$P_{4} + C l_{2} (e x c e s s) ⟶$

P C l_{3}

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P C l_{5}

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Solution

The correct option is B $P C l_{5}$

When white phosphorus reacts with limited $C l_{2}$ , then $P C l_{3}$ is produced as product.

$P_{4} + 6 C l_{2} ⟶ 4 P C l_{3}$

When white phosphorus reacts with excess $C l_{2}$ then $P C l_{5}$ is produced as product. $P_{4} + 10 C l_{2} ⟶ 4 P C l_{5}$

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