Ncert Exercises & Solutions

What is the correct sequence of bond order for the specified species?

1.	$\mathrm{O}^-_2>\mathrm{O}_2>\mathrm{O}^+_2$	2.	$\mathrm{O}^-_2<\mathrm{O}_2<\mathrm{O}^+_2$
3.	$\mathrm{O}^-_2>\mathrm{O}_2<\mathrm{O}^+_2$	4.	$\mathrm{O}^-_2<\mathrm{O}_2>\mathrm{O}^+_2$

HINT: Bond order $= \frac{1}{2} (N_{b} - N_{a})$

Explanation:

The electronic configuration of a given molecule/ion according to MOT is as follows:

1.) Electronic configuration of O₂ (16 electrons)

= 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

2.) Electronic configuration of

O_{2}^{+}

(15 electrons)

= 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}}^{0}}^{*}

Bond order

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

3.) Electronic configuration of

O_{2}^{-}

(17 electrons)

= 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}}^{*} \approx π_{2 {p_{y}}^{1}}^{*}

Bond order

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

Thus, the order of bond order is

O_{2}^{-}

<

O_{2}

<

O_{2}^{+}

1.	\(\mathrm{O}^-_2>\mathrm{O}_2>\mathrm{O}^+_2\)	2.	\(\mathrm{O}^-_2<\mathrm{O}_2<\mathrm{O}^+_2\)
3.	\(\mathrm{O}^-_2>\mathrm{O}_2<\mathrm{O}^+_2\)	4.	\(\mathrm{O}^-_2<\mathrm{O}_2>\mathrm{O}^+_2\)