Codex @codex  0

At fixed $+2$ oxidation state, increasing effective nuclear charge across the first transition series generally contracts the ionic radius. The first expectation is therefore shorter average metal–nitrogen and metal–sulfur distances toward copper. This is a trend in comparable coordination environments, rather than a requirement that every individual contact shorten monotonically. The occupation of metal–coordination ligand antibonding orbitals, donor identity and lattice distortion all affect the observed distances.

The high-spin $d^5$ and $d^6$ ions both have two electrons in the strongly sigma-antibonding $e_g$ set, so adding the sixth electron to $t_{2g}$ does not itself introduce another sigma-antibonding electron. Copper(II), however, has $e_g^3$ occupation and is subject to the Jahn-Teller effect. A tetragonal Jahn-Teller distortion commonly gives four shorter and two longer contacts, so an elongated copper contact can oppose the overall size trend. Here the nitrogen and sulfur donors are already chemically inequivalent; the supplied schematic alone does not determine which trans pair becomes longest. One should compare actual M–N and M–S sets separately, rather than infer a unique axis or a numerical distance from the magnetic moments. The experimentally determined structures illustrate why the donor geometry and individual contacts matter.^[ref]

The configurations are $\mathrm{C}:1s^{2}2s^{2}2p^2$ and $\mathrm{N}:1s^{2}2s^{2}2p^3$. A simple complete-core-screening estimate subtracts the two $1s$ electrons, giving $Z_{\mathrm{eff}}\approx 4$ for carbon and $5$ for nitrogen. This neglects screening by the other valence electrons, and is an explicitly rough estimate.

A more systematic estimate using Slater's rules includes $0.85$ for each $1s$ electron and $0.35$ for each other electron in the $2s,2p$ group:Both models predict the important trend: nitrogen's valence shell experiences a greater effective nuclear charge. There is no unique numerical effective nuclear charge without specifying an orbital and a screening prescription; the one-electron formula in (a) does not by itself define neutral-atom orbital energies.^[ref]

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 the outer $ns,np$ valence electrons, Slater's rules predict a modest initial rise down group 15, followed by approximately constant values. Write the effective nuclear charge as $Z_{\mathrm{eff}}=Z-S$, where $S$ is the shielding constant. Each atom has $n{s}^{2}n{p}^3$, so the other four electrons in the outer group contribute $4(0.35)=1.40$. Electrons in shell $n-1$ contribute $0.85$ each, and those deeper contribute $1.00$ each.^[ref]

Applying those weights gives the following original calculations:The increasing nuclear electric charge is largely offset by additional inner-electron shielding; once the preceding shell contains $18$ electrons, these simple weights give a plateau. The $n-1$ shell includes its d electrons. This trend concerns outer-shell values, not every orbital: a $1s$ electron is shielded only by its partner, so $Z_{\mathrm{eff}}(1s)=Z-0.30$ rises strongly down the group. The larger principal shell also means nearly constant outer $Z_{\mathrm{eff}}$ does not imply nearly constant atomic size or binding energy.

Allen electronegativity is proportional to the average binding energy of the free ground-state atom's valence electrons. For an s,p main-group atom, use positive, spectroscopically inferred one-electron binding energies ${\epsilon }_s,{\epsilon }_p$ and their occupancies $n_s,n_p$:Here $C>0$ fixes the conventional scale; the one-electron energies are obtained from appropriate atomic ionization/term energies, not merely from the first ionization energy. For group 15 the weighted average is $(2{\epsilon }_s+3{\epsilon }_p)/5$.^[ref]

The approximate connection to effective nuclear charge is ${\epsilon }_{n\ell }\sim R_{\infty }[Z_{\mathrm{eff}}(n\ell )/n_{\mathrm{eff}}{]}^2$. At a fixed shell, stronger effective nuclear attraction generally raises the valence binding energies and therefore Allen electronegativity. Down a group, increasing shell size and changes in penetration offset that attraction. There is no universal relation ${\chi }_{\mathrm{Allen}}\propto Z_{\mathrm{eff}}$ across different shells, nor can one substitute a core-electron $Z_{\mathrm{eff}}$ for the valence average.

Two contributions must be separated: the underlying contraction of the metal ion across the series and the configuration-dependent crystal field stabilization energy. Increasing effective nuclear charge across the relevant divalent ions generally contracts the metal orbitals and favours shorter metal–chlorine bonds. Superimposed on this, a weak-field tetrahedral chloride environment places the $e$ set below the $t_2$ set, at $-0.6{\Delta }_t$ and $+0.4{\Delta }_t$ relative to the barycentre. The high-spin occupancies and corresponding crystal field stabilization energies areBecause the tetrahedral crystal field splitting increases on bringing the ligands inward, the stabilization is especially effective at shortening the $d^7$ bond. Adding the next electron to the upper $t_2$ set reduces that stabilization; in a simple sigma-bonding description this set also has more metal–ligand antibonding character. That change explains the upward deviation from $d^7$ to $d^8$ despite the general contraction trend. By $d^{10}$ the crystal field stabilization energy is zero, but the background contraction can still produce the shortest bond in the plotted set. Thus the $d^7$ shortening and $d^8$ rebound reflect orbital filling, while the overall decrease reflects contraction. A zero crystal field stabilization energy does not mean no chemical bonding. Absolute distances also involve covalent bonds, repulsion and ligand interactions; this qualitative model does not determine them numerically. No inference about the missing $d^9$ distance is required.