Question

Among P$_4$, S$_8$ and N$_2$, the elements which undergo disproportionation when heated with NaOH solution

• A. P$_4$, S$_8$ only
• B. N$_2$, S$_8$ only
• C. N$_2$, P$_4$ only
• D. P$_4$, N$_2$, S$_8$

Hint:

Disproportionation reactions involve a single element undergoing both oxidation and reduction — look for such redox-active nonmetals in alkaline conditions.

Answer 1

The Correct Option is A

Phosphorus (P$_4$) and Sulphur (S$_8$) undergo disproportionation when treated with alkali like NaOH.
P$_4$ reacts with hot concentrated NaOH forming PH$_3$ and NaH$_2$PO$_2$, where P is both oxidized and reduced.
Similarly, S$_8$ reacts with NaOH to give Na$_2$S$_2$O$_3$ and Na$_2$S, also indicating disproportionation.
Nitrogen (N$_2$), being very stable due to its triple bond, does not undergo such disproportionation easily under these conditions.