Which of the following is/are disproportionated by aq. solution of NaOH :-

Question

TestHub · Accepted Answer

All four species undergo disproportionation in aqueous NaOH.

I₂: Disproportionates to iodide and hypoiodite/iodate depending on conditions

$I_{2} + 2 O H^{-} \to I^{-} + I O^{-} + H_{2} O$

P₄: In alkali water, gives phosphine and hypophosphite

$P_{4} + 3 O H^{-} + 3 H_{2} O \to P H_{3} + 3 H_{2} P O_{2}^{-}$

S₈: In hot NaOH, forms sulfide and thiosulfate (or related sulfur oxyanions)

Disproportionation occurs (S⁰ → S²⁻ and higher oxidation states)

HCHO: Undergoes Cannizzaro reaction (classic disproportionation)

$2 H C H O + O H^{-} \to H CO O^{-} + C H_{3} O H$

So the correct option is: (a), (b), (c), and (d) — all of the above.

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