Handbook of Solvents - George Wypych - ChemTech - Ventech!

654 Mati Karelson
magnitude of this redistribution depends on both the dielectric constant of the solvent and
the polarizability of the solute molecule. Within the approximations of the spherical solute
cavity and the point dipole interactions between the solute and solvent, the dynamically
changed Onsager reaction field can be expressed by the following formula 14
R
l
2( −1)
( 2ε+ 1)
( ν) ( ν)
μ a + αa R ε l
=
3
a
0
[11.1.36]
Notably, the use of the macroscopic dielectric constant ε = ε 0 in the last formula is justified
only when the lifetime of the solute molecule in a given (ν-th) state is much longer
than the rotational-vibrational relaxation time of the solvent at given temperature. This is
not a valid assumption in the case of the Franck-Condon states, which have the lifetime
much shorter than the rotational-vibrational relaxation time of the solvent. Therefore, the
solvent is only partially relaxed for these states and the corresponding reaction field is characterized
by the dielectric constant at infinite frequency of external electric field, ε ∞ .Byinserting
the expression for the reaction field [11.1.36] into the equation [11.1.18] and
assuming that the static polarizability of the solute molecule is approximately equal to the
one third of the cube of Onsager’s cavity radius
a
α ν
3
( ) 0
a ≈
3
[11.1.37]
the following semiclassical equation can be obtained for the solvation energy of the ν-th
(Franck-Condon) state of the solute molecule 14
E
s
−
a a a a
=−
+ a
−
0 ν 0 ν)
( ν)
⎛ε0
1 ε∞−1⎞⎡2μ
μ 2μ
μ ⎛ ε0
−1⎞⎤
2 μ a ⎛ ε −1⎞
∞
⎜
⎟
⎝ + ⎟⎢
+ ⎜
3
3
3
ε0
2 ε∞
2⎠0
a ⎜
⎟
0 ⎝ε0
+ 2⎟⎥
− ⎜
⎣
⎠⎦
a ⎜
⎟
0 ⎝ε∞
+ 2⎟
⎠
+
⎡⎛
3 ⎞
+ ⎢ ⎜
⎟
⎝2ε
+ 1⎟
0 ⎣⎢
⎠
⎛ ε∞
−1⎞
2
+ ⎜
⎟
⎝ε
+ ⎟ 3
∞ 2⎠
a0
2
( 2ε0 + 1)
( ε0 −ε∞)
( ε −ε∞)
ε
( ) ( ( )
2
2
kT ⎛2ε0
−2⎞
⎤
+ ⎜ 3
2 0 0 3a⎜
⎟
0 ⎝2ε0
+ ⎟ ⎥
1⎠
⎦⎥
( )
∑ ( μ a )
λ≠ν ⎛ E −E
⎜1−
⎜
⎝ E0 −E
2
λν ν λ
s ps
⎞
⎟
⎠
( μ )
∑
λν ( )
a
λ≠ν E −E
λ ν
2
2
+
[11.1.38]
The solvatochromic shift due to the difference in the electrostatic solvation energy of
the ground state and the excited state of the solute, respectively, is thus given as follows:
2 ⎧⎛ε
− − ⎞⎛
+ ⎞
0 1 ε∞1
2ε0 1
ΔE s = ⎜ −
a ⎜
⎟
⎝ + + ⎟
⎜
⎟
3⎨
2 ⎠⎝
+ ⎟
0 ⎩ ε0
ε∞2
ε0
2 ⎠
⎧
⎪⎛
3 ⎞
+ ⎨ ⎜
⎟
⎝2ε
+ 1⎟
0 ⎩⎪ ⎠
2
( 2ε0 + 1)
( ε0 − ε∞)
( ε ε ) ε
0 ν ε
ν
[ ( μ a) μ aμ a ] ( μ a) ( μ a )
2 0 0 1
( ) ⎛ − ⎞ ∞
( )
− + ⎜ ⎟ −
0 2
( μ ) λ ( ) ( νλ)
a ( μ a )
∑ −∑
⎜
⎝ε
+ 2⎟
∞ ⎠
2
kT ⎡
⎤⎫
⎪
3 ⎢
⎥⎬+
2 0 − ∞ 0 3a0
λ≠
E0
−Eλ
λν ≠ Eν −E
⎣⎢
0
λ⎦⎥⎭⎪
2 2
[ ]
⎫
⎬
⎭