Consider the compounds,$BCl_3$ and $CCl_4$. How will they behave with water? Justify.
(N/A) $BCl_3$ is an electron-deficient molecule. The $B$ atom has an incomplete octet and accepts a lone pair of electrons from water,leading to hydrolysis to form boric acid:
$BCl_3 + 3 H_2O \rightarrow H_3BO_3 + 3 HCl$
On the other hand,$CCl_4$ is an electron-precise molecule. The $C$ atom has a complete octet and lacks vacant $d$-orbitals to expand its coordination number. Therefore,it cannot accept electrons from water,and $CCl_4$ does not undergo hydrolysis.
$BCl_3 + 3 H_2O \rightarrow H_3BO_3 + 3 HCl$
On the other hand,$CCl_4$ is an electron-precise molecule. The $C$ atom has a complete octet and lacks vacant $d$-orbitals to expand its coordination number. Therefore,it cannot accept electrons from water,and $CCl_4$ does not undergo hydrolysis.
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