From molecular orbital theory we can give the electronic configuration of singly positive Nitrogen molecular ion $N_{2}^{+}$ as-

σ (1 s)^{2} σ^{*} (1 s)^{2} σ (2 s)^{2} σ^{*} (2 s)^{2} π (2 p)^{4} σ (2 p)^{1}

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σ (1 s)^{2} σ^{*} (1 s)^{2} σ (2 s)^{2} σ^{*} (2 s)^{2} σ (2 p)^{1} π (2 p)^{3}

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σ (1 s)^{2} σ^{*} (1 s)^{2} σ (2 s)^{2} σ^{*} (2 p)^{2} π (2 p)^{4}

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σ (1 s)^{2} σ^{*} (1 s)^{2} σ (2 s)^{2} σ^{*} (2 s)^{2} σ (2 p)^{2} π (2 p)^{2}

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Solution

The correct option is A $σ (1 s)^{2} σ^{} (1 s)^{2} σ (2 s)^{2} σ^{} (2 s)^{2} π (2 p)^{4} σ (2 p)^{1}$
Only option a contains 13 electrons

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Similar questions

N_{2}^{+}

N_{2}^{+}

N_{2}

(σ 2 s)^{2} (σ^{*} 2 s)^{2} (Π 2 p)^{4} (σ 2 p)^{2} (Π^{*} 2 p)^{1}

O_{2}^{2 -}

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