Question

Explain hardness of ionic bond?

Answer 1

For finding the solution to the question, we should know the nature of the species that are involved in the bond formation and the properties of the species and also should know the type of force that keeps them intact.

Complete step-by-step answer: So in the question, it is asked why the ionic bonds are hard i.e. the compounds formed through ionic bonds will be comparatively harder than the other compounds. For explaining the solution for the question, we should know the type of atoms involved in the bonding.
The type of atoms involved in bonding is a cation and an anion. A metal atom loses its atom and forms a cation i.e. a positively charged species and a nonmetal atom gains its electrons to form an anion i.e. a negatively charged species. A cation loses its electron to obtain its nearest noble gas whereas an anion gains the electrons to complete its octet configuration and obtain the nearest noble gas.
The cation and anion combine to form an ionic bond. The force involved in bonding is the strong electrostatic force of attraction. As the charged species are held strongly by the electrostatic force, the bond formed will be harder. So we could conclude that this electrostatic force is the reason for the hardness.
The oppositely charged ions in the compound are arranged and packed tightly with each other and form a crystalline structure.
An example of an ionic compound is NaCl, which is formed by combining $N{{a}^{+}}$and $C{{l}^{-}}$ions, and the ions are held by the electrostatic force of attraction.
$Na\to N{{a}^{+}}+{{e}^{-}}$
$Cl+{{e}^{-}}\to C{{l}^{-}}$
$Na+C{{l}^{-}}\to NaCl$

Note: The compounds formed by ionic bonding will be hard and brittle in nature and it will have high melting and boiling point. Since a very large amount of energy is required for breaking the ionic bonds. The ionic compounds will behave as non-conductors of electricity in the solid-state but they will behave as conductors of electricity when it is in molten state or solution state.