Give the molecular orbital configuration for the valence electrons in $C_2^{2−}$.

Will this ion be stable?

**Solution:**

Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σ* _{p}* orbital. The valence electron configuration for C

_{2}is $(σ_{2s})^2(σ_{2s}^*)^2(π_{2p_y},π_{2p_z})^4$.

Adding two more electrons to generate the $C_2^{2−}$ anion will give a valence electron configuration of $(σ_{2s})^2(σ_{2s}^*)^2(π_{2p_y},π_{2p_z})^4 (σ_{2p_x})^2$.

Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable.

**Check Your Learning**

How many unpaired electrons would be present on a $Be_2^{2−}$ ion? Would it be paramagnetic or diamagnetic?

**Answer:**

two, paramagnetic

Creating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings of the molecular orbitals for C_{6}H_{6}.