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 1 and diamagnetic

As Hund’s rule is violated, molecular orbital configuration of $B_2$ would be :

${\sigma }_{1s^2}$ $>$ ${\sigma }_{1s^2}^{\ast }$ $>$ ${\sigma }_{2s^2}$ $>$ ${\sigma }_{2s^2}^{\ast }$ $>$ ${\pi }_{2p_y{}^2}$ = ${\pi }_{2p_x{}^0}$

Since all the electrons are paired, hence $B_2$ is diamagnetic.

$\text{Bond order of}{B}_2=\frac{N_b-N_a}{2}$
$\therefore \text{Bond order of}{B}_2=\frac{6-4}{2}=1$

Hence, option (a) is correct.