JEE Main 2023 · 12 April · morning · Single correct · +4 / −1
The bond order and magnetic property of acetylide ion are same as that of
- (a) O₂⁺
- (b) O₂⁻
- (c) N₂⁺
- (d) NO⁺
Answer and solution
Answer (d) NO⁺
Solution
Chemistry ma’am · 3 steps
Acetylide is C₂²⁻ with 14 electrons. Find the option that also has 14.
- Count electrons. The acetylide ion is C₂²⁻. It has 6 + 6 + 2 = 14 electrons, the same count as N₂.
- Fill the MOs. For 14 electrons the configuration ends at σ 2p_z² with every orbital doubly filled. There are 10 bonding and 4 antibonding electrons, so the bond order is 3 and there are no unpaired electrons, which makes the ion diamagnetic.
Bond order = (N_b - Nₐ)/2 = (10 - 4)/2 = 3
Species Electrons Bond order Magnetic nature C₂²⁻ 14 3 diamagnetic O₂⁺ 16 - 1 = 15 2.5 paramagnetic (1 unpaired) O₂⁻ 16 + 1 = 17 1.5 paramagnetic (1 unpaired) N₂⁺ 14 - 1 = 13 2.5 paramagnetic (1 unpaired) NO⁺ 7 + 8 - 1 = 14 3 diamagnetic - So. NO⁺ is isoelectronic with C₂²⁻. Both have bond order 3 and both are diamagnetic, so the answer is (d).
The idea to keep. Species with the same total electron count fill the same MOs, so they share bond order and magnetic nature. Count electrons first, then match.
Answer from the NTA final answer key. Solution by PrepWiser.