Question

(i)

The energy of $\sigma 2p_z$ molecular orbital is greater than $\pi 2p_x\text{and}\pi 2p_y$ molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species.

$N_2$

(ii) $N_2^{+}$

(iii) $N_2^{-}$

(iv) $N_2^{2+}$

Answer 1

Electronic configuration of N-atom (Z = 7) is $1s^{2}2s^{2}2p_x^{1}2p_y^{1}2p_z^1$. Total number of electrons present in $N_2$ molecule is 14, 7 from each N-atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of $N_2$ molecule will be

$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2$

Comparative study of the relative stability and the magnetic behaviour of the following species

(a).

$N_2\text{molecule}\sigma 1s^2,{\sigma }^{\ast }{s}^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2$

$\text{Here},N_b=10,N_a=4.\text{Hence, Bond order}=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-4)=3$

Hence, presence on no unpaired electron indicates it to be diamagnetic.

(b).

$N_2^{+}\text{molecule}\sigma 1s^2,{\sigma }^{\ast }{s}^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^1$

$\text{Here},N_b=9,N_a=4.\text{Hence, Bond order}=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(9-4)=2.5$

Hence, presence on no unpaired electron indicates it to be paramagnetic.

(c).

$N_2^{-}\text{molecule}\sigma 1s^2,{\sigma }^{\ast }{s}^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2,{\pi }^{\ast }2p_x^1$

$\text{Here},N_b=10,N_a=5.\text{Hence, Bond order}=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-5)=2.5$

Hence, presence on no unpaired electron indicates it to be paramagnetic.

(d).

$N_2^{2+}\text{molecule}\sigma 1s^2,{\sigma }^{\ast }{s}^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2.$

$\text{Here},N_b=8,N_a=4.\text{Hence, Bond order}=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(8-4)=2$

Hence, presence on no unpaired electron indicates it to be diamagnetic.

As bond dissociation energies are directly proportional to the bond orders, therefore, the dissociation energies of these molecular species in the order.

$N_2>N_2^{-}=N_2^{+}>N_2^{2+}$

As greater the bond dissociation energy, greater is the stability of these species is also in the above order.