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

Asked by Pragya Singh | 1 year ago |  69

##### Solution :-

Assuming 2 H- atoms X and Y with nuclei NX and NY and electrons eX and eY, respectively.

When X and Y are far for each other then there is no interaction between them. As soon as they come closer, the attractive force and repulsive force becomes active.

The repulsive forces are:

(i) Between electrons of both the atoms i.e. eX and eY.

(ii) Between nuclei of both the atoms i.e. NX and NY.

The attractive forces are:

(i) Between the electron and nucleus of the same atom i.e. NX – eX and NY -eY.

(ii) Between the electron of one atom and nucleus of other atom i.e. NX–eY and NY–eX.

The repulsive force pushes the 2 atoms apart whereas the attractive force tend to bring them together.

The values of repulsive forces are less than that of attractive forces. Thus, 2 atoms approach each other. Thus, there is a decrease in potential energy. At the end a stage is reached when the repulsive forces balance the attractive forces and the system achieves the minimum energy,which leads to formation of H2 molecule.

Answered by Abhisek | 1 year ago

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