Question
On reaction with $$C{l_2},$$ phosphorus forms two types of halides $$'A'$$ and $$'B'.$$ Halide $$'A'$$ is yellowish-white powder but halide $$'B'$$ is colourless oily liquid. What would be the hydrolysis products of $$'A'$$ and $$'B'$$ respectively?
A.
$${H_3}P{O_4},{H_3}P{O_3}$$
B.
$$HOP{O_3},{H_2}P{O_2}$$
C.
$${H_3}P{O_3},{H_3}P{O_4}$$
D.
$$HP{O_3},{H_3}P{O_3}$$
Answer :
$${H_3}P{O_4},{H_3}P{O_3}$$
Solution :
$$A$$ is $$PC{l_5}$$ ( yellowish white powder ).
$${P_4} + 10C{l_2} \to 4PC{l_5}$$
$$B$$ is $$PC{l_3}$$ ( colourless oily liquid ).
$${P_4} + 6C{l_2} \to 4PC{l_3}$$
The products of hydrolysis of $$PC{l_3}$$ and $$PC{l_5}$$ are :
$$\eqalign{
& PC{l_3} + 3{H_2}O \to \mathop {{H_3}P{O_3}}\limits_{\left( B \right)} + 3HCl \cr
& PC{l_5} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{\left( A \right)} + 5HCl \cr} $$