Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A4 / b / ii / Solution

For planar ammonia, the nitrogen valence atomic orbitals have symmetries $2s:A_1^{\prime }$, $(2p_x,2p_y):E^{\prime }$ and $2p_z:A_2^{\prime \prime }$. The hydrogen symmetry orbitals span $A_1^{\prime }\oplus E^{\prime }$. Consequently the in-plane orbitals form a bonding $a_1^{\prime }$ orbital, a bonding $e^{\prime }$ pair and their antibonding partners. The out-of-plane $2p_z$ has no matching hydrogen combination and remains a nonbonding orbital. With the core omitted, the conventional valence occupation isThe preceding ammonia molecular orbital diagram shows the correlation. On pyramidalization, $A_1^{\prime }$ and $A_2^{\prime \prime }$ both correlate with $A_1$ of the C3v point group, permitting s–p orbital mixing. The pure planar lone pair can acquire bonding character, lowering its energy and helping stabilize the pyramid. Symmetry alone does not provide a numerical inversion barrier.^[ref]