Question
Bond order of $$N_2^ + ,N_2^ - $$ and $${N_2}$$ will be
A.
2.5, 2.5 and 3 respectively
B.
2, 2.5 and 3 respectively
C.
3, 2.5 and 3 respectively
D.
2.5, 2.5 and 2.5 respectively
Answer :
2.5, 2.5 and 3 respectively
Solution :
$$N_2^ + \left( {13} \right):\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$ $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\sigma 2p_z^1} \right)$$
$$B.O. = \frac{1}{2} \times \left( {9 - 4} \right) = 2.5$$
$$N_2^ - \left( {15} \right):\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$ $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\sigma 2p_z^2} \right)$$ $$\left( {{\pi ^ * }2p_x^1} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 5} \right) = 2.5$$
$${N_2}\left( {14} \right):\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$ $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\sigma 2p_z^2} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 4} \right) = 3$$