Transcript Slide 1

Electron Transfer Reactions
Electron transfer reactions occur by
one of two fundamental mechanisms
In an inner sphere mechanism, there is
a common bridging ligand, and the
electron is transferred from the reductant
to the oxidant through the bridging ligand
In an outer sphere mechanism, there
is an encounter between the reductant
and the oxidant. The electron is
transferred from one to the other whilst
there is no change in the coordination
sphere of either.
Common bridging ligands
Cl
Common bridging ligands
Oxide and hydroxide.
But water is a very poor
bridging ligand
Common bridging ligands
Ligands which have more than one donor atom (called
ambident nucleophiles)
_
S
C
N
2S
Other examples
_
N
O
O
S
O
O
O
Common bridging ligands
Ligands which have more than one donor atom separated by a
delocalised  electron system
How do we distinguish an inner sphere from an outer sphere mechanism?
Henry Taube’s classic experiment
[CoIII(NH3)5Cl]2+
Inert: d6 Co(III)
[CrII(H2O)6]2+
Labile: d4 Cr(II)
Cl- is a bridging ligand; neither H2O nor NH3 are
Observation: products (in acidic medium) are
[CoII(H2O)6]2+ + [CrIII(H2O)5Cl]2+
and this allowed him to deduce the mechanism
The reaction could have occurred through an inner sphere pathway:
[CoIII(NH3)5Cl]2+ + [CrII(H2O)6]2+  [CoIII(NH3)5ClCrII(H2O)5]4+
electron transfer
[CoII(NH3)5ClCrIII(H2O)5]4+
break apart
Co(II) is labile
Cr(III) is inert
[CoII(NH3)5(H2O)]2+ + [CrIII(H2O)5Cl]2+
hydrolysis, in acid
[CoII(H2O)6]2+ + 5NH4+
Or it could have gone through an outer sphere pathway:
[CoIII(NH3)5Cl]2+ + [CrII(H2O)6]2+
electron transfer
[CoII(NH3)5Cl]2+ + [CrIII(H2O)6]2+
hydrolysis, in acid
[CoII(H2O)6]2+ + 5NH4+ + Cl
Co(II) is labile
Cr(III) is inert
[CoII(H2O)6]2+
[CrIII(H2O)6]2+
Observation: products (in acidic medium) are
[CoII(H2O)6]2+ + [CrIII(H2O)5Cl]2+
So there would have had to be a subsequent anation of [CrIII(H2O)6]2+ by Cl
Rate of [CrIII(H2O)6]2+ by Cl : k = 2.9  10-8 M-1 s-1
Rate of electron transfer: k = 6  10+5 M-1 s-1
Rate of electron transfer is 13 orders to magnitude faster than
rate of anation
Rate of electron transfer is 13 orders to magnitude faster than rate of
anation
The reaction could not have proceeded through an outer sphere
mechanism
Taube’s postulate:
A reaction will have proceeded through an inner sphere mechanism if
one of the products is substitution inert and it retains the bridging
ligand
i.e., the rate of the electron transfer reaction is much faster than the
rate of formation of the product by subsequent anation
Corollary:
If the rate of electron transfer is much faster than the rate of ligand
substitution on either metal ion, the reaction must proceed through
an outer sphere mechanism
Example
V2+ is inert (d3 ion, high LFSE – like Cr3+)
Ru3+ is inert (2nd transition series)
Now
[RuIII(NH3)5Br]2+ + [VII(H2O)6]2+
e transfer, k = 5.1  103 M-1 s-1
[RuII(NH3)5Br]+ + [VIII(H2O)6]3+
But
[VII(H2O)6] + Br  [VII(H2O)5Br]+ + H2O
k = 5.0  101 s-1
Hence cannot form the inner sphere complex fast enough – the anation
reaction is too slow. The reaction must have been outer sphere.
Barriers to electron transfer
• Distance
e-
Donor
Acceptor
ΨD
ΨA
Rate of an electronic transition  HDA>2 where
HDA   A Hˆ DA Dd
(Atkins, 8th ed., Chapter 9; 9th ed., Chapter 8)
Hamiltonian operator
that describes the
coupling of the two
wavefunctions
If the coupling is relatively weak,
o
 HDA 2   HDA
2 e  r
Electron coupling
when A and D are
in direct contact (r
= 0)
edge-to-edge
distance
between D
and A
Parameter that
measures the
sensitivity of the
coupling to
distance
and it turns out that the rate constant for electron transfer between D and A is
kET

2 H
2
o
DA
h
(Atkins, 8th ed., p. 897; not in 9th ed.)
e
 r
1/ 2
  


 4 RT 
3
e
G  / RT
A
Will be a constant
if D and A are the
D same
n
kET

2
e  r   3 


h
4

RT


o
2 H DA
1/ 2
eG

/ RT
ln(kET )    r  constant
ln kET
r
Nature often uses large, conjugated macrocycles to do electron transfer
Examples:
Porphyrins
Chlorophylls
Effectively increases radius of D and A, cutting down separation, and
hence increasing rate of e- transfer
distance between metals
effective distance
Q-cytochrome c oxidoreductase – Complex III or the bc1 complex
kET

2 H
2
o
DA
h
e
 r
1/ 2
  


 4 RT 
3
e
G  / RT
which is often written in simplified form as
kET
  N E e
nuclear
frequency
factor
G / RT
electronic factor
0≤κ≤1
• For fast electron transfer, maximise κE
• For fast electron transfer, maximise κE
E
1/ 2

 
 

 4 RT 
3
 minimise the reorganisation energy, λ, of inner and outer sphere
 use appropriate electronic configurations
When an e- is transferred from the D to the A molecule, it cannot change
its spin.
In many cases this is not a problem:
[Co(phen)3]3+ + [Co(bipy)3]2+ → [Co(phen)3]2+ + [Co(bipy)3]3+
low spin
eg
t2g
low spin
low spin
low spin
But in some cases - especially if there is a change of spin state - this is
a barrier to electron transfer
[Co(NH3)4Cl2]3+ + [Co(OH2)6]2+ → [Co(NH3)4Cl2]2+ + [Co(OH2)6]3+
low spin
S=0
high spin
S = 3/2
high spin
S = 3/2
eg
t2g
This cannot be a single step
low spin
S=0
[Co(NH3)4Cl2]3+ + [Co(OH2)6]2+ → {[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+
low spin
eg
t2g
high spin
excited state1
high spin
{[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+→ {[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+
excited state1
eg
t2g
high spin
excited state2
high spin
{[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+→ [Co(NH3)4Cl2]2+ + [Co(OH2)6]3+
excited state2
high spin
high spin
excited state3
eg
t2g
[Co(OH2)6]3+
low spin
and finally
Faster electron transfer occurs if an electron is removed from and added to a
non-bonding orbital (less reorganisational energy λ)
Recall that in complexes with
σ only ligands the t2g orbitals
are non-bonding and eg
orbitals are antibonding
Compare self-exchange rate constants:
[Cr(OH2)6]3+/2+
t2g3/t2g3eg1
1 × 10-5 M-1s-1
[Fe(OH2)6]3+/2+
t2g3eg2/t2g4eg2
1.1 × 10-5 M-1s-1
[Ru(OH2)6]3+/2+
t2g5/t2g6
20 × 10-5 M-1s-1
electrons
going in and
out of the t2g
orbitals makes
for fast
electron
transfer
The Inner Sphere Mechanism
The rate-determining step could be
• the formation of the bridged complex (i.e., the precursor complex)
• the electron transfer step (most commonly rate determining)
• the break-up of the successor complex
We can often rationalise which step will be rate-determining
[RuIII(NH
3)5
Cl]2+
+
[CrII(H
2O)6
]2+
K
 [RuIII(NH5)5ClCrII(H2O)5]4+ + H2O
e transfer
[RuII(NH5)5ClCrIII(H2O)5]4+ + H2O
k1
[RuII(NH3)5(H2O)]2+ + [CrIII(H2O)5Cl ]2+
Both Ru(II) and Cr(III) are inert – so we would expect the breakup of the
successor complex to be rate-limiting
Changes in mechanism are often accompanied by substantial changes in rate
The following reactions must be outer sphere reactions (why?)
oxidant
reductant
k
[CoIII(NH3)5(H2O)]3+
[RuII(NH3)6]2+
3.0 M-1 s-1
[CoIII(NH3)5(OH)]2+
[RuII(NH3)6]2+
0.04 M-1 s-1
and the hydroxo complex reacts some 100 times slower than the aqua
complex
oxidant
reductant
k
[CoIII(NH3)5(H2O)]3+
[CrII(H2O)6]2+
0.1 M-1 s-1
[CoIII(NH3)5(OH)]2+
[CrII(H2O)6]2+
1.5  106 M-1 s-1
The hydoxo complex in this case must be reacting through an inner
sphere mechanism
For the first row of the d block:
If electron transfer is rate-determining, then the rate depends
markedly on
1. the identity of the metal ion
These are the same factors that controlled rate in the outer
sphere mechanism (see later)
2. the nature of the bridging ligands
The ability of the ligand to act as an electron conductor
2.
The ability of the ligand to act as an electron conductor
[(H3N)5CoIII]
O
N
para
O
k = 100 M-1 s-1
N
[(H3N)5CoIII]
meta
[RuII(NH3)4(H2O)]
[RuII(NH3)4(H2O)]
k = 1.6  10-3 M-1 s-1
O
O
If formation of the precursor complex is rate determining, then the rate is
usually not very sensitive to the nature of the bridging ligand
This is because the ligand substitution reactions of the first row d metals are
usually dissociative
 hence does not depend strongly on the nature of the entering ligand
[L5MoxX] + [L5MredY]
rate limiting
[L5Mox] + X + [L5MredY]
[L5MoxYMredL5]
[L5MredYMoxL5]
etc
Example: V2+(aq) is oxidised to
V3+(aq) by a long series of Co3+
oxidants with different bridging
ligands
Inner sphere mechanism always suspected if good bridging ligands are
available:
Cl- Brpyrazine
N
I-
4,4'-bipyridyl
N
N3-
CNN,N-dimethylaminopyridine
N
N
NMe2
N
The Outer Sphere Mechanism
Rudolph Marcus
Energy changes during electron transfer – the Frank-Condon Principle
 Electron transfer is fast compared to nuclear motion
 Hence the nuclei are essential frozen in space during the electron transfer step
Now consider the following situation:
[FeII(H2O)6]2+ + [*FeIII(H2O)6]3+  [FeIII(H2O)6]3+ + [*FeII(H2O)6]2+
Fe(II)-O = 2.02-2.07 Å
Fe(III)-O = 2.00 Å
So suppose
e
Fe(II)OH2 + *Fe(III)OH2
Fe(III)OH2 + *Fe(II)OH2
bond too long for
Fe(III)
bond too short for
Fe(II)
spontaneous (exothermic)
Fe(III)OH2 + *Fe(II)OH2
getting energy
from nothing –
which would be
a violation of
the First Law
What actually happens:
Fe(II)OH2 + *Fe(III)OH2
shrinks
stretches
ENDOTHERMIC
Fe(II)OH2 + *Fe(III)OH2
bonds now about the same length
G‡
e transfer
Fe(III)OH2 + *Fe(II)OH2
rearrangement
Fe(III)OH2 + *Fe(II)OH2
FrankCondon
Energy
EXOTHERMIC
A given system can (in principle) be represented by a wavefunction, 
Represents hypersurface of the
reactants, reac
represents changes to all structural
parameters (bond lengths, angles,
torsions, etc) during the reaction
Parabolic function because bond
stretching and angle bending terms can
be approximated by Hooke’s law
behaviour:
E  k ( x  x0 )2  E0
E0
x0
For example, [Fe3+(H2O)6]
with a short Fe–O bond, xo
products
For example, [Fe2+(H2O)6]
with a long Fe–O bond
Electron transfer from the reactants to products can occur
when the reactant deforms along the reaction coordinate
until it structurally resembles the product (at )

For example, Fe3+––O must stretch
λ is the reorganisation energy, the energy that would be
expended to reorganise the reactant form to the product
form if no electron transfer took place
λ
For an exothermic (exergonic) reaction:
‡
‡
and rearranging:
Show, using similar reasoning, that for an endothermic (endergonic) reaction
So in general
o




G

G  1 

4
 
    G o 
 

4


2
2
 (  G o ) 2

4
2

1
( 2  2G o  G o2 )
4
So when ΔGo << λ
G  
G  
1 2
(  2G o )
4

(  2G o )
4
1
G   (  2G o )
4
Diamonds:
Circles
1
G  (  2G o )
4

λ = 200 kJ mol-1
70
60
ΔG‡ /kJ mol-1
50
40
30
20
10
0
-40
-30
-20
-10
0
10
ΔGo /kJ mol-1
20
30
40
Diamonds:
Circles
1
G  (  2G o )
4

λ = 200 kJ mol-1
700
600
ΔG‡ /kJ mol-1
500
400
300
200
100
0
-600
-400
-200
-100 0
200
400
600
-200
-300
ΔGo /kJ mol-1
Simplified eqt only
good in this region,
i.e., when |ΔGo| << λ
1
G  (  2G o )
4

G‡  Go
But Go = nFEo
G‡  Eo
G‡  Eo
The height of the activation energy barrier to
electron transfer becomes smaller and the
reaction becomes faster as the redox potential
of the couple increases
Reaction becomes activationless when ΔGo = -λ
As ΔGo becomes very large (ΔGo
< -λ), ΔG‡ increases again. Hence kET
goes through a maximum when
ΔGo = -λ
/2
/2
λ = 200 kJ mol-1
700
ΔG‡ /kJ mol-1
600
500
400
300
200
100
0
-600
-400
-200
0
200
ΔGo /kJ mol-1
ΔG‡ = 0 when Go = -λ (here, -200 kJ mol-1)...
400
600
...so RATE goes through a maximum
Relative rate
Inverted Marcus region
+ve
Go
-ve
Inverted Marcus region demonstrated
experimentally by Harry Gray (Caltech)
1990 for an iridium complex
• For fast electron transfer, minimise the reorganisation
energy, λ, of inner and outer sphere
Reorganisation of solvent often major component
Hexaaqua ions:
λ > 100 kJ mol-1
Redox centres in proteins (buried, shielded from solvent):
λ ~ 25 kJ mol-1
Reorganisation of solvent often major component
Hexaaqua ions: λ > 100 kJ mol-1
Redox centres in proteins (buried, shielded from solvent):
λ ~ 25 kJ mol-1
Fe
porphyrin
cytochrome c
Bulky, hydrophobic chelating ligands shield the metal from solvent,
lowering λ
N
N
bipyridyl (bipy)
[Ru(OH2)6]3+/2+
[Ru(bipy)3]3+/2+
kET = 20 M-1s-1
4 × 108 M-1s-1
The Marcus cross-relationship
The speed of an electron transfer reaction depends on
• the reactivity of the complex
This can be measured by the self-exchange rate
[FeII(H2O)6]2+ + [*FeIII(H2O)6]3+  [FeIII(H2O)6]3+ + [*FeII(H2O)6]2+
• Eo
because G‡  Eo
• other factors such as the collision geometry and the collision
frequency (the pre-exponential factor in the Arrhenius equation)
Suppose, as a special case, we have a redox reaction where the electron
donor, D, and the acceptor A, reversibly form an encounter complex;
where the rate determining step is the electron transfer step; and where
break up of the successor complex is fast
K DA

 DA
D + A 

kET
DA 
D+ A fast
D+ A- 
D+ + A -

 DA
D + A 

K DA
d[ P]
= kET [DA]
dt
d[ P]
= kobs [D]0 [A]0
dt
kET
DA 
D+ A fast
D+ A- 
D+ + A -
Assume [D]t ~ [D]0
[A]t ~ [A]0
K DA
[DA]
[DA]


[D]t [A]t
[D]0 [A]0
[DA] = K DA [D]0 [A]0
d[ P]
= kET K DA [D]0 [A]0
dt
kobs = kET KDA
kobs = KDA N E e

GDA
/ RT
G  =


G 

 
o
1 +
4
    G 


4
 
o
2
2
 1 2
o
o2


2


G


G


2
4
 G o G o 2



4
2
4

If ΔGo « λ then ΔGo2/4λ is negligibly small

o

G
G  = 
4
2
For the reaction
A + A → A+ + AΔGo = 0, so

AA
G
=
AA
4
and similarly

DD
G
=
Hence
k AA = Z AA e AA / 4 RT
kDD = Z DDe DD / 4 RT
where Z JJ =  N(JJ) E(JJ)
We will assume that λDA is the arithmetic mean of λAA and λDD
DA =

GDA
DD + AA
2
o
GDA
DD
AA
=
+
+
2
8
8
DD
4
ln( KDA ) = -G
o
DA
/ RT
or KDA = e
o
GDA
/ RT
Since
kobs = KDA ZDAe
ln( K DA ) =  G
o
GDA
/ 2 RT DD /8 RT AA /8 RT
e
o
DA
kobs = K DA Z DA K
=
/ RT
1/2
DA
or
e
K DA = e
o
GDA
/ RT
1/2
1/2
kDD
kAA
 1/2  1/2
Z DD Z AA
K DA Z DA 1/2 1/2 1/2
K DA kDD kAA
1/2 1/2
Z DD Z AA
kobs =
K DA kDD kAA f
2
2
K DA
Z DA
where f 
Z DD Z AA
Marcus Cross Relationship
kobs =
K DA kDD kAA f
where
2
DA
2
DA
K Z
f 
Z DD Z AA
 1
Example
Predict kobs for
[CoIII(bipy)3]3+ + [CoII(terpy)2]2+  [CoII(bipy)3]2+ + [CoIII(terpy)2]3+
Data (0 oC)
bipy
N
N
Self-exchange rate constants
[CoII/III(bipy)3]2+/3+
kAA = 9.0 M-1s-1
[CoII/III(terpy)2]2+/3+
kDD = 48 M-1s-1
N
N
N
terpy
Redox potentials:
bipy complex
terpy complex
0.34 V
0.31 V
Ecell  0.34 V - 0.31 V = 0.03 V
G   RT ln K12  nFE
 nFE 
K DA  exp 

 RT 
1 96485 C mol-1  0.03 J C-1 
 exp 

-1
-1
8.315
J
K
mol
×
273
K


 3.58
k12  kAA kDD K DA f
 9  48  3.58  1
 39 M -1s -1
Experimental value: 62 M-1 s-1
Example 2
For [CoIII(NH3)5Cl]2+ + [EuII(H2O)8]2+  [CoII(NH3)5Cl]+ + [EuIII(H2O)8]3+
[CoII(H2O)6]2+
apply Marcus equation and find k12 = 2.7 M-1s-1
The experimental value is 390 M-1s-1
The Marcus theory was developed for an outer sphere reaction. The
experimental value is very different to the theoretical value. This
suggests that this reaction did not occur through an outer-sphere
mechanism, but rather through an inner sphere mechanism with Cl
as bridging ligand