The number of species from the following which are paramagnetic and with bond order equal to one is _________.
H₂, He₂⁺, O₂⁺, N₂²⁻, O₂²⁻, F₂, Ne₂⁺, B₂

Solution

Chemistry ma’am · 3 steps

Apply two filters one at a time: bond order exactly 1, then at least one unpaired electron.

1. Bond order. Count the total electrons in each species, fill the molecular orbitals in order, and use the formula below.

   Bond order = (N_b - Nₐ)/2

2. Filling order. Up to 14 electrons, s-p mixing pushes σ 2p_z above the two π 2p orbitals, so π 2p fills first. From 15 electrons on, σ 2p_z comes before π 2p. Paramagnetic means at least one electron is unpaired.

   Species	Electrons	Bond order	Unpaired electrons
   H₂	2	1	0, diamagnetic
   He₂⁺	3	0.5	1
   O₂⁺	15	2.5	1
   N₂²⁻	16	2	2
   O₂²⁻	18	1	0, diamagnetic
   F₂	18	1	0, diamagnetic
   Ne₂⁺	19	0.5	1
   B₂	10	1	2, one in each π 2p

3. So. H₂, O₂²⁻, F₂ and B₂ have bond order 1, but only B₂ also has unpaired electrons. The count is 1.

The common slip: 0. Filling σ 2p before π 2p in B₂ pairs its last two electrons, so B₂ looks diamagnetic and the count becomes 0. The experimentally observed paramagnetism of B₂ is what shows that s-p mixing is real.

The idea to keep. For B₂, C₂ and N₂, s-p mixing puts π 2p below σ 2p. That is why B₂ is paramagnetic, with one electron in each degenerate π orbital.

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