Saturday, 17 January 2015

5.2 Formation of Ionic Bonds

Formation of cation and anion
■ Formation of cation and anion

An atom is neutral because it has an equal number of protons and electrons.

An ion formed when an atom donates or accepts one or more electrons.

■ Formation of cation


When an atom donates electrons, it becomes positively charged as there are more protons than electrons in it.

A positively charged ion or cation is obtained.

M → Mn+ + ne
■ Formation of anion


When an atom accepts electrons, it becomes negatively charged as there are more electrons than protons in it.

A negatively charged ion or anion is obtained

A + ne → An-
Formation of ionic bonds
■ Formation of ionic bonds

Electrons are transferred from a metal to a non-metal.

The metal donates its valence electron(s) to attain the noble gas stability. Positive ions (cation) are formed.

The non-metal received the electron(s) to attain the noble gas stability. Negative ion (anion) are formed.

The positive and negative ions are held together by strong electrostatic forces of attraction.

This strong electrostatic forces of attraction between ions of opposite charges is known as ionic bond.

The animation below shows the formation of sodium chloride (NaCl), magnesium oxide (MgO), and sodium oxide (NaO2).

■ This video contains information on the formation of ionic bonds.

Laboratory Activity 5.2.1: Preparation of Ionic Compounds
Determining the ionic compound formulae
■ Formulae of an ionic compound.

Metal atoms will release their valence electrons to achieve the stable electron arrangement as in the inert gases.

Non-metal atoms will accept electrons in order to achieve the stable electron arrangement of the inert gases.

For cations Mb+ and anions Xa-, the formula of an ionic compound formed between them is written as MaNb.

The total positive charge of the cation must be equal to the negative charge of the anion in an ionic compound.

Hence, the formula of an ionic compound formed between them can also be derived as follows :
aMb+ + bXa- → MaNb

Worked-example 5.2(a)
Complete the following table to show the written ionic compound formulae.

Solution 
Atom Proton number Electron arrangement Ion formula Atom Proton number Electron arrangement Ion formula
A 3 2.1 A+ V 9 2.7 V-
B 11 2.8.1 B+ W 8 2.6 W2-
C 12 2.8.2 C2+ X 17 2.8.7 X-
D 20 2.8.8.2 D2+ Y 9 2.7 Y-
E 19 2.8.8.1 E+ Z 17 2.8.7 Z-


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