Meaning of disproportionation

In a disproportionation reaction, the same species undergoes both:

• oxidation, and
• reduction

So the element must be in an intermediate oxidation state, such that it can both increase and decrease its oxidation number.

Oxidation states of halogens in elemental form

In $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2, \mathrm{I}_2$, each halogen is in oxidation state $0$.

For disproportionation, the halogen at $0$ should be able to:

• get reduced to $-1$, and
• get oxidized to a positive oxidation state such as $+1, +3, +5,$ etc.

Check each halogen

(i) $\mathrm{F}_2$

Fluorine is the most electronegative element and shows only oxidation state $-1$ in its compounds. It cannot be oxidized to a positive oxidation state.

Hence, $\mathrm{F}_2$ cannot undergo disproportionation.

(ii) $\mathrm{Cl}_2$

Chlorine can form $-1$ as well as positive oxidation states. Example: $\mathrm{Cl}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H_2O}$ Here chlorine goes from $0$ to:

• $-1$ in $\mathrm{Cl}^-$
• $+1$ in $\mathrm{ClO}^-$

So $\mathrm{Cl}_2$ does disproportionate.

(iii) $\mathrm{Br}_2$

Bromine also shows $-1$ and positive oxidation states. Example: $\mathrm{Br}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{Br}^- + \mathrm{BrO}^- + \mathrm{H_2O}$ Thus $\mathrm{Br}_2$ does disproportionate.

(iv) $\mathrm{I}_2$

Iodine also forms $-1$ and positive oxidation states. Example: $\mathrm{I}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^- + \mathrm{H_2O}$ So $\mathrm{I}_2$ does disproportionate.

Evaluate the options

• A: $\mathrm{F}_2$ and $\mathrm{Cl}_2$ $\rightarrow$ incorrect, because $\mathrm{F}_2$ does not disproportionate.
• B: $\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$ $\rightarrow$ correct.
• C: Only $\mathrm{I}_2$ $\rightarrow$ incorrect.
• D: $\mathrm{F}_2, \mathrm{Cl}_2$ and $\mathrm{Br}_2$ $\rightarrow$ incorrect, because $\mathrm{F}_2$ does not disproportionate.

Final answer

The halogens which can undergo disproportionation are: $\mathrm{Cl}_2, \mathrm{Br}_2, \mathrm{I}_2$

Therefore, the correct option is B.