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P₂H₄ can be removed from phosphine containing traces of it

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bypassing impure PH₃ gas through a freezing mixture

bypassing the impure PH₃ gas through HI and then its treatment with KOH (aq)

by both (a) and (b)

by none of these

answer is C.

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Detailed Solution

When the PH₃ mixture containing P₂H₄ is passed through the freezing mixture, then pure PH₃ is obtained as P₂H₄ gets condensed.

PH₄I forms when PH₃ is passes HI. PH₄I is later treated with KOH (aq), to obtain free PH₃, which is free from P₂H₄.
$\begin{array}{l} {PH}_{3} + HI ⟶ {PH}_{4} I \\ {PH}_{4} I + KOH (aq) ⟶ KI + H_{2} O + {PH}_{3} ↑ \end{array}$

This concludes that both given methods are used to obtain pure PH₃ (free from P₂H₄).

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