1. Tardigrade
  2. Question
  3. Chemistry
  4. Bond order of N2+,N2- and N2 will be

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Bond order of $N_{2}^{+},N_{2}^{-}$ and $N_{2}$ will be

NEETNEET 2022Chemical Bonding and Molecular Structure

Solution:

$N_{2}^{+} (13):\left(\sigma 1s^{2}\right)\left(\sigma^* 1s^{2}\right)\left(\sigma 2s^{2}\right)\left(\sigma^* 2s^{2}\right)\left(\pi 2p_{x}^{2} = \pi2p_{y}^{2}\right)\left(\sigma 2p_{z}^{1}\right)$
$B .O. = \frac{1}{2} \times (9 - 4) = 2.5$
$N_{2}^{-}(15) : \left(\sigma 1s^{2}\right)\left(\sigma^* 1s^{2}\right)\left(\sigma 2s^{2}\right)\left(\sigma^* 2s^{2}\right) \left(\pi 2p_{x}^{2} = \pi2p_{y}^{2}\right)$
$\left(\sigma 2p_{z}\right)^{2} \left(\pi^ * 2p_{x}^{1}\right)$
$B .O. = \frac{1}{2}\times (10-5 )= 2.5$
$N_{2}(14) : \left(\sigma 1s^{2}\right)\left(\sigma^* 1s^{2}\right)\left(\sigma 2s^{2}\right)\left(\sigma^* 2s^{2}\right) \left(\pi 2p_{x}^{2} = \pi2p_{y}^{2}\right)$
$\left(\sigma 2p_{z}^{2}\right)$
$B .O. = \frac{1}{2}\times (10-4 )= 3.0$