In forming (i) $N_{2} \to N_{2}^{+}$ and (ii) $O_{2} \to O_{2}^{+}$ the electrons respectively are removed from

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$(π 2 p_{y} or π 2 p_{x}) and (π 2 p_{y} or π 2 p_{x})$

$(π^{*} 2 p_{y} or π^{*} 2 p_{x}) and (π 2 p_{y} or π 2 p_{x})$

$(π^{\circ} 2 p_{y} or π^{*} 2 p_{x}) and (π^{*} 2 p_{y} or π^{*} 2 p_{x})$

$(σ 2 p_{z}) and (π^{*} 2 p_{y} or π^{*} 2 p_{x})$

answer is C.

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Detailed Solution

Number of electrons in N₂ is 14 and its molecular orbital electronic configuration is:

$\begin{aligned} (σ 1 s^{2}) (σ^{*} 1 s^{2}) (σ 2 s^{2}) (σ^{*} 2 s^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (σ 2 p_{z}^{2}) \end{aligned}$

For formation of $N_{2}^{+}$ from N₂, one electron is removed from $σ 2 p_{z} .$ Molecular orbital electronic configuration of $N_{2}^{+}$ is:

$(σ 1 s^{2}) (σ^{*} 1 s^{2}) (σ 2 s^{2}) (σ^{*} 2 s^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (σ 2 p_{z}^{1})$

Number of electrons of O₂ is 16 and its molecular orbital electronic configuration is:

$(σ 1 s^{2}) (σ * 1 s^{2}) (σ 2 s^{2}) (σ * 2 s^{2}) (σ 2 p_{z}^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π * 2 p_{x}^{1} = π * 2 p_{y}^{1})$

O₂ forms $O_{2}^{+}$ when one electron either remove from $π * 2 p_{y} or π * 2 p_{x} .$ Molecular orbital electronic configuration of $O_{2}^{+ .}$

$(σ 1 s^{2}) (σ * 1 s^{2}) (σ 2 s^{2}) (σ * 2 s^{2}) (σ 2 p_{z}^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π 2 p_{x}^{1} = π 2 p_{y}^{0}) or (σ 1 s^{2}) (σ * 1 s^{2}) (σ 2 s^{2}) (σ * 2 s^{2}) (σ 2 p_{z}^{2}) (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π * 2 p_{x}^{0} = π * 2 p_{y}^{1})$