Question

The bond order of $\mathrm{CO}$ molecule on the basis of molecular orbital theory is ?

• A. Zero
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is C

$3$

Explanation for the correct option:

Option (C): $3$

• Bond order is defined as, the number of bonds formed between the atoms in the compound.
• We know that molecular orbital theory (MOT) explains the formation of molecules, which also helps in determining the bond order of the molecule.
• The bond order formula derived from MOT can be given as:

$\mathrm{Bond}\mathrm{order}=\frac{(\mathrm{No}\mathrm{.}\mathrm{of}\mathrm{bonding}\mathrm{electrons}-\mathrm{No}\mathrm{.}\mathrm{of}\mathrm{antibonding}\mathrm{electrons})}{\mathrm{2}}$

• The electronic configuration of $\mathrm{CO}$ (14 electrons) can be given as:
  $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$
  (Where,$\sigma$ and $\pi$ represents the bonding molecular orbitals and ${\sigma }^{\ast }$,${\pi }^{\ast }$ represents the antibonding molecular orbitals )
• The MO diagram of $\mathrm{CO}$ can be given as:

• We can see from the diagram that, $\text{No}\text{. of bonding}{\text{e}}^{-}s=10$ and $\text{No}\text{.}\text{of antibonding}{\text{e}}^{-}s=4$
• Therefore, the Bond order of $\mathrm{CO}$ can be calculated as:

$\mathrm{B}.\mathrm{O}=\frac{10-4}{2}=3$

Hence, the bond order of $\mathrm{CO}$ molecule on the basis of the molecular orbital theory is $3$.