Chapter 2, Problem 2.57E

Interpretation Introduction

Interpretation:

The hybridization, bond angle and number of $\sigma$- bonds and $\text{π}$- bonds of ${\text{CH}}_3\text{CN}$ has to be determined.

Concept Introduction:

Hybridization is calculated by steric number given as follows:

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{number of atoms bonded to the central atom}\right)+\\ \left(\text{number of lone pairs on the central atom}\right)\end{array}\right]$

The table that relates the steric number with hybridization is as follows:

$\begin{array}{cc}\text{Steric Number}& \text{Type of Hybridisation}\\ 2& \text{sp}\\ 3& {\text{sp}}^2\\ 4& {\text{sp}}^3\\ 5& {\text{sp}}^3\text{d}\\ 6& {\text{sp}}^3{\text{d}}^2\end{array}$

The table that relates the steric number with geometry and bond angles is as follows:

$\begin{array}{cccc}\text{Steric number}& \text{Number of lone pairs}& \text{Molecular geometry}& \text{Bond angles}\\ 2& 0& \text{Linear}& 180°\\ 3& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}\text{Trigonal planar}\\ \text{Bent}\end{array}& 120°\\ 4& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Tetrahedral}\\ \text{Trigonal pyramidal}\\ \text{Bent}\end{array}& 109.5°\\ 5& \begin{array}{c}0\\ 1\\ 2\\ 3\end{array}& \begin{array}{c}\text{Trigonal Bi-pyramidal}\\ \text{See-Saw}\\ \text{T-shaped}\\ \text{Linear}\end{array}& 90°\text{,120 }°,180°\\ 6& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Octahedral}\\ \text{Square pyramidal}\\ \text{Square planar}\end{array}& 90°\text{,}180°\end{array}$