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The one-electron species are ${\mathrm{C}}^{5+}$ and ${\mathrm{N}}^{6+}$, so there is no electron–electron screening and the nuclear charges remain $Z=6$ and $7$.Within the stated hydrogen-like ion model, $2s$ and all three $2p$ atomic orbitals of a given species have the same energy. The energy zero is a separated electron and the residual nucleus.

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.

For a hydrogen-like ion, the Coulomb potential is stronger by the nuclear atomic number $Z$, and the energy levels scale as $Z^2$:Here $Z$ is the number of protons in the atomic nucleus; the ion has exactly one electron, so its net electric charge is $(Z-1)e$. The zero of energy is a separated stationary electron and atomic nucleus at infinite distance. This is the nonrelativistic hydrogenic approximation using the supplied $R_H$; a more precise isotope-specific result also corrects the reduced mass.