Question

Which of the following compound on hydrolysis form white turbidity?

• A. $NC{l}_3$
• B. $PC{l}_3$
• C. $AsC{l}_3$
• D. $SbC{l}_3$

Answer 1

The correct option is D $SbC{l}_3$

$\Rightarrow SbC{l}_3$ forms white turbidity with water.

It is due to the formation of $SbOCl$

$SbC{l}_3+H_{2}O⟶\underset{\text{Turbidity}}{2SbOCl}+2HCl$

$\Rightarrow$ In case of $NC{l}_3$, because of its nonpolar nature it acts differently with water.

$\Rightarrow PC{l}_3$ reacts with water to give Phosphorus acid

$PC{l}_3+3H_{2}O⟶H_{3}P{O}_3+3HCl$

$\Rightarrow$ Arsenic chloride react with water to produce meta Arsenenic acid and $HCl$.

$AsC{l}_3+H_{2}O\to HAs{O}_2+HCl$

Hence, the correct option is $\text{D}$