p-Block Elements-I - Result Question 8-1
8. The chloride that cannot get hydrolysed is
( 2019 Main, 11 Jan I)
(a) $\mathrm{SnCl}_4$
(b) $\mathrm{CCl}_4$
(c) $\mathrm{PbCl}_4$
(d) $\mathrm{SiCl}_4$
Show Answer
Answer:
Correct Answer: 8. (b)
Solution:
- The compounds given are the tetrahalides $\left(\mathrm{MCl}_4\right)$ of group $14$ elements. For the hydrolysis, (nucleophilic substitution) of $M \mathrm{Cl}_4$ the nature of the $M-\mathrm{Cl}$ bond should be as:
Here, $M$ can be $\mathrm{Si}, \mathrm{Sn}$ and $Pb$ because they have vacant $n d$-orbital. But, carbon is a member of second period $(n=2$, $l=0,1$ ), it does not have $d$-orbital $(l=2)$. So, $\mathrm{CCl}_4$ will not be hydrolysed and correct option is (b).
BETA
