Question

Among \(P_4,\;S_8\) and \(N_2\), the elements which undergo disproportionation when heated with \(NaOH\) solution are

• 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:

In disproportionation, the same element is simultaneously oxidized and reduced. White phosphorus and sulphur commonly show this behavior with hot alkali solutions.

Answer 1

The Correct Option is A

Step 1: Recall disproportionation.
In a disproportionation reaction the same element is simultaneously oxidized and reduced. We must find which of $P_4$, $S_8$ and $N_2$ disproportionate when heated with $NaOH$ solution.
Step 2: Examine white phosphorus $P_4$.
White phosphorus reacts with hot concentrated sodium hydroxide as \[ P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2 \] Here phosphorus goes to $-3$ in $PH_3$ and to $+1$ in $NaH_2PO_2$, so it is both reduced and oxidized. $P_4$ disproportionates.
Step 3: Examine sulphur $S_8$.
Sulphur reacts with hot sodium hydroxide as \[ S_8 + NaOH \rightarrow Na_2S + Na_2S_2O_3 \] Sulphur is reduced to sulphide and oxidized to thiosulphate, so $S_8$ disproportionates.
Step 4: Examine nitrogen $N_2$.
Dinitrogen has a very strong triple bond and is chemically inert under these conditions. It does not react with $NaOH$ solution, so $N_2$ does not disproportionate.
Step 5: Combine the conclusions.
Both $P_4$ and $S_8$ disproportionate, while $N_2$ does not, so the correct set is $P_4$ and $S_8$ only.
Step 6: State the answer.
The elements that disproportionate with $NaOH$ are $P_4, S_8$ only, matching the key.
\[ \boxed{P_4,\; S_8 \text{ only}} \]