Question

The bond order of CO molecule on the basis of molecular orbital theory is:
(a) Zero
(b) 2
(c) 3
(d) 1

Answer 1

Hint: To calculate the bond order of a molecule we use the formulae,$\text{Bond}\text{order = }\dfrac{\left( \text{No}\text{.}\text{of}\text{bonding}\,\text{electrons-No}\text{.}\text{of}\text{antibonding}\text{electrons} \right)}{\text{2}}$
Write the electronic configuration of CO according to molecular orbital theory and then calculate the bond order.

Complete answer:
In the question we are supposed to calculate the bond order of carbon monoxide molecule (CO) according to molecular orbital theory.
We are very much familiar with the term bond order from the lower classes, first let’s briefly discuss a few points regarding bond order of a molecule.
Bond order is defined as, half the difference between the number of electrons present in Bonding Molecular orbitals (orbital with lower energy) and number of electrons present in Antibonding Molecular Orbitals. The obtained number i.e. the bond order represents the number of bonds formed between the atoms in the compound.
From the above definition we could derive an equation for calculating bond order of a molecule and the equation is as follows:
$\text{Bond}\text{order = }\dfrac{\left( \text{No}\text{.}\text{of}\text{bonding}\,\text{electrons-No}\text{.}\text{of}\text{antibonding}\text{electrons} \right)}{\text{2}}$
In molecular orbital theory we split the hybrid orbitals formed into bonding orbital with lower energy and anti-bonding orbital with higher energy level.
So now we have to calculate the total electrons present in CO and then have to distribute them between these bonding and antibonding orbitals.
The total number of electrons present in CO molecule$=\,{{e}^{-}}s\,in\,carbon+{{e}^{-}}s\,in\,oxygen$
$Total\,{{e}^{-}}s\,in\,CO=6+8=14{{e}^{-}}s$
$\text{Bond}\text{order = }\dfrac{\left( \text{No}\text{.}\text{of}\text{bonding}\,\text{electrons-No}\text{.}\text{of}\text{antibonding}\text{electrons} \right)}{\text{2}}$
The electronic configuration of Carbon Monoxide (according to Molecular Orbital Theory) is
$\sigma 1{{s}^{2}}\,{{\sigma }^{*}}1{{s}^{2}}\,\sigma 2{{s}^{2}}\,{{\sigma }^{*}}2{{s}^{2}}\,\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\,\sigma 2p_{z}^{2}$
Where,$\sigma$ and $\pi$ indicate bonding molecular orbitals and ${{\sigma }^{*}}$ and ${{\pi }^{*}}$indicate antibonding molecular orbitals.
As we can see, $\text{No}\text{. of bonding }{{\text{e}}^{-}}s=10$
$\text{No}\text{.}\,\text{of antibonding }{{\text{e}}^{-}}s=4$
Therefore, $\text{Bond}\,\text{order}\,\text{of}\,\text{CO}\,\text{=}\,\dfrac{10-4}{2}=3$

Therefore, the answer is option (c).

Note:
A bond order equal to zero indicates that no bond exists, i.e. the compound doesn’t exist. The stability of compounds increases with increasing bond order.
We should know the correct order of the orbitals to distribute the electrons present accordingly to calculate the bond order and the molecular orbitals also obey Aufbau principle, Hund’s rule etc.