Among the following, the number of compounds that can react with $P C l_{5}$ to give $P O C l_{3}$ is:
$C_{2} H_{5} O H, S O_{2}, H_{2} O, H_{2} S O_{4}, P_{4} O_{10}$

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Solution

$C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$
$S O_{2} + P C l_{5} \to S O C l_{2} + P O C l_{3}$
$H_{2} O + P C l_{5} \to P O C l_{3} + 2 H C l$
$H_{2} S O_{4} + P C l_{5} \to P O C l_{3} + H S O_{3} C l + H C l$
$P_{4} O_{10} + 6 P C l_{5} \to 10 P O C l_{3}$

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