Question

Which of the following molecules is paramagnetic?

• A. ${\mathrm{O}}_3$
• B. ${\mathrm{N}}_2$
• C. $\mathrm{CO}$
• D. $\mathrm{NO}$

Answer 1

The correct option is D

$\mathrm{NO}$

Explanation of the correct option:

D. $\mathrm{NO}$

• A compound is diamagnetic if all its electrons are paired and paramagnetic if any of its electrons are unpaired.
• Nitric oxide$\left(\mathrm{NO}\right)$ contains an unpaired electron(odd electron), because of which $\mathrm{NO}$ is paramagnetic.
• The electronic configuration of $\mathrm{NO}$ according to MOT becomes

${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2{\left(\mathrm{\pi }2\mathrm{px}\right)}^2{\left(\mathrm{\pi }2\mathrm{py}\right)}^2{\left(\mathrm{\sigma }2\mathrm{pz}\right)}^2{(\mathrm{\pi }*2\mathrm{px})}^1$

$$\begin{gathered} \mathrm{Bond}\mathrm{order}=\mathrm{Number}\mathrm{of}\mathrm{bonding}\mathrm{electrons}-\mathrm{Number}\mathrm{of}\mathrm{anti}-\mathrm{bonding}\mathrm{electrons} \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times [10–5] \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times 5 \\ \mathrm{Bond}\mathrm{order}=2.5 \end{gathered}$$

The bond order of $\mathrm{NO}$ is $2.5$ and it has one unpaired electron. Thus, it is paramagnetic in nature.

Explanation of incorrect options:

(A) In ozone $\left({\mathrm{O}}_3\right)$ all electrons are paired, so ${\mathrm{O}}_3$ is diamagnetic.

$$\begin{gathered} \mathrm{Bond}\mathrm{order}=\frac{\mathrm{Number}\mathrm{of}\mathrm{shared}\mathrm{pair}\mathrm{in}\mathrm{all}\mathrm{X}-\mathrm{Y}\mathrm{bonds}}{\mathrm{Number}\mathrm{of}\mathrm{X}-\mathrm{Y}\mathrm{links}\mathrm{in}\mathrm{the}\mathrm{molecule}\mathrm{or}\mathrm{ion}} \\ \mathrm{Bond}\mathrm{order}=\frac{\mathrm{Number}\mathrm{of}\mathrm{bonds}}{\mathrm{Number}\mathrm{of}\mathrm{structure}} \\ \mathrm{Bond}\mathrm{order}=\frac{3}{2} \\ \mathrm{Bond}\mathrm{order}=1.5 \end{gathered}$$

(B) In Nitrogen $\left({\mathrm{N}}_2\right)$ there are no unpaired electrons, so ${\mathrm{N}}_2$ is called diamagnetic.

• The electronic configuration of ${\mathrm{N}}_2$ according to MOT is:

${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2{\left(\mathrm{\pi }2\mathrm{px}\right)}^2{\left(\mathrm{\pi }2\mathrm{py}\right)}^2{\left(\mathrm{\sigma }2\mathrm{pz}\right)}^2$

$$\begin{gathered} \mathrm{Bond}\mathrm{order}=\mathrm{Number}\mathrm{of}\mathrm{bonding}\mathrm{electrons}-\mathrm{Number}\mathrm{of}\mathrm{anti}-\mathrm{bonding}\mathrm{electrons} \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times [10–4] \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times 6 \\ \mathrm{Bond}\mathrm{order}=3 \end{gathered}$$

(C) All of the electrons are paired in carbon monoxide $\left(\mathrm{CO}\right)$, so $\left(\mathrm{CO}\right)$ is known as diamagnetic.

• The electronic configuration of $\mathrm{CO}$ according to MOT is:

${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2{\left(\mathrm{\pi }2\mathrm{px}\right)}^2{\left(\mathrm{\pi }2\mathrm{py}\right)}^2{\left(\mathrm{\sigma }2\mathrm{pz}\right)}^2$

$$\begin{gathered} \mathrm{Bond}\mathrm{order}=\mathrm{Number}\mathrm{of}\mathrm{bonding}\mathrm{electrons}-\mathrm{Number}\mathrm{of}\mathrm{anti}-\mathrm{bonding}\mathrm{electrons} \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times [10–4] \\ \mathrm{Bond}\mathrm{order}=\frac{1}{2}\times 6 \\ \mathrm{Bond}\mathrm{order}=3 \end{gathered}$$

Hence, the correct option D i.e. $\mathrm{NO}$ is paramagnetic.