Which of the following is the correct order of bond dissociation enthalpy?

Options

A. `I_(2) lt Br_(2) lt Cl_(2) lt F_(2)`
B. `I_(2) lt Cl_(2) lt Br_(2) lt F_(2)`
C. `I_(2) lt F_(2) lt Br_(2) lt Cl_(2)`
D. `I_(2) lt Br_(2) lt F_(2) lt Cl_(2)`

Answer

3

Explanation

All the Group `17` elements from diatomic molecules. It would be expected that the bond dissociation enthalpy would decreases as the atoms become larger because increased size result in less effective overlap of orbitals .`Cl_(2), Br_(2)` and `I_(2)` show the expected trend but the bond dissociation enthalpy of `F_(2)` does not fit the expected trend.
The bond dissociation enthalpy of `F_(2)` is abnormally low and this is largely responsible for its very high reactivity. Notice that other element in the second row of the peiodic table also have weaker bonds than the element wich follow in their respective groups . For example ,in group `15`, the `N-N` bond in hydrazine si wealer than `P-P` bond ,similarly in Gropu `16` the `O-O` bond in peroxides is weaKer than `S-S` since `F` atoms are small , the `F-F` disatance is also small `(1.48 Å)` amd hence inter electron -electron replusion is apppreciable. The large electron of the two `F` atoms wealen the bond also .The correct order is:
`Cl_(2) gt Br_(2) gt F_(2) gt I_(2)`
Thus `F_(2)` and `I_(2)` have low values of bond dissociation enthalpies.