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Q.

N₂ accept electron and convert into $N_{2}^{-}$ , where this electron goes

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a

Antibonding $π$ molecular orbital

b

Bonding $π$ molecular orbital

c

$σ$ bonding molecular orbital

d

$σ -$ anti-bonding molecular orbital

answer is A.

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Detailed Solution

Number of electrons of N₂ is 14 and its molecular orbital electronic configuration is:
$(σ 1 s^{2}) (σ * 1 s^{2}) (σ 2 s^{2}) (σ * 2 s^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (σ 2 p_{z}^{2}) (π * 2 p_{x}^{0} = π * 2 p_{y}^{0})$
N₂ forms $N_{2}^{-}$ , when N₂ accepts one electron. Considering the above configuration, it concludes that the additional electron can either goes in $π * 2 p_{y} or π * 2 p_{x},$ as both anti-bonding orbitals are degenerate.

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