White phosphorus, P4 (12.4 g), is…

White phosphorus, P4 (12.4 g), is boiled with excess aqueous NaOH in an inert atmosphere, which gives phosphine and sodium hypophosphite. The solid…

Single correct option

White phosphorus,

\mathrm{P_4}

(

12.4\ \mathrm{g}

), is boiled with excess aqueous

\mathrm{NaOH}

in an inert atmosphere, which gives phosphine and sodium hypophosphite. The solid hypophosphite is then recovered and heated strongly in the absence of air so that it decomposes completely to disodium hydrogenphosphate and phosphine. Taking

1\ \mathrm{mol}

of gas as

22.4\ \mathrm{L}

at STP and

\mathrm{P=31}

, the total volume of phosphine obtained in the two steps at STP is

A. $2.24\ \mathrm{L}$
B. $3.36\ \mathrm{L}$
C. $5.60\ \mathrm{L}$
D. $8.96\ \mathrm{L}$

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Answer: (C) $5.60\ \mathrm{L}$

Step 1 is the disproportionation

\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}

. Moles of

\mathrm{P_4}=12.4/124=0.100

, so

0.100\ \mathrm{mol\ PH_3}

and

0.300\ \mathrm{mol\ NaH_2PO_2}

are formed. Step 2 is the thermal disproportionation of hypophosphite,

\mathrm{2NaH_2PO_2\to Na_2HPO_4+PH_3}

, which balances: both sides carry

2\mathrm{Na}

2\mathrm{P}

4\mathrm{H}

and

4\mathrm{O}

. The

0.300\ \mathrm{mol}

of hypophosphite gives

0.150\ \mathrm{mol\ PH_3}

together with

0.150\ \mathrm{mol\ Na_2HPO_4}

. The total phosphine is

0.100+0.150=0.250\ \mathrm{mol}

, which occupies

0.250\times22.4=5.60\ \mathrm{L}

. Option (A) counts only the first step, (B) counts only the second step (

0.150\ \mathrm{mol}

), and (D) corresponds to

0.400\ \mathrm{mol}

, obtained by taking one phosphine per hypophosphite.

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Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake