The total number of species from the following in which one unpaired electron is present, is _______.
N₂, O₂, C₂⁻, O₂⁻, O₂²⁻, H₂⁺, CN⁻, He₂⁺

Solution

Chemistry ma’am · 3 steps

Count the electrons, fill the MO ladder, then look at the last orbital filled.

1. Count electrons. Add the atomic numbers, then add one electron for each negative charge and remove one for each positive charge. Same totals give the same MO filling.
2. Use the right order. Up to 14 electrons, π 2p comes before σ 2p. From O₂ onwards, σ 2p comes first, and the next electrons go into the two degenerate π^* 2p orbitals one at a time (Hund's rule).

   Species	Electrons	Last orbitals filled	Unpaired
   N₂, CN⁻	14 each	σ 2p², all paired	0
   C₂⁻	13	π 2p⁴ σ 2p¹	1
   O₂	16	π^* 2p¹ π^* 2p¹	2
   O₂⁻	17	π^* 2p² π^* 2p¹	1
   O₂²⁻	18	π^* 2p² π^* 2p²	0
   H₂⁺	1	σ 1s¹	1
   He₂⁺	3	σ 1s² σ^* 1s¹	1

3. So. Exactly one unpaired electron is present in C₂⁻, O₂⁻, H₂⁺ and He₂⁺, so the answer is 4.

The common slip: 5. Counting O₂ as well gives 5. O₂ is paramagnetic, but it has two unpaired electrons, not one.

The idea to keep. Any species with an odd electron count has exactly one unpaired electron in these simple diatomics. For even counts, check O₂-type cases, where Hund's rule leaves two unpaired in π^*.

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