In presence of small amount of water, $P C l_{5}$ hydrolyzes to form :

P C l_{3}

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POCl

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P O C l_{3}

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H_{3} P O_{3}

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Solution

The correct option is B $P O C l_{3}$
In limited amount of water,
$P C l_{5} (s) + H_{2} O (l) \to P O C l_{3} (l) + 2 H C l (a q)$
In large amount of water,
$P C l_{5} (s) + 4 H_{2} O (l) \to H_{3} P O_{4} + 5 H C l$
Hence, the correct option is (C).

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Similar questions

Q. In the presence of a small amount of water,

P C l_{5}

hydrolyses to form:

Q. The product formed by complete hydrolysis of

P C l_{3}

are: -

From the following, no. of compounds which do not hydrolyze
$N F_{3}, N C l_{3}, P C l_{3}, P C l_{5}, C C l_{4}, S i C l_{4}, B i C l_{3}, N a C l, F e C l_{3}$ ___

Q. The reaction of

C l_{2}

gas with cold dilute and hot concentrated NaOH in water give sodium salt of two (different) oxoacids of chlorine P and Q respectively. The

C l_{2}

gas reacts with

S O_{2}

gas, in presence of charcoal to give a product R. R reacts with white phosphorous to give a compound S. On hydrolysis, S gives an oxoacid of phosphorous T.
R, S and T, respectively are:

Among the following, the number of compounds which do not hydrolyze

$N F_{3}, N C l_{3}, P C l_{3}, P C l_{5}, C C l_{4}, S i C l_{4}, B i C l_{3}, N a C l, F e C l_{3}$

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In presence of small amount of water, PCl5 hydrolyzes to form :

The correct option is B POCl3In limited amount of water,PCl5(s)+H2O(l)→POCl3(l)+2HCl(aq)In large amount of water,PCl5(s)+4H2O(l)→H3PO4+5HClHence, the correct option is (C).

In presence of small amount of water, $P C l_{5}$ hydrolyzes to form :

The correct option is B $P O C l_{3}$
In limited amount of water,
$P C l_{5} (s) + H_{2} O (l) \to P O C l_{3} (l) + 2 H C l (a q)$
In large amount of water,
$P C l_{5} (s) + 4 H_{2} O (l) \to H_{3} P O_{4} + 5 H C l$
Hence, the correct option is (C).