Question
According to molecular orbital theory which of the following lists rank the nitrogen species in terms of increasing bond order?
A.
$$N_2^ - < {N_2} < N_2^{2 - }$$
B.
$$N_2^{2 - } < N_2^ - < {N_2}$$
C.
$${N_2} < N_2^{2 - } < N_2^ - $$
D.
$$N_2^ - < N_2^{2 - } < {N_2}$$
Answer :
$$N_2^{2 - } < N_2^ - < {N_2}$$
Solution :
According to the molecular orbital theory $$\left( {MOT} \right).$$
$${N_2}\left( {7 + 7 = 14} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\pi 2p_x^2 \approx 2p_y^2,\sigma 2p_z^2$$
$${\text{Bond}}\,{\text{order}} = \frac{{10 - 4}}{2} = 3$$
$$N_2^ - \left( {7 + 7 + 1 = 15} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2,\mathop \pi \limits^* 2p_x^1$$
$${\text{BO}} = \frac{{10 - 5}}{2} = 2.5$$
$$N_2^{2 - }\left( {7 + 7 + 2 = 16} \right) = $$ $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_z^2,$$ $$\pi 2p_x^2 \approx \pi 2p_y^2,\mathop \pi \limits^* 2p_x^1 \approx \mathop \pi \limits^* 2p_y^1$$
$${\text{BO}} = \frac{{10 - 6}}{2} = 2$$
Hence, the increasing order of bond order is, $$N_2^{2 - } < N_2^ - < {N_2}$$