Question

Among the following, the number of halide(s) which is/are inert to hydrolysis is ______

• A. ${\mathrm{BF}}_3$
• B. ${\mathrm{SiCl}}_4$
• C. ${\mathrm{PCl}}_5$
• D. ${\mathrm{SF}}_6$

Answer 1

The correct option is D

${\mathrm{SF}}_6$

Explanation to correct option:

(D) ${\mathbf{SF}}_{\mathbf{6}}$

${\mathrm{SF}}_6$ is inert to hydrolysis because of the presence of six fluoride atoms around sulphur as the six fluoride atoms creates appreciable steric hinderance. So the attack by lone pair of oxygen atom of water on ${\mathrm{SF}}_6$ is not possible.

Therefore, ${\mathrm{SF}}_6$ is inert to hydrolysis.

Explanation to incorrect options:

(A) ${\mathbf{BF}}_{\mathbf{3}}$

1. In ${\mathrm{BF}}_3$, only three fluoride atoms are present around boron.
2. Hence three fluoride atoms doesn't creates steric hinderance around boron atom
3. So ${\mathrm{BF}}_3$ is not inert to hydrolysis.

(B) ${\mathbf{SiCl}}_{\mathbf{4}}$

1. In ${\mathrm{SiCl}}_4$, four fluoride atoms doesn't creates enough steric hinderance around large silicon atom
2. So ${\mathrm{SiCl}}_4$ is not inert to hydrolysis.

(C) ${\mathbf{PCl}}_{\mathbf{5}}$

1. In ${\mathrm{PCl}}_5$, five fluoride atoms doesn't creates enough steric hinderance around large phosphorous atom
2. So ${\mathrm{PCl}}_5$ is not inert to hydrolysis.

Hence, option D is correct, ${\mathrm{SF}}_6$ is inert to hydrolysis.