# Charge-transfer absorption complex - solvent sensitivity

This is a quote from Anslyn's Physical Organic Chemistry:

[...] One way to think of these systems is that the ground state is a weak, intermolecular complex, (D • A). Often such molecules are polar and / or quite polarizable, facilitating complex formation. Because one partner is intrinsically a good electron donor (low ionization potential) and the other is a good acceptor (high electron affinity), the excited state will have a large contribution from states such as (D+• A-) that involve electron transfer, a configuration not possible for either isolated molecule. This preferentially stabilizes the excited state, leading to a lower energy (longer wavelength) absorption. Because of this highly polar character, charge-transfer absorptions are extremely sensitive to solvent polarity, moving to longer wavelengths as the solvent polarity increases.

Earlier I read about Franck-Condon principle but I lack QM-mathematical background behind, so that's why this principle still leave me at unease, and hence my question. How is it possible that absorption wavelength is so sensitive to solvent polarity, when significant charge develops not until the complex is excited? I mean how the excited state of the molecule "knows" there is solvent ready to solvate it, since according to Frank-Condon principle solvent rearrangement is much slower than absorption.

• Without the complex A, the excitation of D would move the electron to a higher orbital of D. With the complex A, complex A has an empty orbital lower in energy than the higher orbital of D. So it takes less energy, a longer wavelength, to excite the D-A complex as opposed to the D atom. – MaxW Dec 10 '17 at 17:50

In a charge transfer DA complex the lowest excited state is the charge separated species $D^+ A^-$. The 'normal' lowest excited singlet states of D and A are each at higher energy. As the molecules D and A come together the interaction between them, although small, (the absorption spectrum is only marginally altered) is nevertheless enough to cause electron transfer upon photon absorption.
The dipolar excited state possesses its own electrostatic self-energy E, as does any dipole. This is the sum of the Born charges $\pm q$ at infinity less the Coulomb energy of bringing the charges together to from the dipole. This works out as $E= q^2/(8\pi\epsilon_0 \epsilon a)$ where $\epsilon_0$ is the permitivitty of free space and $\epsilon$ the dielectric constant of the solvent. The charges have a combined radius of $2a$. The equation can be re-written as $E=\mu^2/(4\pi\epsilon_0\; \epsilon\; L^3)$ if the dipole is defined as $\mu=qL$. Thus you can understand why the energy is lowered, and hence CT transition red shifted in a polar solvent, e.g acetoninitrile $\epsilon =37$ vs hexane $\epsilon \approx 2$.
Notes: Electron transfer upon absorption is in contrast to photoexcited electron transfer where the excited state of the donor or acceptor, whichever is excited, undergoes a thermal reaction in the excited state to form $D^+ A^-$. In the CT complex the charge is transferred on absorption between one molecule and the other, rather than from one part of the molecule to another part as in a normal absorption process. The absorption process is so rapid that, there is plenty of time for this to occur whilst the D and A molecules are close to one another and in the correct relative orientation because molecular diffusion is a relatively slow process.