$\mathrm{P}_{2} \mathrm{H}_{4}$ can be removed from phosphine containing traces of it
- by passing impure $\mathrm{PH}_{3}$ gas through a freezing mixture.
- by passing the impure $\mathrm{PH}_{3}$ gas through HI and then its treatment with $\mathrm{KOH}$ (aq).
- by both (a) and (b).
- by none of these.
Solution
When passed through HI, $\mathrm{PH}_{3}$ is absorbed forming $\mathrm{PH}_{4} \mathrm{I} . \mathrm{PH}_{4} \mathrm{I}$ when treated with $\mathrm{KOH}$
(aq) yields pure phosphine.
$\mathrm{PH}_{3}+\mathrm{HI} \longrightarrow \mathrm{PH}_{4} \mathrm{I}$
$\mathrm{PH}_{4} \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \longrightarrow \mathrm{KI}+\mathrm{H}_{2} \mathrm{O}+\mathrm{PH}_{3} \uparrow$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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