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) \(\mathrm{N}_{2}\) (C) \(\mathrm{O}_{2}\) (D) \(\mathrm{C}_{2}\) (E) \(\mathrm{B}_{2}\) Choose the most appropriate answer from the options given below :-

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Chemical Bonding and Molecular Structure

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)

N_{2}

(C)

O_{2}

(D)

C_{2}

(E)

B_{2}

Choose the most appropriate answer from the options given below :-

(A)(A), (B), (C) only
(B)(B), (C), (E) only
(C)(A), (C) only✓ Correct
(D)(D) only

Answer: (C)

Solution

Bond strength is directly proportional to bond order. Removal of an electron from an antibonding molecular orbital increases the bond order, thereby strengthening the bond.

For NO: Electronic configuration ends in

π_{2 p}^{*}

with 1 electron. Removing this electron increases bond order from 2.5 to 3.

For

O_{2}

: Electronic configuration ends in

π_{2 p}^{*}

with 2 electrons. Removing one electron increases bond order from 2 to 2.5.

For

N_{2}

C_{2}

, and

B_{2}

: The highest occupied molecular orbital is bonding, so removing an electron would decrease bond order.

Therefore, only NO and

O_{2}

have valence electrons in antibonding

π^{*}

orbitals, making the answer (A), (C) only.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Chemical Bonding and Molecular Structure

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