Consider the compounds, `BCl_(3)` and `CCl_(4)`.How will they behave with water? Justify.

Explanation

In `BCl_(3)`,`B` is `sp^(2)` hybridsied and has triangular planar geometry.Due to the presence of vacant `2p` orbital in `B`,it accepts a pair of electron from `O`-centre of the `H_(2)O` molecule and then undergoes hydrolysis to form boric acid,`B(OH)_(3)` and `HCI`. The mechanism is as follows: i.e. `BCl_(3)+3H_(2)Oto B(OH)_(3)+HCl`
However, in `C Cl_(4),C` is `sp^(3)` hybridised and has a tetrahedral geometry.

The octet around `C` is complete, and it has no vacant orbital to accept a pair of electron from water molecule. Hence, `C Cl_(4)` does not undergo hydrolysis.
`C Cl_(4)+H+_(2)Oto` No reaction