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. $2.5,2.5and2.5$respectively

Answer 1

Answer: (A)

$2.5,2.5$and 3 respectively

Answer 2

: $13N_{2}^{+}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})$

$(\pi 2p_{x}^{2} = \pi 2p_{x}^{2})(\sigma 2p_{z}^{1})$

$B.O. = \frac{1}{2} \times (9 - 4) = 2.5$

$15N_{2}^{-}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma*2s^{2})(\pi 2p_{x}^{2} = \pi 2p_{y}^{2})$

$(\sigma 2p_{z})^{2}(\pi*2p_{x}^{2})$

$B.O.. = \frac{1}{2} \times (10 - 5) = 2.5$

$14N_{2}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})$

$(\pi 2p_{x}^{2} = \pi 2p_{y}^{2})(\sigma 2p_{z}^{2})$

$B.O. = \frac{1}{2} \times (10 - 4) = 3.0$