How many of the following species have the bo | vedclass

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(C) The bond order can be determined by using the formula:$	ext{Bond order} = \frac{N_b - N_a}{2}$where $N_b$ is the number of bonding electrons and

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$6$

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(C) The bond order can be determined by using the formula:$	ext{Bond order} = \frac{N_b - N_a}{2}$where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$1. C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. $N_b = 8, N_a = 4$. $B.O. = \frac{8-4}{2} = 2$.$2. B_2^{2-}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.$3. N_2^{2+}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.$4. CN^{+}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.$5. NO^{-}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.$6. O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $N_b = 10, N_a = 6$. $B.O. = \frac{10-6}{2} = 2$.$7. C_2^{+}$ ($11$ electrons): $B.O. = \frac{7-4}{2} = 1.5$.Thus,the species with bond order $2$ are $C_2, B_2^{2-}, N_2^{2+}, CN^{+}, NO^{-}, O_2$. Total $6$ species.

How many of the following species have the bond order $2$?
$C_2, B_2^{2-}, N_2^{2+}, CN^{+}, NO^{-}, O_2, C_2^{+}$

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How many of the following species have the bond order $2$?$C_2, B_2^{2-}, N_2^{2+}, CN^{+}, NO^{-}, O_2, C_2^{+}$