Q151CP - Chemistry: Molecular Nature Of Matter And Change

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.

Step-by-Step Solution

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 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, 4 electrons will fill up the  orbital from the 4 chlorine atoms.

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