Question

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

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

Answer 1

The correct option is D 1 and diamagnetic

For molecules lighter than $O_2$ , the in creasing order of energies of molecular orbitals is:
$\sigma 1s{\sigma }^{\ast }1s\sigma 2s{\sigma }^{\ast }2s\pi 2p_y=\pi 2p_z\sigma 2p_x{\sigma }^{\ast }2p_x\pi 2p_y^{\ast }\pi 2p_z^{\ast }$
where, $\pi 2p_y$ and $\pi 2p_z$ 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:
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^2=\pi 2p_z$
or
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y=\pi 2p_z^2$
No unpaired electron-diamagnetic.
$\text{Bond Order}=\frac{\text{Bonding- electrons- Antibonding Electrons}}{2}$
$=\frac{6-4}{2}=1$