Question

The energy of $\sigma 2p_z$ molecular orbital is greater than and 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 behavior of the following species: $N_2,N_2^{+},N_2^{-},N_2^{2+}$

Answer 1

Configuration of $N_2:\sigma 1s^2<\sigma \ast 1s^2<\sigma 2s^2<\sigma \ast 2s^2<[\pi 2p_x^2=\pi 2p_z^2]<\sigma 2p_z^2<[\pi \ast 2p_z^2=\pi \ast 2p_x]<\sigma \ast 2p_z$
Bond order of $N_2=(10-4)/2=3$.
$N_2$ is Diamagnetic, as all orbitals has unpaired electrons.
$N_2^{+}:\sigma 1s^2<\sigma \ast 1s^2<\sigma 2s^2<\sigma \ast 2s^2<[\pi 2p_x^2=\pi 2p_x^2]<\sigma 2p_z^2<[\pi \ast 2p_x^2=\pi \ast 2p_x^1]<\sigma \ast 2p_z^1$
B.O. of $N_2^{+}=(9-4)/2=2.5$.
$N_2^{-}$ is Paramagnetic, as one of the orbitals has unpaired electrons.
$N_2^{+}:\sigma 1s^2<\sigma \ast 1s^2<\sigma 2s^2<\sigma \ast 2s^2<[\pi 2p_z^2=\pi 2p_x^2]<\sigma 2p_z^2<[\pi \ast 2p_x^2=\pi \ast 2p_x^2]<\sigma \ast 2p_z^1$
B.O. of $N_2^{-}=(9-5)/2=2.5$
$N_2^{-}$ is paramagnetic, as one of the orbitals has unpaired electrons.
$N_2^{+}:\sigma 1s^2<\sigma \ast 1s^2<\sigma 2s^2<\sigma \ast 2s^2<[\pi 2p_x^2=\pi 2p_x^2]<\sigma 2p_z^2<[\pi \ast 2p_x^2=\pi \ast 2p_x^1]<\sigma \ast 2p_z$
B.O. of $N_2^{2+}=(8-4)/2=2$
It is Paramagnetic as two, of the orbitals have unpaired electron.
Stability: Stability of species depends on bond order and also depends on number of electrons present in bonding molecular orbital.
Increasing order of stability: $N_2^{2+}<N_2^{-}<N_2^{+}<N_2$
$N_2^{2+}$ = B.o. is lowest.
$N_2^{-}$ and $N_2^{+}$ is same but number of electrons in bonding and in antibonding molecular orbitals are different. In $N_2^{-}$, in antibonding molecular orbitals, electrons are more than $N_2^{+}$.
$N_2$ = B.O. is higher.