Chapter 2, Problem 2.64E

Interpretation Introduction

Interpretation:

The hybridization, bond angle and number of $\sigma$- bonds and $\text{π}$- bonds of ${\text{HONH}}_2$ has to be determined.

Concept Introduction:

Valence Bond Theory gives information about the chemical bonding between two atoms in terms of an atomic orbital. In general, according to VBT (Valence Bond Theory) single bond, double bond and triple bond is defined as follows:

• Single bond contains $\sigma$- bond.
• Double bond contains $\sigma$- bond and a $\text{π}$- bond.
• Triple bond contains one $\sigma$- bond and two $\text{π}$-bonds.

Hybridization is the hypothetical concept of mixing of atomic orbital into hybrid orbitals that are of dissimilar shapes, energies and are appropriate for combination of electrons to form bonds in valence bond theory.

Hybridization is calculated by the hybrid orbitals and to calculate hybrid orbitals we need to know the steric number that is given by,

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

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:

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