Question

Which of the following species is not paramagnetic?

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

Answer 1

The correct option is D

$\mathrm{CO}$

Explanation for the correct option:

Option (d):

• According to Molecular orbital theory, if there are unpaired electrons in orbitals then it is said to be paramagnetic in nature.
• If there are no unpaired electrons then it is called diamagnetic.
• There are 14 electrons in $\mathrm{CO}$ molecule.
• The electronic configuration of $\mathrm{CO}$ is ${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2){\left(\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}\right)}^2$
• Since there is no unpaired electron in bonding and anti-bonding orbitals, $\mathrm{CO}$molecule is diamagnetic.

Explanation for the incorrect option:

Option (a):

• There are 16 electrons in ${\mathrm{O}}_2$molecule.
• The electronic configuration of ${\mathrm{O}}_2$ is ${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2{\left(\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}\right)}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2)(\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}^1=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}^1)$
• There are two unpaired electrons in $\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}^1$ and $\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}^1$ molecular orbitals.
• So, this is paramagnetic in nature.

Option (b):

• There are 10 electrons in ${\mathrm{B}}_2$molecule.
• The electronic configuration of ${\mathrm{B}}_2$ is ${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1)$
• There are two unpaired electrons in $\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1$ and $\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1$ molecular orbitals.
• So, this is paramagnetic in nature.

Option (c):

• There are 15 electrons in $\mathrm{NO}$molecule.
• The electronic configuration of $\mathrm{NO}$ is ${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{(\mathrm{\sigma }*1\mathrm{s})}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{(\mathrm{\sigma }*2\mathrm{s})}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2){\left(\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}\right)}^2(\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}^1)$
• There is one unpaired electron in $\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}^1$ molecular orbitals.
• So, this is paramagnetic in nature.

Final answer:

Hence, option (d) is the correct answer.