Question
Which of the following are arranged in an increasing order of their bond strengths?
A.
$$\,O_2^ - < {O_2} < O_2^ + < O_2^{2 - }$$
B.
$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
C.
$$\,O_2^ - < O_2^{2 - } < {O_2} < O_2^ + $$
D.
$$O_2^ + < {O_2} < O_2^ - < O_2^{2 - }$$
Answer :
$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
Solution :
$$\eqalign{
& O_2^ + \left( {15} \right) = KK\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr
& \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^1 = \pi 2p_z^0} \right.} \right.\, \cr
& {\text{Bond}}\,{\text{order}}\, = \frac{1}{2}\left( {8 - 3} \right) = \frac{5}{2} = 2.5 \cr
& {O_2}\left( {16} \right) = KK\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr
& \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1} \right.} \right.\, \cr
& {\text{Bond}}\,{\text{order}}\, = \frac{1}{2}\left( {8 - 4} \right) = 2 \cr
& O_2^ - \left( {17} \right) = KK\sigma 2{s^2},\,{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr
& \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^1} \right.} \right. \cr
& {\text{Bond}}\,\,{\text{order}}\, = \frac{1}{2}\left( {8 - 5} \right) - 1.5 \cr
& O_2^{2 - }\left( {18} \right) = KK\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_x^2, \cr
& \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^2} \right.} \right. \cr
& {\text{Bond}}\,\,{\text{order}} = \frac{1}{2}\left( {8 - 6} \right) = 1 \cr} $$
NOTE : As we know that as the bond order decreases,
stability also decreases and hence the bond strength
also decreases. Hence the correct order of their increasing bond strength is
$$O_2^{2 - } < \,O_2^ - < {O_2} < O_2^ + $$