Question

Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule $B_2$ is

• A. 1 and diamagnetic
• B. 0 and diamagnetic
• C. 1 and paramagnetic
• D. 0 and paramagnetic

Answer 1

The Correct Option is A

For molecules lighter than $O_2$, the increasing order of
energies of molecular orbitals is
$\sigma1s _{\sigma}^{*} 1s \sigma2s _{\sigma}^{*}2s$$\bigg[\begin{array}
\ \pi2p_y \\
\pi2p_z \\
\end{array}\bigg]$ $\sigma2p_x {\sigma}^{*}2p_x$ $\bigg[\begin{array}
\ {\pi}^{*}2p_y \\
{\pi}^{*}2p_z \\
\end{array}$ ......
where, $\pi2p_y$ and $\pi2p_z$ are degenerate molecular orbitals, first singly occupied and then pairing starts if Hund's rule is obeyed. If Hund's rule is violated in $B_2$, electronic arrangement would be
$\sigma1 s^2\ _{\sigma}^{*} 1s^2 \sigma2s^2 \ _{\sigma}^{*}2s^2 \bigg[\begin{array}
\ \pi2p_y^2 \\
\pi2p_z \\
\end{array}$.....
No unpaired electron-diamagnetic.
Bond order=$\frac{bonding electrons - antibonding electrons}{2}$
$=\frac {6-4}{2}=1$