The number of paramagnetic species among the following is ___________.
B₂, Li₂, C₂, C₂⁻, O₂²⁻, O₂⁺ and He₂⁺

Solution

Chemistry ma’am · 3 steps

Fill the MO diagram for each species and look for a lone electron, but pick the right order for B₂ first.

1. What to look for. A species is paramagnetic if at least one electron is unpaired in its MO configuration. If every electron is paired, it is diamagnetic.
2. Pick the order. With 14 or fewer electrons, s-p mixing pushes σ 2p_z above π 2pₓ and π 2p_y, so the two π orbitals fill first. With more than 14 electrons, as in the O₂ family, σ 2p_z fills before the π orbitals.

   Species	Electrons	Highest filled orbitals	Unpaired
   B₂	10	π 2pₓ¹ π 2p_y¹ (Hund's rule)	2, paramagnetic
   Li₂	6	σ 2s²	0, diamagnetic
   C₂	12	π 2pₓ² π 2p_y²	0, diamagnetic
   C₂⁻	13	π 2p⁴ σ 2p_z¹	1, paramagnetic
   O₂²⁻	18	π^* 2pₓ² π^* 2p_y²	0, diamagnetic
   O₂⁺	15	π^* 2pₓ¹	1, paramagnetic
   He₂⁺	3	σ 1s² σ^* 1s¹	1, paramagnetic

3. So. B₂, C₂⁻, O₂⁺ and He₂⁺ have unpaired electrons, so the answer is 4.

The common slip: 3. Filling σ 2p_z before the π orbitals for B₂ pairs its last two electrons and makes it diamagnetic, which gives 3, not 4.

The idea to keep. For B₂, C₂ and N₂ and their ions, the π 2p pair fills before σ 2p_z. Get the order right first, then count unpaired electrons.

Answer from the NTA final answer key. Solution by PrepWiser.