NEET UG 2023 — Chemistry Inorganic Chemistry

2023mcqeasy

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

Official previous-year question

Held on 30 Apr 2023 · Verified 9 Jul 2026.

Options

A
$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({\sigma }_{2{p}_{z}})<({{\sigma }^{*}}_{2{p}_{z}})$
B
$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({\sigma }_{2{p}_{z}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({{\sigma }^{*}}_{2{p}_{z}})$
C
$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\sigma }_{2{p}_{z}})<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({{\sigma }^{*}}_{2{p}_{z}})$
D
$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\sigma }_{2{p}_{z}})<({{\sigma }^{*}}_{2{p}_{z}})<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})$

Did you get this right?

Solution

The molecular orbitals filled with electrons according Afbaue, Pauli's exclusion and Hund's rules.

$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({\sigma }_{2{p}_{z}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({{\sigma }^{*}}_{2{p}_{z}})$

The important characteristic feature of this order is that the energy of ${\sigma }_{2\mathrm{pz}}$ molecular orbital is higher than that of ${\pi }_{2\mathrm{px}}\mathrm{and}{\pi }_{2\mathrm{py}}$ molecular orbitals, it is due to 2s-2p mixing is allowed in case of nitrogen molecule.

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Options

$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({\sigma }_{2{p}_{z}})<({{\sigma }^{*}}_{2{p}_{z}})$

$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\sigma }_{2{p}_{z}})<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})<({{\sigma }^{*}}_{2{p}_{z}})$

$({\sigma }_{1s})<({{\sigma }^{*}}_{1s})<({\sigma }_{2s})<({\sigma }_{2s}*)<({\sigma }_{2{p}_{z}})<({{\sigma }^{*}}_{2{p}_{z}})<({\pi }_{2{p}_{x}})=({\pi }_{2{p}_{y}})<({{\pi }^{*}}_{2{p}_{x}})=({{\pi }^{*}}_{2{p}_{y}})$

Solution

The molecular orbitals filled with electrons according Afbaue, Pauli's exclusion and Hund's rules.

NEET UG 2023 — Chemistry Inorganic Chemistry

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NEET UG 2023 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

NEET UG Inorganic Chemistry in other years

More NEET UG Inorganic Chemistry questions

Other Chemistry topics for NEET UG

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry