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CHAPTER 2
STRUCTURE OF ATOM
• Atom is the smallest indivisible particle of the matter
...

PARTICLE
ELECTRON
PROTON
NEUTRON
Sir
...
J
...
6 x 10 Coloumb 1
...
11 x 10-31kg
1
...
67492 x10-27kg
Mass
Electrons were discovered using cathode ray discharge tube experiment
...

Cathode ray discharge tube experiment: A cathode ray discharge tube madeof
glass is taken with two electrodes
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These rays were called cathode rays
...
Cathode rays consist of negatively charged electrons
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Cathode rays themselves are not visible but their behavior can be
observed with
help of fluorescent or phosphorescent materials
...
In absence of electrical or magnetic field cathode rays travel in
straight lines
d
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The characteristics of the cathode rays do not depend upon the
material of the electrodes and the nature of the gas present in the cathode ray
tube
...
The chargeto
mass ratio of an electron as 1
...
The value of the charge on an electron is -1
...

The mass on an electron was determined by combining the results
ofThomson’s experiment and Millikan’s oil drop experiment
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1094 x 10-31kg
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Characteristics of positively charged particles:
a
...
The positively charged particles depend upon the nature of gas present in the
cathode ray discharge tube
c
...

d
...
They are electrically neutral particles having a mass
slightly greater than that of the protons
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Mass Number (A) :Sum of the number of protons and neutrons present in
thenucleus
...
An
important feature of Thomson model of an atom was that mass of atom
isconsidered to be evenly spread over the atom
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Observations from α- particles scattering experiment by Rutherford:
a
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A small fraction of α- particles got deflected through small angles
c
...
Since most of the α-particles passed through foil undeflected, it means most
of the space in atom is empty
b
...
Since only some of the α-particles suffered large deflections, the positively
charged mass must be occupying very small space
d
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The nucleus
comprisesof protons and neutrons
...
Electrons and nucleus are held together by electrostatic forces
ofattraction
...
According to Rutherford’s model of atom, electrons which are negatively
charged particles revolve around the nucleus in fixed orbits
...
theelectrons undergo acceleration
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Thus, an electron in an orbit should
emitradiation
...
But this does not happen
...
The model does not give any information about how electrons
aredistributed around nucleus and what are energies of these electrons
Isotopes: These are the atoms of the same element having the same
atomicnumber but different mass number
...
e g 18Ar40 , 20Ca40
Isoelectronic species: These are those species which have the same numberof
electrons
...
When
anelectrically charged particle moves under acceleration, alternating
electricaland magnetic fields are produced and transmitted
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These waves are called
electromagneticwaves or electromagnetic radiations
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Oscillating electric and magnetic field are produced by oscillating charged
particles
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b
...
That means they can even travel in
vacuum
...
Wavelength: It may be defined as the distance between two
neighbouring crests or troughs of wave as shown
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b
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c
...
In vacuum all types of electromagnetic radiations travel with
thesame velocity
...
It is denoted by v
d
...

Velocity = frequency x wavelength c = νλ
Planck's Quantum Theoryo The radiant energy is emitted or absorbed not continuously but
discontinuously in the form of small discrete packets of energy called
‘quantum’
...
e
...
626 x 10 Js
o Energy is always emitted or absorbed as integral multiple of this
quantum
...

19

Black body: An ideal body, which emits and absorbs all frequencies, is calleda
black body
...

Photoelectric effect: The phenomenon of ejection of electrons from thesurface
of metal when light of suitable frequency strikes it is calledphotoelectric effect
...

Experimental results observed for the experiment of Photoelectric effecto When beam of light falls on a metal surface electrons are
ejectedimmediately
...
Thisis called threshold frequency
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Photoelectric work function (Wo): The minimum energy required to
ejectelectrons is called photoelectric work function
...
e
...
whenever radiation
interacts with matter, it displays particle like properties
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Spectrum is of two types: continuous and line spectrum
a
...

b
...
It has bright lines with dark spaces between them
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It consists of a range
ofelectromagnetic radiations arranged in the order of increasing wavelengths
ordecreasing frequencies
...

Spectrum is also classified as emission and line spectrum
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o Absorption spectrum is the spectrum obtained when radiation is
passedthrough a sample of material
...
The wavelengths which are absorbed are missing
and comeas dark lines
...

Spectral Lines for atomic hydrogen:

Rydberg equation

R = Rydberg’s constant = 109677 cm-1
Bohr’s model for hydrogen atom:
a
...
These paths are called orbits orenergy
levels
...

b
...

c
...
An electron can move only in those orbits for which its angularmomentum
is an integral multiple of h/2π

The radius of the nth orbit is given byrn =52
...
Bohr’s model failed to account for the finer details of the hydrogen
spectrum
...
Bohr’s model was also unable to explain spectrum of atoms containing
more than one electron
...
e
...
de Broglie’s
relation is
21

Heisenberg’s uncertainty principle: It states that it is impossible to
determine simultaneously, the exact position and exact momentum (or
velocity) of an electron
...


Heisenberg’s uncertainty principle rules out the existence of definite pathsor
trajectories of electrons and other similar particles
Failure of Bohr’s model:
a
...

b
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Classical mechanics is based on Newton’s laws of motion
...

Reason: Classical mechanics ignores the concept of dual behaviour of matter
especially for sub-atomic particles and the Heisenberg’s uncertainty principle
...

Quantum mechanics is based on a fundamental equation which is
calledSchrodinger equation
...
Out of the
possible values, only certain solutions are permitted
...
Thus, we can say
that energy is quantized
...
The value of ψhas no physicalsignificance
...
It is called probability density
...

Quantum numbers: There are a set of four quantum numbers which specifythe
energy, size, shape and orientation of an orbital
...

Principal quantum number (n):It identifies shell, determines sizes and

energy of orbitals
Azimuthal quantum number (l): Azimuthal quantum number
...
l
...
e
...
For a
given value of n, it can have n values ranging from 0 to n-1
...

Subshell
s p
d f g
notation
Value of ‘l’
0 1
2 3 4
Number of
1 3
5 7 9
orbitals
Magnetic quantum number or Magnetic orbital quantum number (ml):
Itgives information about the spatial orientation of the orbital with respect
tostandard set of co-ordinate axis
...
For each value of l, ml = – l, – (l –1), – (l–2)
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(l – 2), (l–1), l
Electron spin quantum number (ms): It refers to orientation of the spin of
theelectron
...
+1/2 identifies the
clockwisespin and -1/2 identifies the anti- clockwise spin
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Radial nodes: Radial nodes occur when the probability density of wave
functionfor the electron is zero on a spherical surface of a particular radius
...
Number of angular nodes = l
Total number of nodes = n – 1
23

Degenerate orbitals: Orbitals having the same energy are called
degenerateorbitals
...

So, due to the screening effect, the net positive charge experienced by
theelectron from the nucleus is lowered and is known as effective nuclear
charge
...

Aufbau Principle: In the ground state of the atoms, the orbitals are filled
inorder of their increasing energies
24

n+l rule-Orbitals with lower value of (n+l) have lower energy
...

The order in which the orbitals are filled isas follows:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s
...
Only two electrons may exist in the same
orbitaland these electrons must have opposite spin
...
e
...

Electronic configuration of atoms:Arrangement of electrons in different
orbitals of an atom
...

a
...
notation
...
Orbital diagram:, each orbital of the subshell is represented by a box and the
electron is represented by an arrow (↑) a positive spin or an arrow (↓) a negative
spin
...
Symmetrical distribution of electrons- the completely filled or half filled
sub-shells have
symmetrical distribution of electrons in them and are more stable
...
Exchange energy-The two or more electrons with the same spin present in
the degenerate orbitals of a sub-shell can exchange their position and the
energy released due to this exchange is called exchange energy
...
As a result the exchange energy is maximum and so is
the stability
...
Neutrons can be found in all atomic nuclei except in one case
...
Hydrogen atom
...

2
...

Ans
...
72 x 106 m-1
3
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n=2 and l= 1
4
...
1p and 3f are not possible
...
Write the electronic configuration of the element having atomic number 24
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1s2 2s2 2p6 3s2 3p6 3d5 4s1
6
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1s2 2s2 2p1
b
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a
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Scandium
7
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velocity of light = frequency x wavelength
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8
...

Ans
...

9
...

What does it prove?
Ans
...

10
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The smallest packet of energy of any radiation is called a quantum whereas
that of light is called photon
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Write the complete symbol for the atom with the given atomic number (Z) and
mass number(A)
...
(a) 17Cl
(b) 92U
2
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(a) 1s
(b) 3p (c)4d (d) 4f
3
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n=4, ms= -1/2
b
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(a) 16 electrons
(b) 2 electrons
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An element with mass number 81 contains 31
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Assign the atomic symbol
...
Mass number = 81, i
...
, p + n = 81
If protons = x, then neutrons = x + 31
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317 x
100
x+1
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317x = 81
x=35
Thus proton = 35, i
...
, atomic no
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(i) The energy associated with the first orbit in the hydrogen atom is -2
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What is the energy associated with the fifth orbit
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom
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(i) En = -2
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18 x 10-18/ 52 = -8
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529 x n2 r5 = 0
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225 A0= 1
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Explain , giving reasons, which of the following sets of quantum numbers are
not possible
...
(a) Not possible because n≠ 0 (c) Not possible because when n=1, l≠1
26

(b) Possible
(d) Possible
7
...

Ans
...
Calculate the total number of angular nodes and radial nodes present in 30
orbitals
...
For 3p orbitals, n=3, l= 1
Number of angular nodes = l= 1
Number of radial nodes = n-l-1 = 3-1-1= 1
9
...

Ans
...
It could not explain the stability of an atom
...
It could not explain the line spectrum of H- atom
...
State de-Broglie concept of dual nature of matter
...
Just as light has dual nature, every material particle in motion has dual nature
(particle nature and wave nature)
...


THREE MARKS QUESTIONS
1
...
(a) Pairing of electrons in the orbitals belonging to the same subshell (p, d or
f) does not takeplace until each orbital belonging to that subshell has got one
electron each i
...
, it is singly occupied
...
If two orbitals have the
same value of (n+l) then orbital with lower value of n will have lower energy
...
Write down the quantum numbers n and l for the following orbitals
a
...
3d
c
...

a
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n= 3, l=2 c
...
Write the 3 points of difference between orbit and orbital
...

Orbit
orbital
1
...
It represents the planar
motion of an electron around
the nucleus
3
...
An orbital is the three dimensional
space around the nucleus within
which the probability of finding an
electron is maximum(upto 90 %)
2
...
Different orbitals have different
shapes, i
...
, s-orbitals are spherically
symmetrical, p-orbitals are dumb-bell
shaped and so on
...
State Heisenberg’s uncertainty principle
...
7 x 10 5 m/s
...
It states that it is impossible to determine simultaneously, the exact position
and exact momentum (or velocity) of an electron
...

Δx x (m x Δv) = h/4ᴨ
Δx = h/4ᴨ x m x Δv =
6
...
0 x 10-10 m
4 x 3
...
1 x 10-31 x 5
...
Write 3 points of differences between electromagnetic waves and matterwaves
...
These are associated with
1
...

fields
2
...
They require medium for
medium for propagation
...
They travel with the same
3
...
(i) Calculate the number of electrons which will together weigh one gram
...
(i) Mass of one electron = 9
...
10939 × 10–31 kg = 1
Number of electrons that will weigh 1 g = (1 × 10–3kg)

= 0
...
1098 × 1028
= 1
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10939 × 10–31 kg
Mass of one mole of electron = (6
...
10939 ×10–31 kg)
= 5
...
6022 × 10–19coulomb
Charge on one mole of electron = (1
...
022 × 1023)
= 9
...
Find energy of each of the photons which
(i) correspond to light of frequency 3× 1015Hz
...
50 Å
...
(i) Energy (E) of a photon is given by the expression,
E=
Where,
h = Planck’s constant = 6
...
626 × 10–34) (3 × 1015)
E = 1
...
626 × 10–34Js
c = velocity of light in vacuum = 3 × 108m/s
Substituting the values in the given expression of E:

8
...
Theni= 4 to nf= 2 transition will give rise to a spectral line of the Balmer
series
...
0875 × 10–19 J)
The negative sign indicates the energy of emission
...
An atom of an element contains 29 electrons and 35 neutrons
...

Ans
...

∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is 1s22s2 2p6 3s2 3p64s2 3d10
(iii) Copper
10
...
Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2
∴ Number of electrons in
=2–1=1
Number of electrons in H2 = 1 + 1 = 2
Number of electrons present in oxygen molecule (O 2) = 8 + 8 = 16
∴ Number of electrons in

= 16 – 1 = 15

FIVE MARKS QUESTIONS WITH ANSWERS
30

1
...

Ans
...
Bohr’s model failed to account for the finer details of the hydrogen spectrum
...
Bohr’s model was also unable to explain spectrum of atoms containing more
than one electron
...
Bohr’s model was unable to explain Zeeman effect and Stark effect i
4
...

2
...
The work function for caesium atom is 1
...

Calculate (a) the threshold wavelength and (b) the threshold frequency of the
radiation
...

Ans
...
The ejected electrons are called photoelectrons
...
9 eV
...
53 × 10–7 m
Hence, the threshold wavelength
(b) From the expression,

is 653 nm
...
602 × 10–19J)
ν0= 4
...
593 × 1014s–1
...
3149 × 10–20 J
Kinetic energy of the ejected photoelectron = 9
...
E

v = 4
...
52 × 105ms–1
...
(a)The quantum numbers of six electrons are given below
...
If any of these combination(s) has/have the same
energy lists:
32

1
...
n= 3, l = 2, ml= 1 , ms= +1/2
3
...
n = 3, l= 2, ml= –2 , ms= –1/2
5
...
n = 4, l= 1, ml= 0 , ms= +1/2
(b)Among the following pairs of orbitals which orbital will experience the
larger effective nuclearcharge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p
Ans
...

For n = 3 and l = 2, the orbital occupied is 3d
...

Hence, the six electrons i
...
, 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p,
3d, 3p, and 4p orbitals respectively
...

(b)Nuclear charge is defined as the net positive charge experienced by an
electron in the orbital of a multi-electron atom
...

(i) The electron(s) present in the 2s orbital will experience greater nuclear
charge (being closer to the nucleus) than the electron(s) in the 3s orbital
...

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus
than 3f
...
(i) The unpaired electrons in Al and Si are present in 3p orbital
...
(i) the electrons in the 3p orbital of silicon will experience a more
effective nuclear charge than aluminium
...

(b) Silicon (Si):
Atomic number = 14
The electronic configuration of Si is:1s2 2s2 2p6 3s23p2
The orbital picture of Si can be represented as:
From the orbital picture, silicon has two unpaired electrons
...

(d) Iron (Fe):
Atomic number = 26
The electronic configuration is:1s2 2s2 2p6 3s23p6 4s2 3d6
The orbital picture of chromium is:
From the orbital picture, iron has four unpaired electrons
...
Give the name and atomic number of the inert gas atom in which the total
number of d-electrons is equal to the difference between the numbers of total p
and total s electrons
...
electronic configuration of Kr ( atomic no
...
of s-electrons = 8, total no
...
Difference = 10
No
...
What is the minimum product of uncertainty in position and momentum of an
electron?
Ans
...
Which orbital is non-directional ?
Ans
...
What is the difference between the notations l and L ?
Ans
...

5
...
18
6
...
Write the atomic number of A
...
15
7
...

Ans

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