Question

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

Assuming that Hund's rule is violated, the bond order and magnitude of diatomic molecule of $B_2$ will be $1$ and diamagnetic.$B_2$ has 10 electrons. Its M.O diagram is:

$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^1\pi 2p_y^1$
If Hund's rule is not beyond than M.O diagram will be
$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2$

Bond order $=1/2[6-4]=1$ and as there are no lone electrons it is diamagnetic.