NCERT Class XI Chemistry Solutions

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Question : 33
Total: 40
Explain the formation of H2 molecule on the basis of Valence Bond Theory.
Solution:  
We can consider two hydrogen atoms A and B approaching each other having nuclei NA and NB and electrons present in them are represented by eA and eB. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms begin to approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between :
(i) nucleus of one atom and its own electron that is NAeA and NBeB.
(ii) nucleus of one atom and electron of other atom i.e., NAeB, NBeA.
Similarly repulsive forces arise between :
(i) electrons of two atoms like eAeB.
(ii) nuclei of two atoms NANB.

Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
Experimentally it has been found that the magnitude of new attractive forces is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm.
Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than isolated hydrogen atoms. The energy so released is called bond enthalpy, which corresponds to the minimum in the curve depicted in fig. Conversely, 435.8 kJ of energy is required to dissociate one mole of H2 molecule.
H2(g) + 435.8 kJmol1H(g)+H(g)
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