NEET 2023 — Chemistry solution

NEET2023Chemistry

The correct order of energies of molecular orbitals of N $_{2}$ molecule, is

A
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < σ 2 p_{z} < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$
B
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < σ 2 p_{z} < σ^{*} 2 p_{z} < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y})$
C
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ 2 p_{z} < σ^{*} 2 p_{z}$
D
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} = π 2 p_{y}) < σ 2 p_{z} < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$

Solution

For diatomics of atomic number $\leq 7$ ( $Li_{2}$ through $N_{2}$ ), $s$ - $p$ mixing pushes the $σ 2 p_{z}$ MO above the degenerate $π 2 p_{x}$ and $π 2 p_{y}$ MOs.

So for $N_{2}$ the correct sequence is:

$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} = π 2 p_{y}) < σ 2 p_{z} < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$ .

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