CHAPTER 3 - PERIODIC TABLE

• Second ionization energy (IE2) is the minimum
energy required to remove 1 mole electrons
from 1 mole gaseous ion with a charge of 1+.
X+(g) → X2+(g) + e-

• When an electron is removed from a neutral
atom, the repulsion among the remaining
electrons decrease.

• Since the nuclear charge remain constant, the
electron are held tightly to the nucleus.

• Therefore more energy is needed to remove
another electron from the positively charged
ion. 51

• Thus, ionization energies always increase in
the following order :

IE1< IE2< IE3< IE4<…..

• Although the removal of a subsequent
electron from an atom requires an increment
amount of energy but it may not be
consistence.

• We can determine the electronic
configuration of the valence electron for an
element by using the ionization energy.

52

Example 1

• 4Be
• The ionization energies (kJmol-1) of Beryllium

are shown below.

IE1 IE2 IE3 IE4
899 1757 14850 21005

53

• The ratio between the ionization energies are:

IE2  1757  1.95
IE1 899
IE3  14850  8.45
IE2 1757
IE4  21005  1.41
IE3 14850

54

• A sharp increase in ionization energy occurs
when an inner-shell electron is removed.

• The highest ratio is IE3 / IE2. This indicates a
sharp increase of ionization energy for the third
electron. Thus, the third electron is from the
inner-shell,1s.

• The first and second electrons are removed
from the same energy subshell (2s) .

• 2 valence electrons are present. This element is
in Group 2

• Valence electron configuration : ns2 55

Example 2

• Five successive ionization energies (kJmol-1)
of atom M is shown below:

IE1 IE2 IE3 IE4 IE5
800 1580 3230 4360 16000

• Determine
i) the electronic configuration of the valence
electron.
ii) the group number in the periodic table.

56

Solution

• By determining the IE ratios:

IE2 = 1580 = 1.98
IE1 800
IE3 = 3230 = 2.04
IE2 1580
IE4 = 4360 = 1.35
IE3 3230
IE5 = 16000 = 3.67
IE4 4360

57

• Since the ratio of IE5 / IE4 is the highest, the
fifth electron is removed from the inner shell.

• So, there are 4 valence electrons at the
valence shell.

• Therefore, M is in Group 14 in the periodic
table

• Valence electron configuration for M is ns2 np2

58

Exercise 1

The five successive ionisation energies of an element X
are as follows

IE1 IE2 IE3 IE4 IE5

IE (kJmol-1) 800 2400 3700 25000 31800

i) State the group to which X belongs to in the Periodic
Table. Explain.

ii) If X is an element of Period 3, write its electronic
configuration.

iii) Deduce the empirical formula of its simplest oxide.

59

Solution
i) By determining the IE ratios:

IE2  2400  3.00
IE1 800

IE3  3700  1.54
IE2 2400

IE4  25000  6.76
IE3 3700

IE5  31800  1.27
IE4 25000

60

• Since the ratio of IE4 / IE3 is the highest, the
fourth electron is removed from the inner
shell.

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

• Therefore, X is in Group 13 in the periodic
table

ii) Valence electron configuration for X is 3s2 3p1

iii) x3+ + O2- → X2O3

61

Successive ionization energy)Exercise 2
A graph of successive ionization energy against number of
electrons removed of an element, M is shown below.

6

5

4

3

2

1

0 2 4 6 8 10 12 14 16 18 20

Number of electron removed

62

a. Define the term “ first ionization energy”.

b. State the difference between the first and
second ionization energy (of any element).
Explain.

c. Predict which group and period M belongs to
in the Periodic Table

d. Write the electronic configuration of M

63

Solution

a. The ionization energy (IE1) is the minimum
energy required to remove one mole electron
from one mole gaseous atom in its ground state.

b. IE1 is energy that absorbed when first electron
that has been removed from a gaseous atom

while IE2 is energy that released when second
electron is removed from gaseous ion.

Second ionisation energy is higher than first
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.
64

Solution
c. - Since the sharp increment occur between

second and third ionisation energy, hence
third electron is removed from the inner
shell.

• So, there are 2 valence electrons at the
valence shell.

• Therefore, M is in Group 13 in the periodic
table.

• Since M has 4 shell, so it belong to period 4.

d. M has 20 electrons. 65
1s2 2s2 2p6 3s2 3p64s2

3.2.4 Trends in the electronegativity

• Electronegativity is the relative tendency of
the atom to attract the shared electrons to
itself when chemically combined with
another atom.

• Electronegativity increases up a group and
across a period. This follows the trends for
ionization energy and electron affinity.

66

a) Across a period
• The nuclear charge increase, effective nuclear

charge increase
• The atomic size decrease
• Hence, the nucleus attraction stronger towards

the shared electron stronger.
• Therefore, the electronegativity (ability to

attract electron) increase

67

b) Down a group

• The principle quantum number increase,
shielding effect increase

• The atomic size increase, the shared electrons
farther from nucleus

• Hence, weaker nuclear attraction
• Therefore electronegativity decrease

68

3.2.5 Acid-base character of oxides of Period 3

Period 3:

Na Mg Al Si P S Cl

• When react with oxygen :
(a) Na & Mg will form basic oxide

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

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

69

• Na reacts with oxygen to form a basic oxide, Na2O

4Na (s) + O2 (g)→2Na2O (s)

• The oxide will produce base solution when react with
water.

Na2O (s) + H2O (l) → 2NaOH (aq)

• Mg burns in oxygen to form a basic oxide, MgO.
2Mg (s) + O2 (g) →2MgO (s)

MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l) 70
base acid

• Al forms amphoteric oxide, can react either with
an acid or a base.

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

base acid

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

acid base sodium aluminate

71

Si, P, S & Cl burn in oxygen to form acidic oxide.
Si :
Si (s) + O2 (g) → SiO2 (s)
SiO2 (l) + NaOH (aq) → Na2SiO3(aq) + H2O (l)
acid base

72

P: (limited O2)
P4(s) + 3O2(g) → P4O6(s)

P4O6(g) + 6H2O(l) → 4H3PO3(aq)
phosphorus acid

P4(s) + 5O2(g) → P4O10(s) (excess O2)

P4O10(g) + 6H2O(l) → 4H3PO4(aq)
phosphoric acid

73

S:
S(s) + O2(g) → SO2(g)
SO2(g) + H2O(l) → H2SO3(aq)

sulfurous acid

Cl :
Cl2O7(g) + H2O(l) → 2HClO4(aq)

perchloric acid

74