= Solution
Take the bond axis as $z$. The heavy-atom $2s$ and $2p_z$ <atomic orbitals> can mix with <hydrogen> $1s$ to make three sigma-symmetry <molecular orbitals>. In the qualitative valence ordering, these are a low predominantly heavy-atom $2s$ level, a bonding $\sigma$ level and a higher $\sigma^*$ level. The perpendicular $2p_x,2p_y$ orbitals form a degenerate approximately nonbonding $\pi$ pair because <hydrogen> has no valence orbitals of the matching sideways symmetry. The filled heavy-atom $1s$ core is omitted from the diagram.
Counting valence <electrons> gives
$$
\mathrm{CH}:\quad(\sigma_s)^2(\sigma_b)^2(\pi_x,\pi_y)^1,\qquad
\mathrm{NH}:\quad(\sigma_s)^2(\sigma_b)^2(\pi_x)^1(\pi_y)^1.
$$
Thus \b[both exhibit <paramagnetism>: CH has one unpaired <electron> and NH has two in its ground state.] For NH, <Hund's rules> favour separate parallel-spin occupation of the two degenerate $\pi$ orbitals. The corresponding observed ground terms are $X\,{}^2\Pi$ and $X\,{}^3\Sigma^-$; the total <bond order> is approximately one in this simple sigma-bond/nonbonding-pi description.<NIST Chemistry WebBook, SRD 69>[Methylidyne, CAS 3315-37-5, and Imidogen, CAS 13774-92-0: diatomic constants, ground-state rows]{ref}
The nitrogen-derived orbital energies lie lower than the corresponding carbon-derived ones on an approximate common energy scale, while the isolated <hydrogen> $1s$ reference is unchanged. The bonding orbitals consequently have different polarizations and mixing coefficients; drawing identical energies for C and N would miss the screening trend. The figure specifies ordering and occupancy, not measured gaps or a numerical prediction from (a).
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-ch-nh-orbitals.png]
{title=Original qualitative valence molecular-orbital diagrams of CH and NH with one and two unpaired electrons}
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