The correct order of energies of molecular orbitals of N _ 2...

The correct order of energies of molecular orbitals of

{N}_{2}

molecule, is:

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

Correct answer

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

Step-by-step solution

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

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

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

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