 Solution

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.