Molecular Imaging. Markus Rudin

state of the system just after the jump. Therefore, the energy of the system just before the jump, which is equal to this, would also have to be greater. This simply means that there is a greater energy barrier to forming from the isolated reactants a suitable collision complex in which the electron can jump. Conversely, the more negative ΔF⁰, the less the energy barrier.”

      This is pictorially described in Fig. 5^* (c) and (d) of Chapter 1 in terms of PESs.

      “In spite of the extremely favorable value of ΔF⁰, the redox step of the ferric ion–hydroquinone ion reaction is seen from Table I not to proceed at every collision.”

As a matter of fact the most rapid reactions in solution are those encounter controlled or microscopic diffusion controlled such as the recombination reaction of H⁺ and OH⁻ in water with a rate constant of 1.4 × 10¹¹ liter mole⁻¹ sec⁻¹ at 25^∘C [75] while the k for the Fe⁺³ — benzo-hydroquinone anion is only 3.3 × 10⁹ liter mole⁻¹ sec⁻¹.

      “According to the theory, this is because of the preliminary solvent reorganization prior to the electron jump. However, it is of interest that the rate constant of the redox step in the ferric ion–hydroquinone reaction is much larger, on the average, than those of the isotopic exchange electron transfer reactions having zero standard free energy change considered in the preceding paper [1]. The major reasons for this difference lie in:

      (i)The large negative value of ΔF⁰ in the former reaction as compared with the zero value of the latter.

      (ii)The Coulombic attraction of the Fe⁺³–QH⁻ reactants, as compared with the Coulombic repulsion of the reactants in those isotopic exchange reactions, and moreover:

      “While the general agreement between the calculated and experimental results is satisfactory, the type of agreement obtained in Table II for the absolute value of ΔF^* in the reaction of ferric and hydroquinone ions is partly fortuitous. Two compensating approximations were employed: the a value chosen for the iron ion assumed complete dielectric saturation in the innermost hydration layer of this ion and as in [1] tends to make too small.”

      The reason for this statement is that if the dielectric saturation in the innermost hydration layer is not complete, a reorganization energy contribution of this layer to should also be considered.

      “The a value for the oxygen group correctly assumed no dielectric saturation around the uncharged reactant but made the same assumption when it was charged. This tends to make too large.”

      Because the innermost hydration layer of the ion is supposed completely unsaturated and so it contributes to

      “In the oxygen-leucoindophenol reaction, only the second of these approximations was involved and therefore there is no compensation. This may be the reason why the absolute value of ΔF^* is somewhat larger than in Table II.

In Table I, it is observed that the rate constant for the electron transfer step of the durohydroquinone is very high, namely 2.7×10¹¹ liter mole⁻¹ sec⁻¹. This value is about the maximum value that a rate constant can have in solution. The maximum corresponds to the situation in which the probability of reaction per collision is so high that the slow process in the reaction becomes the diffusion of the reactants toward each other” (vide supra). “Using a formula of Debye [76], we estimate that the rate constant for a diffusion-controlled reaction between two ions of charges +3 and −1 is about 5 × 10¹⁰ liter mole⁻¹ sec⁻¹. Within the error of the various determinations, this is about equal to the rate constant k of the electron transfer step for the durohydroquinone reaction.”

       3.34. Ionic Radii

      “The negative charge on a hydroquinone ion such as HOC₆H₄O⁻ or HOC₆H₄NHC₆H₄O⁻ is largely on the oxygen. Thus it is this atom which polarizes the dielectric. Accordingly, the appropriate polarizing radius a to be used for this charged center may be the same as that for another negatively charged oxygen, such as the OH⁻ ion. It is true that the organic residue will prevent the close approach of some of the solvent molecules and hence reduce their polarization. On the other hand, this residue is itself polarized by the charged oxygen, atomic polarization being induced, although it is less strongly polarized than the solvent. In this way, the organic residue and the solvent play analogous roles.

      A similarity between the hydroquinone ion and the hydroxyl ion in their extent of solvation and therefore in their effective polarizing radius a, can be inferred from the standard entropy change of reaction:

      In such a reaction, the translational and rotational entropies of each of the two products would be expected to be about the same as those of the corresponding two reactants. Moreover, the sum of the vibrational entropies of the products should be about equal to the sum of those of the reactants.

      If there is an appreciable entropy change in the reaction, it would be expected to arise from differences in the ability of the OH⁻ and the HQ⁻ ions to polarize the solvent molecules and therefore to vary in their entropy of solvation”

      There is of course a close relation between the ability of solvent polarization by an ion and its entropy of solvation.

“Now the standard entropy change of reaction (3.105) is readily shown to equal the difference in entropy of ionization of water and of QH₂. The entropy of ionization of QH₂ for the various hydroquinones in Table I is [70], in the respective order in which they occur in the table, −26, −32, −29, and −25 entropy units, with an average value of −28. The entropy of ionization of water is [77] −26.7eu. Accordingly, the entropy change of reaction (3.105) is seen to be zero within the experimental error. This therefore provides some basis for assuming that QH⁻ and OH⁻ have about the same polarizing radius a.

Another question discussed in the text concerned the relative abilities of QH⁻ and OH⁻ to dielectrically saturate the neighboring water molecules. The QH molecule, being uncharged, cannot saturate the dielectric. The former, being charged, could. Thus a would apparently be