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In pyramidal ammonia, the hydrogen symmetry orbitals span $A_1\oplus E$ of the C3v molecular point group. Nitrogen $2s$ and $2p_z$ both have $A_1$ symmetry and $(2p_x,2p_y)$ span $E$. In the planar D3h molecular point group, these become $A_1^{\prime }$, $A_2^{\prime \prime }$ and $E^{\prime }$, respectively, while the hydrogen combinations span $A_1^{\prime }\oplus E^{\prime }$. Thus the planar out-of-plane lone pair is a symmetry-isolated nonbonding orbital; pyramidalization permits mixing with the other $a_1$ functions.^[ref]

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]

Use the stated local axes: $z$ is radial along O–Ru, $x$ is tangential within the Ru–O plane, and $y$ is perpendicular to that plane. The oxygen's three $s{p}^2$ hybrids make the in-plane sigma framework. Its remaining $p_y$ orbital is perpendicular to the plane and can overlap with one $d_{yz}$ orbital on each Ru.

Choose the local orbital phases so their O-facing lobes agree. Of the three metal orbitals, onlyhas the appropriate combination to interact with oxygen $p_y$. The other two independent combinations have coefficients summing to zero and are nonbonding with respect to the central oxygen. Mixing $D$ with $p_y$ gives one lower pi-bonding orbital and one upper pi-antibonding orbital. Each Ru also retains its $d_{xy}$ and $d_{xz}$ orbitals as nonbonding with respect to this central-oxo pi interaction. Thus the complete low-energy diagram contains one bonding level, eight metal-nonbonding levels, and one antibonding level. The eight consist of two $d_{yz}$ combinations, three $d_{xy}$ and three $d_{xz}$ orbitals. They need not be exactly degenerate once other ligands are considered.

For the neutral formula as printed, oxide contributes $-2$ and six acetates contribute $-6$. The Ru oxidation numbers therefore sum to $+8$, giving sixteen metal d electrons overall, formally two Ru(III) and one Ru(II), possibly delocalized. Including the oxygen pi pair gives eighteen electrons, filling the bonding orbital and all eight nonbonding orbitals, leaving pi* empty. If the often encountered singly charged cation were intended instead, there would be seventeen electrons and one singly occupied nonbonding level. This charge distinction changes occupancy, not the orbital connectivity.