Molecular orbital configuration of B₂ (10 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$

Here in B₂, 2 unpaired electrons present.

Since the Hund's rule is violated, two electrons are placed in $\pi 2 p_x$ molecular orbital, so

$$ \mathrm{B}_2(10): {\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} $$

Thus, bond order $=\frac{6-4}{2}=1$.

As there are no unpaired electrons, the nature is diamagnetic.