Explain the formation of H2 molecule on the basis of valence bond theory.
When hydrogen atoms combine to form molecules of hydrogen, 433 KJ/mol. energy is released.
Whenever atoms combine to form a molecule, there is always decrease in the energy. Actually, when two H atoms are far separated, they do not have any force of interaction (attractive or repulsive). As they tend to come closer, two different forces operates:
(a) The nucleus of one H is attracted towards the electrons of the other H atoms and Vice-versa. The energy is released in attraction.
(b) The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion.
Now in case, if the magnitude of the attractive forces is more than that of the repulsive forces, a stable molecule will be formed.
However, if the repulsive forces are more than the attractive forces, then the atoms will not combine.
If for example HA & HB are two hydrogen atoms & eA & eB are their respective electrons, then attractive & repulsive forces may be shown as follows:
------------ old attractive forces
_________ new attractive forces
The number of new attractive and repulsive forces is same, but the magnitude of attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy decreases, hence a stable hydrogen molecule is formed.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 33: Explain the formation of H2 molecule on the basis of valence bond theory.....
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