Question
Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order?
A.
$$O_2^ - < NO < C_2^{2 - } < He_2^ + $$
B.
$$NO < C_2^{2 - } < O_2^ - < He_2^ + $$
C.
$$C_2^{2 - } < He_2^ + < NO < O_2^ - $$
D.
$$He_2^ + < O_2^ - < NO < C_2^{2 - }$$
Answer :
$$He_2^ + < O_2^ - < NO < C_2^{2 - }$$
Solution :
The molecular orbital configuration of
$$O_2^ - \left( {8 + 8 + 1 = 17} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,$$ $$\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^2 \approx \mathop \pi \limits^* 2p_z^1$$
$$\eqalign{
& {\text{Bond}}\,{\text{order}}\left( {{\text{BO}}} \right) \cr
& = \frac{{{N_b} - {N_a}}}{2} \cr
& = \frac{{10 - 7}}{2} \cr
& = 1.5 \cr} $$
$$NO\left( {7 + 8 = 15} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,$$ $$\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^1 \approx \mathop \pi \limits^* 2p_z^0$$
$$\eqalign{
& {\text{BO}} = \frac{{10 - 5}}{2} \cr
& \,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
$$C_2^{2 - }\left( {6 + 6 + 2 = 14} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\pi 2p_y^2 \approx \pi 2p_z^2,\sigma 2p_x^2$$
$$\eqalign{
& {\text{BO}} = \frac{{10 - 4}}{2} \cr
& \,\,\,\,\,\,\,\,\, = 3 \cr} $$
$$\eqalign{
& He_2^ + \left( {2 + 2 - 1 = 3} \right) = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^1} \cr
& {\text{BO}} = \frac{{2 - 1}}{2} \cr
& \,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr
& \,\,\,\,\,\,\,\,\, = 0.5 \cr} $$
Hence, order of increasing bond order is $$He_2^ + < O_2^ - < NO < C_2^{2 - }$$