MY CHEMISTRY 1 MODULE, KMPh SDS : SK015

CHAPTER 3

PERIODIC
TABLE
Prepared by:
Faizah bt Mohd Yasin
Arinawati bt Mohd Hasman
Hazlima bt Hamzah
Alis Suhairin bt Abd Ghani
Mohd Nor Hisham bin Ahmad

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JK Bahan PdPc Tutorial Sesi 2021/ 2022, Unit Kimia, KMPh

MY CHEMISTRY 1 MODULE SDS : SK015

3.0 : PERIODIC TABLE

3.1 : CLASSIFICATION OF ELEMENTS
At the end of this topic, students should be able to:
(a) Describe period, group and block (s,p,d,f).
(b) Deduce the position of elements in the periodic table from its electronics
configuration.

CLASSIFICATION OF ELEMENTS

1) PERIOD 2) GROUP

• Horizontal rows in periodic • Vertical columns in periodic

table. table.
• Numbered from 1 to 7. • Numbered from 1 to 18.

3) BLOCK

• Divided into 4 major regions

(s, p, d, f block)

ns1
ns2 np1 to ns2 np6
ns2

(n-1)d1 ns2 to (n-1)d10 ns2

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EXAMPLE 1

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

A: 1s2 2s2 2p6 3s2 3p6

B: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
C: 1s2 2s2 2p6 3s2 3p6 4s2
D: 1s2 2s2 2p6 3s2 3p6 3d3 4s2
E: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

Solution

Element Group Period Block

A 18 3 p
B 17 4 p
C 2 4 s
D 5 4 d
E 18 4 p

PRACTICE QUESTION 1

The following table shows the proton number for four elements F, G, H and I.

Element Proton number

F 8
G 16
H 19
I 21

(a) State the position of element H in the periodic table.

(b) Choose two elements which belong to the same group.
(c) Choose two elements which belong to the same period.
(d) Which element is a d-block element?
Answer

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3.0 : PERIODIC TABLE

3.2 : PERIODICITY
At the end of this topic, students should be able to:
(b) Analyse the variation in atomic radii
i. across a period.
ii. Across the first row of transition elements.
iii. down a group.

FACTORS AFFECTING ATOMIC RADII

EFFECTIVE NUCLEAR CHARGE SHIELDING/SCREENING EFFECT

(Zeff)

The net nuclear charge actually The effect caused by mutual repulsion of
experiences by an electron as a results of electrons within the same orbital as well
shielding/screening effects by other as between inner and outer electrons
electrons

Zeff = Z – S

Z = proton number
S = number of inner or core electrons

Example :
Mg: 1s2 2s2 2p6 3s2

inner electron

• The inner electron shields the outer

- -
ee electrons from full effect of the nuclear
-
charge more effectively than electrons
e- e- -
e- e e-
e- e - e of the same orbital.
e- e-
• Shielding effect reduce the attraction
forces between an outer electron and
the nucleus in any atom with more than
Zeff = Z – S one electron shell.
= 12 – 10
= +2 • The more electron shells there are, the
greater the shielding effect is.

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VARIATION IN ATOMIC RADII

1
ACROSS PERIOD
• Proton number increase.
• Effective nuclear charge increase.
• Nucleus attraction towards
valence electron stronger.
• Atomic radius becomes smaller.

DOWN GROUP
• Number of shells increase.
• Shielding effect increase.
• Nucleus attraction
towards valence electron
weaker.
• Atomic radius becomes
larger.

ACROSS THE FIRST ROW OF TRANSITION ELEMENT

Atomic radii decrease but does not change significantly (approximately constant)

• Electrons are added to the inner 3d orbitals.

• The 3d electrons shield the outer 4s electrons from the increasing proton
number in the nucleus.
• The increase in Zeff effects is cancelled by the increase in shielding effect.
• The effective nuclear charge is about the same.
• Thus, atomic radii of the transition elements are approximately constant.

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EXAMPLE 2

The electronic configuration of element J and K is 1s2 2s2 2p6 3s2 3p4 and 1s2 2s2 2p6
3s2 3p3 respectively. Which atom has smaller size? Explain your answer.

Solution

TIPS:
Element J Element K
Zeff = Z – S Zeff = Z – S
= 16 – 10 = 15 – 10
= +6 = +5

• Proton number of element J is higher than element K.

• Effective nuclear charge of element J is higher than element K.
• Nucleus attraction towards valence electron of element J is stronger than
element K.
• Thus, atomic radius of element J is smaller than element K.

EXAMPLE 3

The table below shows the proton numbers of element L, M, N, P.

Element Proton number

L 11
M 3
N 1
P 19

Arrange the elements in order of increasing atomic radius. Explain your answer.

Solution

TIPS:
Element Proton number Electronic
configuration
L 11 1s2 2s2 2p6 3s1
M 3 1s2 2s1
N 1 1s1
P 19 1s2 2s2 2p6 3s2 3p6 4s1

Atomic radius: N < M < L < P

• From element N to element P, number of shells increase.
• Shielding effect increase.
• Nucleus attraction towards valence electron weaker
• Thus, atomic radius becomes larger from element N to element P.

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PRACTICE QUESTION 2

The proton number of element Q and R is 20 and 12 respectively. Compare the atomic
radius of atom Q and R. Explain your answer.

Answer

PRACTICE QUESTION 3

The elements below are found in the periodic table.

24 23 32
12𝑋 11𝑌 16𝑍

Arrange the elements in descending order of atomic size. Explain your answer.

Solution

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PRACTICE QUESTION 4

Consider the following elements P, Q, R and S in the table below:

Element Proton Number

P 3
Q 7
R 11
S 12

Choose the element that has the smallest atomic radius. Explain.

Solution

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3.0 : PERIODIC TABLE

3.2 : PERIODICITY
At the end of this topic, students should be able to:
(c) Compare the atomic radius of an element and its corresponding ionic radius.

(e) Compare the radius of isoelectronic species

(f) Analyse the variation in the ionic radii across period 2 and period 3.

IONIC RADII

ION vs ITS ATOM ION vs ION

Cation Anion Cation

Isoelectronic
vs vs vs
species
its atom its atom Anion

Across Across
period 2 period 3

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COMPARISON OF IONIC RADIUS AND

ITS CORRESPONDING ATOMIC RADIUS

RADIUS OF CATION vs ITS CORRESPONDING ATOM

• Cation is formed when atom removes electron.

• The effective nuclear charge of cation is higher than its atom.
• Nucleus attraction towards valence electrons of cation is stronger than its
atom.
• The cation formed has smaller radius compare to its atom

RADIUS OF ANION vs ITS CORRESPONDING ATOM

• Anion is formed when atom receives electron.

• The mutual repulsion among electrons of anion is greater than its atom.
• Nucleus attraction towards valence electrons of anion is weaker than its
atom.
• The anion formed has bigger radius compare to its atom.

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EXAMPLE 4
Explain the difference in radius between the ion of Mg2+ and its neutral atom, Mg.

Solution

TIPS:
Magnesium atom Magnesium ion
Mg: 1s2 2s2 2p6 3s2 Mg2+: 1s2 2s2 2p6
Zeff = Z – S Zeff = Z – S
= 12 – 10 = 12 – 2
= +2 = +10

• Magnesium ion, Mg2+ is formed when magnesium atom, Mg removes electrons.

• The effective nuclear charge of Mg2+ is higher than Mg.
• Nucleus attraction towards valence electrons of Mg2+ is stronger than Mg.
• Thus, Mg2+ has smaller radius compare to Mg.

PRACTICE QUESTION 5

Removal and addition of electron results in change of atomic radii. Explain the difference
in radius between the ions and their respective neutral atoms, based on the table below:

Species Na Na+ Cl Cl-

Radius/nm 0.156 0.095 0.099 0.181

Answer

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RADIUS OF ISOELECTRONIC SPECIES

Isoelectronic species are groups of atoms or ions which have the same electronic
configuration (same number of electrons)

Ions Electronic Configuration Ionic Radii (pm)

+
Na 1s2 2s2 2p6 95
2+
Mg 1s2 2s2 2p6 65
3+
Al 2 2
1s 2s 2p 6
50
The more positive the charge, the smaller the species.

Ions Electronic Configuration Ionic Radii (pm)

N3- 1s2 2s2 2p6 171
2-
O 2 2
1s 2s 2p 6
140
F- 1s2 2s2 2p6 136
The more negative the charge, the larger the species.

EXAMPLE 5

Arrange Na+, Mg2+, Al3+, Si4+ in order of increasing ionic radius.

Solution

TIPS:
Ions Proton number Electronic Configuration Zeff = Z-S
Na+ 11 1s2 2s2 2p6 11 – 2 = +9
Mg2+ 12 1s2 2s2 2p6 12 – 2 = +10
Al3+ 13 1s2 2s2 2p6 13– 2 = +11
Si4+ 14 1s2 2s2 2p6 14– 2 = +12

Ionic radius: Si4+ < Al3+ < Mg2+ < Na+

• Na+, Mg2+, Al3+, Si4+ are isoelectronic species because they have the
same electronic configuration
• Proton number increases from Na+ to Si4+.
• Effective nuclear charge increases.
• Nucleus attraction towards valence electrons stronger.
• Thus, ionic radii decrease from Na+ to Si4+.

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EXAMPLE 6

Arrange N3-, O2- , F- in order of increasing ionic radius.

Solution

TIPS:
Ions Proton number Electronic Configuration Zeff = Z-S
N3- 7 1s2 2s2 2p6 7 – 2 = +5
O2- 8 1s2 2s2 2p6 8 – 2 = +6
F- 9 1s2 2s2 2p6 9 – 2 = +7

Ionic radius: F- < O2- < N3-

• N3-, O2- , F- are isoelectronic species because they have the same
electronic configuration
• Proton number increases from N3- to F-.
• Effective nuclear charge increases.
• Nucleus attraction towards valence electrons stronger.
• Thus, ionic radii decrease from N3- to F-.

EXAMPLE 7

Arrange N3-, F-, Mg2+, Si4+ in order of increasing ionic radius.

Solution

TIPS:
Ions Proton number Electronic Configuration Zeff = Z-S
N3- 7 1s2 2s2 2p6 7 – 2 = +5
F- 9 1s2 2s2 2p6 9 – 2 = +7
Mg2+ 12 1s2 2s2 2p6 12 – 2 = +10
Si4+ 14 1s2 2s2 2p6 14– 2 = +12

Ionic radius: Si4+ < Mg2+ < F- < N3-

• N3-, F-, Mg2+, Si4+ are isoelectronic species because they have the same
electronic configuration
• Proton number increases from N3- to Si4+.
• Effective nuclear charge increases.
• Nucleus attraction towards valence electrons stronger.
• Thus, ionic radii decrease from N3- to Si4+.

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PRACTICE QUESTION 6

Given ionic radius of Na+ and Mg2+ ion are 0.095 nm and 0.065 nm respectively. Ionic
radius of Mg2+ smaller than Na+, explain.

Answer

PRACTICE QUESTION 7

S2- ion and Cl- ion are formed from elements located in period 3 of Periodic Table.
Compare the ionic radius of S2- ion and Cl- ion. Explain your answer.

Answer

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PRACTICE QUESTION 8

Arrange the ions below in order of increasing ionic radius. Explain.

Mg2+, Si4+, N3- , F-, Al3+, Na+, O2-

Answer

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VARIATION OF IONIC RADII ACROSS PERIOD 2 & PERIOD 3

IONIC RADII ACROSS PERIOD 2

• Li+, Be2+, B3+ and C4+ are

isoelectronic species because they
Proton Electronic have same electronic configurations.
Ions Zeff
number Configuration • Proton number increases from Li+
Li+ 3 1s2 +3 to C4+.
Be2+ 4 1s2 +4 • Effective nuclear charge also
B3+ 5 1s2 +5 increases from Li+ to C4+.
C4+ 6 1s2 +6 • Nucleus attraction towards
valence electron stronger.
• Ionic radii decrease from Li+ to
C4+.

❖ N3- ion has extra shells compare to

Proton Electronic C4+ ion.
Ions
number Configuration ❖ N3- ion has larger shielding effect.
C4+ 6 1s2 ❖ Nucleus attraction towards
N3- 7 1s2 2s2 2p3 valence electron weaker.
❖ N3- ion has larger radius compare to
C4+ ion.

➢ N3- , O2- and F- ions are

Proton Electronic isoelectronic species because they
Ions Zeff
number Configuration have same electronic configurations.
N3- 7 1s2 2s2 2p3 +5 ➢ Proton number increases from N3-
O2- 8 2 2
1s 2s 2p 4
+6 to F-.
F- 9 1s2 2s2 2p5 +7 ➢ Effective nuclear charge also
increases from N3- to F-.
➢ Nucleus attraction towards
valence electron stronger.
➢ Ionic radii decreases from N3- to F-.

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IONIC RADII ACROSS PERIOD 3

• Na+, Mg2+, Al3+, Si4+ are

isoelectronic species because they
Proton Electronic have same electronic configurations.
Ions Zeff
number Configuration • Proton number increases from Na+
+
Na 11 1s2 2s2 2p6 +9 to Si4+.
Mg2+ 12 1s2 2s2 2p6 +10 • Effective nuclear charge also
Al3+ 13 1s2 2s2 2p6 +11 increases from Na+ to Si4+.
Si4+ 14 1s2 2s2 2p6 +12 • Nucleus attraction towards
valence electron stronger.
• Ionic radii decrease from Na+ to
Si4+.

❖ P3- ion has extra shells compare to

Proton Electronic Si4+ ion.
Ions
number Configuration ❖ P3- ion has larger shielding effect.
Si4+ 14 1s2 2s2 2p6 ❖ Nucleus attraction towards
P3- 15 1s2 2s2 2p6 3s2 3p6 valence electron weaker.
❖ P3- ion has larger radius compare to
Si4+ ion.

➢ P3- , S2- and Cl- ions are

Proton Electronic isoelectronic species because they
Ions Zeff
number Configuration have same electronic configurations.
1s2 2s2 2p6 +5 ➢ Proton number increases from P3-
P3- 15 2 6 to Cl-.
3s 3p
1s2 2s2 2p6 +6 ➢ Effective nuclear charge also
S2- 16 2 6 increases from P3- to Cl-.
3s 3p
1s2 2s2 2p6 +7 ➢ Nucleus attraction towards
Cl- 17 2 6 valence electron stronger.
3s 3p
➢ Ionic radii decreases from P3- to Cl-
.

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EXAMPLE 8

Li+, Be2+, B3+, C4+, N3-, O2- and F- are ions for elements in period 2. Arrange the ions in
increasing order of ionic radius. Explain.

Solution

Ionic radius: C4+ < B3+ < Be2+ < Li+ < F- < O2- < N3

• Li+, Be2+, B3+ and C4+ are isoelectronic species because they have same
electronic configurations.
• Proton number increases from Li+ to C4+.
• Effective nuclear charge also increases from Li+ to C4+.
• Nucleus attraction towards valence electron stronger.
• Ionic radii decrease from Li+ to C4+.

• N3- ion has extra shells compare to C4+ ion.

• N3- ion has larger shielding effect.
• Nucleus attraction towards valence electron weaker.
• N3- ion has larger radius compare to C4+ ion.

• N3- , O2- and F- ions are isoelectronic species because they have same electronic
configurations.
• Proton number increases from N3- to F-.
• Effective nuclear charge also increases from N3- to F-.
• Nucleus attraction towards valence electron stronger.
• Ionic radii decrease from N3- to F-.

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PRACTICE QUESTION 9

Na+, Mg2+, Al3+, Si4+, P3-, S2- and Cl- are ions for elements in period 3. Arrange the ions
in decreasing order of ionic radius. Explain.

Answer

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3.0 : PERIODIC TABLE

3.2 : PERIODICITY
At the end of this topic, students should be able to:
(h) Analyse the variations in the first ionisation energy across a period and down a
group

IONISATION ENERGY

First Ionisation Energy, IE1 Second Ionisation Energy, IE2

The minimum energy required to The energy required to remove 1 mole

remove 1 mole of electron from 1 mole of electron from 1 mole its gaseous
its gaseous atom to form 1 mole ion.
gaseous ion from its ground state.
Mg+(g) Mg2+(g) + e
Mg(s) +
Mg (g) + e ∆H = +ve
∆H = +ve

1 ACROSS PERIOD:
• Proton number increases.
• Effective nuclear charge increase.
• Nucleus attraction towards valence
electron stronger.
• Atomic radius becomes smaller.
• More energy is needed to remove the
electron.
• First ionisation energy, IE1 increases.

2
DOWN GROUP:
• Number of shells increase.
• Shielding effect increase.
• Nucleus attraction towards
valence electron weaker.
• Atomic radius becomes
larger.
• Less energy is needed to
remove the electron.
• First ionisation energy, IE1
decreases.

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ANOMALOUS BEHAVIOUR OF FIRST IONISATION ENERGY

GROUP 2 & 13 GROUP 15 & 16

Be > B N>O
Be : 1s2 2s2 N : 1s2 2s2 2p3
P (completely filled 2s orbital) (half filled 2p orbital)
E B : 1s2 2s2 2p1 O : 1s2 2s2 2p4
R (partially filled 2p orbital) (partially filled 2p orbital)
I
O • Completely filled 2s orbital is more • Half filled 2p orbital is more stable
D stable than partially filled 2p orbital. than partially filled 2p orbital.
• More energy required to remove • More energy required to remove
2 the first electron in 2s orbital of Be the first electron in 2p orbital of
than 2p orbital of B. N than 2p orbital of O.
• Thus, first ionisation energy of Be is • Thus, first ionisation energy of N
higher than B. is higher than O.

Mg > Al P>S
Mg: 1s2 2s2 2p6 3s2 P : 1s2 2s2 2p6 3s2 3p3
(completely filled 3s orbital) (half filled 3p orbital)
P Al: 1s2 2s2 2p6 3s2 3p1 S : 1s2 2s2 2p6 3s2 3p4
E (partially filled 3p orbital) (partially filled 3p orbital)
R
I • Completely filled 3s orbital is more • Half filled 3p orbital is more stable
O stable than partially filled 3p orbital. than partially filled 3p orbital.
D • More energy required to remove • More energy required to remove
the first electron in 3s orbital of Mg the first electron in 3p orbital of P
3 than 3p orbital of Al. than 3p orbital of S.
• Thus, first ionisation energy of Mg • Thus, first ionisation energy of P
is higher than Al. is higher than S.

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EXAMPLE 9

Element R has electronic configuration as follows.

1s2 2s2 2p6 3s2 3p4

The electronic configuration of element Q is 1s2 2s2 2p6 3s2 3p3. Between Q and R, which
element has higher first ionization energy? Give reason for each element.

Solution

Q : 1s2 2s2 2p6 3s2 3p3 (half filled 3p orbital)

R: 1s2 2s2 2p6 3s2 3p4 (partially filled 3p orbital)

• Half filled 3p orbital is more stable than partially filled 3p orbital.

• More energy required to remove the first electron in 3p orbital of Q than 3p
orbital of R.
• Thus, first ionisation energy of Q is higher than R.

PRACTICE QUESTION 10
Ionisation energy for nitrogen atom is greater than that of oxygen atom but the ionisation
energy for N+ is lower than that of O+. Explain.

Answer

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PRACTICE QUESTION 11

Among the element of magnesium, sodium and silicon, which has the highest second
ionisation energy? Explain.

Answer

PRACTICE QUESTION 12

The table shows the electronic configuration of elements W, X, andY

Element Electronic configuration

W [Ne]3s1
X [Ne] 3s23p4
Y [Ar]4s2

(a) Explain:
i. the difference in the first ionisation energy between elements W and X
ii. the difference in ionic radius between X 2- and Y 2+

Answer

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3.0 : PERIODIC TABLE

3.2 : PERIODICITY
At the end of this topic, students should be able to:
i) Explain the increase in the successive ionisation energies of an element.
j) Deduce the electronic configuration of an element and its position in the
periodic table based on successive ionization energy data.

SUCCESSIVE IONIZATION ENERGY

Successive ionization energy of Be in kJmol-1

Electronic configuration of Be : 1s2 2s2

- -
e e -
e
-
e
n=1
n=2

Ionisation 1st 2nd 3rd 4th

energy, IE
(kJ mol-1) 899 1757 14845 21000

Generally, successive ionisation energies always increase because each subsequent

electron is being pulled away from an increasingly more positive ion, and that requires
more energy.

SMALL CHANGE OF IE
• Electrons are removed from the same
valence shell

Based on successive
ionization energy data

DRASTIC INCREASE/ LARGER

DIFFERENCES/ SUDDEN INCREASE
OF IE

• Electron is removed from an inner shell

• More energy required to remove
electron from inner shell

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DEDUCE ELECTRONIC CONFIGURATION OF AN ELEMENT & ITS

POSITION IN THE PERIODIC TABLE BASED ON SUCCESSIVE
IONIZATION ENERGY DATA

1. Successive ionisation energy, IE data:

a. Table of data
(group, block and valance electronic configuration)
b. Graph
(group, block, period and electronic configurations)

2. TWO methods to deduce electronic configuration:

a. IE ratio
b. IE difference

KEY CONCEPT Drastic increase in

successive IE data

indicates:

Removal of electron from an inner Number of valence electrons that

shell (which has a stability of full- the particular element has
filled orbital)

deduces:

Electronic configuration of the Position of the element in periodic

element table

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EXAMPLE 10

Six successive ionisation energies (kJmol-1) for atom X is shown below. Deduce its
valence electronic configuration and position in periodic table.

IE First Second Third Fourth Fifth Sixth

(kJ mol-1) 1086 2352 4619 6221 37820 47260

Solution

METHOD 1: IE RATIO METHOD 2: IE DIFFERENCE

Step 1: Calculate the IE ratio Step 1: Calculate the IE difference

IE2 / E1 = 2.17 IE2 - E1 = 1266
IE3 / E2 = 1.96 IE3 - E2 = 2267
IE4 / E3 = 1.35 IE4 - E3 = 1602
IE5 / E4 = 6.08 IE5 - E4 = 31599
IE6 / E5 = 1.25 IE6 - E5 = 9440

Step 2: Making deduction Step 2: Making deduction

IE5 / IE4 has the highest ratio, There is sudden increase from IE4 to
indicates very large energy is required IE5, indicates the 5th electron is very
to remove the 5th electron. Hence it can dificult to be removed. It can be
be deduced that: deduced that:
➢ Element X has 4 valence ➢ The 5th electron is removed
electrons and the 5th electron is from an inner shell and
removed from an inner shell element X has 4 valence
➢ Valence electronic electron.
configuration: ns2 np2 ➢ Valence electronic
➢ Position: Group 14 configuration: ns2 np2
➢ Position: Group 14

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EXAMPLE 11

Based on the ionisation energy curve below, predict the group for Element Y.

70000

60000

50000
Ionisation Energy

40000

30000

20000

10000

0
1 2 3 4 5
No of electron removed

Solution

There is sudden increase from IE3 to IE4, indicates the 4th electron is very
dificult to be removed. It can be deduced that:

❖ The 4th electron is removed from an inner shell.

❖ The element Y has 3 valence electron.
❖ Thus, element Y is in group 13

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PRACTICE QUESTION 13

The table shows the successive ionisation energies (IE / kJ mol-1) for elements K and
Answer
L,
Element IE1 IE2 IE3 IE4 IE5
K 1086 2350 4620 6200 38000
L 786 1580 3230 4360 16000
M 418 3052 4410 5900 8000

a) i. Identify the element that will form ion with the charge of +1, and give
the reason.
ii. Choose two elements which belong to the same group. Explain the
difference in the first ionization energy values for the two elements.

b) Determine the group for element L.

Answer

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PRACTICE QUESTION 14

The graph shows the variation of ionisation energies for element Z.

log ionisation energy
X
X

X
X
X
X
X
X
X
X

X
X
X

No. of electron
removed
(a) Distinguish the group and period for element Z.
(b) Explain why a higher energy is needed to remove the fourth electron?
(c) What happen to the atomic radius of Z when an electron is removed?
(d) State the oxidation number of element Z.

Answer

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PRACTICE QUESTION 15

Sketch a plot of the ionisation energies of carbon atom against the number of ionisations.
Explain the trend shown in the plot.

Answer

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3.0 : PERIODIC TABLE

3.2 : PERIODICITY
At the end of this topic, students should be able to:
(l) Explain the variation in electronegativity of elements.
(m) Explain the acid-base character of oxides of elements in Period 3.

ELECTRONEGATIVITY

The relative tendency of an atom to attract electrons to itself when chemically

combined with another atom.

ACROSS PERIOD:

• Proton number increases.

• Effective nuclear charge
increase.
• Nucleus attraction towards
valence electron stronger.
• Atomic radius becomes smaller.
• Tendency to attract bonding
electron increases.
• Electronegativity increases.

2
DOWN GROUP:

• Number of shells increase.

• Shielding effect increase.
• Nucleus attraction towards
valence electron weaker.
• Atomic radius becomes
larger.
• Tendency to attract bonding
electron decreases.
• Electronegativity decreases.

121 | P a g e Chemistry Unit, KMPh

MY CHEMISTRY 1 MODULE SDS : SK015

EXAMPLE 12

Fluorine and chlorine are in the same group. State and explain which is more
electronegative.

Solution

• Fluorine has less number of shell than chlorine.

• Shielding effect in fluorine is less than chlorine
• Fluorine has stronger nucleus attraction towards valence electron and smaller
atomic radius than chlorine.
• Tendency to attract bonding electron is higher in fluorine.
• Thus, fluorine is more electronegative than chlorine

PRACTICE QUESTION 16

The proton number of several elements are given in the following table.
Element Proton
number
V 17
W 14
X 12

(a) Compare the electronegativity of element X and W. Explain your answer.

(b) Arrange the elements in order of increasing electronegativity. Explain your
answer.

Answer

122 | P a g e Chemistry Unit, KMPh

MY CHEMISTRY 1 MODULE SDS : SK015

ACID-BASE CHARACTER OF OXIDES OF ELEMENTS IN PERIOD 3

Reaction of elements in period 3 with oxygen will form oxides. Metal will form basic
oxide and nonmetal will form acidic oxide.; metalloid will form amphoteric (both
acidic and basic) oxide.

Element
Na Mg Al Si P S Cl
period 3
Metallic
Metal Metalloid Non-metal
character
P4O10 (or SO3 (or Cl2O7 (or
Oxides Na2O MgO Al2O3 SiO2
P4O6) SO2 Cl2O
Acid-
Basic Basic Amphoteric Acidic Acidic Acidic Acidic
base
Oxide Oxide Oxide Oxide Oxide Oxide Oxide
character

P4O10 + Cl2O7 +
SO3 +
6H2O H2O
H2 O
→ →
→ H2SO4
Na2O 4H3PO4 2HClO4
Adding + H2O
Insoluble Insoluble Insoluble
H2O →
2NaOH P4O6 + SO2 + Cl2O +
6H2O H2 O H2O
→ → →
4H3PO3 H2SO3 2HOCl

Na2O + MgO + Al2O3 +

Adding 2HCl→ 2HCl→ 6HCl→ No No No No
HCl 2NaCl MgCl2 + 2AlCl3 + reaction reaction reaction reaction
+ H2O H2O 3H2O

Cl2O7 +
P4O10 + SO3 +
2NaOH
12NaOH→ 2NaOH→
→
4Na3PO4 + Na2SO4 +
2NaClO4
Al2O3 + SiO2 + 6H2O H2 O
+ H2O
Adding No No 2NaOH + 2NaOH→
NaOH reaction reaction 3H2O → Na2SiO3
2NaAl(OH)4 + H2O P4O6 + Cl2O +
SO2 +
12NaOH→ 2NaOH
2NaOH→
4Na3PO3 + →
Na2SO3 +
6H2O 2NaOCl
H2 O
+ H2O

123 | P a g e Chemistry Unit, KMPh

MY CHEMISTRY 1 MODULE SDS : SK015

EXAMPLE 13

X is an element from Period 3 of the Periodic Table with molecular formula XO3 as an
oxide.
(a) Write the chemical reaction when oxide X reacts with water.
(b) State the acid-base property of oxide X. Write an equation to show the stated
property

Solution

(a) XO3 + H2O → H2XO4

(b) Acidic oxide

XO3 + 2NaOH → Na2XO4 + H2O

PRACTICE QUESTION 17

(a) Classify the following oxides as basic, acidic or amphoteric.

Al2O3, SiO2, MgO, P4O10, SO3, Na2O, Cl2O
(b) Write the equations for the reaction between the following oxides with acid and
base.
i. Na2O
ii. SO3
iii. Al2O3

Answer

124 | P a g e Chemistry Unit, KMPh