The B₂ molecule has 10 electrons. Write its configuration, find its bond order, and find the number…

Step 1. Configuration. $\sigma 1s^2$ $\sigma^* 1s^2$ $\sigma 2s^2$ $\sigma^* 2s^2$ $\pi 2p_x^1$ $\pi 2p_y^1$.

Step 2. Bond order. Bonding $2 + 2 + 2 = 6$, antibonding $4$: $\dfrac{6 - 4}{2} = 1$.

Step 3. Unpaired. The two $\pi 2p$ electrons are in separate orbitals: $2$ unpaired, so $\text{B}_2$ is paramagnetic.

Step 4. The other order. The last two electrons would pair in $\sigma 2p$, giving $0$ unpaired. The real paramagnetism of $\text{B}_2$ shows that $\pi 2p$ is lower for boron.