Chapter 7, Harder Questions, Answers
Q6
Electronegativity is the ability of an atom or molecule to attract electrons. Why is it
then (from one definition) the average of the ionization energy and the electron
affinity (Eq. 7.32), rather than simply the electron affinity?
Equation 7.32 is
w¼IþA
2We can call this the Mulliken electronegativity. Why is electronegativity not
defined simply as the electron affinity (A)? First, we saw two derivations of
Eq. 7.32. In the first, electronegativity(w) was intuitively taken as the negative of
electronic chemical potential (the more electronegative a species, the more its
energy should drop when it acquires electrons). This led to approximating the
derivative of energy with respect to number of electrons at a point corresponding
to a species M as the energy difference of M+and M#divided by 2. In the second,
Mulliken, derivation, a simple argument equated electron transfer from X to Y to
transfer from Y to X. Both derivations clearly invoke ionization energy (I). It is no
surprise thatwshould be connected withA, but the intrusion ofImay be puzzling;
however, our surprise diminishes if we note that the more electronegative a species,
the more readily it should gain an electronandthe less readily it should part
with one.
But could we alternatively reasonably define electronegativity quantitatively just
as electron affinity? Let’s compare with the popular Pauling electronegativity scale
[1] electronegativities calculated from Eq. 7.32 and calculated simply asA. (The
Pauling scale has been criticised by Murphy et al.[2] and their criticisms were
acknowledged and improvements to the scale suggested, by Smith [3]; Matsunaga
et al., provided a long defence of Pauling’s scale [4]). Below are some electro-
negativities (preceded by a table of the calculated needed energies, at the MP2/6-
311 þG* level) by these three methods.
Energies in hartrees
Li Ca Cb F
Neutral #7.43202 #37.61744 #37.74587 #99.55959
Cation #7.23584 #37.16839 #37.33742 #98.79398
Anion #7.44251 #37.78458 #37.78458 #99.67869
aStarting from a neutral quintet 1s (^2) , 2s (^1) , 2px (^1) , 2py (^1) , 2pz 1
bStarting from a neutral triplet 1s (^2) , 2s (^2) , 2px (^1) , 2py (^1) , 2pz 0
I,A, and Mullikenw, in eV, Paulingwin kJ mol#^1. Hartrees were converted to eV
by multiplying by 27.212.
642 Answers