Question

Two $\pi$ bonds and half $\sigma$ bonds are present in:

• A. ${\mathrm{N}}_2^{+}$
• B. ${\mathrm{N}}_2$
• C. ${\mathrm{O}}_2^{+}$
• D. ${\mathrm{O}}_2$

Answer 1

The correct option is A

${\mathrm{N}}_2^{+}$

Explanation for correct option

(A) ${\mathbf{N}}_{\mathbf{2}}^{\mathbf{+}}$

Bond order

• The number of covalent bonds shared by two atoms is known as bond order.
• Bond order indicates the stability of a bond.
• The formula to calculate bond order is as follows:

$\mathrm{BO}=\frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)$

• $\mathrm{BO}$ is the bond order.
• ${\mathrm{N}}_{\mathrm{b}}$ is the number of electrons present in the bonding orbital.
• ${\mathrm{N}}_{\mathrm{a}}$ is the number of electrons present in the antibonding orbital.

${\mathbf{N}}_{\mathbf{2}}^{\mathbf{+}}$

• Total number of electrons ${\mathrm{N}}_2^{+}=13$
• The electronic configuration of ${\mathrm{N}}_2^{+}$ molecules is as follows:

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

• The bond order is as follows:

$\begin{array}{rcl}\mathrm{BO}& =& \frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)\\ & =& \frac{1}{2}\left(9-4\right)\\ & =& \frac{5}{2}\\ & =& 2.5\end{array}$

• ${\mathrm{N}}_2^{+}$ forms half $\sigma$ and two $\pi$ bonds.

Explanation for incorrect options

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

• Total number of electrons ${\mathrm{N}}_2=14$
• The electronic configuration of ${\mathrm{N}}_2$ molecules is as follows:

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

• The bond order is as follows:

$\begin{array}{rcl}\mathrm{BO}& =& \frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)\\ & =& \frac{1}{2}\left(10-4\right)\\ & =& \frac{6}{2}\\ & =& 3\end{array}$

• ${\mathrm{N}}_2$ forms one $\sigma$ and two $\pi$ bonds.

(C) ${\mathbf{O}}_{\mathbf{2}}^{\mathbf{+}}$

• Total number of electrons ${\mathrm{O}}_2^{+}=15$
• The electronic configuration of ${\mathrm{O}}_2^{+}$ molecules is as follows:

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

• The bond order is as follows:

$\begin{array}{rcl}\mathrm{BO}& =& \frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)\\ & =& \frac{1}{2}\left(10-5\right)\\ & =& \frac{5}{2}\\ & =& 2.5\end{array}$

• ${\mathrm{O}}_2^{+}$ forms one $\sigma$ and half $\pi$ bonds.

(D) ${\mathrm{O}}_2$

• Total number of electrons ${\mathrm{O}}_2=16$
• The electronic configuration of ${\mathrm{O}}_2$ molecules is as follows:

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

• The bond order is as follows:

$\begin{array}{rcl}\mathrm{BO}& =& \frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)\\ & =& \frac{1}{2}\left(10-6\right)\\ & =& \frac{4}{2}\\ & =& 2\end{array}$

• ${\mathrm{O}}_2$ forms one $\sigma$ and one$\pi$ bonds.

Therefore the correct option is (A) ${\mathbf{N}}_{\mathbf{2}}^{\mathbf{+}}$.