Ncert Exercises & Solutions

What is the effect of the following processes on the bond order in N₂ and O₂?

According to molecular orbital theory, electronic configurations and bond order of

N_{2}, N_{2}^{+}, O_{2} and O_{2}^{+}

species are as follows

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} (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}}^{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

(a)

\underset{B . O . = 3}{N_{2}} \to \underset{B . O . = 2.5}{N_{2}^{+}} + e^{-}

Thus, bond order decreases.

(b)

\underset{B . O . = 2}{O_{2}} \to \underset{B . O . = 2.5}{O_{2}^{+}} + e^{-}