What products are expected from the disproportionation reaction of hypochlorous acid?

For any disproportionation reaction, the same element is simultaneously oxidised as well as reduced. Hence we must identify the oxidation state of chlorine in each species and then search for two products in which chlorine appears in a lower and a higher oxidation state than in the reactant.

In hypochlorous acid $$HClO$$, the oxidation state of chlorine is $$+1$$ because
$$+1\;(\text{for }H) + x\;(\text{for }Cl) + (-2)\;(\text{for }O) = 0 \; \Longrightarrow \; x = +1.$$

Therefore, in the disproportionation of $$HClO$$, part of the chlorine must be reduced to an oxidation state lower than $$+1$$, and the remaining chlorine must be oxidised to an oxidation state higher than $$+1$$.

Common reduced species of chlorine (lower than $$+1$$) formed in aqueous medium are $$Cl^{-}$$ in $$HCl$$ ($$-1$$ oxidation state) or $$Cl_2$$ ($$0$$).
Common oxidised species (higher than $$+1$$) are $$ClO_3^{-}$$ in $$HClO_3$$ ($$+5$$) or $$ClO_4^{-}$$ in $$HClO_4$$ ($$+7$$).

A well-known disproportionation equation for hypochlorous acid in water is

$$3\,HClO \;\longrightarrow\; 2\,HCl \;+\; HClO_3$$

• Chlorine in $$HClO$$ (initial $$+1$$) is reduced to $$-1$$ in $$HCl$$.
• Chlorine in $$HClO$$ (initial $$+1$$) is oxidised to $$+5$$ in $$HClO_3$$.
Hence both reduction and oxidation occur, satisfying the definition of disproportionation.

Balancing check:
Cl: $$3 \rightarrow 2+1$$ ✔️  H: $$3 \rightarrow 2+1$$ ✔️  O: $$3 \rightarrow 3$$ ✔️

No other offered pair of products simultaneously places chlorine both below and above the $$+1$$ oxidation state while keeping mass and charge balanced.

Therefore the expected products are $$HCl$$ and $$HClO_3$$.

Answer: Option D which is: $$HCl$$ and $$HClO_3$$