Question

Give in brief ${\mathrm{O}}_2$is paramagnetic and N₂ is diamagnetic.

Answer 1

${\mathbf{N}}_{\mathbf{2}}$ is diamagnetic:-

• The total number of electrons present in the Nitrogen (${\mathrm{N}}_2$) molecule is 14, 7 from each Nitrogen-atom.
• The electronic configuration of the Nitrogen (${\mathrm{N}}_2$) molecule will be:

$\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

• As we can see that all the electrons are paired therefore Nitrogen (${\mathrm{N}}_2$ is diamagnetic in nature).

${\mathbf{O}}_{\mathbf{2}}$is paramagnetic:-

• The total number of electrons present in the Oxygen (${\mathrm{O}}_2$) molecule is 16, 8 from each Oxygen atom.
• The electronic configuration of the Oxygen (${\mathrm{O}}_2$) molecule will be:

$\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2{\mathrm{\pi }}^{\ast }2{\mathrm{p}}_{\mathrm{x}}^1\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1$

• The presence of two unpaired electrons in the ${\mathrm{\pi }}^{*}\left(2{\mathrm{p}}_{\mathrm{x}}\right)$ and ${\mathrm{\pi }}^{*}\left(2{\mathrm{p}}_{\mathrm{y}}\right)$ molecular orbitals results in the paramagnetic nature of the Oxygen molecule.

Therefore, ${\mathrm{O}}_2$ is paramagnetic and ${\mathrm{N}}_2$ is diamagnetic in nature.