Question

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

$N_2,N_2^{+},N_2^{-},N_2^{2+}$

Answer 1

The general sequence of the energy levels of the molecular orbital

(If the number of electrons in the molecule is less than or equal to fourteen)

$\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<\pi 2p_x=\pi 2p_y<\sigma 2p_z<{\pi }^{\ast }2p_x={\pi }^{\ast }2p_y$

Molecular orbital configuration and magnetic behaviour of given species
* $N_2(14e-):\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2_y^2\sigma 2p_z^2$

Bond order = 1/2[electrons in BMO-electrons in ABMO]

BMO = Bonding molecular orbital

ABMO = Antibonding molecular orbital

For $N_2=1/2(10-4)=3$

Since, the molecular orbitals has no unpaired electrons.So, molecule $N_2$ is diamagnetic in nature.

$N_2^{+}(13e-):\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2_y^2\sigma 2p_z^1$

Bond order for $N_2^{+}=1/2(9-4)=2.5$

Since, the molecular orbitals has $1$unpaired electrons.So, molecule $N_2^{+}$ is paramagnetic in nature.

$N_2^{-}(15e-):\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2_y^2\sigma 2p_z^1{\pi }^{\ast }2p_x^1$

Bond order for $N_2^{-}=1/2(10-5)=2.5$

Since, the molecular orbitals has $1$unpaired electrons.So, molecule $\left(N_2^{-}\right)$ is paramagnetic in nature.

$N_2^{2+}(12e-):\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2_y^2$

Bond order for $N_2^{2+}=1/2(8-4)=2$

Since, the molecular orbitals has $1$unpaired electrons.So, molecule $\left(N_2^{2+}\right)$ is diamagnetic in nature.

Relative stability order among given specise

* The compound with higher bond order is more stable.

* If the bond order is same the species with more bonding electrons and less antibonding electrons will be more stable

Thus, the order of stability is :

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