Question

Write molecular orbital configuration for the following molecular ions. Comment on the paramagnetic properties of these ions: (i) N2^+ (ii) N2^2-

Answer 1

1)Diamagnetic
2) Paramagnetic

Answer 2

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration hydrogen to nitrogen will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 7 electrons present in nitrogen.

(i) The number of electrons present in $N_2^+$ molecule = 2(7) - 1 = 13

The molecular orbital configuration of $N_2^+$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^1,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in $N_2^+$ molecule is, 1. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic

(ii) The number of electrons present in $N_2^{2-}$ molecule = 2(7) + 2 = 16

The molecular orbital configuration of $N_2^{2-}$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^2,[(\pi_{2p_x}^*)^1=(\pi_{2p_y}^*)^1],(\sigma_{2p_z}^*)^0$

The number of unpaired electrons in $N_2^{2-]$ molecule are, 2. So, this is paramagnetic.