Question

The energy of σ2p_z molecular orbital is greater than π2p_x and π2p_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 :

\[\ce{N2, N^{+}2, N^{-}2, N^{2+}2}\]

Answer 1

The general molecular energy levels for nitrogen atom is:

`σ1s < σ^∗1s < σ2s < σ^∗2s < π2p_x = π2p_y < σ2p_z`

\[\ce{N2}\] molecule has 14 electrons in its molecular structure. Its electronic configuration will be:

`σ1s^2 < σ^∗1s^2 <σ2s^2 < σ^∗2s^2 < π2px^2 = π2py^2 < σ2pz^2`

\[\ce{N^{+}2}\] has 13 electrons in its molecular structure. Its electronic configuration will be:

`σ1s^2 < σ^∗1s^2 < σ2s^2< σ^∗2s^2 < π2px^2 = π2py^2 < σ2pz^1`

\[\ce{N^{-}2}\] has 15 electrons in its molecular structure. Its electronic configuration will be:

`σ1s^2 < σ^∗1s^2 < σ2s^2 < σ^∗2s^2 < π2px^2 = π2py^2 < σ2pz^2 < π^∗2px^1`

\[\ce{N^{2+}2}\] has 12 electrons in its molecular structure. Its electronic configuration will be:

`σ1s^2 < σ^∗1s^2 < σ2s^2 < σ^∗2s^2 < π2px^2 = π2py^2`

The formula for finding the bond order (B. O) for any molecular species is:

BO = `1/2[N_b - N_a]`

Hence, the bond orders of the given molecular species are:

N₂ = `(10 - 4)/2` = 3

\[\ce{N^{+}2}\] = `(9 - 4)/2` = 2.5

\[\ce{N^{-}2}\] = `(10 - 5)/2` = 2.5

\[\ce{N^{2+}2}\] = `(8 - 4)/2` = 2

The order for the stability of the given molecular species thus will be:

\[\ce{N2 > N^{-}2 > N^{+}2 > N^{2+}2}\]