According to MO theory, number of species/ions from the following having identical bond order is ________.
CN⁻, NO⁺, O₂, O₂⁺, O₂²⁺

Solution

Chemistry ma’am · 3 steps

Count the electrons first: every 14-electron species has bond order 3.

1. Count electrons. Add the atomic numbers and adjust for charge. Subtract one electron for each + charge and add one for each − charge. Species with the same total electron count fill the same MOs, so they have the same bond order.

   Bond order = (N_b - Nₐ)/2

   Species	Total electrons	N_b, Nₐ	Bond order
   CN⁻	6 + 7 + 1 = 14	10, 4	3
   NO⁺	7 + 8 - 1 = 14	10, 4	3
   O₂²⁺	16 - 2 = 14	10, 4	3
   O₂⁺	16 - 1 = 15	10, 5	2.5
   O₂	16	10, 6	2

2. Why 14 gives 3. With 14 electrons, the bonding orbitals up to π 2p are full. That gives 10 bonding electrons. The antibonding σ^* 1s and σ^* 2s hold 4. The π^* 2p orbitals stay empty, so the bond order is (10 - 4)/2 = 3, the same as in N₂.
3. So. CN⁻, NO⁺ and O₂²⁺ share bond order 3, so the answer is 3.

The common slip: 2. Reading O₂²⁺ as the peroxide ion O₂²⁻ gives it 18 electrons and bond order 1. That leaves only CN⁻ and NO⁺, so the count becomes 2.

The idea to keep. Isoelectronic diatomics have the same bond order. Memorise the ladder: 14 electrons give 3, 15 give 2.5, 16 give 2, 17 give 1.5, 18 give 1.

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