Question

Complete this molecular orbital diagram for CN^– then determine the bond order. Note that the 1s orbital is not shown in this problem. To add arrows to the MO diagram, click on the blue boxes.

Bond order of CN^-

• 0

• 0.5

• 1

• 1.5

• 2

• 2.5

• 3

Answer 1

Concepts and reason

• The molecular orbital theory explains the bonding in terms of the combination and organization of atomic orbitals of an atom which are associated in the molecule to form molecular orbitals

• The number of molecular orbitals is always equal to the number of atomic orbitals from which they are formed. Two types of orbitals are bonding and anti-bonding orbitals.

• The molecules are said to be stable if the number of electrons in the bonding molecular orbitals is greater than that in anti-bonding molecular orbitals.

• Bond order is defined as one half the differences between the number of electrons present in the bonding and the antibonding orbitals.

Fundamentals

When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital and the other is called an anti-bonding molecular orbital.

$\begin{array}{l}\\{\rm{Antibonding}}\,{\rm{MO: }}\,{\rm{\sigma *2s}}\,\,\,{\rm{\sigma *2}}{{\rm{p}}_{\rm{z}}}\,{\rm{\pi *2}}{{\rm{p}}_{\rm{x}}}\,{\rm{\pi *2}}{{\rm{p}}_{\rm{y}}}\\\\{\rm{Bonding}}\,\,{\rm{MO: }}\,\,{\rm{\sigma 2s}}\,\,\,{\rm{\sigma 2}}{{\rm{p}}_{\rm{z}}}\,{\rm{\pi 2}}{{\rm{p}}_{\rm{x}}}\,{\rm{\pi 2}}{{\rm{p}}_{\rm{y}}}\\\end{array}$

The energies of various molecular orbitals of ${\rm{C}}{{\rm{N}}^ - }$ in ascending order is:

${\rm{\sigma 1s}}\,{\rm{ < }}\,{\rm{\sigma *1s}}\,{\rm{ < \sigma 2s}}\,{\rm{ < \sigma *2s}}\,{\rm{ < }}\left( {{\rm{\pi 2}}{{\rm{p}}_{\rm{x}}}{\rm{ = \pi 2}}{{\rm{p}}_{\rm{y}}}} \right)\,{\rm{ < }}\,{\rm{\sigma 2}}{{\rm{p}}_{\rm{z}}}\,{\rm{ < }}\,\left( {{\rm{\pi *2}}{{\rm{p}}_{\rm{x}}}{\rm{ = \pi *2}}{{\rm{p}}_{\rm{y}}}} \right)\,{\rm{ < }}\,{\rm{\sigma *2}}{{\rm{p}}_{\rm{z}}}$ .

${\rm{Bond}}\,{\rm{order}}\,{\rm{ = }}\,\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{number}}\,{\rm{of}}\,{\rm{bonding}}\,{\rm{electron}}\,{\rm{ - }}\,{\rm{number}}\,{\rm{ofantibonding}}\,{\rm{electron}}} \right)$

Electronic configuration of ${\rm{C}}{{\rm{N}}^{\rm{ - }}}$ is

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}\,{\rm{,}}\,{\rm{\sigma *1}}{{\rm{s}}^{\rm{2}}}\,{\rm{,\sigma 2}}{{\rm{s}}^{\rm{2}}}\,{\rm{,\sigma *2}}{{\rm{s}}^{\rm{2}}}\,{\rm{,}}\left( {{\rm{\pi 2p}}_{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}_{\rm{y}}^{\rm{2}}} \right)\,{\rm{,}}\,{\rm{\sigma 2p}}_{\rm{z}}^{\rm{2}}\,{\rm{,}}\,\left( {{\rm{\pi *2}}{{\rm{p}}_{\rm{x}}}{\rm{ = \pi *2}}{{\rm{p}}_{\rm{y}}}} \right)\,{\rm{,}}\,{\rm{\sigma *2}}{{\rm{p}}_{\rm{z}}}$ .

The molecular orbital diagram of ${\rm{C}}{{\rm{N}}^{\rm{ - }}}$ .

Number of electrons present in the anti-bonding orbitals = 2

Number of electrons present in the bonding orbitals = 8

$\begin{array}{c}\\{\rm{therefore}}\,{\rm{bond}}\,{\rm{order}}\,{\rm{ = }}\,\frac{{{\rm{8 - 2}}}}{{\rm{2}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,{\rm{3}}\\\end{array}$

Ans:

The completed molecular orbital diagram of ${\rm{C}}{{\rm{N}}^{\rm{ - }}}$ .