Question
The pair of species with the same bond order is
A.
$$O_2^{2 - },{B_2}$$
B.
$$O_2^ + ,N{O^ + }$$
C.
$$NO,CO$$
D.
$${N_2},{O_2}$$
Answer :
$$O_2^{2 - },{B_2}$$
Solution :
According to molecular orbital theory,
$$O_2^{2 - }\left( {8 + 8 + 2 = 18} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\sigma 2p_z^2,\pi 2p_x^2 \approx \pi 2p_y^2,$$ $$\mathop \pi \limits^* 2p_x^2 \approx \mathop \pi \limits^* 2p_y^2$$
$$\eqalign{
& {\text{Bond}}\,\,{\text{order}}\left( {BO} \right) \cr
& = \frac{{{N_b} - {N_a}}}{2} \cr
& = \frac{{10 - 8}}{2} \cr
& = 1 \cr
& {B_2}\left( {5 + 5 = 10} \right) \cr} $$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\pi 2p_x^1 \approx \pi 2p_y^1$$
$$\eqalign{
& BO = \frac{{6 - 4}}{2} \cr
& \,\,\,\,\,\,\,\,\,\, = 1 \cr} $$
Thus, $$O_2^{2 - }$$ and $${B_2}$$ have the same bond order.
NOTE
$$BO\,\,{\text{of}}\,\,O_2^ + = 2.5,\,N{O^ + } = 3,$$ $$NO = 2.5,\,CO = 3,\,{N_2} = 3$$ $${\text{and}}\,\,{O_2} = 2$$