Arrange CCl_4, AlCl_3, PCl_5 and SiCl_4 in th | vedclass

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Question

(A) The ease of hydrolysis depends on the availability of vacant $d$-orbitals and the polarity of the $M-Cl$ bond. $CCl_4$ does not undergo hydrolysis

Vedclass · Accepted Answer

$CCl_4 < SiCl_4 < PCl_5 < AlCl_3$

Vedclass · Answer

(A) The ease of hydrolysis depends on the availability of vacant $d$-orbitals and the polarity of the $M-Cl$ bond. $CCl_4$ does not undergo hydrolysis because carbon lacks vacant $d$-orbitals. $SiCl_4$ undergoes hydrolysis due to the presence of vacant $d$-orbitals on the $Si$ atom. $PCl_5$ is highly reactive and undergoes rapid hydrolysis. $AlCl_3$ is ionic in nature and undergoes hydrolysis readily. The order of ease of hydrolysis is $CCl_4 < SiCl_4 < PCl_5 < AlCl_3$ is incorrect based on standard reactivity; the correct order is $CCl_4$ (no hydrolysis) $< SiCl_4 < PCl_5 < AlCl_3$ (ionic). However,comparing the covalent chlorides,the order is $CCl_4 < SiCl_4 < PCl_5$. Considering the ionic nature of $AlCl_3$,it hydrolyzes very easily. The correct sequence is $CCl_4 < SiCl_4 < PCl_5 < AlCl_3$.

Arrange $CCl_4, AlCl_3, PCl_5$ and $SiCl_4$ in the order of their ease of hydrolysis.

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