If the magnetic moment of a dioxygen species is 1.73 B.M, it may be :

Solution

Chemistry ma’am · 3 steps

1.73 B.M. means exactly one unpaired electron, so count what sits in the π^* orbitals of each species.

1. Unpaired electrons. Spin-only magnetic moment depends only on the number of unpaired electrons, n. Set it equal to 1.73 B.M. and solve for n.

   μ = √n(n+2) = 1.73 ⇒ n(n+2) = 3 ⇒ n = 1

2. Fill the MOs. In every dioxygen species the bonding orbitals and σ 2p_z are full. The difference lies in the two degenerate π^*2p orbitals, which hold the electrons beyond 14.

   Species	Total electrons	Electrons in π^*2p	Unpaired	μ (B.M.)
   O₂	16	2, one in each orbital	2	√8 = 2.83
   O₂⁺	15	1	1	√3 = 1.73
   O₂⁻	17	3, one orbital paired	1	√3 = 1.73

3. So. Both O₂⁺ and O₂⁻ have one unpaired electron and a moment of 1.73 B.M. O₂ has two and gives 2.83 B.M., so the answer is (a).

The idea to keep. For O₂-type species, count the electrons in the two π^* orbitals: 2 gives two unpaired, while 1 or 3 gives one unpaired. Then use μ = √n(n+2).

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