= Solution
For planar <ammonia>, the <nitrogen> valence <atomic orbitals> have symmetries $2s:A_1'$, $(2p_x,2p_y):E'$ and $2p_z:A_2''$. The hydrogen <symmetry orbitals> span $A_1'\oplus E'$. Consequently the in-plane orbitals form a bonding $a_1'$ orbital, a bonding $e'$ 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 is
$$
\boxed{(2a_1')^2(1e')^4(1a_2'')^2.}
$$
The preceding <ammonia molecular orbital diagram> shows the correlation. On pyramidalization, $A_1'$ and $A_2''$ 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.<Explaining the geometry of simple molecules using molecular orbital energy-level diagrams built by using symmetry principles by Sérgio P. Machado and Roberto B. Faria>[The ammonia molecule, Tables 5–6 and Figures 7–9: planar nonbonding $p_z$ and the pyramidal comparison]{ref}
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