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“mixed” products in reactions, Eqs. (79) and (81), would be favored over
the hard acid/hard base alternative[69].
Because the electron-transfer energy is inversely proportional to the
hardness of the reagents, Eq. (41), electron-transfer effects have a propen-
sity to be insignifi cant when hard acids react with hard bases. Hard acids
and hard bases tend to be small and highly charged; therefore, the electro-
static interactions between a hard acid and a hard base have a propensity
to dominate.
Now the electrostatic contribution to the reaction energy is given by
the following equation:
ΔE qq = q A q B /( r A + r B )
(86)
here, q A , q B , r A , and r B represent the charges and radii of the acidic and
basic reactive sites.
Just as the electron-transfer contribution to the reaction energy is de-
termined using the fundamental variational principle for the density, the
electrostatic contribution to the reaction energy is determined using the
fundamental variational principle for the external potential:
when two reactants come together, their nuclei arrange themselves so
that the energy of the 'supermolecule' is minimized [71].
Hence to proceed further Ayers [3] considered the double-exchange re-
action including the electrostatic contribution to the reaction energy and by
omitting the electron-transfer contribution, he showed that the hard-hard
and soft-soft adducts are still favored. Moreover, in the single-exchange
reactions with a hard reagents, Eqs. (79) and (81), the hard acid-hard base
products are now favored over their mixed alternatives.
In the exchange reactions with a soft reagent, “mixed” compounds such
as the hard acid/soft base are favored over the soft acid/soft base product
that would be predicted based on the HSAB principle. This anomaly of
mixed product formation was explained [3] in terms of the stability of
electron-transfer contributions to the reaction energy, which favored over
the soft acid/soft base product formation.
1.2.20.2 POLARIZABILITY EFFECTS
Ayers [3] noted that the soft reagents are not small or highly charged, but
they are polarizable. Thus, he [3] further commented that “determining
 
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