List - I List - II (A) O 2 → O 2 + (P) An electron is removed from the π ∗ molecular orbital. (B) N 2 → N 2 + (Q) An electron is removed from the π molecular orbital. (C) C 2 → C 2 + (R) F

Question

List - I List - II (A) O 2 ​ → O 2 + ​ (P) An electron is removed from the π ∗ molecular orbital. (B) N 2 ​ → N 2 + ​ (Q) An electron is removed from the π molecular orbital. (C) C 2 ​ → C 2 + ​ (R) Fractional bond order of molecular ion (D) NO → NO + (S) Bond order decreases

TestHub · Accepted Answer

(A): Electron removed from π* orbital → bond order decreases → A → PR

(B): Electron removed from π orbital → bond order decreases slightly → B → RS

(C): Molecular ion has fractional bond order → affected by π and π* electrons → C → QRS

(D): Bond order decreases when electron is removed from antibonding orbital → D → P

List - I	List - II
(A) $O_{2} \to O_{2}^{+}$	(P) An electron is removed from the $π^{*}$ molecular orbital.
(B) $N_{2} \to N_{2}^{+}$	(Q) An electron is removed from the $π$ molecular orbital.
(C) $C_{2} \to C_{2}^{+}$	(R) Fractional bond order of molecular ion
(D) $NO \to NO^{+}$	(S) Bond order decreases

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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JEE Advanced 2027 (Class 12)