Question

Many molecules with central atoms from Period 3 or higher take part in Lewis acid-base reactions in which the central atom expands its valence shell. SnCl4 reacts with (CH₃)₃N as follows:

(a) Identify the Lewis acid and the Lewis base in the reaction.

(b) Give the nl designation of the sublevel of the central atom in the acid before it accepts the lone pair.

Short answer

(a)$\text{(}{\left.{\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$is the Lewis base and${\text{SnCl}}_4$ is the Lewis acid.

(b)$\text{Sn}$will accept the lone pair on the 5 d orbital.

Step-by-step solution

Step 1(a): Identify the Lewis acid and the Lewis base

A Lewis acid is an electron acceptor and a Lewis base is an electron donor. As seen in the reaction in the book, the nitrogen in the$\text{(}{\left.{\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$have a lone pair of electrons, which can be donated to the${\text{SnCl}}_4$to form an adduct.

Therefore,$\text{(}{\left.{\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ is the Lewis base and${\text{SnCl}}_4$ is the Lewis acid.

Step 2(b): Give the nl designation of the sublevel

Sn has an electron configuration of${\text{[Kr]4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{2}}.$Then,4electrons will fill up theorbital from the4chlorine atoms.

Therefore, the will accept the lone pair on the 5 d orbital.