Arrange the following  H–H, D–D and F–F in order of increasing bond dissociation enthalpy.

Asked by Abhisek | 1 year ago |  63

1 Answer

Solution :-

The bond pair in D–D bond is more strongly attracted by the nucleus than the bond pair in H–H bond. This is because of the higher nuclear mass of D2. The stronger the attraction the greater will be the bond strength and the higher is the bond dissociation enthalpy.

Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the repulsive and attractive forces present in a molecule.

Hence, the bond dissociation enthalpy of D–D is higher than H–H. However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre.

Therefore, the increasing order of bond dissociation enthalpy is as follows:

F–F < H–H < D–D

Answered by Pragya Singh | 1 year ago

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