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J.J.
Thomson’s
Model
Rutherford’s
Model
Bohr’s
Model
Planck's
Quantum
Theory
de Broglie’s
hypothesis
SWE
Heisenberg’s
Uncertainty
principle
Thomson’s Model
An atom has a spherical shape
(radius~10–10 m).
Positive charge is uniformly distributed
throughout the sphere.
Negatively charged electrons are
embedded in it like raisins in a
pudding.
Thomson’s Model
Mass of the atom is assumed to be
uniformly distributed all over the
atom.
Plum pudding Watermelon
Thomson’s Model
Electrons are embedded in an atom in such a way that
the most stable electrostatic arrangement is achieved.
Positively-
charged matter
Electrons
Limitation of Thomson’s Model
Not consistent
with the results
of later
experiments
A stream of high energy 𝞪–particles was
directed at a thin gold foil (thickness ∼ 100 nm).
Rutherford’s Experiment
Radioactive
source
ZnS
screen
Gold foil
Rutherford’s Experiment: Observations
Gold atoms
Nucleus
Observation Conclusion
Most 𝞪-particles passed
through the foil without
deflection
Presence of large
empty space
in the atom
Observation Conclusion
Few 𝞪-particles
were deflected by small
angles
Positive charge is
concentrated in a
very small region
Observation Conclusion
Very few 𝞪-particles
(∼1 of 20,000)
deflected at 180°
Small positively -
charged core at
the centre
Extranuclear Part
Nucleus is surrounded by electrons
revolving in circular paths called orbits.
Electrostatic Forces
Electrons and the nucleus are held together by
electrostatic forces of attraction in an atom.
FCentripetal
α particles can be represented by :
B) He2+
C) He
A) Li+
D) None of the above
Imagine that in any atom about 50% of the space is occupied
by the atomic nucleus. If a silver foil is bombarded with α-
particles, majority of the α-particles would :
B) be absorbed by the nuclei
C) pass through the foil undeflected
A) be scattered
D) get converted into photons
Nucleus
An atom consists of
a small positively -
charged core at the
centre which
carries almost the
entire mass
of the atom.
It has negligible
volume as
compared to the
volume
of the atom.
Nucleus
Radius of the atom
Radius of the nucleus
~10–10 m
~10–15 m
Both protons and neutrons present in the
nucleus are collectively called nucleons.
Nucleus
R R0 (A)
=
R = Radius of nucleus of an element
A = Mass number of element
R0 = 1.11 x 10-15 m to 1.44 x 10-15 m
1
3
Extranuclear part
Nucleus is surrounded by
revolving electrons.
Electrons and
nucleus are held
together by
electrostatic forces
of attraction.
Extranuclear part
FCentripetal
FElectrostatic FCentripetal
=
Rutherford’s Model: Drawbacks
Electron
Nucleus
It could not explain stability of the atom.
Rutherford’s Model: Drawbacks
It could not explain the line spectrum of the H atom.
It could not explain the electronic structure of the atom.
What is the ratio of specific charge of a proton to that of an
𝜶-particle?
B) 1 : 2
C) 1 : 4
A) 2 : 1
D) 4 : 1
How much time does light take to travel 1.0 m?
B) 3.3 microsecs
C) 3.3 nanosecs
A) 3.3 millisecs
D) 3.3 seconds
Smallest quantity of energy that can
be emitted or absorbed in the
form of EM radiation
Packet or bundle of energy
Quantum of radiation
Photon
Planck’s Quantum Theory
E h𝝂
= = h h = Planck’s constant
= 6.626 × 10-34 Js
Quantum Theory of Radiation
c
λ
E nh𝝂
=
n = Number of photons
= 0, 1, 2, 3, ….
_
×
)
Important Conversions
1 eV 1.6 × 10-19 J
=
E (eV) = 12,400
λ (Å)
=
eV
particle
E (
kJ
mole
E
( ) 96.48
An electric bulb marked as 60 watt emits light of wavelength
3000 Å. If 25% of the energy is emitted as light, what is the
number of photons emitted in one second?
B) 7.32 x 1020
C) 2.27 x 1019
A) 2.27 x 1020
D) 7.22 x 1019
31
What is the equation E = h𝝂 indicate?
B) Photons are waves.
C) Photons are stream of particles.
A) Photons have both particle and wave nature.
D) No such inference can be drawn from the given
equation.
Photoelectric Effect
Observations
Electrons are ejected as soon as the beam of light
of sufficient frequency strikes the metal surface.
mevmax
2
2
Photoelectric Effect
h𝝂 h𝝂0
= +
me = Mass of the electron
vmax = Maximum velocity of the electron
From the law of energy conservation
1
Which of the following statements is/are true in the context of
photoelectric effect?
A) The kinetic energy of ejected electron is independent of the
intensity of radiation.
B) It provided evidence for the quantum nature of light.
C) The number of photoelectrons ejected depends upon the
intensity of the incident radiation.
D) The kinetic energy of the emitted electrons depends on the
frequency of the incident radiation.
Electromagnetic radiation having λ = 310 Å is subjected to a
metal sheet whose work function = 12.8 eV. What will be the
velocity of the photoelectron having the maximum kinetic
energy?
B) 4.352 ✕ 106 m/s
C) 3.09 ✕ 106 m/s
A) No emission will occur
D) 8.72 ✕ 106 m/s
Postulates
Stationary orbits
Concentric circular orbits around the nucleus
These orbits have fixed value of energy
Electrons revolve without radiating energy
Postulates
Energy states/levels
Stationary orbits or
L M
n = 1
n = 2
n = 3
K
Postulates
mvr
nh
2π
n = 1, 2, 3...
Planck’s constant
h
Velocity of electron
v
Mass of electron
m
Radius of orbit
r
=
Bohr’s
Model
rn = 0.529 Å
n2
Z
vn = 2.18 x 106 m/s
Z
n
En = -13.6 eV/atom
Z2
n2
Bohr’s
Model
Tn = 1.5 x 10-16 s
n3
Z2
fn = 6.6 x 1015 Hz
Z2
n3
If the radius of the 2nd orbit of Li2+ ion is x, then find the radius
of the 1st orbit of He+ in terms of x.
B)
C)
A)
D)
5x
8
7x
8
3x
8
x
8
Calculate the energy ratio for the 3rd orbit of Li2+ ion and the
2nd orbit of Be3+ ion.
B)
C)
A)
D)
1
2
1
4
1
3
1
9
Emission Spectral Lines/
De-Excitation Series
Line Spectrum of Hydrogen
Line
Spectrum
Emission spectral lines/
De-excitation series
Line
Spectrum
n = 1
n = 2
n = 3
n = 4
n = 5
n = ∞
13.6
eV
12.09
eV
-13.6 eV
-3.4 eV
-1.51 eV
-0.85 eV
-0.54 eV
0 eV
Energy Level Diagram for H atom
10.2
eV
3.4
eV
12.75
eV
13.06
eV
Rydberg’s Formula
RH Z2 x
1
λ
1, 2, 3, ...
n1
n1 + 1, n1 + 2, ...
n2
For any atom
1
n1
2
_ 1
n2
2
=
λ
For any atom
= 1
n2
2
1
n1
2
_
912
Z2
Å
Line Spectra is a characteristic of :
B) atoms
C) radicals
A) molecules
D) none of these
The wavelength of a spectral line for an electronic transition is
inversely proportional to :
B) the nuclear charge of the atom
C) the velocity of an electron undergoing transition
A) the number of electrons undergoing transition
D) the difference in the energy involved in the transition
de Broglie Wavelength (λ)
h
p
h
mv
λ = =
Momentum
of particle
p
Mass
of particle
m
Velocity
of particle
v
Planck’s
constant
h
p
0
λ
A beam of helium atoms move with a velocity of 2 × 104 m s–1.
Find the wavelength (in pm) of the particles constituting the
beam.
B) 3
C) 6
A) 4
D) 5
de Broglie’s Equation and Kinetic Energy
Multiplying both sides by m and rearranging
K.E. =
m2v2 2 K.E. × m
=
mv
mv2
1
2
√2 K.E. x m
=
de Broglie’s Equation and Kinetic Energy
h
λ =
Since, mv 2 K.E. x m
h
p
h
mv
λ
2 K.E. × m
de Broglie
equation
= √
√
= =
If the kinetic energy of a proton is increased to nine times, the
wavelength of the de Broglie wave associated with it would
become :
times
times
1
3
1
9
B) 9 times
C)
A) 3 times
D)
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is
3.0 × 10–25 J, calculate its wavelength.
B) 0.89 × 10-7 m
C) 0.89 × 10-5 m
A) 0.89 × 10-6 m
D) 0.89 × 10-8 m
Heisenberg’s Uncertainty Principle
Werner Heisenberg
Heisenberg’s Uncertainty Principle
Exact position and momentum of a microscopic particle
cannot be determined simultaneously
Heisenberg’s Uncertainty Principle
Δx . Δp
4𝜋
h
≥
Δx . m . Δv
Uncertainty in
position
Δx
Mass of
particle
m
Uncertainty in
momentum
Δp
Uncertainty in
velocity
Δv
≥
4𝜋
h
Principle of Optics
λ
Minimum error in the
position measurement (Δx) =
If a light (wavelength ‘λ’) is used to locate the
position of a particle, then
+
-
Heisenberg’s Uncertainty Principle
=
Since
For accurate
position
Δx 0 λ 0
Δx λ
Heisenberg’s Uncertainty Principle
E
λ
hc
=
For a photon
λ 0 E
Heisenberg’s Uncertainty Principle
High energy photon
strikes particle
Δp
For accurate momentum Δx
Similarly
Heisenberg’s Uncertainty Principle
For an electron
≥
Δx . m . Δv
4𝜋
h
≥
Δx . Δv
4𝜋m
h
Heisenberg’s Uncertainty Principle
=
Δx . Δv
4 × 3.14 × 9.1 × 10-31 kg
6.626 × 10-34 Js
≅
Δx . Δv 10-4 m2s-1
Considering minimum error
Position
High
accuracy
Δx is small
Velocity Uncertain Δv is large
Heisenberg’s Uncertainty Principle
If Δx = 10-8 m then Δv = 104 ms-1
Velocity
High
accuracy
Δv is small
Position Uncertain Δx is large
Heisenberg’s Uncertainty Principle
If Δv = 10-8 ms-1 then Δx = 104 m
Uncertainty Principle for
Large Particles
A golf ball has a mass of 40 g, and a speed of 45 m s-1. If the
speed can be measured within accuracy of 2 %, calculate the
uncertainty in the position.
Conclusion
Heisenberg
Uncertainty Principle
Meaningless for large
particles
Significance of Uncertainty Principle
2
Rules out the existence of
definite paths of electrons
3
Introduced concept of probability
of finding the electrons
1
Not an instrumental error, rather
conceptual error
Heisenberg’s Uncertainty Principle
The uncertainty in position of a particle of 25 g in space is 10-5 m.
What will be the uncertainty in velocity (m s-1) ?
B) 2.1 × 10-34
C) 0.5 × 10-34
A) 2.1 × 10-28
D) 5 × 10--24
The Heisenberg’s uncertainty principle is significant for :
B) a football
C) a jet aeroplane
A) a cricket ball
D) an electron
Quantum Numbers
Set of four numbers
required to define
an electron in an
atom completely
Quantum Numbers
Principal Quantum Number (n)
1
Azimuthal Quantum number (l)
2
Magnetic Quantum Number (ml )
3
Spin Quantum Number (s)
4
Principal Quantum Number (n)
Describes the size
of electron wave &
the total energy of
the electron
n = 1, 2, 3,...
Represented as K, L, M, N,...
Azimuthal Quantum Number (l )
2
Energy of the orbital in a
multielectron species (both n & l)
1
Designates the subshell to which
the electron belongs
Describes the 3-D
shape of the orbital
or the electron
cloud
Boundary Surface Diagram
Encloses the 3-D region, where the
probability of finding electrons is maximum
Shape: Spherical
Example
s - orbital
Shape: Spherical
p - orbital
Shape: Dumb bell
d - orbital
3:4
Shape: Double
dumb bell
Subshell
Collection of
similar shaped
orbitals of the
same n
Azimuthal Quantum Number (l )
For a given value of principal quantum number (n)
0 to (n - 1)
l =
=
Subshell
l
0
Subshell
s
1 p
2 d
3 f
Description
Sharp
Principal
Diffused
Fundamental
Subshell Representations
n
1
l
0
2 0, 1
3 0, 1, 2
4 0, 1, 2, 3
Subshell
notation
1s
2s, 2p
3s, 3p, 3d
4s, 4p, 4d, 4f
Number of subshells in the nth shell n
Azimuthal Quantum Number (l )
Orbital angular
momentum (L)
l (l + 1) ħ
Subshell
s
Orbital angular
momentum
0
p 2 ħ
d 6 ħ
ħ =
h
2π
=
Magnetic Quantum Number (ml )
2 Describes the orientation of orbitals
3
Accounts for the splitting of lines of
atomic spectrum in magnetic field
1
Designates the orbital to which
the electron belongs
Magnetic Quantum Number (ml )
Subshell
s
Number of orbitals
1
p 3 (px, py, pz)
d 5 (dxy, dyz, dzx, dx - y ,dz )
2
2 2
f 7
2l + 1
Maximum number of
orbitals in a subshell =
2
s - orbital
Shape: Spherical
Non-directional
in nature
x
y
z
p - orbital
Shape: Dumb bell
Directional in
nature
z
y
x
z
y
x
z
y
x
pz py px
d - orbital
3:4
x
y
z
dxy
x
y
z
dxz
x
y
z
dyz
z
x
y
z
y
x
dz
dx - y
Axial
d-orbitals
Non-axial
d-orbitals
2
2 2
Remember!
Maximum number of
electrons in a subshell
2 (2l + 1)
Subshell
l
s
0
Number of electrons 2
p
1
6
d
2
10
f
3
14
=
An orbital can accommodate maximum of 2 electrons.
Spin Quantum Number (s)
Spin of an electron
s = +
1
2
s =
1
2
Presence of two closely-spaced lines in
atomic spectrum
√n(n+2) B.M.
Spin Quantum Number (s)
Spin magnetic
moment (μ)
n = Number of unpaired
electrons
=
n2
Remember!
Number of orbitals in nth shell =
2n2
Maximum electrons in nth shell =
What is the total number of atomic orbitals in fourth energy
level of an atom?
B) 16
C) 32
A) 8
D) 4
Which of the following shell can contain maximum of 72
electrons?
B) 6th
C) 4th
A) 5th
D) 3rd
Which of the following statements concerning the four
quantum numbers is false?
B) l gives the shape of an orbital.
C) m gives the energy of the electron in the orbital.
A) n gives an idea of the size of an orbital.
D) s describes the spin of the electron.
Rules for Filling of Electrons in Orbitals
Rules
Aufbau principle
Hund’s rule of maximum
multiplicity
Pauli’s exclusion principle
Aufbau Principle
Electrons are
filled in various orbitals
in order of their
increasing energies
Energies of Subshells of H-like Species
1s < 2s = 2p < 3s = 3p = 3d
< 4s = 4p = 4d = 4f < …
Order of
energy
Energy of single electron species depends only on the
Principal quantum number
What are Degenerate Orbitals?
Orbitals
that have
same energy
Energies of Subshells of H-like Species
4s
3s 3p 3d
2s 2p
1s
Energy
4p 4d 4f
Energy of Subshells of Multielectron Species
Different subshells have different energy
which depends on:
Principal quantum
number
Azimuthal quantum
number
( n + l ) rule or Bohr-Bury’s Rule
Lower the
energy of subshell
Lower value of
(n + l )
Example
3d n + l 3+2 5
= = =
4s n + l 4+0 4
= = =
4s 4s
>
Value of n + l for :
4s will be filled before 3d
( n + l ) rule
Subshell with lower ‘n’
value has lower energy
Two subshells with
same (n + l) value
Example
3d n + l 3 + 2 5
= = =
4p n + l 4 + 1 5
= = =
3d 4p
=
Value of n + l for :
3d will be filled first due to
the lower ‘n’ value
Comparision of orbital energy
2s
1s
3d
2p
4d
4f
4f
5d
4p
3d
Energies of Subshells of Multielectron Species
Energy
4p
4s
3d
3p
3s
2s
2p
1s
Energies of Subshells of Multielectron Species
Energies of Subshells of Multielectron Species
Energies of Subshells of Multielectron Species
In a 3d subshell, all the five orbitals are degenerate. What does
it mean?
B) All the orbitals have the same shape.
C) All the orbitals have the same energy.
A) All the orbitals have the same orientation.
D) All the orbitals are unoccupied.
Degeneracy of the second excited state of H is y. Find the
value of y.
B) 9
C) 5
A) 3
D) 4
Pauli’s Exclusion Principle
Wolfgang Pauli
Pauli’s Exclusion Principle
No two electrons in
an atom can have
the same set of all
four quantum
numbers
Pauli’s Exclusion Principle
Restrict the filling of number of electrons
in an orbital
⥮ ⥮
⥮
⥮
⥮ ⥮
s2
⥮
⥮
⥮ ⥮
p6
⥮
d10
⥮ ⥮ ⥮
f14 ⥮ ⥮
Subshell electron capacity
Friedrich Hund
Hund’s Rule of Maximum Multiplicity
Hund’s Rule of Maximum Multiplicity
Deals with the filling of electrons
In the degenerate orbitals of
the same subshell
Hund’s Rule of Maximum Multiplicity
No electron pairing takes place in
the orbitals in a subshell
Until each orbital is occupied
by 1 electron with parallel spin
Hund’s Rule of Maximum Multiplicity
Hund’s rule is an empirical rule
Determines the lowest energy
arrangement of electrons
Why Maximum multiplicity?
Maximum spin of
an atom (S)
1
2
Spin Multiplicity (S.M.) 2S + 1
=
= x n
Spin Multiplicity (S.M.) Stability
Electronic Configuration
Electronic Configuration
Distribution
of electrons in
orbitals of
an atom
Electronic Configuration
Two ways of Electronic Configuration
Representation
sa pb dc ... notation Orbital diagram
sa pb dc ... Notation
Similar subshell represented for
different shells is differentiated by
Writing the principal
quantum number before the
respective subshell
sa pb dc ... Notation
2 p
5
Subshell ‘p’ belongs
to 2nd shell (n = 2)
Subshell ‘p’, l = 1 Number of
electrons = 5
F
19
9
Fluorine
1s2 2s2 2p5
Orbital Diagram Notation
1s2 2s2 2p5
Electron is Represented By
Upward arrow ( or ↿) Downward arrow ( or ⇂)
Represent all 4 Quantum numbers
⥮ ⥮ ⥮ ↿
⥮
F
H
1s1
He
1s2
Li
⥮
⥮
1s2 2s1
↿
↿
Electronic Configuration of Various Elements
C ⥮ ⥮ ↿ ↿
B ⥮ ⥮ ↿
2p2
1s2 2s2
1s2 2s2 2p1
Be
1s2 2s2
⥮ ⥮
Electronic Configuration of Various Elements
F
1s2 2s2 2p5
⥮ ⥮ ⥮ ⥮ ↿
O ⥮ ⥮ ↿
⥮ ↿
2p4
1s2 2s2
N ↿ ↿
↿
2p3
1s2 2s2
⥮ ⥮
Electronic Configuration of Various Elements
Ne
2p6
1s2 2s2
⥮ ⥮ ⥮ ⥮ ⥮
Mg
2p6
1s2 2s2
Na
⥮ ⥮
⥮ ⥮
⥮ ⥮ ⥮
⥮ ⥮ ⥮ ↿
3s1
3s2
⥮
Electronic Configuration of Various Elements
2p6
1s2 2s2
Simplified Electronic Configuration
Na
1s2 2s2 2p6
⥮ ⥮ ⥮ ⥮ ⥮ ↿
3s1
Ne
Na Ne
3s1
Simplified configuration: ↿
Configuration of Sodium:
Simplified Electronic Configuration
1s2 2s2 2p6
⥮ ⥮ ⥮ ⥮ ⥮
3s2
Ne
Simplified
configuration:
Configuration of Argon:
Ar
3p6
⥮ ⥮ ⥮
⥮
Ar Ne
3s2
⥮
3p6
⥮ ⥮ ⥮
Write the electronic configuration and find the number of
unpaired electrons as well as total spin for the following atoms.
21Sc
26Fe
B)
A)
a) 21Sc 1s2 2s2 2p6 3s2 3p6 4s2
3d1
[Ar] 4s2 3d1
[Ar] 3d1 4s2
Number of unpaired electrons 1
Total spin
1
2
1
2
or
3d1
↿
+ -
4s2
⥮
Sc
45
21
Scandium
Exceptions
[Ar] 4s2 3d4
d5 is more stable than d4 configuration
24Cr
[Ar] 4s1 3d5
24Cr
Cr
52
24
Not
Correct
Correct
Exceptions
[Ar] 4s2 3d9
29C
u
[Ar] 4s1 3d10
29C
u
Cu
63
29
Not
Correct
Correct
d10 is more stable than d9 configuration
What is special about
half filled & fully filled orbitals?
Half-filled & Fully Filled Orbitals
Exactly half filled &
fully filled orbitals make the
configuration more stable
p3, p6, d5, d10, f7 & f14
configurations are stable
Half-filled & Fully Filled Orbitals
Stability of half filled & fully filled orbitals
Symmetry
Exchange
Energy
Symmetry
Symmetrical
distribution of
electrons
Symmetry leads
to stability
Electrons in
the same
subshell
Different
spatial
distribution
Equal
energy
Symmetry
Consequently, their shielding of one
another is relatively small
Electrons are more strongly
attracted by the nucleus
Have less energy
and more stability
Exchange Energy
Tends to exchange their positions
Energy released when two or more electrons
with the same spin in the degenerate orbitals
Exchange Energy
Number of exchanges that
can take place is maximum
When subshell is either half
filled or fully filled.
Exchange Energy
↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿
↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿
4 exchange by electron ‘a’ 3 exchange by electron ‘b’
2 exchange by electron ‘c’ 1 exchange by electron ‘d’
a b
c d
The orbital diagram in which both Hund’s rule and Aufbau
principle are violated is:
⥮ ⥮
↿
↿
⥮
⥮ ↿ ↿
⥮ ↿
↿↿ ↿ ↿
↿
2p 2p
2p 2p
2s
2s
2s 2s
Which rule is/are violated in the given electronic configurations?
⥮ ↿
⥮
↿ ↿
⥮ ↿
↿↿ ↿ ↿
↿
↿ ⥮ ⥮
2p
2s
2p
2s
2p
2s
2p
2s
Aufbau Pauli’s Hund’s
THANK YOU
“Stay Positive. Work Hard. Make It Happen!”

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