Question

Which of the following has the highest bond order?

• A. ${\mathrm{H}}_2$
• B. ${\mathrm{O}}_2$
• C. ${\mathrm{N}}_2$
• D. ${\mathrm{He}}_2$

Answer 1

The correct option is C

${\mathrm{N}}_2$

Explanation for the correct option

(C) ${\mathbf{N}}_{\mathbf{2}}$

Bond order

• The bond order represents the number of chemical bonds between two atoms.
• The bond order describes the stability of the bond.
• The formula to calculate bond order is as follows:

$\mathrm{Bond}\mathrm{order}=\frac{1}{2}\left(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbital}-\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbital}\right)$

Calculate bond order of ${\mathbf{N}}_{\mathbf{2}}$ molecule

• Number of electrons of ${\mathrm{N}}_2$ (Nitrogen gas) is $14$.
• Molecular orbital configuration is as follows:

${\mathrm{N}}_2={\left(\sigma 1s\right)}^2{\left({\sigma }^{*}1s\right)}^2{\left(\sigma 2s\right)}^2{\left({\sigma }^{*}2s\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left(\sigma 2p_z\right)}^2$

The bond order of ${\mathrm{N}}_2$ is as follows:

$\begin{array}{rcl}\mathrm{Bond}\mathrm{order}& =& \frac{1}{2}\left(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbital}-\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbital}\right)\\ & =& \frac{1}{2}\left(10-4\right)\\ & =& \frac{1}{2}\times 6\\ & =& 3\end{array}$

Explanation for incorrect options

(A) ${\mathbf{H}}_{\mathbf{2}}$

Calculate bond order of ${\mathbf{H}}_{\mathbf{2}}$ molecule

• Number of electrons of ${\mathrm{H}}_2$ (Hydrogen gas) is $2$.
• Molecular orbital configuration is as follows:

${\mathrm{H}}_2={\left(\sigma 1s\right)}^2$

The bond order of ${\mathrm{H}}_2$ is as follows:

$\begin{array}{rcl}\mathrm{Bond}\mathrm{order}& =& \frac{1}{2}\left(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbital}-\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbital}\right)\\ & =& \frac{1}{2}\left(2-0\right)\\ & =& \frac{1}{2}\times 2\\ & =& 1\end{array}$

(B) ${\mathbf{O}}_{\mathbf{2}}$

Calculate bond order of ${\mathbf{O}}_{\mathbf{2}}$ molecule

• Number of electrons of ${\mathrm{O}}_2$ (Oxygen gas) is $16$.
• Molecular orbital configuration is as follows:

${\mathrm{O}}_2={\left(\sigma 1s\right)}^2{\left({\sigma }^{*}1s\right)}^2{\left(\sigma 2s\right)}^2{\left({\sigma }^{*}2s\right)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left({\pi }^{\mathit{*}}2p_x\right)}^1{\left({\pi }^{\mathit{*}}2p_y\right)}^1$

The bond order of ${\mathrm{O}}_2$ is as follows:

$\begin{array}{rcl}\mathrm{Bond}\mathrm{order}& =& \frac{1}{2}\left(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbital}-\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbital}\right)\\ & =& \frac{1}{2}\left(10-6\right)\\ & =& \frac{1}{2}\times 4\\ & =& 2\end{array}$

(D) ${\mathbf{He}}_{\mathbf{2}}$

Calculate bond order of ${\mathbf{He}}_{\mathbf{2}}$ molecule

• Number of electrons of ${\mathrm{He}}_2$ is $4$.
• Molecular orbital configuration is as follows:

${\mathrm{He}}_2={\left(\sigma 1s\right)}^2{\left({\sigma }^{*}1s\right)}^2$

The bond order of ${\mathrm{He}}_2$ is as follows:

$\begin{array}{rcl}\mathrm{Bond}\mathrm{order}& =& \frac{1}{2}\left(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbital}-\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbital}\right)\\ & =& \frac{1}{2}\left(2-2\right)\\ & =& \frac{1}{2}\times 0\\ & =& 0\end{array}$

Therefore, the correct option is (C) ${\mathbf{N}}_{\mathbf{2}}$.