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CHAPTER

8

The d- and
f- Block Elements

8
...
2 Electronic Configurations of the d-Block
Elements
8
...
4 Some Important Compounds of Transition
Elements
8
...
6 The Actinoids
8
...
1

8
...
3

8
...

8 Maximum VSA type questions were asked from
General Properties of the Transition Elements
(d-Block)
...
5

8
...
7

8 Maximum SA and LA I type questions
were asked from General Properties of the
Transition Elements (d-Block)
...

8 Their general electronic configuration is
(n – 1)d1 – 10ns0 – 2
...

Form coloured compounds due to d-d transitions
Transition metal ions and their compounds are paramagnetic due to presence
of unpaired electrons in the (n – 1)d-orbitals and it is calculated by using the
formula, m =

Catalytic behaviour
Interstitial compounds
Alloy formation

n(n + 2) where, n is the no
...


Due to variable oxidation states and ability to form complexes
Due to empty spaces in their lattices, small atoms can be easily accommodated
Due to similar atomic sizes

8 Some important compounds :
Compounds
Potassium dichromate
(K2Cr2O7)

O 179 pm O
O Cr 126°
O

2–

O
Cr
O
16
3p O
m

Potassium permanganate
(KMnO4)

O

O

Mn
O

O–

Preparation

Properties

Uses

From sodium dichromate
(obtained from chromite
ore)
4FeCr2O4 + 8Na2CO3 + 7O2
8Na2CrO4 + 2Fe2O3
+ 8CO2
2Na2CrO4 + 2H+
Na2Cr2O7 + 2Na+ + H2O
Na2Cr2O7 + 2KCl
K2Cr2O7 + 2NaCl

Orange red crystalline solid,
oxidising agent having melting
point 398°C
...


From potassium
manganate (obtained from
pyrolusite)
2MnO2 + 4KOH + O2
2K2MnO4 + 2H2O
2K2MnO4 + Cl2
2KMnO4 + 2KCl

Deep purple crystalline solid,
oxidising agent, having melting
point 240°C
...


X

Volumetric titrations involving KMnO4 are carried out only in the presence of dil
...
H2SO4 does not give any oxygen of its own to oxidise the
reducing agent
...
Cerium (at
...
58) to lutetium (at
...
71)
...
Thorium (at
...
90) to lawrencium (at
...
103)
...


Oxidation states

Most common oxidation state of lanthanoids is +3
...
g
...


Action of air

All the lanthanoids are silvery white soft metals and tarnish readily in
moist air
...


Coloured ions

They form coloured trivalent metal ions due to f-f transitions of unpaired
electrons
...


Magnetic properties

La3+, Lu3+ are diamagnetic while trivalent ions of the rest of lanthanoids
are paramagnetic
...


Electropositive character Highly electropositive because of low ionisation energies
...


Tendency to form
complexes

Lanthanoids do not have much tendency to form complexes due to low
charge density because of their large size
...


8 Lanthanoid contraction : In lanthanoid
series, with increasing atomic number, there
is progressive decrease in atomic/ionic radii
(M3+ ions) from La3+ to Lu3+
...


X

Consequences : Their separation is difficult,
they have small differences in properties
and 4d and 5d transition series have almost
same atomic radii (Zr and Hf have similar
properties due to same size)
...
It is used to make tracer bullets,
shells and lighter flints
...
Actinoid contraction
is greater due to poor sheilding effect of the 5f-electrons
...


Oxidation states

Like lanthanoids, most common oxidation state is +3
...
g
...

They show a large number of oxidation states because of very small energy
gap between 5f, 6d and 7s subshells
...

Coloured ions

Coloured due to f-f transition except Ac3+(5f 0), Cm3+(5f 7) and Th4+(5f 0)
which are colourless
...


Density

All actinoids except thorium and americium have high densities
...

Ionisation energy

They have low ionisation energies than lanthanoides
...


Electropositive
character

Highly electropositive metals
...
U and Pu are used as fuel
in nuclear reactor
...

and +7 besides +3
...


All actinoids are radioactive
...


They form oxo-cations like UO2, PuO2, UO 2, etc
...


Actinoid compounds are more basic
...


They have greater tendency of complex
formation
...
2 Electronic Configurations of
the d-Block Elements
VSA

(1 mark)

1
...

(1/5, 2020)
2
...

(1/5, AI 2014)
(2
marks)
SA
3
...

(Delhi 2015)

8
...
In general, the electronic
configuration of these elements is (n – 1)d1 – 10
ns1 – 2
...
e
...
However, Zn, Cd and Hg are
not regarded as transition elements
...
Transition metals are hard (except
Zn, Cd and Hg) and have a high melting point
...
Why are Zn, Cd and Hg non-transition
elements?
5
...
Why do transition metals and their
compounds show catalytic activity?

7
...


Why is Cu2+ ion coloured while Zn2+ ion is
colourless in aqueous solution?

9
...
Out of the following transition elements, the
maximum number of oxidation states are
shown by
(a) Sc (Z = 21)
(b) Cr (Z = 24)
(c) Mn (Z = 25) (d) Fe (Z = 26)
...
Assertion (A) : Transition metals have high
melting point
...

(a) Both Assertion (A) and Reason (R) are
correct statements, and Reason (R) is
the correct explanation of the Assertion
(A)
...

(c) Assertion (A) is correct, but Reason (R)
is incorrect statement
...

(2020)
12
...


(1/5, 2020)
13
...

(Delhi 2017)
14
...


(Delhi 2017)
15
...
Zn2+ salts are white while Cu2+ salts are
coloured
...
Why do transition elements show variable
oxidation states?
(Delhi 2014C)
18
...

(1/2, Delhi 2011)
19
...


(1/2, AI 2011)

SA

(2 marks)

20
...
 (2/5, 2020)
21
...
Give reasons for the following :

(i) Transition metals form alloys
...


(2/3, Delhi 2019)
23
...
91 –1
...
44

–0
...
41 +1
...
77

+1
...
(i) How is the variability in oxidation states
of transition metals different from that
of the p-block elements?
(ii) Out of Cu+ and Cu2+, which ion is
unstable in aqueous solution and why?

(2/5, Delhi 2017)
25
...


(ii) Zn, Cd and Hg are soft metals
...
Give reasons :
(i) Mn shows the highest oxidation state
of +7 with oxygen but with fluorine it
shows the highest oxidation state of +4
...

(2/3, Delhi 2016)
27
...

(ii) Cu+ ion is unstable in aqueous solution
...
Why do transition elements show variable
oxidation states? In 3d series (Sc to Zn),
which element shows the maximum number
of oxidation states and why? (Foreign 2015)
29
...

(ii) Transition metals and their compounds
show catalytic properties
...
What is meant by ‘disproportionation’? Give
an example of a disproportionation reaction
in aqueous solution
...
Account for the following :
(i) Mn2+ is more stable than Fe2+ towards
oxidation to +3 state
...

(Delhi 2014)
32
...

(ii)
Which of the 3d-block elements may
not be regarded as the transition
elements and why? (2/3, Foreign 2014)
33
...


(ii) The ionisation enthalpies (first and
second) in the first series of the
transition elements are found to vary
irregularly
...
Assign reasons for the following :
(i) Copper(I) ion is not known to exist in
aqueous solutions
...

(AI 2014C)
35
...

(ii) Cobalt (II) is very stable in aqueous
solutions but gets easily oxidised in the
presence of strong ligands
...
Assign reasons for the following :
(i) Cu(I) ion is not known to exist in
aqueous solutions
...

(2/3, AI 2014C)
37
...

(ii) Transition metals generally form
coloured compounds
...
(i) 
Which metal in the first transition
series (3d-series) exhibits +1 oxidation
state most frequently and why?

(ii) Which of following cations are coloured
in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+

(At
...
Sc = 21, V = 23, Ti = 22, Mn = 25)
(Delhi 2013)
39
...

(ii) Transition metals and their compounds
act as catalyst
...
How would you account for the following?
(i) Many of the transition elements are
known to form interstitial compounds
...


(2/3, Delhi 2012)
41
...


43
...


45
...


47
...


(d4) Cr2+ is a reducing agent while Mn3+
is an oxidising agent
...

(2/3, Delhi 2012)
How would you account for the following :
(i) The E°M2+/M for copper is positive
(+0
...
Copper is the only metal in
the first series of transition elements
showing this behaviour
...
(2/3, AI 2012)
Explain the following observations :
(i) Many of the transition elements are
known to form interstitial compounds
...

(2/3, AI 2012)
Explain the following observations
...

(ii) There is hardly any increase in atomic
size with increasing atomic numbers in
a series of transition metals
...

(ii) The transition metals and many of their
compounds act as good catalysts
...

(ii) The Fe2+ is much more easily oxidised to
Fe3+ than Mn2+ to Mn3+
...

(2/5, AI 2012C)
Account for the following :
(i) Mn2+ compounds are more stable than
Fe2+ compounds towards oxidation to
their +3 state
...


(2/5, AI 2012C)
49
...

(ii)
Manganese exhibits the highest
oxidation state of +7 among the 3rd
series of transition elements
...
How would you account for the following :
(i) Cr2+ is reducing in nature while with
the same d-orbital configuration (d4)
Mn3+ is an oxidising agent
...

(AI 2011)
51
...

(ii) Unlike Cr3+, Mn2+, Fe3+ and the
subsequent other V2+ ions of the 3d
series of elements, the 4d and the 5d
series metals generally do not form
stable cationic species
...
Explain giving a suitable reason for each of
the following :
(i) Transition metals and their compounds
are generally found to be good catalysts
...


(AI 2011)

LA I

(3 marks)

53
...

(ii) unstable in aqueous solution
...

Give suitable reason in each
...
Give reasons :
(i) Eº value for Mn3+/Mn2+ couple is much
more positive than that for Fe3+/Fe2+
...


(iii) Sc3+ is colourless in aqueous solution
whereas Ti3+ is coloured
...
Account the following :
(i) Transition metals form large number of
complex compounds
...

(iii) E° value for the Mn3+/Mn2+ couple is
highly positive (+1
...

(3/5, Delhi 2017)
56
...
Account for the following :
(i) Mn shows the highest oxidation state
of +7 with oxygen but with fluorine, it
shows the highest oxidation state of +4
...

(iii) Cu2+ salts are coloured, while Zn2+ salts
are white
...
E°(M2+/M)

Cr Mn
Fe
Co
Ni
Cu
–0
...
18 –0
...
28 –0
...
34

From the given data of E° values, answer the
following questions :
(i) Why is E°(Cu2+/Cu) value exceptionally
positive?
(ii) Why is E°(Mn2+/Mn) value highly negative
as compared to other elements?
(iii) Which is a stronger reducing agent Cr2+
or Fe2+? Give reason
...
Assign suitable reasons for the following :
(i) The Mn2+ compounds are more stable
than Fe2+ towards oxidation to their +3
state
...



60
...


62
...


64
...

(Foreign 2014)
Give reasons for the following :
(i) Mn3+ is a good oxidising agent
...

(iii) Although ‘F’ is more electronegative
than ‘O’, the highest Mn fluoride is
MnF4, whereas the highest oxide is
Mn2O7
...

(ii) The enthalpies of atomization of the
transition metals are high
...


(3/5, AI 2013C)
Explain the following :
(i) Copper (I) ion is not stable in an
aqueous solution
...

(iii) Transition metals in general act as good
catalysts
...

(ii) Mn (III) undergoes disproportionation
reaction easily
...
 (3/5, Foreign 2011)
Account for the following :
(i) The transition metals and their
compounds act as good catalysts
...

(iii) A transition metal exhibits higher
oxidation states in oxides and fluorides
...
The elements of 3d transition series are given
as
Sc Ti V Cr Mn Fe Co Ni Cu Zn

Answer the following :
(i) Write the element which shows
maximum number of oxidation states
...

(ii) Which element has the highest melting
point?
(iii) Which element shows only +3 oxidation
state?
(iv) Which element is a strong oxidising
agent in +3 oxidation state and why?

(AI 2016)

8
...
Account for the following :

Chromates change their colour when kept
in an acidic solution
...
Orange colour of Cr2O72– ion changes to
yellow when treated with an alkali
...
Complete the following equation :
2MnO4– + 6H+ + 5NO2–


(1/2, Delhi 2015)
69
...
Complete the following equation :
MnO4– + 8H+ + 5e–

(1/5, Delhi 2014)
71
...
Give reason :
Orange solution of potassium dichromate
turns yellow on adding sodium hydroxide to
it
...
Write the balanced chemical equations
involved in the preparation of KMnO4 from
pyrolusite ore (MnO2)
...
Write the balanced ionic equations showing
the oxidising action of acidified dichromate
(Cr2O2–
7 ) solutions with (i) iron (II) ion and
(ii) tin (II) ion
...
When MnO2 is fused with KOH in the
presence of KNO3 as an oxidizing agent,

it gives a dark green compound (A)
...
An
alkaline solution of compound (B) oxidises
KI to compound (C) whereas an acidified
solution of compound (B) oxidises KI to (D)
...
 (Delhi 2019)
76
...
Complete the following equations :
(i) 2MnO–4 + 16H+ + 5S2– →

(2018)

Heat

(ii) KMnO4 → 
(2/5, AI 2017)
78
...
Compound (B) on
reaction with KCl forms an orange coloured
crystalline compound (C)
...

(ii) Write one use of compound (C)
...
Complete the following chemical equations :
(i) 8MnO4– + 3S2O32– + H2O →
(ii) Cr2O72– + 3Sn2+ + 14H+ →

(Delhi 2016)
80
...



(i) 2MnO2 + 4KOH + O2
(ii) Cr2O72– + 14H+ + 6I–

(2/5, AI 2016)
81
...
How does the acidified
permanganate solution react with oxalic
acid? Write the ionic equations for the
reaction
...
Describe the oxidising action of potassium
dichromate and write the ionic equations for
its reaction with (i) an iodide (ii) H2S
...
How do you prepare :
(i) K2MnO4 from MnO2?
(ii) Na2Cr2O7 from Na2CrO4?

(2/5, Delhi 2014)

84
...
Complete the following equations :
(i) 2MnO4– + 5S2– + 16H+ →
(ii) Cr2O72– + 2OH– →
(Foreign 2014)
86
...
Complete the following chemical equations :
(i) Cr2O72– + H+ + I–
(ii) MnO4– + NO2– + H+
(Delhi 2012)
88
...
Complete the following reactions in an
aqueous medium :
(i) MnO4– + C2O42– + H+ →
(ii) Cr2O72– + H2S + H+ →

(2/5, Foreign 2011)
90
...
Describe the reactions involved in the
preparation of K2Cr2O7 from chromite ore
...
Explain the method of preparation of sodium
dichromate from chromite ore
...

(AI 2019)
93
...

(i) MnO2 + KOH + O2 →
(ii) I– + MnO4– + H+ →

(iii) Cr2O72– + Sn2+ + H+ →
(AI 2019)
94
...
Describe the preparation of potassium
permanganate from pyrolusite ore
...

(3/5, AI 2013C)

96
...
What is the
effect of change of pH on dichromate ion?
(3/5, AI 2012C)
97
...

What happens when acidified potassium
permanganate solution reacts with ferrous
sulphate solution? Write balanced chemical
equations
...
Complete the following chemical equations :
(i) MnO4– + C2O42– + H+
Heated

(ii) KMnO4
(iii) Cr2O72– + H2S + H+

 (Delhi 2011)

8
...
Give reasons for the following :
Eu2+ is a strong reducing agent
...
Account for the following :
Zr and Hf have almost similar atomic radii
...
Name a member of the lanthanoid series
which is well known to exhibit +2 oxidation
state
...
Name a member of the lanthanoid series
which is well known to exhibit +4 oxidation
state
...
What are the different oxidation states
exhibited by the lanthanoids?
(1/3, Foreign 2014)
104
...

(1/3, Delhi 2013, 1/5, AI 2013C)
105
...
However, occasionally in
solutions or in solid compounds, +2 and +4
ions are also obtained
...
Give reason :

There is a gradual decrease in the size of
atoms with increasing atomic number in the
series of lanthanoids
...
What is meant by ‘lanthanoid contraction’?
(AI 2011)

SA

(2 marks)

108
...
What is lanthanoid contraction and what
is it due to? Write two consequences of
lanthanoid contraction
...
What is lanthanoid contraction? Mention its
main consequences
...
What is lanthanoid contraction? What are
the consequences of lanthanoid contraction?
(Delhi 2015C)
112
...
6 The Actinoids
VSA

(1 mark)

113
...

(1/5, Delhi 2017)
114
...
Give two reasons
...
Give reasons :
Actinoids show irregularities in their
electronic configurations
...
How would you account for the following :
Actinoid contraction is greater than
lanthanoid contraction? (1/3, Delhi 2015)
117
...


(1/3, AI 2014C, Delhi, AI 2012,

1/2, Delhi 2011)
118
...

(1/3, AI 2014C)
119
...

(1/3, Delhi 2012)

between the chemistry of lanthanoids and
actinoids
...
Explain giving reasons :
The chemistry of actinoids is not as smooth
as that of lanthanoids
...
With reference to structural variability and
chemical reactivity, write the differences
between lanthanoids and actinoids
...
Why do actinoids show a wide range of
oxidation states? Write one similarity

123
...

(3/5, 2020)

Detailed Solutions
1
...
e
...

2
...

3
...

Characteristics of transition elements :
(i) They show variable oxidation states
...

4
...

5
...

6
...
This activity
is ascribed to their ability to adopt multiple
oxidation states, ability to adsorb the reactant(s)
and ability to form complexes
...

Catalysis involves the formation of bonds between
reactant molecules and atoms at the surface of the
catalyst
...
Strong metallic bonds between its atoms of
transition elements are responsible for the high
melting points
...
Zn2+ ion has completely filled d-subshell and no
d-d transition is possible
...


Configuration of Cu2+ is [Ar] 3d9
...

9
...
(c)
11
...
Transition elements show
variable oxidation states
...
Cu(I) compounds have completely filled
d-orbitals and there are no vacant d-orbitals
for promotion of electrons whereas in Cu(II)
compounds have one unpaired electron which is
responsible for colour formation
...
Oxoanion of chromium in which it shows
+6 oxidation state equal to its group number is
Cr2O72– (dichromate ion)
...
Formula of oxoanion of manganese is MnO–4
...

15
...

When an electron from a lower energy d-orbital
is excited to a higher energy d-orbital, the energy
of excitation corresponds to the frequency which
generally lies in the visible region
...
The frequency of the
light absorbed is determined by the nature of the
ligand
...
Refer to answer 8
...
(i) Cr2+ is a stronger reducing agent than Fe2+
...
Transition elements can use their ns and
(n – 1)d orbital electrons for bond formation
therefore, they show variable oxidation states
...

It utilizes two electrons from its ns subshell then
its oxidation state = +2
...


E°Cr3+/Cr2+ is negative (–0
...
77 V)
...

Hence, Cr2+ is stronger reducing agent than Fe2+
...

As E°Co3+/Co2+ is maximum, thus Co2+ ion is most
stable
...
(i) Variable oxidation states of transition
metals arise due to incomplete filling of d-orbitals
and it differs from each other by unity e
...
, V(V),
V(IV), V(III), V(II)
...
e
...
, Sn(II),
Sn(IV), PCl3, PCl5, etc
...

25
...

(ii) In Zn, Cd and Hg, all the electrons in
d-subshell are paired
...
That is why they are soft metals with low
melting and boiling points
...
(i) Manganese can form pp - dp bond with
oxygen by utilising 2p-orbital of oxygen and
3d-orbital of manganese due to which it can show
highest oxidation state of +7
...

(ii) Refer to answer 17
...
(i) Refer to answer 17
...

28
...

Among the 3d series manganese (Mn) exhibits
the largest number of oxidation states from +2 to
+7 because it has maximum number of unpaired
electrons
...
(i) Refer to answer 26(i)
...

30
...

The examples of disproportionation reaction are :
(i) Aqueous NH3 when treated with Hg2Cl2 (solid)
forms mercury aminochloride disproportionatively
...
In aqueous solutions, Cu+ undergoes
disproportionation to form a more stable Cu2+
ion
...
It compensates the
second ionisation enthalpy of Cu involved in the
formation of Cu2+ ions
...
Transition metals and most of their compounds
contain unpaired electrons in the (n – 1)d orbitals
hence show paramagnetic behaviour
...
(b) Co = [Ar]3d 74s2
Co2+ = [Ar]3d7
i
...
, there are three unpaired electrons (n = 3)
...
M
...
87 B
...

21
...
So,
it is paramagnetic due to presence of unpaired
electrons
...
e
...

After Cr, the number of unpaired electrons goes
on decreasing
...
pt and b
...

decrease after middle (Cr) because of increasing
pairing of electrons
...
(i) Transition metals form alloys because
they have similar atomic radii
...
Oxidation state of Mn in Mn2O3 is
+3 while in Mn2O7 is +7
...
(i) Electronic configuration of Mn2+ is 3d5
which is half filled and hence stable
...
e
...
In case of Fe2+,
electronic configuration is 3d6
...

(ii) Zinc (Z = 30) has completely filled d-orbital
(3d10), so d-orbitals do not take part in interatomic
bonding
...

This is why it has very low enthalpy of atomisation
(126 kJ mol–1)
...
(i) Refer to answer 3
...

33
...
e
...
Hence,
the atomic volume decreases
...
Hence, the density from
titanium (Ti) to copper (Cu) increases
...

(ii) Irregular variation of ionisation enthalpies
is mainly attributed to varying degree of stability
of different 3d-configurations (e
...
, d0, d5, d10 are
exceptionally stable)
...
(i) Refer to answer 18
...

35
...

(ii) The tendency to form complexes is high for
Co(III) as compared to Co(II)
...
Co3+ ions are less
stable and are reduced by water
...
g
...

36
...

(ii) This is attributed to the involvement of greater
number of electrons from (n–1)d in addition
to the ns electrons in the interatomic metallic
bonding
...
(i) Refer to answer 6
...

38
...
Electronic configuration of Cu in the
ground state is 3d10 4s1
...
Thus,
it shows +1 oxidation state
...

Ions with d0 and d10 will be colourless
...
Ti4+(3d0) and
Sc3+(3d0) are colourless
...
(i) Refer to answer 17
...

40
...
) get
trapped in voids or vacant spaces of lattices of the
transition metals
...

(ii) This is due to lanthanoid contraction
...
(i) E° values for the Cr3+/Cr2+ and Mn3+/Mn2+
couples are
–
Cr3+
Cr2+
(aq) + e
(aq); E° = –0
...
551 V
These E° values indicate that Cr2+ is strongly
reducing while Mn3+ is strongly oxidising agent
...
This is called d-d transition
...
The mixture of the wavelength
which is not absorbed is transmitted out
...

42
...

This is why E°M2+/M for Cu is positive
...

43
...

(ii) As we move along transition metal series
from left to right (i
...
, Ti to Cu), the atomic radii
decrease due to increase in nuclear charge
...
At the same time,
atomic mass increases
...

44
...

(ii) As one proceeds along a transition series, the
nuclear charge increases which tends to decrease
the size but the addition of electrons in the
d-subshell increases the screening effect which
counterbalances the effect of increased nuclear
charge
...

45
...
Hence, they have high enthalpy of
atomization
...

46
...

(ii) Refer to answer 31(i)
...
The variability in oxidation states of transition
metals is due to the incomplete filling of d-orbitals
...

For example, Fe and Fe , Cu and Cu , etc
...
For example,
Pb2+ and Pb4+, Sn2+ and Sn4+, etc
...

3+

2+

2+

+

48
...

(ii) Refer to answer 41(i)
...
(i) Refer to answer 15
...
Thus, it exhibits
+7 oxidation state
...
(i) Refer to answer 41(i)
...


51
...

(ii) This is because due to lanthanoid contraction
the expected increase in size does not occur hence
they have very high value of ionisation enthalpies
...
(i) Refer to answer 6
...
So
the valence electrons are less tightly held and form
metal-metal bonding more frequently
...
(i) Cr2+ is reducing since its configuration
is converted to d3 from d4
...

(ii) Cu+ is unstable in aqueous solution
In
aqueous
solutions,
Cu+
undergoes
disproportionation to form a more stable Cu2+
ion
...
Therefore, third
ionization enthalpy is very high, i
...
, 3rd electron
cannot be lost easily
...
(i) From the relation, DG° = –nFE°
More positive is the value of E°, reaction will be
feasible
...

(ii) Greater the number of unpaired electrons,
stronger is the metallic bond and therefore, higher
is the enthalpy of atomisation
...

(iii) Only those ions are coloured which have
partially filled d-orbitals facilitating d-d transitions
...

55
...

– High ionic charges
...


(ii) Lowest oxidation compounds of transition
metals are basic due to their ability to get oxidised
to higher oxidation states
...

e
...
,  MnO is basic whereas Mn2O7 is acidic
...
This also explains that +3 state of Mn is of
little importance
...
(i) Ti4+ has highest oxidation state among the
given ions
...

On the other hand, V2+, Mn3+, Cr3+ have unstable
electronic configuration and hence, are less stable
...

(iii) Due to absence of unpaired electron in Ti4+, it
is a colourless ion
...
C
...
(i) Refer to answer 26(i)
...
d3 has half filled t2g
configuration with higher stability
...

58
...

(ii) Mn2+ ion has stable half-filled (3d5) electronic
configuration
...
Hence
E°Mn2+/Mn is more negative
...

59
...

(ii) Refer to answer 31(ii)
...

Sc3+ with 3d0 configuration is colourless while Ti3+
(3d1) is coloured
...
(i) Mn is more stable due to half filled d
configuration and Mn3+ easily changes to Mn2+
hence, it is a good oxidising agent
...
e
...

(iii) Refer to answer 26(i)
...
(i) All transition elements except the first and
the last member in each series show a large number of
variable oxidation states
...

Hence, electrons from both the energy levels can
be used for bond formation
...

(iii) Refer to answer 47
...
(i) Refer to answer 18
...

(iii) Refer to answer 6
...
(i) Refer to answer 52(ii)
...
That is why, Mn3+ undergoes
disproportionation reaction
...

64
...

(ii) Refer to answer 55(ii)
...
For
example, vanadium shows as oxidation state of +5
in V2O5
...

65
...
of oxidation
states from +2 to +7 because Mn has maximum
number of unpaired electrons in 3d sub-shell
...

(iii) Sc shows only +3 oxidation state because
after losing 3 electrons, it has noble gas electronic
configuration
...
5 V
...
In aqueous solution,
Cr2O72– + H2O

Dichromate ion
(Orange red)

2CrO42– + 2H+

Chromate ion
(Yellow)

When an acid is added (i
...
, pH of solution
decreased), the concentration of H+ ions is
increased and the reaction proceeds in the
backward direction producing an orange red
dichromate solution
...
Orange colour of Cr2O72– ion changes to yellow
when an alkali such as NaOH is added because
on addition of an alkali, the concentration of H+
ions decreases and hence, the reaction proceeds in
the forward direction producing yellow solution
2–
containing CrO4 ions
...
2MnO4– + 6H+ + 5NO2–
2Mn2+ + 5NO3– +

3H2O
+
69
...
MnO4– + 8H+ + 5e–
71
...
Refer to answer 67
...
Preparation of potassium permanganate :
Potassium permanganate is prepared by the fusion
of MnO2 (pyrolusite) with potassium hydroxide
and an oxidising agent like KNO3 to form
potassium manganate which disproportionates in
a neutral or acidic solution to form permanganate
...
Cr2O2–
7 + 14H + 6Fe

2Cr3++ 6Fe3++ 7H2O

Cr2O72– + 3Sn2+ + 14H+ → 2Cr3+ + 3Sn4+ + 7H2O
...
(A) is K2MnO4, (B) is KMnO4, (C) is KIO3
and (D) is I2
...
(i) 5Fe2+ + MnO4– + 8H+
(ii) 2MnO4– + H2O

Mn2+ + 4H2O + 5Fe3+
+ I → 2OH– + 2MnO2 + IO3–
–

77
...
(i) 4FeCr2O4 + 16NaOH + 7O2
8Na2CrO4 + 2Fe2O3 + 8H2O


(A)

2Na2CrO4 + H2SO4


Na2Cr2O7 + 2KCl


Na2Cr2O7 + Na2SO4 + H2O
(B)

K2Cr2O7 + 2NaCl
(C)

(ii) Potassium dichromate is used as a powerful
oxidising agent in industries and for staining and
tanning of leather
...
(i) 8MnO4– (aq) + 3S2O2–
3 (aq) + H2O(l) →
–

8MnO2(aq) + 6 SO2–
4 (aq) + 2OH (aq)
2+
+
(ii) Cr2O2–
2Cr3+ + 3Sn4+
7 + 3Sn + 14H
+ 7H2O

80
...
Refer to answer 73
...
Potassium dichromate is a strong oxidising
agent
...

(i) Refer to answer 80(ii)
...
(i) Refer to answer 80(i)
...

2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na++H2O
–

84
...
(i) H2S
2H + S
5S2– + 2MnO4– + 16H+
2Mn2+ + 8H2O + 5S
2–
–
(ii) Cr2O7 + 2OH
2CrO42– + H2O
2–
+
86
...
(i) Refer to answer 80(ii)
...

88
...


+
2+
(ii) Cr2O2–
2Cr3+
7(aq) + 14H (aq) + 6Fe (aq)
(aq)

+ 6Fe3+
+
7H
(aq)
2O(l)
+
89
...
(i) 2Fe3+ + 2I– → 2Fe2+ + I2
(ii) 2CrO42– + 2H+ → Cr2O72– + H2O
91
...

Cr2O72–

OH–
H+

CrO42–

92
...


101
...

102
...

103
...
However, some of
the lanthanoids also show +2 and +4 oxidation states
in solution or in solid compounds
...

e
...
, Ce4+ : 4f 0 , Eu2+ : 4f 7
Tb4+ : 4f 7 , Yb2+ : 4f 14
105
...


The solution of sodium dichromate is treated
with potassium chloride to obtain potassium
dichromate
...

2Mn2+ + 5I2
+ 8H2O

94
...

(ii) Refer to answer 90(ii)
...

95
...

Oxidising nature of KMnO4
2MnO4– + 5C2O42– + 16H+ → 2Mn2+ + 10CO2 + 8H2O
96
...

97
...

5Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5Fe3+
98
...

(ii) Refer to answer 86(ii)
...


100
...
g
...


104
...




93
...

(ii) 2MnO4– + 10I– + 16H+

(iii) Refer to answer 79(ii)
...
Eu2+ has a strong tendency to loose electrons
to attain the more stable +3 oxidation state of
lanthanoids hence, it is a strong reducing agent
...
As the atomic number increases, each
succeeding element contains one more electron in
the 4f orbital and one extra proton in the nucleus
...
As a result,
there is gradual increase in the nuclear attraction
for the outer electrons
...
This is called lanthanoid
contraction
...
The steady decrease in the atomic and ionic
radii (having the same charge) with increase in
atomic number across the series from lanthanum
to lutetium is known as lanthanoid contraction
...

Lanthanoid contraction : The steady
decrease in the atomic and ionic radii of
lanthanoid elements with increase in atomic
number is called lanthanoid contraction
...

Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides
of lanthanoids decrease with increasing atomic
number
...
e
...
,
atomic radii of zirconium(Zr) is same as that of
hafnium Hf
...
Refer to answer 108
...
Refer to answer 108
...
Refer to answer 108
...
Refer to answer 108
...
Similarity : The elements of both the series
are electropositive in nature
...

Difference : Lanthanoids except promethium are
non-radioactive elements, while all actinoids are
radioactive elements
...
Chemistry of actinoids is more complicated
than lanthanoids because
(i) actinoids show greater number of oxidation
states due to the comparable energies of 5f, 6d and
7s orbitals
...

115
...

116
...

117
...
This is because there is less
energy difference between 5f and 6d orbitals in
actinoids than the energy difference between 4f
and 5d orbitals in case of lanthanoids
...
Refer to answer 116
...
Refer to answer 117
...
The chemistry of actinoids is not as smooth as
lanthanoid because they show greater number of

oxidation states due to comparable energies of 5f,
6d and 7s orbitals
...
Refer to answer 117
...

122
...
Hardness of Lanthanoids
increases with increasing atomic number
...
The structural
variability is due to irregularities in metallic radii
which are greater than that of lanthanoids
...

The actinoids are highly reactive in finely divided
state
...
Difference between lanthanoids and actinoids
are following :
(i) Electronic configuration : The general
electronic configuration of lanthanoids is
[Xe]4f 1–14 5d0–1 6s2 whereas, that of actinoids is
[Rn] 5f 1–14 6d0–1 7s2
...

(ii) Oxidation states : Lanthanoids have principal
oxidation state of +3
...
On the other hand, actinoids show a
large number of oxidation states because of small
energy gap between 5f and 6d subshells
...
Lanthanoids
react with dilute acids to liberate H2 gas whereas
actinoids react with boiling water to give a mixture
of oxide and hydride

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