CHAPTER 3.0

Variation In First Ionisation Energy Down The Group

01 Shielding Effect increases. Atomic size increases. 02

03 Attraction between 04Less energy is required
nucleus and valence
electrons become weaker. to remove first electron.

05 Frist Ionisation energy
decreases

Example:

Explain the variation of first ionisation energy in descending group.

Going down a group,
 Proton number increases
 Number of shells filled by electrons increases
 Shielding effect increases
 Atomic size increases
 Nucleus attraction toward valence electrons becomes weaker.
 First ionisation energy decreases.

Variation In First Ionisation Energy Across A Period 53

 Across a period, IE1 generally INCREASES (Reason: Zeff , atomic size )
 Irregularities occur between Group 2 & 13 and Group 15 & 16.

First Ionisation Energy 2500 Ne
(kJ mol-1)
2000 N F The anomaly can
1500 O only be explained
1000 Be C by inspecting the
500 Li B valence
electronic
configurations.

3 4 5 6 7 8 9 10
Proton Number

Variation In First Ionisation Energy Across Period 2 54

Across period 2, Atomic size decrease. 02

01 effective nuclear charge,

Zeff increases.
Shielding effect constant.

03 Attraction between nucleus
and valence electrons

become stronger. 04More energy is required to

05 First ionisation energy remove the first electron.
increases. However, 1st IE of

B and O are lower than

expected.

 Although the pattern of ionisation energy increase across a period, there are exceptions

to the trend of first ionisation energy between group 2 and 13 and group 15 and 16.
 This exceptions also known as irregularities or anomaly.

Irregularities Between Group 2 (Be) and Group 13 (B) In Period 2

55

 Be is found to have a higher first ionisation energy than B.
 This irregularity can be explained by the stability of the completely filled orbital.

4Be: 1s2 2s2 First electron of Be is removed from a completely filled 2s orbital
which is more stable and therefore requires higher energy.

5B: 1s2 2s2 2p1 First electron of B is removed from the 2p orbital which is higher
energy orbital and shielded from nucleus by two inner 2s
electrons@ partially-filled 2p orbital less stable. Thus, require less
energy.

Irregularities Between Group 15 (N) and Group 16 (O) In Period 2 56

 N is found to have a higher first ionisation energy than O.
 This irregularity can be explained by the stability of the half-filled orbital.

7N: 1s2 2s2 2p3 The first electron of N is removed from half-filled 2p orbital.

8O: 1s2 2s2 2p4 The first electron of O is removed from the partially- filled 2p
orbital.

 Half-filled 2p orbital is more stable than partially-filled 2p orbital, therefore more
energy is required to remove the first electron in N atom than O atom.

Irregularities In Period 3 Elements Between Group 2 & 13 and Group 15 & 16

 The same reason applied in Period 3.

12Mg: 1s2 2s2 2p6 3s2  Mg has completely filled 3s orbital.

13Al: 1s2 2s2 2p6 3s2 3p1  Al has partially-filled 3p orbital.

 More energy is required to remove the first electron in more stable completely
filled 3s orbital in Mg atom than partially-filled 3p orbital in Al atom.

15P: 1s2 2s2 2p6 3s2 3p3  P has half-filled 3p orbital.

16S: 1s2 2s2 2p6 3s2 3p4  S has partially-filled 3p orbital.

 More energy is required to remove the first electron in more stable half-filled 3p
orbital in P atom than partially-filled 3p orbital in S atom.

57

Example: 58
Explain the variation of first ionisation energy across period 3

Ar
First Ionisation Energy
(kJ mol-1) P Cl
S
Mg Si
Na Al

11 12 13 14 15 16 17 18
Proton Number

01 Across period 3, Atomic size decrease. 02
effective nuclear charge, Zeff
increases. 04More energy is required to
Screening effect constant.
remove the first electron.
03 Attraction between nucleus
and valence electrons
become stronger.

05 First ionisation energy
increases. However,
1st IE of Al and S are lower
than expected.

12Mg: 1s2 2s2 2p6 3s2 13Al: 1s2 2s2 2p6 3s2 3p1

 The graphs shows Mg has higher first ionisation energy, IE1 than Al because the first
valence electron of Mg is removed from completely filled 3s orbital which is

more stable. Therefore requires higher energy.

 However, the first electron for Al is removed from partially-filled 3p orbital which
is higher energy orbital and is shielded from nucleus by two inner 3s
electrons@ partially-filled 3p orbital less stable. Less energy required to remove
first electron in Al atom.

15P: 1s2 2s2 2p6 3s2 3p3 16S: 1s2 2s2 2p6 3s2 3p4

 P has higher first ionisation energy, IE1 than S because the first electron of P is
removed from half-filled 3p orbital which is more stable. Therefore requires more
energy.

 However, the first electron for S is removed from partially-filled 3p orbital which
is less stable. Less energy required to remove first electron of S.

Variation in Successive Ionisation Energies

 If an atom has n electrons, its successive ionisation energies show increasing trend:

IE1 < IE2 < IE3 …..< IEn Inner electrons

Group

1

2 Ionisation
13 energy
14 (kJ mol-1)

15
16

Valence electrons

 Successive IE values can be used to deduce the position of elements in the Periodic

Table. 62

Variation in Successive Ionisation Energies

 Analysing the data of the successive ionisation energy, IE will enable us to figure out

the valence electronic configuration and the position of element in the Periodic
table.

 Generally, successive ionisation energy will increase for each of the electron
being removed.

 Increases gradually when the electrons are removed from an inner shell which is
closer to the nucleus.

 The position of a particular element can be determined by looking for the drastic
increase in the IE value.

 The drastic increase in the IE indicates the difficulty to remove electron
from an inner shell which is closer to nucleus or completely@
fully filled orbital.

63

Example: The following graph shows all the successive energies of an element Z.

Drastic increase in ionisation
energy from IE3 to IE4 because
the 4th e- removed from inner
shell/ different shell which is
closer to nucleus.
Indicate Z → 3 valence e-
(Group 13, p-block)
Valence electronic
configuration: 3s2 3p1

64

Example: 65
An element X has the following successive ionisation energies, IE :

i. Determine the group of X is located in the periodic table. Explain

 Group 14

 Drastic increase in ionisation energy from

IE4 to IE5@ IE5 highest ratio, indicate more
energy is IEr4equired to remove the 5th

electron.

 5th electron is removed from inner shell@
different shell which is closer to nucleus.

 X has 4 valence electrons.

ii. Write the valence electronic configuration of an atom of X.

 Valence electronic configuration of X: ns2 np2

Example: The first five ionisation energy of two elements are given below:

Element Ionisation Energy (kJ mol-1) Determine the position of
A first second third fourth fifth each element in the periodic
B 786 1540 3230 4360 16000 table. Explain
418 3070 4600 5860 7990

Element A: Element B:

drastic increase in ionisation energy drastic increase in ionisation energy
from IE4 to IE5@ IE5/IE4 highest ratio
indicates that the 5th electron is from IE1 to IE2@ IE2/IE1 highest ratio
removed from inner shell which is indicates that the 2nd electron is
closer to nucleus.
removed from inner shell which is
A has 4 valence electrons (ns2 np2)
closer to nucleus.
Group 14, p-block.
A has 1 valence electrons (ns1)

Group 1, s-block. 66

Electronegativity

 Electronegativity is the relative tendency of an atom to attract bonding electrons
to itself when chemically combined with another atom.

 The greater the electronegativity of an atom in a molecule, the more strongly it
attracts the bonding electrons.

HH

Same electronegativity Cl is more electronegative

67

Down a Group  Electronegativity Decreases Variation In Electronegativity Of Elements

Across Period  Electronegativity Increases

The most electronegative atom

68

Across a Period,
Proton number increases
Effective nuclear charge increases
Atomic size@ atomic radius decreases
Ability@ tendency of atom to attract bonding electrons increases
Electronegativity increases

Down a Group,
Number of shells filled by electrons

increases
Shielding effect increases
Atomic size@ atomic radius increases
Ability@ tendency of atom to attract

bonding electrons decreases
Electronegativity decreases

69

Example:

Arrange Oxygen, sulphur and Selenium in the order of increasing electronegativity.
Explain your answer.

 8O: 1s2 2s2 2p4
 16S: 1s2 2s2 2p6 3s2 3p4
 34Se: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

 Electronegativity: Se < S < O
 Se, S and O are in the same group@ Group 17 elements.
 From Se, S to O,
 Number of shells filled by electrons decreases
 Shielding effect decreases
 Atomic size@ atomic radius decreases
 Ability@ tendency of atom to attract bonding electrons increases

70

Example:

Given the proton number of elements P, Q and R is 12, 15 and 17 respectively. Arrange
the elements P, Q and R in the order of increasing electronegativity. Give explanation for
your arrangement.
 12P: 1s2 2s2 2p6 3s2
 15Q: 1s2 2s2 2p6 3s2 3p3

 17R: 1s2 2s2 2p6 3s2 3p5

 Electronegativity: P < Q < R

 P, Q and R are elements of the same period@ period 3.

 From P, Q to R. 71
 Proton number increases
 Effective nuclear charge increases
 Atomic size@ atomic radius decreases
 Ability@ tendency of atom to attract bonding electrons increases

Acid-Base Character Of Oxides Of Period 3 Elements

Element Na Mg Al Si P S Cl
Na2O MgO Al2O3 SiO2 P4O6@ Cl2O7
Oxide Basic Basic Amphoteric Acidic P4O10 SO2@ Acidic
Formulae SO3
Acidic
Acid-base Acidic
Properties

1. Basic oxide

 Sodium oxide react with water to form a basic solution, NaOH.

Na2O(s) + H2O(l) → 2NaOH(aq)
(basic solution)

 Sodium oxide react with acid to form salt and water.
Na2O(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l)
(base) (acid)

72

1. Basic oxide 73

 It is slightly soluble in water to form an basic solution. 73

2MgO(s) + H2O(l) → Mg(OH)2(aq)
(basic solution)

 Magnesium oxide react with acid to form salt and water.

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

(base) (acid) (salt)

2. Amphoteric oxide

 Aluminium oxide is insoluble in water but it is amphoteric oxide because it has
properties of both acid and base.

 It can react with either an acid or a base.

i. Al2O3 act as base – react with acid

Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O (l)

(base) (acid) (salt)

ii. Al2O3 act as acid – react with base

Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq)

(acid) (base) (sodium aluminate)

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3. Acidic oxide

 Silicone oxide is insoluble in water, but react with base to form salt and water.

SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l)

(Acid) (Base) (sodium silicate)

 The oxides of nonmetals are acidic because they react with water to produce

acidic solutions: Cl2O7(l) + H2O(l) → 2HClO4(aq)
(Perchloric acid)
P4O6 (s) + 6H2O(l) → 4H3PO3(aq)
(Phosphorous acid)

P4O10(s) + 6H2O(l) → 4H3PO4(aq)
(Phosphoric acid)

SO2(g) + H2O(l) → H2SO3(aq)
(Sulphurous acid)

SO3(g) + H2O(l) → H2SO4(aq)
(Sulphuric acid)

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CHEMISTRY UNIT, KMNS

THE END