CHAPTER 2 - ATOMIC STRUCTURE

Example 3

Line ED C B A
spectrum
Paschen series

Which of the line in the Paschen series
corresponds to the longest wavelength of photon?

Describe the transition that gives rise to the line.

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Solution

Line A.
Line A formed when the electron absorbed the
energy, and promoted from n=3 to n=4 (Electron
is excited). At excited states (n=4), the
electron is unstable. It will fall back to a lower
energy level (n=3) and releases a specific
amount of energy in the form of light (radiation),
called a photon. The energy released is detected
by detector and line A formed on the spectrum.

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Exercise
Calculate the
a) wavelength in nm
b) frequency
c) energy

that associated with the second line in the
Balmer series of the hydrogen spectrum.

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a) Second line of Balmer series:

Solution :
The transition of electron is from n2=4 to n1=2

n1 < n2 11 1
 = RH n12 n22

1 = (1.097x107 m1) 1 1
 22 42

 = 4.86x107 m
= 486 nm

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b) Frequency = 6.17 X 1014 s-1

 c

 
3.00 x 108 ms 1
4.86 x 10 7 m

 6.1728 x 1014 s-1

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c) Energy = 4.09 X 10-19 J

  h
 6.63 x 1034 Js x 6.1728 x1014 s1
 4.0926 x1019 J

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Exercise

Use the Rydberg equation to calculate the
wavelength of the spectral line of hydrogen atom
when an electron drops from the fourth orbit to the
first orbit.
Identified the series the line would be found.

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Solution:

1 R H 1 1
λ 2 n 22
n 1

n1  4 to n2 1

1  1.097  10 7 m1 1  1 
  12 42 

  9.7235 108 m → Lyman series

 9.7235 108 109 nm 58

 97.23nm

EXERCISE

Calculate the;
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen

spectrum in Lyman series
Wave number = 1/wavelength

For Lyman series; n1 = 1 & n2 = ∞
Ans:
i. 9.116 x10-8m
ii. 3.29 x1015 s-1
iii. 1.0970 X 107 m-1

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Solution

i) ∆E = RH (1/n12 – 1/n22) – 1/ ∞
= 2.18 X 10 -18 (1/12
= 2.18 X 10 -18 (1 – 0) 2)

= 2.18 X 10 -18 J

  hc

  
6.63 x 10 34 Js 3.00 x 108 ms 1
2.18 x 10 18 J

 9.12 x 10-8 m

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Solution

ii) ∆E = RH (1/n12 – 1/n22) – 1/ ∞
= 2.18 X 10 -18 (1/12
= 2.18 X 10 -18 (1 – 0) 2)

= 2.18 X 10 -18 J

  h

 2.18 x 10 18 J
6.63 x 1034 Js

  3.29 x 1015 s1

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Solution

iii)   9.12 x 10-8 m

1  9.12 1
 x 10-8 m

1.096 x107 m1

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Ionization Energy

• Ionization energy is the minimum energy
required to remove 1 mol of electron in its
ground state from an atom (or an ion) in
gaseous state
M (g)  M+ (g) + e H = +ve

• The hydrogen atom is ionised when electron is
removed from its ground state (n = 1) to n = .

• At n = , the potential energy of electron is zero,
here the nucleus attractive force has no effect on
the electron (electron is free from nucleus).

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Example 1

n1 = 1, n2 = ∞

∆E = RH (1/n12 – 1/n22)
= 2.18 X 10 -18 (1/12 – 1/ ∞ 2)
= 2.18 X 10 -18 (1 – 0)
= 2.18 X 10 -18 J

Ionisation energy for 1 mol atom

= 2.18 X 10 -18x 6.02 X 1023J mol-1
=1.312 x 106 J mol-1

= 1312 kJ mol-1

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Example 2:
Calculate the energy to ionized
(a) a hydrogen atom
(b) 1 mol of hydrogen atom

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Solution:

(a) ΔE = RH  1  1 
 n12 n22 

= 2.18  1018 J  1  1 
 12 2 

= 2.18  1018 J

• The energy to ionized a hydrogen atom is 2.18  1018 J

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(b) 1 H atom needs 2.18  1018 J
6.022  1023 H atoms need 2.18  1018J  6.022 x1023

= 1.31  106 J
The energy to ionized 1 mol of hydrogen atom

= 1.31 X 106 J

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Finding ionisation energy experimentally:

Convergent limit 1 st line

 1
Ionisation energy is determined by detecting
the wavelength of the convergence point

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Example 1

10.97 10.66 10.52 10.27 9.74 8.22 ( x 106 )

wavenumber, m-1

The Lyman series of the spectrum of hydrogen is
shown above. Calculate the ionisation energy of
hydrogen from the spectrum.

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Solution

ΔE = hc/λ ; where 1/ λ is the wave number

= h x c x wave no.

= 6.626 x 10-34 J s x 3 x 108 m s-1 x 10.97 x 106 m-1
= 218.06 x 10-20 J

= 2.18 x 10-18J

Ionisation energy
= 2.18 X 10 -18x 6.02 X 1023 J mol-1
=1.312 x 106 J mol-1
= 1312 kJ mol-1

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Example

9.12 9.38 9.51 9.73 ( x10-8 m)

wavelength, 

The Lyman series of the spectrum of hydrogen is
shown above. Calculate the ionisation energy of
hydrogen from the spectrum.

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Solution
ΔE = hc/λ

= (6.626 x 10-34 J s x 3 x 108 m s-1 ) / 9.12 x 10-8 m
= 2.18 x 10-18J

Ionisation energy
= 2.18 X 10 -18x 6.02 X 1023 J mol-1
=1.312 x 106 J mol-1
= 1312 kJ mol-1

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The limitation of Bohr’s Atomic Model

• It can only explain the hydrogen spectrum or
any spectrum of ions contain one electron

eg: He+, Li2+

• Unable to account for the emission spectrum of
atom containing more than 1 electron

• Electrons are wavelike, we can’t define the
precise location of a wave because a wave
extends in space.

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de Broglie’s Postulate

• In 1924 Louis de Broglie proposed that not only
light but all matter has a dual nature and
possesses both wave and particle properties.

• Tiny particle such as an electron have wave

properties.

• An electron is both particle and wave.  hm=

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• de Broglie deduced that the particle and
wave properties are related by the expression

λ= h
mv

h = Planck constant (J s)
m = particle mass (kg)
v = velocity (m/s)
 = wavelength of a matter wave

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Heisenberg’s Uncertainty Principle

It is impossible to know at the same time, both
momentum p (defined as mass times velocity) and
position of a particle with certain.

Stated mathematically, x p  h
4

where x = uncertainty in measuring the position
p = uncertainty in measuring the momentum
= mv
h = Planck constant

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2.2 Quantum Mechanical
Model

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2.2 Quantum Mechanical Model

At the end of this topic students should be able to:-
• Define the term orbital.
• Explain the four quantum numbers.

i. principle quantum number, n
ii. angular momentum quantum number, l
iii. magnetic quantum number , m
iv. electron spin quantum number, s
• Sketch the 3-D shapes of s, p and d orbitals.

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Atomic Orbital

Definition

An orbital is a three-
dimensional region
in space around the
nucleus where there
is a high probability of
finding an electron.

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Quantum Numbers
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers :
i. principal quantum number, n

ii. angular momentum quantum number, 

iii. magnetic quantum number, m
iv. electron spin quantum number, s.

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i) Principal Quantum Number, n
• n determines the energy level (electron shell)

and size of an orbital.

• The principal quantum number n, starting from
n =1, 2, 3, …, .

• As n increase : i) the orbital become larger

ii) electron has higher energy

n1 2 3 4

Orbital size

Energy increases

81r

ii) Angular Momentum Quantum Number, 

Alternative name: Subsidiary / Azimuthal /
Orbital / Quantum Number

 The value of  indicates the shape of the atomic

orbital.

 The allowed values of  are 0, 1, 2,…, (n1)
 Letters are assigned to different numerical values of .

Numerical value of  Symbol Orbital shape

0 s spherical
1 p dumbbell
2 d cloverleaf
3 f

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• A collection of orbitals with the same value of n
is called a shell.

• One or more orbitals with the same n and 

values are called sub-shell.

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-  value is depend on n, i.e, 0, 1, …. (n -1).

Example:
If n = 1,  = 0 (s-orbital) - only one shell

If n = 2,  = 0 (s-orbital) two subshells
= 1 (p-orbital) (s and p-orbitals)

If n = 3,  = 0 (s-orbital) three subshells
= 1 (p-orbital) (s, p, and d-orbitals)
= 2 (d-orbital)

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iii) Magnetic Quantum Number, m
• Describe the orientation of orbitals in space.
• Possible values of m depend on the value of  that

is m = , …, 0, …, +
Example:
If  = 0, m = 0  1 orientation of s orbital
If  = 1, m = 1, 0, +1 means 3 values of m

 3 orientations of p orbital (px, py, pz)

If  = 2, m = 2, 1, 0, +1, +2 means 5 values of m
 5 orientations of d orbital ( dxy,dxz,dyz,dx2-y2,dz2)

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iv) Electron Spin Quantum Number, s

• The value of s represent the direction of
an electron rotation on its own axis.

• either clockwise or anticlockwise
• It has 2 value : + 1 or - 1

22

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Shape of Atomic Orbitals

a) s orbital
 Spherical shape with the nucleus at the

centre.
 When l = 0 , m = 0 , only 1 orientation of s

orbital
 The larger value of n, the larger the size of s

orbital.

Shape of s orbital
with different n

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b) p orbital
 is represented as a pair of dumb-bell shaped
 When l = 1, m = -1,0,+1
 3 orientation of p-orbitals are px, py, and pz.
 p-orbitals lie along x, y and z axis
 As n increases, the p-orbitals get larger.

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c) d orbitals
• All d orbitals are not look alike.

• When  = 2 , m = -2, -1, 0, +1, +2.

There are five orientations of d orbitals.

• The orbitals are labeled as
dxy, dxz, dyz, dx2-y2, dz2

• Orbitals of dxy, dxz, dyz lie between the axis while
dx2-y2,dz2 lie along the axis.

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90

• The 4 quantum numbers n,l,m and s
enable us to locate an electron in any
orbital of an atom.

Eg:
• 4 quantum numbers of 2s orbital electron

are
n = 2 , l = 0 , m = 0 and s = +1/2 or -1/2
• Written as (n = 2, l = 0, m = 0, s = +1/2)
or (n = 2, l = 0, m = 0, s = -1/2)

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Summary to determine an orbital

1) What is the size and energy level of the orbital?

n (1,2,3…) – size and energy level
e.g: n = 3 > n = 2

2) What is the shape of the orbital?
ℓ (0,1,2…. n-1) – 0 (spherical)

1 (dumbell)

2 (cloverleaf)

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3) Where is the orientation (direction/ point of
reference/ axis) of the orbital?

m (-ℓ…0…+ℓ) – m = 0 : center
m = -1 : x axis  px
0 : y axis  py
+1 : z axis  pz
m = -2 : between axis  dxy
-1 : between axis  dxz
0 : between axis  dyz
+1 : x and y axis  dx2- y2
+2 : z axis  dz2

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Answer Example 1

n/  Orbital m No. of No of
Shell designation orbitals electrons

10 1s 0 1 2
2s 0 1 2
0 2p -1, 0, +1 3 6
2 3s 0 1 2
3p -1, 0, +1 3 6
1 -2, -1, 0, +1, +2
3d 5 10
0
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31

2

Example 2

Predict the following quantum numbers
whether they are allowed or not
(a) (n=1, l=0, m=0, s= -1/2)
(b) (n=2, l=0, m=1, s= 1)
(c) (n=0, l=1, m=1, s= +1/2)
(d) (n=4, l=1, m=0, s= -1/2)

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Solution:

(a) (n=1, l=0, m=0, s= -1/2)
allowed

(b) (n=2, l=0, m=1, s= 1)
not allowed because when l=0, the possible value
of m=0 and s= +1/2 or -1/2

(c) (n=0, l=1, m=1, s= +1/2)
not allowed because the possible value of n=1 or
2 or 3 or 4….

(d) (n=4, l=1, m=0, s= -1/2)
allowed

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Exercise

1. Give the values of the four quantum numbers of an

electron in the following orbitals

a) 2p b) 4p c) 3d

2. What are the similarities and differences between
a) a 1s and a 2s orbital
b) a 2px and a 2py

3. Calculate the total number of electrons that can occupy
a) a one s orbital
b) three p orbitals
c) 5dxz orbital

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Solution:

1. a) (n=2, l=1, m= +1, s=-1/2) or
(n=2, l=1, m= +1, s=+1/2) or
(n=2, l=1, m= 0, s=-1/2) or
(n=2, l=1, m= 0, s=+1/2) or
(n=2, l=1, m= -1, s=-1/2) or
(n=2, l=1, m= -1, s=+1/2)

b) (n=4, l=1, m= +1, s=-1/2) or
(n=4, l=1, m= +1, s=+1/2) or
(n=4, l=1, m= 0, s=-1/2) or
(n=4, l=1, m= 0, s=+1/2) or
(n=4, l=1, m= -1, s=-1/2) or
(n=4, l=1, m= -1, s=+1/2)

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Solution:

c) (n=3, l=2, m= +1, s=-1/2) or
(n=3, l=2, m= +1, s=+1/2) or
(n=3, l=2, m= 0, s=-1/2) or
(n=3, l=2, m= 0, s=+1/2) or
(n=3, l=2, m= -1, s=-1/2) or
(n=3, l=2, m= -1, s=+1/2) or
(n=3, l=2, m= +2, s=-1/2) or
(n=3, l=2, m= +2, s=+1/2) or
(n=3, l=2, m= -2, s=-1/2) or
(n=3, l=2, m= -2, s=+1/2)

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Solution:

2. a) 1s and 2s orbitals have same shape but 1s orbital
has smaller size than 2s orbital. 1s orbitals also has
lower energy than 2s orbitals.

b) 2px and 2py orbitals have same size, energy and

shape but 2px and 2py orbitals are located

at different axis.

3. a) 2
b) 6
c) 2

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