Question

Which of the following is paramagnetic in nature?

• A. $N_2$
• B. $F_2$
• C. $K{O}_2$
• D. None of these

Answer 1

The correct option is C $K{O}_2$

As we know, $K{O}_2$ is paramagnetic, due to presence of one unpaired electron in $O_2^{-}$.

In case of $N_2$ molecule, all electrons are paired. That means half of the electrons spin clockwise and half of the electrons spin anticlockwise. Because of their opposite spins they produce magnetic field in opposite direction thus the resultant magnetism becomes zero.

In case of $F_2$, all the electrons are paired. Thus, it is diamagnetic in nature.

$O_2^{-}B.O.=1.5$ stable paramagnetic

$\begin{array}{ccc}& \underset{–}{}{\sigma }_{2p}^{\ast }& \\ \underset{–}{\uparrow \downarrow }{\pi }_{2p}^{\ast }& & \underset{–}{\uparrow }{\pi }_{2p}^{\ast }\\ \underset{–}{\uparrow \downarrow }{\pi }_{2p}& & \underset{–}{\uparrow \downarrow }{\pi }_{2p}\\ & \underset{–}{\uparrow \downarrow }{\sigma }_{2p}& \\ & \underset{–}{\uparrow \downarrow }{\sigma }_{2s}^{\ast }& \\ & \underset{–}{\uparrow \downarrow }{\sigma }_{2s}& \\ & \underset{–}{\uparrow \downarrow }{\sigma }_{1s}^{\ast }& \\ & \underset{–}{\uparrow \downarrow }{\sigma }_{1s}& \end{array}$