One mole of $\mathrm{P}_{4}$ reacts with 8 moles of $\mathrm{SOCl}_{2}$ to give 4 moles of A , $x$ mole of $\mathrm{SO}_{2}$ and 2 moles of B. A, B and $x$ respectively are

(A) $\mathrm{PCl}_{3}, \mathrm{S}_{2} \mathrm{Cl}_{2}$ and 4

(B) $\mathrm{POCl}_{3}, \mathrm{S}_{2} \mathrm{Cl}_{2}$ and 4

(C) $\mathrm{PCl}_{3}, \mathrm{S}_{2} \mathrm{Cl}_{2}$ and 2

(D) $\mathrm{POCl}_{3}, \mathrm{S}_{2} \mathrm{Cl}_{2}$ and 2

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · p-Block Elements

One mole of

P_{4}

reacts with 8 moles of

SOCl_{2}

to give 4 moles of A ,

x

mole of

SO_{2}

and 2 moles of B. A, B and

x

respectively are

(A) $PCl_{3}, S_{2} Cl_{2}$ and 4✓ Correct
(B) $POCl_{3}, S_{2} Cl_{2}$ and 4
(C) $PCl_{3}, S_{2} Cl_{2}$ and 2
(D) $POCl_{3}, S_{2} Cl_{2}$ and 2

Answer: (A)

Solution

The balanced reaction is:

P_{4} + 8 SOCl_{2} \to 4 PCl_{3} + 2 S_{2} Cl_{2} + 4 SO_{2}

From the stoichiometry, A =

PCl_{3}

, B =

S_{2} Cl_{2}

, and

x = 4

moles of

SO_{2}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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