= Solution
\b[<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 $\varepsilon_s,\varepsilon_p$ and their occupancies $n_s,n_p$:
$$
\chi_{\rm Allen}=C\,\frac{n_s\varepsilon_s+n_p\varepsilon_p}{n_s+n_p}.
$$
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 $(2\varepsilon_s+3\varepsilon_p)/5$.<Electronegativity is the average one-electron energy of the valence-shell electrons in ground-state free atoms by Leland C. Allen>[1989, definition and spectroscopic construction]{ref}
The approximate connection to <effective nuclear charge> is $\varepsilon_{n\ell}\sim R_\infty[Z_{\rm eff}(n\ell)/n_{\rm eff}]^2$. 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. \b[There is no universal relation $\chi_{\rm Allen}\propto Z_{\rm eff}$ across different shells], nor can one substitute a core-electron $Z_{\rm eff}$ for the valence average.
Back to article page
