Question

Explain the formation of H₂ molecule on the basis of valence bond theory.

Answer 1

Let us consider the combination between atoms of hydrogen H_A and H_B and eA and eB be their respective electrons.

As they tend to come closer, two different forces operate between the nucleus and the electron of the other and vice versa. The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion. Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy of the system decreases. Thus, a stable molecule of hydrogen is formed.

Answer 2

Let us assume that two hydrogen atoms (A and B) with nuclei (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 interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating.

• Attractive force arises between:

1. Nucleus of one atom and its own electron i.e., N_A – e_A and N_B – e_B.
2. Nucleus of one atom and electron of another atom i.e., N_A – e_B and N_B – e_A.

• Repulsive force arises between:

1. Electrons of two atoms i.e., e_A – e_B.
2. 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 attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.