The electronegativity of potassium and rubidium is reckoned at 0.82 for both. Why is it same for both of them? Shouldn't it be less for rubidium as compared to potassium owing to the addition of one extra shell in rubidium?
1
-
3$\begingroup$ $0.82$ is the Pauling electronegativity. Several other ways exist for calculating the electronegativity, for example according to Mulliken, to Allred-Rochow, and to Sanderson. On those scales, Rubidium is a little bit smaller than Potassium. See Electronegativity scales on Wikipedia. $\endgroup$– MauriceSep 27, 2020 at 13:20
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
There are a few ways that electronegativity can be measured/calculated, the original being that of Pauling, followed by others scales such as the Allred-Rochow and Mulliken electronegativity. When looking at the values obtained by Pauling, one must keep in mind that these values aren't grounded in quantum mechanics (i.e. electronegativity does not have an observable that can be measured). However, these values seen on the Pauling electronegativity charts are relative values estimated from bond energies.
Anyways, defining Pauling's electronegativity as the "ability of an atom to attract electrons" (Matthew Salem, Chem Libre Texts), one can look at two factors to estimate these values.
1. Nuclear Charge/Atomic radius - As we know from Coulombs law,
$\large F=k\frac{q_{1}q_{2}}{r^2}$
the attraction between to oppositely charged points is a function of the magnitude of each charge (q) and the distance between them ($r^2)$. So the number of protons in the nucleus along with the distance of the electron from the nucleus determine the attractive force it feels from the nucleus.
2. Electron Shielding – The electrons that are closer to the nucleus can shield bonding electrons from the attractive forces of the nucleus. The ability to shield outer electrons is greatest in the s orbitals, a little weaker in the p orbitals, and poor in the d/f orbitals.
With these factors in mind, one can approximate the effective nuclear charge $(Z_{eff})$ of an atom, which is an estimation of the attraction certain electrons will feel. Effective nuclear charge is estimated with the equation below,
$(Z_{eff}) = Z - S$
where Z is the atomic number and S is the shielding constant (number of non-valence electrons).
If you calculate the $(Z_{eff})$ for rubidium and potassium, one would find that they are equal, suggesting that the electronegativity of each atom is are approximately the same.
This link discusses $(Z_{eff})$ in more depth if you are interested.