Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?
(A) NO
(B) N₂
(C) O₂
(D) C₂
(E) B₂
Choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Ask where the electron leaves from: an antibonding orbital raises bond order, a bonding one lowers it.

1. The test. Bond order is 1/2(N_b - Nₐ). Removing an electron from an antibonding orbital raises the bond order and strengthens the bond. Removing one from a bonding orbital lowers it.
2. Find the HOMO. The electron leaves from the highest occupied MO. For 14 or fewer electrons (N₂, C₂, B₂), that HOMO is bonding. With 15 or 16 electrons (NO, O₂), the outermost electrons sit in π^*2p, which is antibonding.

   Molecule	Electrons	Electron removed from	Bond order change	Bond
   (A) NO	15	π^*2p (antibonding)	2.5 → 3	stronger
   (B) N₂	14	σ 2p_z (bonding)	3 → 2.5	weaker
   (C) O₂	16	π^*2p (antibonding)	2 → 2.5	stronger
   (D) C₂	12	π 2p (bonding)	2 → 1.5	weaker
   (E) B₂	10	π 2p (bonding)	1 → 0.5	weaker

3. So. Only NO and O₂ get stronger bonds when they lose an electron, so the answer is (c), (A) and (C) only.

The idea to keep. Losing an antibonding electron strengthens the bond; losing a bonding electron weakens it. Among diatomics, only those with more than 14 electrons have an antibonding HOMO.

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