JEE Chemistry 2023 Question with Solution

Given: The options are $\mathrm{Cl_2}$, $\mathrm{I_2}$, $\mathrm{Br_2}$ and $\mathrm{F_2}$.

Find: Which molecule has the highest bond dissociation energy.

Bond dissociation energy is the energy required to break a bond in a molecule.

From the solution, the bond energy order for halogens is:

$$\mathrm{Cl_2 > Br_2 > F_2 > I_2}$$

Although $\mathrm{F_2}$ has a shorter bond length than $\mathrm{Cl_2}$, its bond energy is lower because of lone pair-lone pair repulsions.

Thus $\mathrm{Cl_2}$ has the highest bond dissociation energy.

Therefore, the correct option is A.

Given: Compare bond dissociation energies of halogen molecules.

Find: The molecule with maximum bond strength.

In general, shorter bonds tend to have higher bond dissociation energy. However, this trend is not followed perfectly for $\mathrm{F_2}$ because strong lone pair-lone pair repulsions weaken the $\mathrm{F-F}$ bond.

So the comparative order becomes:

$$\mathrm{Cl_2 > Br_2 > F_2 > I_2}$$

Hence the molecule with highest bond dissociation energy is $\mathrm{Cl_2}$.

Therefore, the correct option is A.