What is the effect of the following processes on the bond order in $N_{2}$ and $O_{2}$ ?

(i). $N_{2} \to N_{2}^{+} + e^{-}$

(ii) $O_{2} \to O_{2}^{+} + e^{-}$

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Solution

According to molecular orbital theory, electronic configurations, and bond order of $N_{2}, N_{2}^{+}, O_{2} and O_{2}^{+}$ species are as follos

$N_{2} (14 e^{-}) = σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, (π 2 p_{x}^{2} \approx π 2 p_{y}^{2}), σ 2 p_{z}^{2} Bond order = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 4) = 3 N_{2}^{+} (13 e^{-}) = σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, (π 2 p_{x}^{2} \approx π 2 p_{y}^{2}), σ 2 p_{z}^{1} Bond order = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (9 - 4) = 2.5 O_{2} (16 e^{-}) = σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2} (π 2 p_{x}^{2} \approx π 2 p_{y}^{2}), (π 2 p_{x}^{1} \approx π^{} 2 p_{y}^{1}) Bond order = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 6) = 2 O_{2}^{-} (15 e^{-}) = σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2} (π 2 p_{x}^{2} \approx π 2 p_{y}^{2}), (π 2 p_{x}^{1} \approx π^{} 2 p_{y}) Bond order = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 5) = 2.5$

(i). $N_{2} B . O . = 3 \to N_{2}^{+} B . O . = 2.5 + e^{-}$ . Thus, the bond order decreases.

(ii). $O_{2} B . O . = 2 \to O_{2}^{+} B . O . = 2.5 + e^{-}$ . Thus, the bond order increases.

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Similar questions

$N_{2}$ and $O_{2}$ are converted into monocations, $N_{2}^{+}$ and $O_{2}^{+}$ respectively. Which of the following is wrong ?

N_{2}

O_{2}

N_{2}^{+}

O_{2}^{+}

N_{2}^{+}

N_{2}

O_{2}^{+}

O_{2}

What is meant by the term bond order? Calculate the bond order of: $N_{2}$ , $O_{2}$ , $O_{2}^{+}$ and $O_{2}^{-}$ .

N_{2} ⟶ N_{2}^{+}

O_{2} ⟶ O_{2}^{+}

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