The bond order and magnetic behaviour of O₂ - ion are respectively :

Solution

Chemistry ma’am · 4 steps

Fill 17 electrons into the O₂ MO diagram and look at the π^* level: three electrons, one unpaired.

1. Count electrons. Each O atom has 8 electrons, so O₂ has 16. The extra negative charge adds one more, so O₂⁻ has 17 electrons.
2. Fill the orbitals. For O₂ and heavier, σ 2p_z lies below π 2p. The 17 electrons go in as σ 1s² σ^* 1s² σ 2s² σ^* 2s² σ 2p_z² π 2pₓ² = π 2p_y² π^* 2pₓ² = π^* 2p_y¹.

   Bond order = (N_b - Nₐ)/2 = (10 - 7)/2 = 1.5

3. Magnetic behaviour. The two π^* orbitals have the same energy and hold 3 electrons between them. One of these electrons has no partner, so the ion is paramagnetic.
4. So. O₂⁻ has bond order 1.5 and is paramagnetic, which is option (a).

The idea to keep. In the O₂ family, every electron you add goes into π^* and lowers the bond order by 0.5. An odd total electron count always means at least one unpaired electron, so the species is paramagnetic.

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