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

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

Vedclass · Accepted Answer

$CCl_4 < SiCl_4 < PCl_5 < AlCl_3$

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(A) The ease of hydrolysis depends on the availability of empty $d$-orbitals and the polarity of the $M-Cl$ bond.$CCl_4$ does not undergo hydrolysis because carbon lacks $d$-orbitals.$SiCl_4$ undergoes hydrolysis due to the presence of empty $d$-orbitals.$PCl_5$ is highly reactive towards water due to the high positive charge on $P$ and empty $d$-orbitals.$AlCl_3$ is ionic in nature and undergoes rapid hydrolysis.Thus,the increasing order of ease of hydrolysis is $CCl_4 < SiCl_4 < PCl_5 < AlCl_3$.

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

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