$\mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} ightarrow \mathrm{A}+3 \mathrm{NaH}_{2}…
- $\mathrm{NH}_{3}$
- $\mathrm{PH}_{3}$
- $\mathrm{H}_{3} \mathrm{PO}_{4}$
- $\mathrm{H}_{3} \mathrm{PO}_{3}$
Solution
$$
\mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} ightarrow \mathrm{PH}_{3}+3 \mathrm{NaH}_{2} \mathrm{PO}_{2}
$$
In this reaction, phosphorus disproportionate into phosphine and sodium hydrogen phosphite. ^
Asked in: JEE-TOPICTESTS-CHEMISTRY
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