Question

Construct the molecular orbital diagram for H2^- and then identify the bond order.

Bond order:

• 0

• 0.5

• 1

• 1.5

• 2

Answer 1

Concepts and reason

The concept used to solve this problem is based on molecular orbital diagram.

A molecular orbital diagram is used to explain chemical bonding in a molecule. This diagram is based upon the molecular orbital theory.

Fundamentals

According to the molecular orbital theory, molecular orbitals of a molecule is formed by combination of its atomic orbitals and electrons are distributed among the molecular orbitals.

The bond order of any molecule is calculated by using electrons which are distributed in molecular orbital diagram.

${B_O}{\rm{ = }}\frac{1}{2}\left( {{n_{\rm{b}}} - {n_{\rm{a}}}} \right)$ …… (1)

Here, ${B_O}$ is bond order, ${n_{\rm{b}}}$ is total number of electrons present in bonding orbital and ${n_{\rm{a}}}$ is total number of electrons present in antibonding orbital.

(1)

Construct molecular orbital diagram and fill $3$ electrons in bonding and antibonding $1{\rm{s}}$ orbitals.

(2)

Total number of electrons in bonding orbitals is $2$ .

Total number of electrons in antibonding orbital is $1$ .

The formula for calculating bond order is as follow.

${B_O}{\rm{ = }}\frac{1}{2}\left( {{n_{\rm{b}}} - {n_{\rm{a}}}} \right)$

Substitute $2$ for ${n_{\rm{b}}}$ and $1$ for ${n_{\rm{a}}}$ in the equation (1) to calculate the bond order.

$\begin{array}{c}\\{B_O} = \frac{1}{2}\left( {2 - 1} \right)\\\\ = 0.5\\\end{array}$

Bond order for ${{\rm{H}}_2}^ -$ is $0.5$ .

Ans: Part 1

The molecular orbital diagram of ${{\rm{H}}_2}^ -$ is shown as follows:

Part 2

The bond order for ${{\rm{H}}_2}^ -$ is $0.5$ .