p-Block Elements-I 2 Question 8
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8. The chloride that cannot get hydrolysed is (2019 Main, 11 Jan I)
======= ####8. The chloride that cannot get hydrolysed is (2019 Main, 11 Jan I)
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $\mathrm{SnCl}_{4}$
(b) $\mathrm{CCl}_{4}$
(c) $\mathrm{PbCl}_{4}$
(d) $\mathrm{SiCl}_{4}$
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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 $\mathrm{Pb}$ 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).
