Assuming that Hund's rule is violated and electrons are forced to pair up before occupying degenerat — Chemical Bonding Chemistry Question

💡 Solution & Explanation

The $B_2$ molecule ($10 e^-$) normally has the valence configuration $\sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$ (Bond Order 1, Paramagnetic). If Hund's rule is violated, both electrons will pair up in the lower of the two degenerate orbitals: $\pi 2p_x^2$. The molecule becomes diamagnetic, but the bond order remains $\frac{6 - 4}{2} = 1$.