SK015

Rydberg Equation

ii) frequency, Ʋ • Wavelength emitted by the electron between two
energy levels can be calculated using Rydberg
  E equation:
h

= 1 .0 6 x 1 0 -19 J 1  RH 1 1
6 .6 3 x 1 0 -34 J s λ n 2 n 22
1

= 1 .60 x 1 0 J where
6 .6 3 x 1 0 -34 J s
RH = 1.097  107 m-1
= 1 .6 0 x 1 0 14 s -1  = wavelength
Since  should have a positive value thus n1 < n2

25 26

Example Solution:
Calculate the wavelength, in nanometers of the
spectrum of hydrogen corresponding to n = 2 and 1  RH 11
n = 4 in the Rydberg equation. λ 2 n 22
n 1
27
1  1.097  10 7 m1 1  1 
  22 42 

  4.86  107 m

 4.86  107  109nm

 486nm

28

Exercise: Answer:

1) Calculate the energy of hydrogen electron in the: 1) Calculate the energy of hydrogen electron in the:
(a) 1st orbit
(b) 3rd orbit (a) 1st orbit (b) 3rd orbit
(b) 8th orbit
E1  RH  1  E1  RH  1 
2) Calculate the energy change (J), that occurs 12 12
when an electron falls from n = 5 to n = 3 energy
level in a hydrogen atom.  2.18 x 10 18  1    2.18 x 10 18  1 
12 32
3) Calculate the frequency and wavelength (nm) of   2.18 x 1018 J
the radiation emitted in question 2.   2.42 x 1019 J

29 (c) 8th orbit

E1  RH  1 
 12 

  2.18 x 10 18  1 
 82 

  3.41 x 1020 J 30

48

Answer: Answer:
2) Calculate the energy change (J), that occurs
3) Calculate the frequency and wavelength (nm) of
when an electron falls from n = 5 to n = 3 the radiation emitted in question 2.
energy level in a hydrogen atom.
  h

E  RH ( 1  1 )  1.55 x 10 19 J
52 32 6.63 x 1034 Js

 2.18 X 10 18 ( 1  1 )   2.34 x 1014 s1
52 32
  hc
 1.55 X10-19 J 

  
6.63 x 1034 Js 3.00 x108 ms1
1.55 x 10 19 J

1.28 x 10-6 m

31 32

Emission Spectra i) Continuous Spectrum
• A spectrum consists of radiation with
i) Continuous ii) Line
Spectra Spectra unbroken sequence of frequencies over a
relatively wide range of wavelength.
Emission and absorption spectra may show a continuous spectrum, a line spectrum or a band spectrum • Example : electromagnetic spectrum, rainbow

33 34

Regions of the Electromagnetic Spectrum FORMATION OF CONTINUOUS SPECTRUM

Continuous spectrum is produced by white light
(sunlight or incandescent lamp) that passed through
a prism

35 36

49

FORMATION OF ATOMIC / LINE SPECTRUM

ii) Line Spectrum (atomic spectrum) Spectral lines

A spectrum consists of discontinuous or discrete prism film
lines with specific wavelengths and frequencies
in which each line is produced by the transition gas discharge tube
of electron between two energy levels.
• It is composed when the light from a gas discharge
37 tube containing a particular element is passed
through a prism.

• The emitted light (photons) is then separated into its
components by a prism.

38

• Each component is focused at a definite position, Example :
according to its wavelength and forms as an
image on the film of the photographic plate. • The line emission spectrum of hydrogen atom

• The images are called spectral lines. • Line spectrum are composed a few wavelengths
giving a series of discrete line separated by blank
39 areas

• It means each line corresponds to a specific
wavelength and frequency.

line emission spectrum of hydrogen atom

40

FORMATION OF ATOMIC / LINE SPECTRUM FORMATION OF ATOMIC / LINE SPECTRUM

• Energy absorbed

n= by the atom causes n= When the electrons
n=5 the electron n=6 fall to lower energy
n=4 n=5 levels, nf, photons
Energy n=4 energies are
Energy n=3 emitted in the form
n=3 promoted from a of light with specific
lower to a higher- n=2 wavelength can be
detected as a line
n = 2 energy level, ni. spectrum

• The electron is at Emission of photon
excited state (very

n = 1 unstable) and will
fall to lower energy
level.

41 42

50

Line spectrum series Exercise: Complete the following table

• For the emission line spectra, Series nf ni Spectrum
nf is the energy level to where the electron falls region
ni is the energy level from where the electron falls
Lyman 1 2,3,4,… ultraviolet
• If an electron falls to energy level 1 (nf = 1), it forms Lyman Balmer 2 3,4,5,… Visible
series. Paschen 3 4,5,6,… Infrared
Brackett 4 5,6,7,… Infrared
• If an electron falls to energy level 2 (nf = 2), it forms Balmer Pfund 5 6,7,8,… Infrared
series.
44
• If an electron falls to energy level 3 (nf = 3), it forms
Paschen series.

• If an electron falls to energy level 4 (nf = 4), it forms
Brackett series.

• If an electron falls to energy level 5 (nf = 5), it forms Pfund
series.

43

energy The transition of FORMATION OF ATOMIC / LINE SPECTRUM
electrons to form
n=5 series of spectrum n=
n=4 n=5
n=3 n=4

n=2 Energy n=3
The fall from any excited state to n=2
produces Balmer series

n=1 n=2

The fall from any excited state to n=1
produces a line in Lyman series

First line in the Balmer ‫ג‬1 n=1
series, electron falls from
n=3 to n=2 Emission of photon

Line spectrum of Balmer series E

Line spectrum of Lyman series
46

FORMATION OF ATOMIC / LINE SPECTRUM Emission series of hydrogen atom

n= n=
n=5
n=4 n=4
n=3
Energy n=3 n=2 Pfund series
n=2
emission of photons Brackett series E4   RH
42
Paschen series RH
E3   32

Balmer series E2   RH
22
n=1
Line spectrum of Lyman Series Balmer Series E n=1 Lyman series E1   RH
 12

47 48

51

Example 1 Example 2

The following diagram is the line spectrum of W Y
hydrogen atom. Line A is the first line of the Lyman
series. Line spectrum of Balmer series

A B C DE E State the transitions of electrons that lead to the
Line spectrum of Lyman series  lines W, and Y, respectively.

Solution
Specify the increasing order of the radiant energy,
frequency and wavelength of the emitted photon. Line W: transition of electron is from n=4 to n=2
Line Y: electron shifts from n=7 to n=2
Which of the line that corresponds to
50
i) the shortest wavelength? Line E

ii) the lowest frequency? Line A

49

Example 3 A Solution

ED C B a) Line A.

Line spectrum of Paschen series b) Line A formed when the electron absorbed the
energy, and promoted from n=3 to n=4
a) Consider the line spectrum above. Which of the (Electron is excited). At excited states (n=4),
line in the Paschen series corresponds to the the electron is unstable. It will fall back to a
longest wavelength of photon? lower energy level (n=3) and releases a
specific amount of energy in the form of light
b) Describe the transition that gives rise to the (radiation), called a photon. The energy
line. released is detected by detector and line A
formed on the spectrum.
51
52

Exercise a) Second line of Balmer series:
Calculate the
a) wavelength in nm Solution :
b) frequency The transition of electron is from n2=4 to n1=2
c) energy
n1 < n2 1 = RH 1 1
that associated to the second line in the  n12 n22
Balmer series of the hydrogen spectrum.
1 = (1.097x107 m1) 1 1
53  22 42

 = 4.86x107 m
= 486 nm

54

52

b) Frequency = 6.17 X 1014 s-1 c) Energy = 4.09 X 10-19 J
 c   h
  6.63 x 1034 Js x 6.1728 x1014 s1
 4.0926 x 1019 J
 3.00 x108 ms1
56
 4.86 x107 m
 6.1728 x 1014 s-1

55

Exercise Solution:

a) Calculate the wavelength, λ of the spectral line a) 1 11 b) Lyman series
λ
of hydrogen atom when an electron drops from  RH n 2 n 22
the fourth to the first orbit. 1
b) Identified the spectral line series.
n1  4 to n2 1
57
1  1.097  10 7 m1  1  1 
  12 42 

  9.7235 108 m

 9.7235  108  109 nm

 97.23nm

58

EXERCISE Solution

Calculate the; i) ∆E = RH (1/n12 – 1/n22)
i ) Wavelength = 2.18 X 10 -18 (1/12 – 1/ ∞ 2)
ii ) Frequency = 2.18 X 10 -18 (1 – 0)
iii ) Wave number of the last line of hydrogen = 2.18 X 10 -18 J

spectrum in Lyman series   hc
Wave number = 1/wavelength 

For Lyman series; n1 = 1 & n2 = ∞  6.63 x 1034 Js 3.00 x 108 ms1
Ans:
i. 9.116 x10-8m   2.18 x 10 18 J
ii. 3.29 x1015 s-1  9.12 x 10-8 m
iii. 1.0970 X 107 m-1

59 60

53

Solution Solution

ii) ∆E = RH (1/n12 – 1/n22) iii)   9.12 x 10-8 m
= 2.18 X 10 -18 (1/12 – 1/ ∞ 2) 11
= 2.18 X 10 -18 (1 – 0)   9.12 x 10-8 m
= 2.18 X 10 -18 J 1.096 x107 m1

  h
2.18 x 10 18 J

  6.63 x 1034 Js
  3.29 x 1015 s1

61 62

Ionization Energy Example 1
• Ionization energy is the minimum energy
n1 = 1, n2 = ∞
required to remove 1 mol of electron in its
ground state from an atom (or an ion) in ∆E = RH (1/n12 – 1/n22)
gaseous state = 2.18 X 10 -18 (1/12 – 1/ ∞ 2)
= 2.18 X 10 -18 (1 – 0)
M (g)  M+ (g) + e H = +ve = 2.18 X 10 -18 J

• The hydrogen atom is ionised when electron is Ionisation energy for 1 mol atom
removed from its ground state (n = 1) to n = . = 2.18 X 10 -18x 6.02 X 1023J mol-1
=1.312 x 106 J mol-1
• At n = , the potential energy of electron is zero, = 1312 kJ mol-1
here the nucleus attractive force has no effect on
the electron (electron is free from nucleus). 64

63

Example 2: Solution:
Calculate the energy to ionized
(a) a hydrogen atom (a) ΔE = RH  1  1 
(b) 1 mol of hydrogen atom  n12 n22 

65 = 2.18  1018 J 112  1 
2 

= 2.18  1018 J

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

66

54

(b) 1 H atom needs 2.18  1018 J Finding ionisation energy experimentally:
6.022  1023 H atoms need 2.18  1018J  6.022 x1023
Convergent limit 1 st line
= 1.31  106 J
The energy to ionized 1 mol of hydrogen atom  1
Ionisation energy is determined by detecting
= 1.31 X 106 J the wavelength of the convergence point

67 68

Example 1 Solution

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

10.97 10.66 10.52 10.27 9.74 8.22 ( x 106 ) = 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

wavenumber, m-1 = 2.18 x 10-18J

The Lyman series of the spectrum of hydrogen is Ionisation energy
shown above. Calculate the ionisation energy of = 2.18 X 10 -18x 6.02 X 1023 J mol-1
hydrogen from the spectrum. =1.312 x 106 J mol-1
= 1312 kJ mol-1

69 70

Example Solution
ΔE = hc/λ
9.12 9.38 9.51 9.73 ( x10-8 m)
= (6.626 x 10-34 J s x 3 x 108 m s-1 ) / 9.12 x 10-8 m
wavelength,  = 2.18 x 10-18J

The Lyman series of the spectrum of hydrogen is Ionisation energy
shown above. Calculate the ionisation energy of = 2.18 X 10 -18x 6.02 X 1023 J mol-1
hydrogen from the spectrum. =1.312 x 106 J mol-1
= 1312 kJ mol-1
71
72

55

The limitation of Bohr’s Atomic Model de Broglie’s Postulate

• It can only explain the hydrogen spectrum or • In 1924 Louis de Broglie proposed that not only
any spectrum of ions contain one electron light but all matter has a dual nature and
possesses both wave and particle properties.
eg: He+, Li2+
• Tiny particle such as an electron have wave
• Unable to account for the emission spectrum of properties.
atom containing more than 1 electron
• An electron is both particle and wave.
• Electrons are wavelike, we can’t define the  hm=
precise location of a wave because a wave
extends in space.

73 74

• de Broglie deduced that the particle and Heisenberg’s Uncertainty Principle
wave properties are related by the expression
It is impossible to know at the same time, both
λ= h momentum p (defined as mass times velocity) and
mv position of a particle with certain.

h = Planck constant (J s) Stated mathematically, x p  h
m = particle mass (kg) 4
v = velocity (m/s)
 = wavelength of a matter wave where x = uncertainty in measuring the position
p = uncertainty in measuring the momentum
75 = mv
h = Planck constant

76

56

2.2 Quantum mechanics 2.2 Quantum mechanics

79 At the end of this topic students should be able to:-

a) define the term orbital.

b) explain all four quantum numbers of an electron
in an orbital
i. principle quantum number, n
ii. angular momentum quantum number, ℓ
iii. magnetic quantum number , m
iv. electron spin quantum number, s

c) sketch the 3-D shapes of s, p and d orbitals.

80

Atomic Orbital Quantum Numbers

Definition Each electron that occupy an orbital in an atom is
described and characterised by a set of four
An orbital is a three- quantum numbers :
dimensional region in i. principal quantum number, n
space around the nucleus ii. angular momentum quantum number, 
where there is a high iii. magnetic quantum number, m
probability of finding an iv. electron spin quantum number, s.
electron.
82
81

i) Principal Quantum Number, n ii) Angular Momentum Quantum Number, 
• n determines the energy level (principle electronic
Alternative name: Subsidiary / Azimuthal /
shell) and size of an orbital. [n = energy level = shell] Orbital / Quantum Number

• The principal quantum number n, starting from n =1,  The value of  indicates the shape of the atomic
2, 3, …, . orbital.

• As n increase : i) the orbital become larger  The allowed values of  are 0, 1, 2,…, (n1)
 Letters are assigned to different numerical values of .
ii) electron has higher energy
 subshell Orbital shape
n1 2 3 4
0s spherical
Orbital size
1 p dumbbell

Energy 2 d cloverleaf

increases 83r 3f 84

57

• All orbitals with the same value of n are in the -  value is depend on n, i.e, 0, 1, …. (n -1).
same principle electronic shell or principle level
called shell. Example:

• All orbitals with the same n and  values are in the If n = 1,  = 0 (s subshell) one s subshell
same subshell or sublevel
If n = 2,  = 0 (s subshell) two subshells ; s and p
85 = 1 (p subshell)

If n = 3,  = 0 (s subshell)

= 1 (p subshell) three subshells ; s, p, and d

= 2 (d subshell)

86

iii) Magnetic Quantum Number, m iv) Electron Spin Quantum Number, s
• Describe the orientation of orbitals in space.
• Possible values of m depend on the value of  that • The value of s represent the direction of
an electron rotation on its own axis.
is m = , …, 0, …, +
Example: • either clockwise or anticlockwise
If  = 0, m = 0 , 1 value of m
 1 orientation of s orbital • It has 2 value : + 1 or - 1
If  = 1, m = 1, 0, +1 means 3 values of m 2 2
 3 orientations of p orbital (px, py, pz)
88
If  = 2, m = 2, 1, 0, +1, +2 means 5 values of m
 5 orientations of d orbital ( dxy,dxz,dyz,dx2-y2,dz2)

87

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

 When l = 0 , m = 0 , only 1 orientation of s 90
orbital

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

Shape of s orbital
with different n value

89

58

c) d orbitals Orbitals of dxy , dxz , dyz lie between the x,y and z
• All d orbitals are not look alike. axes
92
• When  = 2 , m = -2, -1, 0, +1, +2 (5 values of m)

Five orientations of d orbitals, there are 5 d
orbitals

• These orbitals are dxy, dxz, dyz, dx2-y2, dz2

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

91

Orbitals dx2-y2 and dz2 lie along the x,y and axes • The 4 quantum numbers n,l,m and s
enable us to locate an electron in any
93 orbital of an atom.

Eg:
• 4 quantum numbers of 2 electrons that

occupy the 2s orbital 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)

94

Important facts concerning an orbital 3) The orientation (direction/ point of reference/
axis) of the orbital.
1) The size and energy level of the orbital.
m = (-ℓ…0…+ℓ) – m = 0 : center
n = (1,2,3…) – size and energy level m = -1 : x axis  px
e.g: n = 3 > n = 2 0 : y axis  py
+1 : z axis  pz
2) The shape of the orbital. m = -2 : between axis  dxy
ℓ = (0,1,2…. n-1) – 0 (spherical) -1 : between axis  dxz
0 : between axis  dyz
1 (dumbell) +1 : x and y axis  dx2- y2
+2 : z axis  dz2
2 (cloverleaf)
96
95

59

Example 1 Solution

Complete the table below for n =1 to n=3. n/ shell = Orbital No. of No of
n= (0,1,2 designation orbitals electrons
n/ shell  = Orbital m = (-ℓ…0…+ℓ) No. of No of (1,2,3..) …n-1) (subshell) m = (-ℓ…0…+ℓ)
0 orbitals electrons 12
n = (0,1,2 designation 1 0 1s 0

(1,2,3..) …n-1) (subshell)

10 1s 0 2s 0 12
2 2p -1, 0, +1 36

1

0 3s 0 12

31 3p -1, 0, +1 36

2 3d -2, -1, 0, +1, +2 5 10

97 98

Example 2 Solution
(a) (n=1, l=0, m=0, s= -1/2)
Predict the following quantum numbers as
allowed or not allowed. allowed

(a) (n=1, l=0, m=0, s= -1/2) (b) (n=2, l=0, m=1, s= 1)
(b) (n=2, l=0, m=1, s= 1) not allowed because when l=0, the possible
(c) (n=0, l=1, m=1, s= +1/2) value of m=0 and s= +1/2 or -1/2
(d) (n=4, l=1, m=0, s= -1/2)
(c) (n=0, l=1, m=1, s= +1/2)
99 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

100

Exercise Solution

1. Give the set of four quantum numbers of an 1. a) (n=2, l=1, m= +1, s=-1/2) or
(n=2, l=1, m= +1, s=+1/2) or
electron that occupies each the following orbitals (n=2, l=1, m= 0, s=-1/2) or
(n=2, l=1, m= 0, s=+1/2) or
a) 2p b) 4p c) 3d (n=2, l=1, m= -1, s=-1/2) or
(n=2, l=1, m= -1, s=+1/2)
2. What are the similarities and differences between
a) a 1s and a 2s orbital b) (n=4, l=1, m= +1, s=-1/2) or
b) a 2px and a 2py (n=4, l=1, m= +1, s=+1/2) or
(n=4, l=1, m= 0, s=-1/2) or
3. Give the total number of electrons that can occupy (n=4, l=1, m= 0, s=+1/2) or
a) a one s orbital (n=4, l=1, m= -1, s=-1/2) or
b) three p orbitals (n=4, l=1, m= -1, s=+1/2)
c) 5dxz orbital
102
101

60

Solution Solution:

c) (n=3, l=2, m= +1, s=-1/2) or 2. a) 1s and 2s orbitals have same shape but 1s
(n=3, l=2, m= +1, s=+1/2) or orbital has smaller size than 2s orbital. 1s
(n=3, l=2, m= 0, s=-1/2) or orbitals also has lower energy than 2s orbitals.
(n=3, l=2, m= 0, s=+1/2) or
(n=3, l=2, m= -1, s=-1/2) or b) 2px and 2py orbitals have same size, energy
(n=3, l=2, m= -1, s=+1/2) or and shape but 2px and 2py orbitals are located
(n=3, l=2, m= +2, s=-1/2) or at different axis.
(n=3, l=2, m= +2, s=+1/2) or
(n=3, l=2, m= -2, s=-1/2) or 3. a) 2
(n=3, l=2, m= -2, s=+1/2) b) 6
c) 2

103 104

61

2.3 Electronic configuration 2.3 Electronic configuration

At the end of this topic students should be able to:-
a) explain Aufbau Principle, Hund’s Rule and

Pauli Exclusion Principle.
b) predict the electronic configuration of atoms

and monatomic ions using spdf notation and
orbital diagram.
c) explain the anomalous electronic
configurations of chromium and copper.

106

What is an electronic configuration? Example:
Draw orbital diagram and write spdf notation for
• An electronic configuration : oxygen atom.

 describes the arrangement of electrons and i) orbital diagram magnetic quantum
electrons filling in the orbitals of an atom.
number , m
• Electronic configuration can be expressed as

i) orbital diagram 2p 8O : 1s 2s 2p
e.g 9F :
1s 2s principle quantum angular momentum
number, n quantum number, l
ii) spdf notation
e.g 9F : 1s2 2s2 2p5

107 108

ii) spdf notation Principles and rule of filling electrons in
orbitals
e.g 8O: 1s2 2s2 2p4
To write an electronic configuration, the following
Principal quantum Azimuthal quantum rule and principles need to be obeyed :
number, 
number, n a) The Aufbau Principle
1s2 2s2 2p4 b) The Pauli Exclusion Principle
8 O: c) The Hund’s rule

Number of electrons in 110
the subshells

109

62

Relative Energy Level of Atomic Orbitals

a) Aufbau Principle energy n=5 5s 4d
4p 3d
• States that electrons are filled in the orbitals in n=4
order of increasing energy. 4s
energy
• Electrons should occupy the orbital that has the n=3 3p
lowest energy, 1s before occupying the higher 3s
energy orbitals.
n=2 2p
111 2s

n=1
1s

Orbital energy levels in a many-electrons atom

112

Order of filling electrons in orbitals for a multi-electrons Example
atom
Write the electronic configuration as the spdf notation
• It starts with the 1s for
orbital and moves a) beryllium (4 electrons)
downward, then b) boron (5 electrons)
2s, 2p and
follows the Answer
arrows. a) 4Be : 1s2 2s2
b) 5B : 1s2 2s2 2p1
• The order of filling
orbitals is: 114

1s < 2s < 2p < 3s <
3p < 4s < 3d < 4p <
5s

113

Example: Answer:

Write the electronic configurations for following 1) Carbon (6 electrons)
atoms: 1s22s22p2
1) Carbon (6 electrons)
2) Nitrogen (7 electrons) 2) Nitrogen (7 electrons)
3) Oxygen (8 electrons) 1s22s22p3

115 3) Oxygen (8 electrons)
1s22s22p4

116

63

b) Hund’s Rule Example

 States that when electrons are added to the Draw the electronic configuration as the orbital
degenerate orbitals (equivalent energy diagram for
orbitals i.e the p and d orbitals), each orbital a) Carbon (6 electrons)
are filled singly with electron of the same spin b) Oxygen (8 electrons)
before it is paired.
118
 The electrons in half-filled or partially orbitals
have the same spins, that is, parallel spins.

This can be shown in orbital diagram as the electronic
configuration.

117

Answer c) Pauli Exclusion Principle
No two electrons in an atom can have the same
a) 6C : • p - the four quantum numbers (n, , m, s)
1s Eg : Li (3 electrons)
2s 2p degenerate (n, , m, s)
orbitals, each (n=1,  = 0, m= 0, s= +½) - first electron
electron is (n =1,  = 0, m = 0, s= -½) – second electron
(n =2,  = 0, m = 0, s= +½) – third electron
6C : 2s 2p filled singly of
1s the same spin * First and second electron - same value of n, , m but
as in carbon, different value of s

then paired as 120

b) 8O : in oxygen
1s
2s 2p

8O : 2s 2p
1s

119

Stability of orbital.

The Anomalous Electronic Configurations of Cr  For degenerate orbitals, half- filled orbitals is
and Cu more stable than the partially-filled.

• Cr and Cu have electronic configurations which are Half-filled orbital (stable) nd
inconsistent with the Aufbau principle.
or
• The anomalous are explained on the basis that a np
filled or half-filled orbital is more stable.
Partially-filled orbital ( less stable)
* a filled orbital IS a completely filled orbital
a half-filled orbital is NOT a partially filled orbital or nd
np
121
122

64

Based on the Aufbau principle, the electronic
configuration for 24Cr is 1s22s22p63s23p6 4s2 3d4

or 24Cr :
1s 2s
24Cr : 2p 3s 3p 4s 3d
1s 2s
2p 3s 3p 4s 3d • The actual spdf electronic configuration of 24Cr is
1s22s22p63s23p6 4s1 3d5
• The 3d orbital is partially-filled, so it is less stable .
• Half-filled 3d orbital arrangement increase the
• Thus, to achieve stability, one electron from the stability of Cr atom.
4s orbital occupies one of the 3d orbital in order to
have a half -filled orbital arrangement as follow :

123 124

The electronic configuration of 29Cu based on Aufbau As a conclusion, the electronic configuration of
principle is 1s2 2s2 2p6 3s2 3p6 3d9 4s2 or chromium and copper are :

29Cu :

1s 2s 2p 3s 3p 4s 3d Element Expected (Aufbau Observed/actual
Principle)
• To achieve stability 1 electron from 4s orbital 1s22s22p63s23p6 4s1
occupies 1 of the 3d orbital in order to have a 24Cr 1s22s22p63s23p6 4s2 3d5
filled orbital arrangement. 3d4
1s22s22p63s23p6 3d10
29Cu : 29Cu 1s22s22p63s23p6 3d9 4s1
1s 2s 4s2
2p 3s 3p 4s 3d

Or 1s2 2s2 2p6 3s2 3p6 3d10 4s1 125 126

* To fill electron(s) z = 21
- follow the Aufbau principle, Pauli Exclusion z = 30
principle and Hund’s rule except for
anamolous cases (Cr and Cu) 128

* To remove / eliminate electron(s)
- remove electron(s) from the orbital with the
largest n value

e.g :
Co ( Z = 27 ), write the electronic configuration
of Co3+
Co : 1s2 2s2 2p6 3s2 3p6 4s2 3d7
Co3+ : 1s2 2s2 2p6 3s2 3p6 3d6

127

65

Exercise: Answer: 5) 17Cl

Write the electronic configuration of the following 1) 6C 1s2 2s2 2p6 3s2 3p5
atoms and ions: 1s2 2s2 2p2
1) 6C
2) 10Ne 2) 10Ne 6) 17Cl-
3) 13Al 1s2 2s2 2p6 1s2 2s2 2p6 3s2 3p6
4) 13Al3+
5) 17Cl 3) 13Al
6) 17Cl- 1s2 2s2 2p6 3s2 3p1

129 4) 13Al3+ 130
1s2 2s2 2p6

66

CHAPTER 2: ATOMIC
STRUCTURE

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

67

CHEMISTRY
TUTORIAL TOPIC 2: ATOMIC STRUCTURE

2.1: Bohr’s Atomic Model

1. Based on Bohr’s atomic model of the hydrogen atom,
a) explain the existence of energy levels in an atom.
b) state the difference between ground state and excited state.
c) distinguish a continuous spectrum and a line spectrum.

2. a) Describe the formation of a line in the hydrogen atom emission spectrum.
b) Compare and contrast the Lyman and the Balmer series.

3. The emission spectrum of hydrogen atom in the visible region is shown below:
λ (nm)

line spectrum X YZ
W

a) State the transition of electrons that produce line W and Y respectively.
b) Which of the above lines has the lowest frequency?
c) Calculate the wavelength that corresponds to line X.

4. One of the lines in the Balmer has a frequency of 4.10 x 1014 Hz
a) Calculate its wavelength.
b) Calculate the energy emitted.
c) Determine the transition of electron involved.

5. Calculate the wavelength (in nm) and frequency of the:
a) third line in the Paschen series.
b) second line in the Brackett series.

6. Calculate the ionization energy of hydrogen in kJ mol−1.

7. If an emission spectrum of hydrogen atom emit photons with the average kinetic
energy of 10370 J mol−1 is in the infrared region, calculate the frequency of IR wave

formed.

68

CHEMISTRY
TUTORIAL TOPIC 2: ATOMIC STRUCTURE

8. Two important concepts that relate to the behavior of electrons in an atom are the
Heisenberg uncertainty principle and the wave-particle duality of matter.

(a) State the Heisenberg uncertainty principle as it related to the determining the
position and momentum of an object.

(b) What aspect of the Bohr theory of the atom is considered unsatisfactory as a
result of the Heisenberg uncertainty principle?

(c) Explain why the uncertainty principle or the wave nature of particles is not
significant when describing the behavior of macroscopic objects, but it is very
significant when describing the behavior of electrons.

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1. a) Define orbit and orbital.
b) What do these quantum numbers represent?
i. n
ii. l
iii. m
iv. s
c) Give one set of possible quantum numbers for an electron that occupy the 3p,
4d and 5s orbitals respectively.

2. For the following quantum numbers of n and l below,
nl

I 31

II 4 2

III 5 0

a) name the subshell and number of orbitals involved for each subshell.
b) determine the maximum number of electrons that occupy each subshell.

3. Explain the following subshell as allowed or not allowed.

a) 6s b) 4p c) 2d d) 3f

4. Draw the shape of the following orbitals:
1s, 2s, 3py, 3dxz, 3 d x2 −y2 and 3 d z2

5. a) State Aufbau principle.
b) Arrange the following orbitals in the order of increasing energy.
4dxy, 3dxy, 3dyz, 4pz, 3pz, 3py, 2py, 3s, 2s, 1s, 4s

69

CHEMISTRY
TUTORIAL TOPIC 2: ATOMIC STRUCTURE

6. Atom X has 15 electrons.
a) State Hund’s rule and Pauli Exclusion Principle
b) Draw the orbital diagram for X.
c) Give a set of quantum numbers for the highest energy electron.

7. Give the electronic configuration of the following species as orbital diagram and spdf
notation.
a) O2−
b) Al3+
c) Ni2+

8. a) Write the electronic configuration of iron(II) and iron(III) ions.
b) Which of these two species is more stable? Explain.
c) Write the sets of quantum numbers for the electrons in the outermost orbital of
iron(III) ions.

70

CHEMISTRY
MEKA TOPIC 2: ATOMIC STRUCTURE

2.1: Bohr’s Atomic Model

1. Explain how the lines in the Lyman series are formed.

2. The Lyman series of the hydrogen emission spectrum is shown below

zy x

line spectrum

a) What is the transition that produces line z?
b) In what region does the line exist?
c) Calculate the energy that corresponds to line y.
d) Determine the wavelength and frequency of line y.

3. Calculate the wavelength of the fourth line in Balmer series.

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1. The valence electrons of an element X has the orbital diagram of,

3s 3p

a) Write a set of four quantum numbers for each valence electron.
b) Write the electronic configuration for X.
c) Draw the shape of orbitals of the valence electrons.

2. Vanadium, V is a d-block element in the Periodic Table with a proton number 23.
a) Draw the orbital diagram for V2+ and V3+ ions.
b) Give a set of possible quantum number for one of the electrons with the
highest energy in V2+ ion.

3. Chromium, Cr and copper, Cu are transition elements in the d-block of the Periodic
Table. The proton number of chromium is 24 while copper is 29.
a) Write the expected and actual electronic configuration of chromium and
copper.
b) State the reasons for the anomaly.

71

CHEMISTRY
KUMBE TOPIC 2: ATOMIC STRUCTURE

2.1: Bohr’s Atomic Model

1. Draw a labeled energy diagram of the hydrogen atom. By using arrows, indicate the
electronic transitions corresponding to:
a) the first line in the Lyman series.
b) the third line in the Balmer series.
c) the second line in the Paschen series.
d) the last line in the Brackett series.

2. A B C D E F

The above diagram shows the lines of emission spectrum in the Paschen series.

a) Describe the formation of line spectrum.

b) Calculate the wavelength (in nm) of the third line, C. [1094 nm]

c) Mark the direction of energy increase in the spectral lines.

d) Why do the lines become increasingly closer together from left to right?

3. The diagram below shows the line spectrum of hydrogen atom in the visible region.

red cyan blue violet

frequency increase

a) Name the series above.

b) On the energy levels diagram, show the electronic transitions which produce

the above lines. [−4.58×10−19 J]
c) Calculate the energy of blue line.

4. An electron of a hydrogen atom is excited to an energy level of n=7 before it falls to a

lower energy level to produce a line in the Paschen series.

a) State the energy level to which the electron falls.

b) Calculate the energy of the electron in the excited state before it falls to

produce the line. [−4.45×10−20 J]

5. A light with a wavelength of 486.3 nm was discovered to be one of the line that falls
in the visible region of the hydrogen emission spectrum.
a) Name the series of the line spectrum.
b) Determine the transition that produces this line.

72

CHEMISTRY
KUMBE TOPIC 2: ATOMIC STRUCTURE

6. Calculate the frequency of the light emitted when an electron transits from the fourth

to the second orbit of the hydrogen atom. [6.17×1014 s−1]

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1. Which of the following quantum numbers is permissible for an electron in a hydrogen
atom. Explain why others are incorrect.
a) n=2, l=1, m=+1, s= +½
b) n=1, l=0, m= -1, s= -½
c) n=3, l=3, m= -1, s= +½
d) n=1, l=0, m= -1, s= -½

2. The table below shows elements A and B with their respective proton numbers.

Element Proton Number
A 15
B 23

a) Write the electronic configuration for elements A and B.
b) State the number of unpaired electrons for elements A and B.
c) Draw the shapes of the orbital and give sets of the four quantum numbers for

each of the valence electrons in element A.

3. The given sets of quantum numbers represent the four outermost electrons of element
X, at the ground state.
n=3, l=1, m= -1, s= +½
n=3, l=1, m= -1, s= -½
n=3, l=1, m= 0, s= +½
n=3, l=1, m= +1, s= +½

a) Draw the orbital diagram for this element.
b) Suggest the charge of an ion that can be formed by this element and write its

electronic configuration.

73

CHEMISTRY
KUMBE TOPIC 2: ATOMIC STRUCTURE

4. a) Name the quantum number that represents the shape of an orbital.
b) The following sets of quantum numbers are the quantum numbers for the
valence electrons in atom J
n=3, l=2, m=0, s= -½
n=3, l=2, m= +1, s= -½
n=3, l=2, m= -1, s= -½
n=4, l=0, m=0, s= -½
n=4, l=0, m=0, s= +½
i. Draw the shapes of the respective orbitals and suggest its name.
ii. Write the electronic configuration of J.
iii. If five electrons are removed from J, write the electronic configuration
of the ion formed.

74

CHAPTER 3: PERIODIC TABLE

LECTURE NOTE

lecture notes & questions

75

3.0 3.1
PERIODIC TABLE CLASSIFICATION OF ELEMENTS

3.1 Classification of elements At the end of the lesson the students should be
3.2 Periodicity able to:

1 (a) Describe period, group and block (s, p, d, f).
(b) Deduce the position of element in the periodic

table from its electronic configuration.

2

3.1 Classification of element • In the periodic table, a vertical column of
elements is called a group while a horizontal
• The periodic table is a table that arranges all row is known as a period.
the known elements in order of increasing
proton number. • Elements in the same group have the same
number of valence electrons.
• This order generally coincides with increasing
atomic mass. For elements in s and d block
Group number = number of valence electrons
3 For elements in p block
Group number = number of valence electron + 10

4

GROUP

The groups in Periodic Table are

e.g: oxygen and sulphur are both found in group 16 – Group 1 : alkali metals (except H)
since they both have 6 valence electrons. – Group 2 : alkaline earth metals
– Group 3-12 : transition elements
5 – Group 15 : pnictogens
– Group 16 : chalcogens
– Group 17 : halogens
– Group 18 : inert/ noble gases

6

76

• The periods in the Periodic Table are Blocks
numbered 1 to 7
• Elements in the Periodic Table are classified into 4 main
E.g : hydrogen and helium are in Row 1 or blocks according to their valence electron configuration.
Period 1 because their principal quantum
number, n, of the main electrons shell is 1. • These main blocks are s, p, d and f block.
(H: 1s1; He: 1s2) • s block and p block are the main group elements or

Period number = Principal quantum number (n) known as the representative elements

7 8

s block p block
• Group 1 and 2
• The filling of valence electrons involves only s • Groups 13 to 18
• Valence electrons configuration :
orbital
• Valence electrons configuration : ns2 np1 to ns2 np6

ns1 to ns2 Eg.

e.g 11Na: 1s2 2s2 2p6 3s1 13Al:1s2 2s2 2p6 3s2 3p1
20Ca: 1s2 2s2 2p6 3s2 3p6 4s2
52Te:1s2 2s2 2p6 3s2 3p6 3d104s2 4p6 4d10 5s2 5p4
9
10

d block f block

• Also known as transition elements • The elements in the series of lanthanides (Ce
• Groups 3 to 12 to Lu) and actinides (Th to Lr)
• Valence electrons configuration :
• The valence electrons are fill in the subshell
(n-1)d1 ns2 to (n-1)d10 ns2 of 4f and 5f.

eg. 12
23V : 1s2 2s2 2p6 3s2 3p6 3d3 4s2

28Ni : 1s2 2s2 2p6 3s2 3p6 3d8 4s2

11

77

Example

Classify the following elements into its appropriate
group, period and block.

P ………1s2 2s2 2p6 3s2 3p6
Q ……… 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
R ……… 1s2 2s2 2p6 3s2 3p6 4s2
S ……… 1s2 2s2 2p6 3s2 3p6 3d3 4s2
T ……… 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

13 14

Answer : Class/block Exercise
Element Group Period State the group, period and block of the
followings :
P 18 3 Inert gas / p block 1. 20X
2. 28Y2+
Q 17 4 halogen / p block 3. 16Z2-

R 2 4 alkaline earth metal / s 16
block

S 5 4 Transition element
/d block

T 18 4 Inert gas / p block

15

Answer :

Species group period block
2 4 s
20X 10 4 d
28Y2+ 16 3 p
16Z2-

17

78

3.2 LEARNING OUTCOMES
PERIODICITY
At the end of this lesson, students should be able
a. Periodic trends in the size (atomic radii) of to :
atom a) describe the variation in atomic radii :

b. Trends in the ionic radii i. across a period
c. Isoelectronic series ii. across the first row of transition elements
d. Trends in the ionization energies iii. down a group
e. Trends in the electronegativity
f. Acid-base character of oxides b) analyse the variation in atomic radii :
i. across a period
18 ii. across the first row of transition elements
iii. down a group
19

LEARNING OUTCOMES LEARNING OUTCOMES

c) compare the atomic radius of an element and i) explain the increase in the successive ionisation
its corresponding ionic radius energies of an element
d) define the term isoelectronic
e) compare the radius isoelectronic species j) deduce the electronic configuration of an
f) analyse the variations in the ionic radii across element and its position in the periodic table
period 2 and period 3 based on successive ionisation energy data.
g) define the first and second ionisation energy
h) analyse the variations in the first ionisation k) define electronegativity
energy l) explain the variation in electronegativity of
i. across a period ii. down a group
elements
20 m) explain the acid-base character of oxides of

elements in Period 3.

21

a) Periodic trends in the size of atom Radius, r  a
(atomic radii) 2

• The size /radius of atom is difficult to be a
defined exactly because the electron cloud has
no clear boundary. • Down the group, atomic radii increases
• Across the period, atomic radii decreases
• Therefore, the distance between the 2 nuclei
in a molecule represent the radius of an atom. 23

22

79

• Across the period of d block (transition Two factors that influence the changes of
element) the change in atomic radii is atomic radii in the Periodic Table are:
small as valence electron are filled in the
3d degenerate orbitals. i. Effective nuclear charge (Zeff)
experience by the valence electrons
24
ii. The principal quantum number, n of
the valence electrons

25

i. Effective nuclear charge (Zeff) Zeff = Z –S
• Electrons around the nucleus experience
where :
difference nuclear attraction. Z = no. of proton
• Electrons that are closer to the nucleus S = no. of electrons filled at the inner
shell
experience a greater force than those that are
farther away. 27

effective nuclear charge, Zeff
The actual/nett nuclear charge experience by an
electron as a result of shielding effects due to the
presence of other electrons

26

• Effective nuclear charge ↑, nucleus a rac on ii. The principal quantum number of the
↑, atomic radii ↓ valence electrons

• Across the period, the effective nuclear charge • Moving down a group, the number of shells
increases as proton number increases. increases, more inner electrons are present to
shield the valence electrons from the nucleus
• e.g : 11Na : 1s2 2s2 2p6 3s1 Zeff= +1 (shielding effect @ screening effect increases).
Zeff= +2
12Mg : 1s2 2s2 2p6 3s2 Zeff= +3 • The valence electrons are farther from the
13Al : 1s2 2s2 2p6 3s2 3p1 Zeff= +4 nucleus.
14Si : 1s2 2s2 2p6 3s2 3p2
• Thus, the attraction between the nucleus and
• As a result, the attraction between the nucleus valence electrons decreases, therefore, the
and valence electrons become stronger, causing atomic radius increase.
the atomic radius to decrease.
28 29

80

• Down a group, the atomic radius increases Across
because of the increasing principal quantum period 3
number (n) of the valence electron. Across
Period 2

30 31

The graph shows that : Example
•Atomic radius decreases when : Arrange the following atoms, P,Si and N in order of increasing radius
* Across a period (from left to right)
* Moving up a group in the periodic table. Solution Group Group
•Atomic radius increases when : 14 15
* Going down the group • N and P are in the same group and N is above
• The greater the nucleus attraction, the P. Si N

smaller the atomic radius. • Atomic radius increases as we go down the P
group.
32 33
• Therefore, the radius of N is smaller than that
of P

• Both Si and P are in the third period and Si is to
the left of P.

• Atomic radius decreases as we move from left
to right.

• Therefore, the radius of P is smaller than Si.
• Thus the order of increasing radius :

N < P < Si

b) Trends in the ionic radii Ionic radii for cation
Ionic radii for anion • When electrons are removed from the valence
• When electrons are added to an atom, the
shell, the electron-electron repulsions
mutual repulsions between them increase. decrease but the nuclear charge remains the
• This causes the electrons to push apart and same.
• So the remaining electrons are to be pulled
enlarges the domain of electron cloud. closer together around the nucleus.
• Therefore, negative ions (anions) are larger • Therefore, cations are smaller than the atoms
from which they are formed.
than the atoms from which they are formed.
35
34

81

Trend of ionic radius across a period 3

• cations in the same period (Na+ to Si4+) show a
decrease in the radius

• ionic radius increases from Si4+ to P3- while
the anions formed show a decrease in size
going from P3- to Cl-

• Discussion should involve:
a) isoelectronic factor
b) the effect of varying nuclear charge

36 37

c) Isoelectronic species
Isoelectronic species are groups of atoms
and ions which have the same electronic

configuration.

E.g :
Na+, Mg2+, Al3+and Si4+ ions are isoelectronic

(10 e) with the electronic configurations as 1s2 2s2 2p6.

38 39

Within isoelectronic species: b) the more negative the charge, the larger the species.

a) the more positive the charge, the smaller the E.g :
species. • Cl-, S2- and P3- ions are isoelectronic (18 e) with the

• Na+, Mg2+, Al3+ and Si4+ ions are isoelectronic (10 e) electronic configurations as 1s2 2s2 2p6 3s2 3p6.
with the electronic configurations as 1s2 2s2 2p6 . • It is due to the proton number of Cl- > S2- > P3-
• Thus, nuclear charge of Cl- > S2- > P3-
• It is due to the proton number increases from Na+ < • Electrons are attracted tightly to the nucleus from P3-
Mg2+ < Al3+ < Si4+.
to Cl-.
• Thus, nuclear charge of Na+ < Mg2+ < Al3+ < Si4+. • Thus, the size of Cl- < S2- < P3- ion
• Electrons are attracted tightly to the nucleus from Na+
41
to Si4+.
• Therefore the size of Na+ > Mg2+ > Al3+ > Si4+ ion.

40

82

Exercise Answer :
Na+, Si4+ ,Mg2+, N3-, O2- , Al3+ and F- are isoelectronic with N3-> O2- > F-> Na+ > Mg2+ > Al3+ > Si4+
the electronic configuration as 1s2 2s2 2p6.
Arrange in a descending order the size of those • Proton number increase from N3- to Si4+
isoelectronic species. Explain. • Hence the nuclear charge increase from N3- to Si4+
Answer : • Nucleus attractions towards electrons are stronger

42 from N3- to Si4+.
• So the size of N3-> O2- > F- > Na+ > Mg2+ > Al3+ > Si4+

43

d) Trends in the ionization energies i) Ionization energy across a period :
The minFiimrstuimoneinzaetrigoynreenqeurirgeyd(ItEo1)remove  The effective nuclear charge increases, the atomic size

first electron from gaseous atom in its decreases.
ground state.  Electrons are held tightly to the nucleus thus it is

E.g: difficult to remove the first electron.
energy + X(g) → X+(g) + e- H = IE1  Therefore the first ionisation energy is high.

44 It can be said that the first ionization energy increases
from left to right.
However, there are some irregularities in the trend.

45

Anomalous cases in Period 2

a) Between group 2 and 13

‫٭‬Be ■N • 25Bs2: 1s2 2s2 2p1 in group 13 has a lower IE1 than 4Be: 1s2
▲O in group 2.
♦B

■ • Be loses a 2s electron while B loses a 2p electron.
▲
• Less energy is needed to remove an electron from a
‫♦٭‬ higher energy of partially-filled 2p orbital which is
highly shielded from the nucleus attraction by 1s and
2s orbital as in B than in Be.

• In Be the electrons in 2s orbital is shielded from the
nucleus attraction by 1s orbital only.

46 47

83

b) Between group 15 and 16 ii) Ionization energy going down the group

• O (group16) has lower IE1 than N (group 15). • Going down the group, the atomic size increases
7N :1s2 2s2 2p3 (the half-filled 2p orbital ) as the energy level, n increases.
8O :1s2 2s2 2p4( the partially-filled 2p orbital)
• Therefore the outer electrons are farther from
• Although the 2p subshells are degenerate, the repulsion the nucleus and are held less tightly (weaker
between electrons in the partially-filled 2p orbital makes attraction) by the nucleus.
it easier to remove one of those electrons than that of
2p half-filled orbital. • Thus, it is easy to remove the first electron.
• Hence the Ionization Energy decreases down the
• N loses an electron from the half-filled 2p orbital which
is more stable than that of electron of the partially-filled group.
orbital in O.
49
• As a result, the first ionization energy of N is higher
than of O.

48

Second ionization energy (IE2) • Thus, ionization energies always increase in
the following order :
The minimum energy required to remove the second
electrons from gaseous ion with a charge of 1+.

X+(g) → X2+(g) + e- IE1< IE2< IE3< IE4<…..
• Although the removal of a subsequent
• When an electron is removed from a neutral atom,
the repulsion among the remaining electrons electron from an atom requires an increment
decrease. amount of energy but it may not be
consistence.
• Since the nuclear charge remain constant, the
electron are held tightly to the nucleus. • We can determine the electronic

• Therefore more energy is needed to remove another configuration of the valence electron for an
element by using the ionization energy.
electron from the positively charged ion. 50 51

Example 1 • The ratio between the ionization energies are:

4Be IE2  1757 1.95
The ionization energies (kJmol-1) of Beryllium IE1 899
are shown below.
IE3  114785570 8.45
IE1 IE2 IE3 IE4 IE2
899 1757 14850 21005
IE4  1241805005 1.41
52 IE3

53

84

• A sharp increase in ionization energy occurs Example 2
when an inner-shell electron is removed.
Five successive ionization energies (kJmol-1) of
• sThhaerhpiginhcersetarsaetioofiisoInEi3z/atIiEo2n. This indicates a atom M is shown below:
energy for the third
electron. Thus, the third electron is from the IE1 IE2 IE3 IE4 IE5
inner-shell,1s. 800 1580 3230 4360 16000

• The first and second electrons are removed from Determine
the same energy subshell (2s) . i) the electronic configuration of the valence
electron.
• 2 valence electrons are present. This element is ii) the group number in the periodic table.
in Group 2
55
• Valence electron configuration : ns2 54

Solution • Since the ratio of IE5 / IE4 is the highest, the
fifth electron is removed from the inner shell.
By determining the IE ratios:
• So, there are 4 valence electrons at the
IE2 = 1580 = 1.98 valence shell.
IE1 800
IE3 = 3230 = 2.04 • Therefore, M is in Group 14 in the periodic
IE2 1580 table
IE4 = 4360 = 1.35
IE3 3230 • Valence electron configuration for M is ns2 np2
IE5 = 16000 = 3.67
IE4 4360 56 57

Exercise 1 Solution

The five successive ionisation energies of an element X i) By determining the IE ratios:
are as follows
IE2  2400  3.00
IE1 800

IE1 IE2 IE3 IE4 IE5

IE (kJmol-1) 800 2400 3700 25000 31800 IE3  3700  1.54
IE2 2400

i) State the group to which X belongs to in the Periodic IE 4  25000  6.76
Table. Explain. IE3 3700

ii) If X is an element of Period 3, write its electronic IE5  31800  1.27
configuration. IE 4 25000

iii) Deduce the empirical formula of its simplest oxide. 58 59

85

• Since the ratio of IE4 / IE3 is the highest, the Exercise 2Successive ionization energy)
fourth electron is removed from the inner A graph of successive ionization energy against number of
shell. electrons removed of an element, M is shown below.

• So, there are 3 valence electrons at the 6
valence shell. 5

• Therefore, X is in Group 13 in the periodic 4
table 3
2
ii) Valence electron configuration for X is 3s2 3p1 1

iii) x3+ + O2- → X2O3 0 2 4 6 8 10 12 14 16 18 20
Number of electron removed
60
61

Solution

a. Define the term “ first ionization energy”. a. Tehneerigoynrizeaqtuioirnedentoerrgeym(oIEv1e) is the minimum
b. State the difference between the first and one mole electron
from one mole gaseous atom in its ground state.
second ionization energy (of any element).
Explain. b. tIEh1aits energy that absorbed when first electron
c. Predict which group and period M belongs to has been removed from a gaseous atom
in the Periodic Table welheiclteroIEn2 is energy that released when second
d. Write the electronic configuration of M is removed from gaseous ion.
Second ionisation energy is higher than first
62 ionisation energy because the nuclear charge
remain constant and the electron are held tightly
to the nucleus. Therefore more energy is needed
to remove another electron from the positively
charged ion.
63

Solution e) Trends in the electronegativity
c. - Since the sharp increment occur between Electronegativity

second and third ionisation energy, hence The relative tendency of the atom to
third electron is removed from the inner attract the shared electrons to itself when
shell. chemically combined with another atom.
• So, there are 2 valence electrons at the
valence shell. • Electronegativity increases up a group and
• Therefore, M is in Group 13 in the periodic across a period. This follows the trends for
table. ionization energy and electron affinity.
• Since M has 4 shell, so it belong to period 4.
65
d. M has 20 electrons. 64
1s2 2s2 2p6 3s2 3p64s2

86

a) Across a period b) Down a group
• The nuclear charge increase, effective nuclear
• The principle quantum number increase,
charge increase shielding effect increase
• The atomic size decrease
• Hence, the nucleus attraction stronger towards • The atomic size increase, the shared electrons
farther from nucleus
the shared electron stronger.
• Therefore, the electronegativity (ability to • Hence, weaker nuclear attraction
• Therefore electronegativity decrease
attract electron) increase
67
66

f) Acid-base character of oxides of Period 3 • Na reacts with oxygen to form a basic oxide, Na2O
4Na (s) + O2 (g)→2Na2O (s)
Period 3:
• The oxide will produce base solution sodium hydroxide
Na Mg Al Si P S Cl when react with water.
Na2O (s) + H2O (l) → 2NaOH (aq)
• When react with oxygen :
(a) Na & Mg will form basic oxide 69

(b) Al will form amphoteric (both acidic and basic)
oxide.

(c) Si, P, S & Cl will form acidic oxide

68

• Magnesium, Mg burns in oxygen to form MgO a basic • Al forms amphoteric oxide, can react either
oxide with an acid or a base.
2Mg (s) + O2 (g) →2MgO (s)
Al2O3 (s) + 6HCl (aq) →2AlCl3 (aq) + 3H2O (l)
• The white magnesium oxide powder is dissolved in base acid
water, the pH of the liquidis around pH 9 - showing
that it is slightly alkaline. Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2NaAl(OH)4 (aq)

• Magnesium oxide reacts with warm dilute hydrochloric acid base sodium aluminate
acid to give magnesium chloride solution.

MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l)
base acid

70 71

87

Si, P, S & Cl burn in oxygen to form acidic oxide. P: (limited O2)
Si : P4(s) + 3O2(g) → P4O6(s)
Si (s) + O2 (g) → SiO2 (s)
P4O6(g) + 6H2O(l) → 4H3PO3(aq)
SiO2 (l) + NaOH (aq) → Na2SiO3(aq) + H2O (l) phosphorus acid (weak acid)
• acid base
P4(s) + 5O2(g) → P4O10(s) (excess O2)
Silicon dioxide doesn't react with water
• Silicon dioxide reacts with sodium hydroxide P4O10(g) + 6H2O(l) → 4H3PO4(aq)
phosphoric acid (weak acid)
solution, but only if it is hot and concentrated. A
colourless solution of sodium silicate is formed. 73

72

S: As a conclusion…..
S(s) + O2(g) → SO2(g)
SO2(g) + H2O(l) → H2SO3(aq) • From left to right on the periodic table, acid-base
character of oxides go from basic to acidic.
sulfurous acid
• Ionic oxides are usually basic (element act as a
Cl : base when reacting with H2O) (Na2O, MgO)
Cl2O7(g) + H2O(l) → 2HClO4(aq)
• Aluminium oxide, Al2O3 are amphoteric
perchloric acid (elements acts as an acid and/or base when
reacting depending on pH of solution)
74
75

• Covalent oxides are usually acidic (elements
act as an acid when reacts with H2O)
(Si2O,P4O6,P4O10,SO2,Cl2O7)

Na2O, MgO, Al2O3, Si2O, P4O6, SO2, Cl2O7
More basic amphoteric More acidic

76

88

CHAPTER 3: PERIODIC TABLE

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

89

CHEMISTRY
TUTORIAL TOPIC 3: PERIODIC TABLE

3.1: Classification of elements

1. By referring to the electronic configuration of each element below:
A : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
B : 1s2 2s2 2p6 3s2 3p6 3d3 4s2
C : 1s2 2s2 2p6 3s2 3p6 4s2
D : 1s2 2s2 2p6 3s2
E : 1s2 2s1

a) State the period, group and block for each element.

b) State how the elements A to E are arranged in the periodic table?

c) Why are elements C and D in the same group?

d) Why are elements A and C in the same period?

2. The table below shows eight elements Q to X with consecutive proton number.

Element Q R S TUVW X

Valence 3s23p1 3s23p3 4s2
electronic

configuration

From the table,
a) Write the ground state valence electronic configuration of elements U and W

and state their most stable oxidation numbers.
b) Choose one

i. noble gas iv. element that forms acidic oxide

ii. alkaline earth metal v. element that forms basic oxides

iii. group 14 element vi. element that forms amphoteric oxides

c) Deduce the group and period for element S

3.2: Periodicity

1. Explain the trend of atomic and ionic radius down group 1 of the periodic table.

2. Removal or addition of electron(s) to an atom results in changes of atomic radii.

Species Na Na+ Cl Cl-

Radius (nm) 0.156 0.095 0.099 0.181

Explain the difference in radius between the ions and their respective neutral atoms.

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CHEMISTRY
TUTORIAL TOPIC 3: PERIODIC TABLE

3. Cations Mg2+, Al3+, Na+, Si4+ are isoelectronic with the electronic configuration,
1s2 2s2 2p6. Arrange the cations in the ascending order of ionic radius. Explain.

4. a) Define first ionisation energy.
b) Sketch the graph of the ionization energies versus the number of electrons of
carbon atom. Explain the trend shown in the graph.
c) Explain why the first ionisation energy of aluminium is larger than that of
sodium.
d) Why is the first ionization energy of Be higher than B?
e) Explain the trend of ionization energy down group 1.

5. The table shows the successive ionisation energies, IE (kJ mol−1) for elements J, K, L,
M and N.

Element IE1 IE2 IE3 IE4 IE5

J 801 2430 3660 25000 32820

K 1086 2350 4620 6200 38000

L 786 1580 3230 4360 16000

M 418 3052 4410 5900 8000

N 738 1450 7730 10500 13600

a. What is meant by the first and second ionisation energies?
b. Determine the element that forms a stable +1 ion.
c. Choose two elements which belong to the same group.

i. Choose two elements which belong to the same group.
ii. Explain the difference in the first ionization energy values for the two

elements.
d. Determine the group for element J.
e. Write the molecular formula of oxide J
f. The first ionisation energy of nitrogen is greater than that of oxygen. Explain.

6. Explain qualitatively the variation of electronegativity across a period 3 (Na to Ar)
and down a group 2 (Be to Ba)

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CHEMISTRY
TUTORIAL TOPIC 3: PERIODIC TABLE

7. a) Classify the following oxides as basic, acidic or amphoteric.
Al2O3, SiO2, MgO, P4O10, SO3, Na2O, Cl2O

b) Write the equations to show the acidic or basic properties of the following
oxides:
i. Na2O
ii. SO3
iii. Al2O3

92

CHEMISTRY
MEKA TOPIC 3: PERIODIC TABLE

3.1: Classification of elements

1. Fill in the blank:

ELEMENT ELECTRONIC GROUP PERIOD BLOCK
CONFIGURATION

F 1s2 2s2 2p6

G 1s2 2s2 2p6 3s1

H 1s2 2s2 2p6 3s2 3p6 3d5 4s2

I 13 4 p

J 3 4d

2. The table below shows elements and their respective proton numbers.

Elements KLMN

Proton number 8 10 11 13

Answer the following questions based on the information above.
a) State the group and period of each element in the Periodic Table
b) Arrange these elements in the ascending order of:

i. atomic radius
ii. first ionization energy
c) Which element is inert?
d) Write the formula of compound formed when K react with M.

3.2: Periodicity

1. An element Z has the following successive ionization energy:

Ionisation IE1 IE2 IE3 IE4 IE5 IE6
energy 800 1820 2750 11600 12500 13500
(kJ/mol)

Determine the group and block of element Z.

2.

Element K L MN

Proton number 12 11 17 8

a) Define electronegativity.

b) Arrange in ascending order the electronegativity of the above elements.

Explain.

93

CHEMISTRY
KUMBE TOPIC 3: PERIODIC TABLE

3.1: Classification of elements

1. Electronic configuration for elements O and P are as follows:
O : 1s22s22p63s23p1
P : 1s22s22p63s23p5

a) State the period and group for O and P in the periodic table.
b) What is the molecular formula for chloride compound of O?
c) What is the molecular formula and type of bond for hydride compound of P?

d) Write the molecular formula of the compound formed when O reacts with P.

3.2: Periodicity

1. Sodium and Sulphur are elements of period 3. The electronic configurations for the ions

are as follows:

11Na+ : 1s2 2s2 2p6

16S2− : 1s2 2s2 2p63s2 3p6

Compare and explain the radius of Sodium ion, Na+ with that of Sulphur ion, S2−.

2. Based on the following graph,

IE 1 (kJ/mol)

0 Na Mg Al Si P S Cl

a) Explain the trend of the first ionization energy.
b) Explain the variation of atomic radii.

3. The table below shows the first, second, third and fourth ionisation energies, IE in kJ
mol−1 of elements X, Y and Z.

Element IE1 IE2 IE3 IE4

X 738 1450 7730 10500

Y 800 2427 3658 25024

Z 495 4663 6912 9540

a) State the element(s) with the oxidation number +2.
b) Determine the group of Y in the periodic table. Explain.
c) Write the formula of oxide for Y.
d) The second ionisation energy of Z is higher than its first ionisation energy.

Explain.

94

log (ionization energy) CHEMISTRY
KUMBE TOPIC 3: PERIODIC TABLE

4. The graph below shows the successive ionization energies of all electrons of element
Z.

18
16
14
12
10

8
6
4
2
0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
number of electrons removed

a) State the group and period of element Z. Explain.
b) Write the electronic configuration of Z.
c) State the oxidation number of Z.

5. Elements D and E are in Period 4 of the Periodic Table. D is a p block element and E
is a d block element. Atom D and E2+ion, each has three valence electrons. Atom D has
one unpaired electron, while a E2+ion has three unpaired electrons. Identify D and E.
Explain.

6. Consider the periodic table in FIGURE 1.

FIGURE 1
a) i. Which elements are located in the same group? Explain your choice.

ii. Name the group.
b) i. State the character of the element S.

ii. Write the stable electronic configuration of S. Justify your choice.
c) i. What is the product when P react with Q?

ii. What is the acid-base property of this product?
iii. Write the chemical equations to represent this property.
d) Which element will generally not react with other elements? Explain.

95

CHAPTER 4: CHEMICAL
BONDING

LECTURE NOTE

lecture notes & questions

96

CHAPTER 4 4.1 LEWIS STRUCTURE
CHEMICAL BONDING
2
4.1 Lewis Structure
4.2 Molecular Shape and Polarity
4.3 Orbital Overlap and Hybridisation
4.4 Intermolecular Forces
4.5 Metallic Bond

1

At the end of this lesson, students should be (d) draw Lewis structures of molecules and
able to : polyatomic ions with single, double and triple
bonds.
(a) state the octet rule.
(e) compare the bond length between single, double
(b) describe how atoms achieve stability by and triple bonds.
attaining stable configuration of
(i) noble gas (f) determine the formal charge and the most plausible
(ii) pseudo-noble gas Lewis structures.
(iii) half-filled orbital
(g) explain the exception to the octet rule : incomplete
(c) describe the formation of the following octet, expanded octet and odd number electrons.
bonds using Lewis dot symbol.
i. ionic or electrovalent bond 3 (h) Illustrate the concept of resonance using 4
ii. covalent bond appropriate examples.
iii. dative or coordinate bond

 Chemical bond is the force that 4.1.1 Lewis Symbol
holds two atoms together in a A Lewis symbol consists of:
molecule or compound
 the symbol of an element
 Valence electrons play an important  dots or cross to represent the valence
role in the formation of chemical
bonds electrons in an atom of the element.

5 6

97