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 1

Answer: (A)

$N_2^{+}(13):\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)\left(\sigma 2p_z^1\right)$
$B.O.=\frac{1}{2}\times (9-4)=2.5$
$N_2^{-}(15):\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)$
${\left(\sigma 2p_z\right)}^2\left({\pi }^{\ast }2p_x^1\right)$
$B.O.=\frac{1}{2}\times (10-5)=2.5$
$N_2(14):\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)$
$\left(\sigma 2p_z^2\right)$
$B.O.=\frac{1}{2}\times (10-4)=3.0$