Ncert Exercises & Solutions

Match the species in Column I with the bond order in Column II.

Codes

A. $\to$ (3) B.

\to

(4) C. $\to$ (1) D. $\to$ (2)

NO (7 + 8 = 15) = 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}

Bond order

= \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 5}{2} = 2.5

CO (6 + 8 = 14) = 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}

Bond order

= \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 4}{2} = 3

{O_{2}}^{-} (8 + 8 + 1 = 17) = 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}}^{2} ≃ π * 2 {p_{y}}^{1}

Bond order

= \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 7}{2} = 1.5

O_{2} (8 + 8 = 16) = 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} ≃ π * 2 {p_{y}}^{1}

Bond order

= \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 6}{2} = 2