Chemical Bonding: Numerical Problems Set (1)

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Chemical Bonding: numerical problems set for JEE+NEET

doubts.ai · Accepted Answer

Solution — Step by Step where N b is electrons in bonding MOs and N a in antibonding MOs. The MOs in order: {1s}, ^ {1s}, {2s}, ^ {2s}, {2p z}, {2p x} = {2p y}, ^ {2p x} = ^ {2p y}. For O 2: N b = 10, N a = 6. Bond order = (10-6)/2 = 2. Lose one antibonding electron. N b = 10, N a = 5. Bond order = 5/2 = 2.5. Gain one antibonding electron. N b = 10, N a = 7. Bond order = 3/2 = 1.5. Bond orders: O 2 = 2, O 2^+ = 2.5, O 2^- = 1.5 . Why This Works Adding an electron to an antibonding orbital weakens the bond; removing one strengthens it. That's why O 2^+ has a higher bond order than O 2, despite having one fewer electron overall. The bond length follows the inverse trend: shorter bonds for higher bond orders. The key insight is that bonding electrons stabilise the molecule, antibonding electrons destabilise it, and the difference (divided by 2) is the bond order. Alternative Method Just memorise the trend for second-period diatomics: N 2 (3) NO (2.5) O 2 (2). Add or remove from antibonding to shift in steps of 0.5. JEE Main loves this exact comparison. Bond order bond strength \propto 1/bond length. Higher bond order means shorter, stronger bonds. Use this to rank species in MCQ traps. Common Mistake Forgetting that O 2 has unpaired electrons in ^ (paramagnetic, two unpaired). 	ext{O} 2^+ has one unpaired electron, O 2^- has one unpaired electron, O 2^{2-} has zero. NEET 2023 had a paramagnetism trap on O 2^{2-}.

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