Question

$P_4\left(excess\right)+C{l}_2\to X$
$P_4+C{l}_2\left(excess\right)\to Y$
Then the correct statement is:

• A. Hydrolysis of $X$ and $Y$ forms oxyacids of phosphorous having equal reducing nature
• B. Hydrolysis of $X$and $Y$ forms oxyacids of phosphorus contains equal number of $\pi$ bonds
• C. Hydrolysis of $X$ and $Y$ form oxyacids of phosphorous which have equal basicity
• D. Hydrolysis of $X$ and $Y$ form oxyacids of phosphorous which have equal oxidation state of phosphorus atoms.

Answer 1

The correct option is A Hydrolysis of $X$ and $Y$ forms oxyacids of phosphorous having equal reducing nature

Hydrolysis of $X$ and $Y$ forms oxyaids of Phosphorous contains equal number of $\pi$ bonds

$P_4\left(excess\right)+C{l}_2\to PC{l}_3\left(X\right)$

$P_4+C{l}_2\left(excess\right)\to PC{l}_5\left(Y\right)$

$PC{l}_3+H_{2}O\to H_{3}P{O}_3+HCl\left(Notbalanced\right)$

$PC{l}_5+H_{2}O\to H_{3}P{O}_4+HCl$

$H-\stackrel{O}{\stackrel{||}{\underset{OH}{\underset{|}{P}}}}-OH$ $HO-\stackrel{O}{\stackrel{||}{\underset{OH}{\underset{|}{O}}}}-OH$

Both contains $1\pi -bond$