting in positive deviation from Raoults
law
...
056 mol
180 g mol -1
Mass of solution
Volume of solution =
Density of solution
100 g
=
= 83
...
2 g mL-1
83
...
083 L
1000
Moles of solute
Molarity =
Volume of solution in litre
0
...
67 mol L–1
0
...
09 kg
1000 g kg -1
Moles of glucose
Molality =
Mass of water in kg
0
...
62 mol kg–1
0
...
(a)
2
(i) Cr2O7 – + 2OH– ¾ ¾ ® 2CrO2 – + H2O
¾
4
(ii) MnO– + 4H+ + 3e 4
(b)
Heat
¾ ¾ ¾ ® MnO2 + 2H2O
¾
(i) As zinc atom has completely filled d orbitals (3d 10 4s 2 ) in its ground state as well as in its
oxidised state, therefore zinc is not considered as transition element
...
= Presence of vacant orbitals of appropriate energy which can accept lone pairs of
electrons donated by ligand
...
Due to this the third ionisation energy of Mn is very high
...
OR
(i)
S
...
Lanthanoids
Actinoids
(i)
Atomic or ionic radii does not show much Atomic or ionic radii show much variation
...
(ii)
Besides +3 oxidation state, they show +2 and Besides +3 oxidation states, they show higher
+4 oxidation states in few cases
...
(ii) Cerium, Ce ([Xe] 4f 2 5d0 6s2)
...
(iii) MnO– + 8H+ + 5e 4
¾¾®
¾
Mn 2 + + 4H2O
(iv) Mn 3 + (3d4 4s0) has 4 unpaired electrons while Cr3+(3d3 4s0) has 3 unpaired electrons therefore
Mn3+ is more paramagnetic than Cr3+
...
(a) (i) CH3
pH 9-10
C—O
+ HCN
C
H
H
CN
Ethanal
Hydrogen cyanide
(ii) CH3
C—O
H
+ H2N—OH
Ethanal
O
||
(iii) 2CH3 — C— H
Ethanal
Ethanal cyanohydrin
H
+
pH 3
...
6C6H5OH + FeCl3 ¾ ¾ ® [Fe(OC6H5)6]3– + 3H+ + 3HCl
¾
Violet coloured complex
3C6H5COOH + FeCl3 ¾ ¾ ® (C6H5COO)3Fe + 3HCl
¾
Benzoic acid
Ferric benzoate
(Buff coloured ppt
...
CH3 CH2 CHO + 2[Ag(NH3 ) 2 ] + + 3OH- ® CH3 CH2 COO- + 2Ag ¯ + 4NH3 + 2H2 O
Silver
mirror
Propanal
CH3 COCH3
Tollens' reagent
No silver mirror
¾¾¾¾¾¾ ®
¾
Propanone
OR
(a)
(i) Because of –I effect of Cl atom in ClCH2COOH and +I effect of CH3 group in CH3COOH
the electron density in the O—H bond in ClCH2COOH is much lower than CH3COOH
...
Hence ClCH2COOH acid is stronger acid
than CH3COOH
...
+
C
O
III
–
C—O
IV
Because of contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is
electrophilic
...
As carbonyl carbon of carboxyl group is less
electropositive than carbonyl carbon in aldehydes and ketones therefore carboxylic acids
do not give nucleophilic addition reactions of aldehydes and ketones
...
KOH
¾¾¾¾¾ ®
¾
Formaldehyde
CH3 ¾ OH + HCOO- K +
Methyl alcohol
CHO
CH2
Benzaldehyde
–+
COO Na
OH
conc
...
Adsorption occurs with decrease in entropy i
...
, DS is –ve
...
This can be possible only when DH is –ve for the process i
...
,
adsorption is exothermic
...
Mond process (Vapour phase refining)
3
...
To become more stable by acquiring noble gas configuration having 8 electrons in the valence
shell of N, it undergoes dimerisation to form N2O4
...
Elastomer
5
...
C12 H22 O11
H+
+ H2 O ¾ ¾¾®
Maltose
C 6 H12 O6
D -(+) - Glucose
+
C 6 H12 O6
D -(+) - Glucose
Cl
O
|
||
6
...
10
...
Silica acts as a flux in the extraction of
copper from copper matte to remove ferrous oxide as ferrous silicate slag
...
The role of a depressant is to prevent
one type of sulphide particle from forming the froth with air bubble
...
Under these
conditions, only PbS forms froth and hence is separated from ZnS
...
(i) 2Ag
+ PCl5
(ii) CaF2
11
...
F
F
Xe
F
F
(XeF4) Square planar
(ii) Perchloric acid
H
O
Cl
O
O
O
HClO4
13
...
(i) A fuel cell is a device which converts the energy produced during the combustion of fuels like
hydrogen, methanol, methane etc
...
One of the most successful
fuel cell is H2—O2 fuel cell
...
It is represented by L°m
...
® 0
(i) The linkage between two monosaccharides through oxygen atom in an oligosaccharide or a
polysaccharide is known as glycosidic linkage
...
5°) but after hydrolysis it gives an equimolar mixture of
D-(+)-glucose and D-(–)-fructose, which is laevorotatory
...
Examination Papers 365
(iii) Carbohydrates which on hydrolysis give two to ten molecules of monosaccharides are called
oligosaccharides e
...
, sucrose
...
In milk liquid fat acts as dispersed phase and water acts as dispersion medium
...
Electrolytic refining
3
...
Therefore it acts as a
Lewis base
...
Nylon
6
...
C 6 H12 O6
+
Glucose
C 6 H12 O6
Galactose
COOH
OH
2-hydroxy-benzoic acid
9
...
) ¾ ¾ ® CO2 + 2SO2 + 2H2O
¾
(ii) 2XeF2(s) + 2H2O(l) ¾ ¾ ® 2Xe(g) + 4HF(aq) + O2(g)
¾
10
...
(i) Tetrafluoroethene
(ii) 1, 3–butadiene and acrylonitrile
13
...
(i) Molar conductivity, L m of a solution at a dilution V is defined as the conductance of all the
ions produced from one gram mole of the electrolyte dissolved in V cm3 of the solution when
the electrodes are one centimetre apart and the area of the electrodes is so large that the whole
of the solution is contained between them
...
(ii) Secondary batteries are those batteries which can be recharged by passing electric current
through them and hence can be used over again e
...
lead storage battery
...
366 Xam idea Chemistry—XII
This method of concentration of ore is based upon the principle that the surface of sulphide ore is
preferentially wetted by oils while that of gangue is preferentially wetted by water
...
g
...
23
...
e
...
When nucleoside is linked to phosphoric acid at 5’
position of sugar moiety, we get a nucleotide
...
(iii) The amino acids which cannot be synthesised in our body and must be obtained through diet
are known as essential amino acids e
...
, valine, lysine, histidine
...
SET–I
1
...
What type of forces are responsible for the occurrence of physisorption?
3
...
What type of isomerism is shown by the following complex:
[Co(NH3)6] [Cr(CN)6]
5
...
c
...
unit cell
...
Write the IUPAC name of the following compound:
OH
CHO
7
...
Name the two components of a-glucose which constitute starch
...
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5
...
What mass of nickel will be deposited at the cathode?
(Given: At
...
7 g mol–1, 1 F = 96500 C mol–1)
10
...
Write the expression of half-life for
(i) zero order reaction and
(ii) first order reaction
...
Write the chemical reactions involved in the extraction of silver from silver ore
...
Name the two most important allotropes of sulphur
...
(ii) Why is K a2 << K a1 for H2 SO4 in water?
13
...
Write the state of hybridisation, shape and IUPAC name of the complex [CoF6 ] 3 -
...
of Co = 27)
15
...
(ii) chlorobenzene is treated with CH3 COCl in presence of anhydrous AlCl 3
...
(a) Which alkyl halide from the following pairs would you expect to react more rapidly by S N 2
mechanism and why?
CH3 — CH2 — CH— CH3
CH3 — CH2 — CH2 — CH2 — Br
|
Br
(b) Racemisation occurs in SN1 reactions
...
Write the mechanism of the following reaction:
HBr
CH3 CH2 OH ¾ ¾ ¾ ® CH3 CH2 Br + H2 O
¾
18
...
20
...
(ii) Equilibrium constant (Kc) for the given cell reaction is 10
...
cell
A(s) + B2+(aq)
A2+(aq) + B(s)
21
...
4
2
100
0
...
(Given: log 4 = 0
...
3010)
22
...
(b) What are the dispersed phase and dispersion medium of butter?
(c) A delta is formed at the meeting place of sea and river water
...
(a) What are the different oxidation states exhibited by the lanthanoids?
(b) Write two characteristics of the transition elements
...
(b) In the 3d series from Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of Zn is the
lowest
...
24
...
Define the following terms:
(a) Anomers
(b) Denaturation of proteins
(c) Essential amino acids
26
...
(ii) Which one of the following drugs is an antibiotic:
Morphine, Equanil, Chloramphenicol, Aspirin
(iii) Why is use of aspartame limited to cold food and drink?
27
...
To make it
more impactful, they organised a rally by joining hands with other schools and distributed paper
bags to vegetable vendors, shopkeepers and departmental stores
...
After reading the above passage, answer the following questions:
(i) What values are shown by the students?
(ii) What are biodegradable polymers? Give one example
...
(a) State Raoult’s law for a solution containing volatile components
...
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g of CaCl2 to 200 g
of water
...
512 K kg mol–1, Molar mass of CaCl2 = 111 g mol–1)
370 Xam idea Chemistry—XII
OR
(a) Define the following terms:
(i) Azeotrope
(ii) Osmotic pressure
(iii) Colligative properties
(b) Calculate the molarity of 9
...
02 g mL–1
...
(a) Account for the following:
(i) Bi is a strong oxidizing agent in the +5 state
...
(iii) Iron dissolves in HCl to form FeCl2 and not FeCl3
...
(ii) Unlike xenon, no distinct chemical compound of helium is known
...
30
...
HCl
(iii) C6H5CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzoic acid and Ethyl benzoate
(ii) Propanal and Butan-2-one
OR
(a) Account for the following:
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN
...
However, only one is
involved in the formation of semicarbazone
...
On what principle is the method of zone refining of metals based?
2
...
3
...
What type of isomerism is exhibited by the following complex:
[Co(NH3 ) 5 SO4 ]Cl
5
...
c
...
unit cell
...
Which of the two is more basic and why?
CH3NH2 or NH3
7
...
9
...
(ii) Graphite rod in the electrometallurgy of aluminium
...
Complete the following equations:
(i) 2MnO4– + 5NO2– + 6H+ ¾ ¾ ®
¾
(ii) Cr2O72– + 14H+ + 6e– ¾ ¾ ®
¾
11
...
(Atomic no
...
Write chemical equations when
(i) methyl chloride is treated with AgNO2
...
13
...
Define the following terms:
(i) Polysaccharides
(ii) Amino acids
(iii) Enzymes
SET–III (Questions Uncommon to Set–I and II)
1
...
2
...
What type of isomerism is exhibited by the complex [Co(NH3)5NO2]2+?
5
...
Which of the two is more basic and why?
NH2
NH2
or
CH3
7
...
9
...
(ii) Silica in the extraction of copper from copper matte
...
Write the state of hybridization, shape and IUPAC name of the complex [Co(NH3)6]3+
...
of Co = 27)
13
...
(ii) chlorobenzene is treated with CH3Cl in the presence of anhydrous AlCl3
...
Name the different reagents needed to perform the following reactions:
(i) Phenol to benzene
(ii) Dehydration of propan-2-ol to propene
(iv) Dehydrogenation of ethanol to ethanal
24
...
Collectors (e
...
, pine oil, xanthates etc
...
2
...
3
...
4
...
r (atomic radius) =
3
a (edge length of unit cell)
4
6
...
In CH3 NH2 , the +I effect of —CH3 group makes lone pair of electrons on N-atom more available for
donation
...
Hence, CH3NH2 is more basic
than C6H5NH2
...
The two component of starch are amylose and amylopectin
...
Q = I ´ t = 5 A × 20 × 60 s = 6000 C
Ni 2 + + 2e - ¾ ¾ ® Ni
¾
2 × 96500 C deposit Ni = 58
...
7 g mol -1
2 ´ 96500 C mol -1
´ 6000 C = 1
...
The half life (t ½ )of a reaction is the time in which the concentration of a reactant is reduced to one
half of its initial concentration
...
693
(ii) t ½ for a first order reaction =
k
11
...
Two most important allotropes of sulphur are rhombic sulphur and monoclinic sulphur
...
When rhombic sulphur is heated above 370 K it gets
converted into monoclinic sulphur
...
6 kJ mol–1
374 Xam idea Chemistry—XII
The forward reaction is exothermic and proceed with decrease in number of moles
...
(ii) H2SO4 is a dibasic acid, it ionizes in two steps and has two dissociation constants
...
2 × 10–2
K a2 << K a , because the negatively charged HSO- ion has much less tendency to donate a
4
proton to H2 O as compared to neutral H2SO4
...
(i) MnO- + 8H+ + 5e - ¾ ¾ ® Mn 2 + + 4H2 O] × 2
¾
4
S 2 - ¾ ¾ ® S + 2e - ] ´ 5
¾
2MnO4 - + 5S 2 - + 16H+ ¾ ¾ ® 2Mn 2 + + 5S + 8H2 O
¾
2
(ii) Cr2 O7 - + 2OH- ¾ ¾ ® 2CrO2 - + H2 O
¾
4
14
...
Hybridisation = sp d , Shape = Octahedral
CH3—CH2—OH + KCl
(i) CH3—CH2—Cl + KOH(aq)
Ethyl chloride
(ii)
Cl
Ethyl alcohol
Cl
O
+ CH3—C—Cl
Chlorobenzene
Anhyd
...
(a) CH3—CH2—CH2—CH2—Br
...
(b) This is due to the fact that carbocations are the intermediates in S N 1 reactions
...
e
...
17
...
26 (a) Set-I CBSE Delhi 2014
...
(i) Aqueous bromine (Br2/H2O)
(ii) Lithium aluminium hydride (LiAlH4 ) or sodium borohydride (NaBH4 )
...
AlCl3)
Examination Papers 375
(iv) Acidified potassium dichromate solution (K2Cr2O7/H2SO4) or alkaline potassium permanganate
solution (KMnO4/OH–) followed by hydrolysis with dil
...
19
...
(i)
(ii)
(iii)
(iv)
KCl shows schottky defect as the cation, K+ and anion, Cl– are of almost similar sizes
...
CO2
Ferromagnetic
(i) The two main advantages of H2–O2 fuel cell over ordinary cell are as follows:
l It has high efficiency of 60%–70%
...
(ii) A ( s) + B 2 + ( aq) l
A 2 + ( aq) + B ( s)
Here, n = 2,
K = 10
0
...
059
0
...
0295 V
2
2
21
...
22, Set-I CBSE Delhi 2014
...
(a) Freundlich adsorption isotherm equation for adsorption of gases on solids:
x
= kp1 / n ( n > 1)
...
(b) In butter, water acts as disperse phase and oil acts as dispersion medium
...
Sea water contains a variety of electrolytes
...
23
...
Besides this some of the lanthanoid also exhibit
+2 and +4 oxidation states
...
(ii) Transition elements exhibit variable oxidation states
...
(iv) A number of transition elements and their compounds show catalytic properties
...
OR
(a) The electronic configuration of Mn 2 + is [Ar] 3d5 which is half filled and hence stable
...
On the other hand, Fe 2 + has electronic configuration [Ar] 3d6
...
Hence Mn2+ is
more stable than Fe2+ towards oxidation
...
That is why, the enthalpy of atomisation of zinc is the lowest in the
series
...
Hence, it is coloured in aqueous solution
...
Hence, it is colourless in aqueous solution
...
Refer to Ans 27 Set-I CBSE Delhi 2014
...
(a) The carbohydrates which differ in configuration at the glycosidic carbon (i
...
, C1 in aldoses and
C2 in ketoses) are called anomers e
...
, a–D-(+)-glucose and b–D-(+)-glucose
...
Due to this, globules
unfold and helix get uncoiled and protein loses its biological activity
...
During denaturation, 2° and 3° structures are destroyed but 1° structures remain intact,
e
...
, coagulation of egg while on boiling, curdling of milk, etc
...
g
...
26
...
g
...
(ii) Chloramphenicol
(iii) Use of aspartame is limited to cold foods because it is unstable at cooking temperature
...
(i) Concern towards water pollution, concern for environmental protection, team work, socially
aware
...
The required
enzymes are produced by microorganism e
...
, poly-b-hydroxybutyrate-co-b-hydroxyvalerate
(PHBV), nylon-2-nylon-6 etc
...
28
...
Thus, for a solution of volatile liquids
o
o
A and B, PA µ xA and PB µ xB or PA = PA x A and PB = PB x B where PA and PB are partial
o
o
vapour pressures, xA and xB are mole fractions, PA and PB are vapour pressure of pure
components A and B respectively
...
(b) The given quantities are:
WB = 10 g, WA = 200 g, K b = 0
...
512K Kg mol -1 ´ 10 g ´ 1000 g kg –1
111 g mol -1 ´ 200 g
DTb = 0
...
(ii) The excess of pressure which must be applied to the solution side to prevent the passage
of solvent into it through a semipermeable membrane is called osmotic pressure
...
(b) Let the mass of solution = 100 g
Mass of H2 SO4 = 9
...
8 g
=
= 0
...
02
1
...
02
1000 mL L-1
1
L
=
10
...
1 mol
=
1L
10
...
02 mol L–1 or 1
...
(a) (i) Due to inert pair effect Bi in +3 state is much more stable than in +5 state
...
Hence Bi is strong
oxidising in +5 state
...
It does not have d orbitals to expand its
covalency beyond four
...
PCl5 is known as P has vacant
3d orbital to which 3s electrons can be excited to make available five half filled orbitals
needed for the formation of five P—Cl bonds
...
=
Fe + 2HCl ¾ ¾ ® FeCl2 + H2
¾
H2 thus produced prevents the oxidation of FeCl2 to FeCl3
...
(ii) This is due to the following reasons:
He does not have d-orbitals in the valence shell and hence electron cannot be excited to
higher energy levels like in Xe to form bonds
...
(iii) Acids which contains P—H bonds have reducing character
...
O
30
...
HCl
Acetaldehyde
(iii)
O
C H
OH
(b)
Ethanal
O
CH CH CHO
CH CH2 C H
O
+ CH3 C H
Benzaldehyde
Ethane
– H2O
dil
...
Examination Papers 379
– +
COONa
COOH
+ NaHCO3
+ CO2 + H2O
Benzoic acid
Sod
...
No effervescence due to evolution of CO2
Ethyl benzoate
(ii) Propanal being an aldehyde reduces Tollens’ reagent to silver mirror but butan-2-one
being a ketone does not
...
Moreover it also hinder the approach of nucleophile CN-
...
–
–
O
O
O
(ii)
H2N – C – NHNH2
(b)
+
+
H2N = C – NH–NH2
H2N–C = NH – NH2
Semicarbazide has two –NH2 groups but one of them (i
...
, directly attached to C = O) is
involved in resonance as shown above
...
In contrast, the lone pair of electrons on
the other NH2 group (i
...
, attached to —NH) is not involved in resonance and hence is
available for nucleophilic attack on the C = O group of aldehydes and ketones
...
KOH
¾¾¾¾¾ ®
¾
Formaldehyde
CH3 ¾ OH + HCOO- K +
Methyl alcohol
Potassium formate
SET–II
1
...
2
...
Order of basic character
BiH3 < SbH3 < AsH3 < PH3 < NH3
4
...
Atomic radius (r) =
2 2
6
...
CH3 group due to its +ve I effect pushes electron towards nitrogen
··
in CH3 NH2 and this makes the unshared electron pair more available for sharing with the proton of
the acid
...
Chloroprene (CH2 = C - CH = CH2 )
9
...
330 - 350K
Ni + 4CO ¾ ¾ ¾ ¾ ¾ ® Ni(CO) 4
¾
The nickel tetracarbonyl complex thus obtained is then heated to a higher temperature so that it
is decomposed to give pure metal
...
Carbon reacts with oxygen liberated at anode producing CO and CO2 otherwise oxygen
liberated at the anode may oxidise some of the liberated aluminium back to Al2O3
...
(i) 2MnO4– + 5NO2– + 6H+ ¾ ¾ ® 2Mn2+ + 5NO3– + 3H2O
¾
(ii) Cr2O72– + 14H+ + 6e– ¾ ¾ ® 2Cr3+ + 7H2O
¾
11
...
2
Examination Papers 381
12
...
AlCl3
o-bromo toluene
(Minor)
13
...
HNO3/conc
...
+
-
(iv) Pyridinium chlorochromate, PCC (C5 H5 N HCrO3 C l in CH2 Cl 2 )
...
(i) Carbohydrates that yield a large number of monosaccharide units on hydrolysis are called
polysaccharides e
...
, starch, cellulose, gums etc
...
(iii) Enzyme: Enzymes are complex nitrogenous organic compounds produced in living cells of
plants and animals
...
SET–III
1
...
Associated colloids
4
...
Propane–1, 2, 3-triol
NH2
NH2
is stronger base than
effect
...
as CH3 group is electron releasing by +I effect and hyperconjugation
CH3
7
...
g
...
9
...
It is then decomposed by heating over a tungsten
filament at 2075K to give pure zirconium
...
In the converter FeS gets converted into FeO
...
2FeS
+
3O2
¾ ¾ ® 2FeO + 2SO2
¾
382 Xam idea Chemistry—XII
FeO
+
Impurity
11
SiO2
¾ ¾ ® FeSiO3
¾
Flux
Slag
(i) 2MnO- + 5SO2 – + 6H+ ¾ ¾ ® 2Mn2+ + 5SO42– + 3H2O
¾
4
3
(ii) 2CrO2 - + 2H+ ¾ ¾ ® Cr2O72– + H2O
¾
4
12
...
13
...
KOH ¾ ¾ ® CH2 — CH2 + KCl + H2O
¾
Ethyl chloride
Ethene
Cl
(ii)
Cl
+
CH3Cl
Anhyd
...
Cl
CH3
1-chloro-2-methylbenzene
(Minor)
CH3
1-chloro-4-methylbenzene
(Major)
(i) Zinc dust
(ii) 85% H2SO4 at 440 K
(iv) Heat copper at 573 K (Cu/573 K)
24
...
5°) but after hydrolysis it gives an equimolar mixture of
D-(+)-glucose and D-(–)-fructose, which is laevorotatory
...
(b) Vitamins are generally regarded as organic compounds required in the diet in small amounts to
perform specific biological functions for normal maintenance of optimum growth and health of
the organism
...
HO––H 2C 5'
4'
H
O
H
3'
OH
Base
H1'
2'
H
OH
zzz
Title: CBSE Chemistry Class 12 Examination Paper upto 2015
Description: This PDF contains a full set of CBSE class 12 question paper from 2007 till 2015 for practice purpose.
Description: This PDF contains a full set of CBSE class 12 question paper from 2007 till 2015 for practice purpose.