Question

Both molecular orbitals are gerade or ungerade, In which of the following pairs?
A.${\sigma _{2s}},{\pi _{2{p_x}}}$
B.$\sigma {*_{2s}},\pi {*_{2{p_x}}}$
C.$\sigma {*_{2s}},{\pi _{2{p_z}}}$
D.${\pi _{2{p_x}}},\pi {*_{2{p_x}}}$

Answer 1

Hint: If a molecular orbital is symmetrical with respect to inversion, it has a subscript g (gerade, for even). If it is asymmetrical with respect to inversion, it is given a subscript u (ungerade, for uneven).

Complete Step by step answer: cular orbital, then the MO is said to have gerade (g) symmetry, from the German word for even.
 If inversion through the center of symmetry in a molecule results in a phase change for the molecular orbital, then the MO is said to have ungerade (u) symmetry, from the German word for odd.
For a bonding MO with $\sigma $-symmetry, the orbital is ${\sigma _g}$ (s' + s'' is symmetric), while an antibonding MO with $\sigma $ -symmetry the orbital is ${\sigma _u}$, because inversion of s' – s'' is antisymmetric. For a bonding MO with π-symmetry the orbital is πu because inversion through the center of symmetry for would produce a sign change (the two p atomic orbitals are in phase with each other but the two lobes have opposite signs), while an antibonding MO with $\pi $-symmetry is ${\pi _g}$ because inversion through the center of symmetry for would not produce a sign change (the two p orbitals are antisymmetric by phase).
-A→​${\sigma _{2s}}$ gerade , ​​ ${\pi _{2{p_x}}}$ - ungerade
-B→σ∗2s​$\sigma {*_{2s}}$ ungerade ,$\pi {*_{2{p_x}}}$-​​ gerade
-C→$\sigma {*_{2s}}$ungerade ,${\pi _{2{p_z}}}$​​ ungerade
-D→​​${\pi _{2{p_x}}}$ ungerade ,$\pi {*_{2{p_x}}}$ gerade.

Hence the correct option is (c).

Note: An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes remains the same, the orbital is gerade, and if the sign changes, the orbital is ungerade.