Explain the formation of `H_(2)` molecule on the basis of valence bond theory.

Explanation

Let us assume that two hydrogen atoms (A and B) with nucleil `(N_(A) " and " N_(B))` and electrons `(e_(A)" and "e_(B))` are taken to undergo a reaction to form a hydrogen molecule.
When A and B are at a large distance, there is no interation between them. As they being to approach each other, the attractive and repulsive forces starts operating.
Attractive force arises between:
(a) Nucleus of one atom and its own electron i.e., `N_(A)-e_(A) " and " N_(B)-e_(B)`.
(b) Nucleus of one atom adn electron of another atom i.e., `N_(A)-e_(B) " and " N_(B)-e_(A)`.
Repulsive force arises between :
(a) Electrons of two atoms i.e., `e_(A)-e_(B)`.
(b) Nuclei of two atoms i.e. , `N_(A)-N_(B)`.
The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attactive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.