CH307

Inorganic Kinetics

Dr. Andrea ErxlebenRoom C150Andrea.Erxleben@nuigalway.ie

Textbook: Inorganic ChemistryC. E. Housecroft and A. G. Sharpe2nd edition: Chapter 253rd edition: Chapter 26

Topics

Kinetically labile and inert complexes

Dissociative, associative and interchange mechanisms

Activation parameters

Substitution in square planar complexes

Substitution and isomerization in octahedral complexes

Electron-transfer reactions

1 Ligand Substitution

[MLxX] + Y [MLxY] + X

X is the leaving group and Y is the entering group.

Metal complexes that undergo substitution reactions with t1/2

1 min. at 25 C are called kinetically labile. If t1/2 > 1 min.,the complex is kinetically inert (H. Taube).

Examples:

Cr(III) complexes are generally inert:

[Cr(en)2(ox)]+ + 4 H2O [Cr(ox)(H2O)4]+ + 2 en slow

(en = ethylendiamine, ox = oxalate, practical 5)

Cu(II) complexes are generally very labile:

[Cu(H2O)4]2+ + 4 NH3 [Cu(NH3)4]

2+ + 4 H2O fast

(practical 6)

There is no connection between the thermodynamicstability of a complex and its lability towardssubstitution!

Example:

[Ni(CN)4]2- is thermodynamically very stable (high complex

formation constant), but kinetically labile:

[Ni(CN)4]2- + 13CN [Ni(CN)3(

13CN)]2- + CN-

t1/2 = 30 s

1.1 Types of Substitution Mechanisms

dissociative (D)

associative (A)

interchange (I)

1. Dissociative Reaction Mechanism

MLxX MLx + X

MLx + Y MLxY

intermediate

two-step pathway formation of an intermediate coordination number of the intermediate is lower than

that in the starting complex corresponds to SN1 mechanism for organic compounds

2. Associative Reaction Mechanism

MLxX + Y MLxXY

MLxXY MLxY + X

intermediate

Example:

two-step pathway

formation of an intermediate

coordination number of the intermediate is higher thanthat in the starting complex

[PtCl4]2- + NH3 {PtCl4(NH3)}

2- [PtCl3(NH3)]- + Cl-

3. Interchange Reaction Mechanism

Bond formation between the metal and entering group

is concurrent with bond cleavage between the metal

and the leaving group.

corresponds to SN2 reaction in organic chemistry

no intermediate

MLxX + Y Y....MLx....X MLxY + X

transition state

1.2 Intermediates and Transition States

intermediate occurs at a local minimum can be detected by spectroscopy and, sometimes, isolated

transition state occurs at an energy maximum cannot be detected or isolated

transitionstate

transitionstate

Gibbsenergy

reaction coordinate

products

reactants

intermediate

reaction profile fordissociative andassociative reactionmechanism

Summary

Dissociative and associative mechanisms

involve two-step pathways and an

intermediate.

An interchange mechanism is a concerted

process where there is no intermediate.

1.3 Activation parameters

Gibbs energy of activation ∆G‡

Gibbsenergy

∆G‡1

∆G‡2

∆G = ∆H - T∆S

Gibbs energy relationship with enthalpy and entropy:

Analogously for Gibbs energy of activation:

∆G‡ = ∆H‡ - T∆S‡∆H‡ = enthalpy of activation∆S‡ = entropy of activation

Relationship between rate constant and enthalpy

and entropy of activation

ln kT

=∆H‡

RT- + ln

k’h

∆S‡

R+

k = rate constant

T = temperature in K

R = molar gas constant = 8.314 J K-1 mol-1

k’ = Boltzmann constant = 1.381.10-23 J K-1

h = Planck constant = 6.626.10-34 J s

The enthalpy of activation and the entropy of activationcan be determined by measuring the rate constant atdifferent temperatures:

Plotting lnkT

against1T

gives a straight line with

slope∆H‡

R- and intercept ln

k’h

∆S‡

R+

1T

ln kT A plot of ln k

Tvs.

1T

is called an Eyring plot.

Example: [Pt(dien)Cl]+ + H2O [Pt(dien)(H2O)]2+ + Cl-

T k [s-1]298 6.01 x 10-6

313 3.19 x 10-5

328 1.47 x 10-4

intercept = lnk’h

∆S‡

R+ = 15.89

23.76

∆S‡ = (15.89 - 23.76) x 8.314 = -65.43 J K-1 mol-1

slope =∆H‡

R- = -10001.37 ∆H‡ = 83.18 kJ mol-1

slope = -10001.37

0.000 0.001 0.002 0.003-20

-15

-10

-5

0

5

10

15

20

ln(k

/T)

1/T

∆G‡ = ∆H‡ - T∆S‡

at r.t: ∆G‡ = 83.18 – 298 x (-0.06543) = 102.68 kJ mol-1

intercept = 15.89

Volume of Activation ∆V‡

associative mechanism:

MLxX + Y {MLxX...Y} MLxXY

transitionstate

The transition state has a greater volume than the initialstate. The volume of activation ∆V‡ is positive.

dissociative mechanism:

MLxX {MLx...X} MLx + X-

transition intermediatestate

The transition state is compressed relative to the reactants,i.e. has a smaller volume. We say the volume of activation∆V‡ is negative.

The value of the volume of activation can be used todistinguish between dissociative and associativemechanisms:

A negative value of ∆V‡ indicates an associativemechanism, a positive value suggests that themechanism is dissociative.

V‡ [cm3 mol-1]

(NH3)5Co3+ +1.2

(NH3)5Rh3+ -4.1

(NH3)5Cr3+ -5.8

[(NH3)5M(H2O)]3+ + H2O* [(NH3)5M(H2O*)]3+ + H2O

Typical values:

associative mechanism:volume of transition state smaller than initial stateincrease in pressure increase in rate constant

dissociative mechanism:volume of transition state greater than initial stateincrease in pressure decrease in rate constant

The value of the volume of activation can be determinedby measuring the rate constant at different pressures.

Example:[Fe(CN)5(NH2Me)]3- + py [Fe(CN)5(py)]3- + MeNH2

p / MPa k / s-1

5 0.02625 0.02250 0.01775 0.013100 0.011

dissociativemechanism

When an associative mechanism isoperative, the rate constant increaseswith increasing pressure.

In the case of an dissociative mechanism,the rate constant decreases withincreasing pressure.

1.4 Substitution in Square Planar Complexes

square planar complexes:

metal ions with d8 configuration (RhI, IrI, PtII, PdII, AuIII) best studied: PtII complexes

rate of ligand substitution relatively slow convenient to measure

Nucleophilic substitution reactions in square planarPtII complexes usually proceed by an associativemechanism.

Evidence: negative values for ∆V‡

[ ] will be used for “concentration of “ in rateequations.

For the sake of clarity, square brackets aroundformulae of complexes are therefore omittedon the following slides.

PtL3X + Y PtL3Y + X

experimental rate law:

rate = -d[PtL3X]

dt= k1 [PtL3X] + k2 [PtL3X] [Y]

under pseudo-first order conditions (excess Y):

rate = -d[PtL3X]

dt= kobs [PtL3X]

kobs = k1 + k2 [Y]

[Y] = const.

Determination of k1 and k2:

kobs

[Y]

k1

slope k2

Plots for different entering groups (but the same solvent)

kobs

[Y]

SCN-

Br-

M–X M–Y

M–S

Y

SY S = Solvent

Origin of the two terms in the rate law:

There are two parallel ways of substitution:

direct substitution: entering group displaces leaving group k2 [Y] term

solvolytic pathway: solvent molecule displaces leavinggroup, then entering group displacessolvent

rate determining step is solvolysis independent on concentration of

entering group k1 term

[(dien)PtCl] + Y [(dien)PtY] + Cl-

kobs

[Y]

in hexane

in H2O or methanol

H2O, methanol:coordinating solvents, solvolytic pathway predominates k2 (slope) = 0 ; kobs = k1

hexane:non-coordinating solvent, only direct substitution k1 (intercept) = 0; kobs = k2 [Y]

Plots for differentsolvents, but thesame entering group

PtX

Y

L

T

L L

T

L

Y

XPt Pt

X

Y

L

T

L

squarepyramide

trigonalbipyramide

squarepyramide

Pt XY

L

T

L

Pt

XY

L

T

L+

Substitution at square planar Pt(II) is stereoretentive.

Trans-Effect

PtCl42- + 2 NH3

Pt(NH3)42+ + 2 Cl-

Pt

Cl

H3N Cl

H3Ncis isomer

PtCl

H3N Cl

NH3

trans isomer

The trans-effect is the ability of ligands to direct trans-substitution.The choice of leaving group in a square planar complex isdetermined by the ligand trans to it.

Order of trans-effect: H2O, OH- < NH3, pyridine < Cl- < Br- < I-

< NO2- < R- < PR3 « CO, CN-

Examples:

Pt

Cl

Cl Cl

OC+ NH3 Pt

Cl

Cl NH3

OC

Pt

I

I I

I

-

2-

+ 2 pyridine Pt

py

I py

I

1.5 Substitution and Isomerization inOctahedral Complexes

Examples: Cr(III), Co(III)

Volumes of activation for water exchange reactions:

metal ion ∆V‡ [cm3 mol-1]

V2+ -4.1Mn2+ -5.4Fe2+ +3.7Co2+ +6.1Ni2+ +7.2Ti3+ -12.1V3+ -8.9Cr3+ -9.6Fe3+ -5.4

associative

dissociative

associative

Substitution rates for aqua ligands in M(H2O)6n+

Examples:

Ni(H2O)62+ + Y Ni(H2O)5Y

2+ + H2O

entering ligand k [s-1]NH3 3 x 10-4

pyridine 3 x 10-4

acetate 3 x 10-4

F- 0.8 x 10-4

SCN- 0.6 x 10-4

Little variation in k consistent with adissociative mechanism

When a dissociative mechanism is operative, the rate of ligandsubstitution depends on the nature of the leaving ligand.

rate: OH- < NH3 ~ NCS- < CH3COO- < Cl- < Br- < I- < NO3-

The stronger the M-X bond, the slower the rate. The ratedetermining step involves bond breaking!

Co(NH3)5X2+ + H2O Co(NH3)5(H2O)2+ + X

The Eigen-Wilkins Mechanism

ML6 + Y ML5Y + L

For substitution reactions of octahedral metal complexes thefollowing is very often observed:

At high concentration of Y, the rate is independent of [Y],suggesting a dissociative mechanism. At low concentrationsof Y, the rate depends on [Y] and [ML6], suggesting anassociative mechanism.

These contradictions can be explained by the Eigen-Wilkinsmechanism:Metal complex and entering ligand form an encountercomplex in a pre-equilibrium step. This is followed by loss ofthe leaving ligand in the rate-determining step.

The Eigen-Wilkins mechanism:

1. Pre-equilibrium step:

ML6 + YKE

{ML6,Y}

weakly boundencounter complex

2. Rate-determining step:

{ML6,Y}k

ML5Y + L

Formation of {ML6,Y} and back reaction to ML6 and Y aremuch faster than conversion to ML5Y.The concentration of {ML6,Y} cannot be measured directlyand usually the equilibrium constant KE can only beestimated using theoretical models.

KE = equilibriumconstant

k = rate constant

KE =[{ML6,Y}]

[ML6] [Y]

[ML6] + [{ML6,Y}] = [M]total

[M]total = [ML6] + KE [ML6] [Y] = [ML6] (1 + KE [Y])

[ML6] =[M]total

1 + KE [Y]

rate = k [{ML6,Y}] = k ([M]total – [ML6])

rate = k[M]total

1 + KE[Y][M]total - =

k KE [M]total [Y]

1 + KE [Y]

rate =

At low concentrations of Y, KE[Y] « 1 can be assumed andthe equation simplifies to

rate = k KE [M]total [Y] = kobs [M]total [Y]

kobs can be measured experimentally.

k KE [M]total [Y]

1 + KE [Y]

k =kobs

KE

measured experimentally

estimated theoretically

At high concentration of Y (e.g. Y is solvent), KE [Y] » 1can be assumed and the equation simplifies to

rate = k [M]total

kobs = k KE

Base-catalysed Hydrolysis

Substitution reactions of CoIII ammine complexes arecatalysed by OH-.

Co(NH3)5X2+ + OH- Co(NH3)5OH2+ + X-

Experimentally determined rate law:

rate = kobs [Co(NH3)5X2+] [OH]

Reaction mechanism:Conjugate-base mechanism (Dcb or SN1cb mechanism)

(1) Co(NH3)5X2+ + OH- Co(NH3)4(NH2)X

+ + H2OK

(2) Co(NH3)4(NH2)X+ Co(NH3)4(NH2)

2+ + X-k

(3) Co(NH3)4(NH2)2+ + H2O Co(NH3)5(OH)2+

fast

rate =K k [Co(NH3)5X

2+] [OH]

1 + K [OH]

if K [OH] « 1, then

rate = K k [Co(NH3)5X2+] [OH]

= kobs [Co(NH3)5X2+] [OH], where kobs = K k

Co

NH3

NH3

NH3

NH3

H2N5-coordinateintermediate

Cis-trans Isomerization in Octahedral Complexes

mechanism:(1) Formation of a 5-coordinate intermediate:

MX4Y2 MX4Y + Y

(2) Berry pseudo-rotation

trans-MX4Y2 trans-MX4Y2 + cis-MX4Y2

(3) Re-formation of the M-Y bond leads to mixture of cisand trans isomer

2 Electron-transfer Processes

Fe(CN)63- + Co(CN)5

3- Fe(CN)64- + Co(CN)5

2-

+3 +2 +2 +3

Two classes of electron-transfer reactions:

outer-sphere mechanism inner-sphere mechanism

2.1 Inner-sphere mechanism

In an inner-sphere mechanism, electron transfer occursvia a covalently bound bridging ligand.

ox. stateof M:

Example:

Mechanism:

(NH3)5CoIIICl2+ + CrII(H2O)62+ (NH3)5CoIII(µ-Cl)CrII(H2O)5

4+ + H2O

(NH3)5CoIII(µ-Cl)CrII(H2O)54+ (NH3)5CoII(µ-Cl)CrIII(H2O)5

4+

(NH3)5CoII(µ-Cl)CrIII(H2O)54+ CoII(NH3)5

2+ + CrIII(H2O)5Cl2+

CoII(NH3)52+ decomposes in water to give CoII(H2O)6

2+ and NH4+

step 1: bridge formation

step 2: electron transfer via bridging ligand

step 3: bridge cleavage

CoIII(NH3)5Cl2+ + CrII(H2O)62+ CoII(NH3)5

2+ + CrIII(H2O)5Cl2+

evidence for this mechanism:

CoIII(NH3)5Cl2+ + CrII(H2O)62+ CoII(NH3)5

2+ + CrIII(H2O)5Cl2+*Cl

*Cl = radioactive Cl

If the reaction is carried out in the presence of free *Cl,labelled Cl is not incorporated into the product complex. The transferred Cl must have been bound to both

metal centres during the reaction.

Common bridging ligands in inner-sphere mechanisms:

halides OH-

CN-

NCS-

The bridging ligand is transferred from Co to Cr.

Transfer of the bridging ligand is often – but not always –observed.

(NH3)5CoII – Cl – CrIII(H2O)54+

bond cleavage

CoII more labile than CrIII

CoIII(NH3)5Cl2+ + CrII(H2O)62+ CoII(NH3)5

2+ + CrIII(H2O)5Cl2+

FeIII(*CN)63- + CoII(CN)5

3- FeII(*CN)64- + CoIII(CN)5

2-

(CN)5FeII – CN – CoIII(CN)56-

bond cleavage

The bridging ligand isnot transferred.

Kinetics:

Most inner-sphere processes exhibit second orderkinetics.

Any of the three steps (bridge formation, electrontransfer, bridge cleavage) can be rate-determining.

Typical rate constants:

Bridging ligand X k / M-1 s-1

F- 2.5 x 105

Cl- 6.0 x 105

Br- 1.4 x 106

N3- 3.0 x 106

OH- 1.5 x 106

H2O 0.1

CoIII(NH3)5X2+ + CrII(H2O)6

2+

2.2 Outer-sphere mechanism

Example:

FeII(CN)64- + FeIII(phen)3

3+ FeIII(CN)63- + FeII(phen)3

2+

In an outer-sphere mechanism, electron transfer occurswithout a covalent linkage being formed between thereactants.

MIIIL6 + MIIY6 MIIL6 + MIIIY6

1. formation of a precursor complex (reductant-oxidant pair;also called encounter complex)

MIIIL6 + MIIY6 (L5MIIIL)(YMIIY5)

2. electron transfer

(L5MIIIL)(YMIIY5) (L5M

IIL)(YMIIIY5)

3. product formation

(L5MIIL)(YMIIIY5) MIIL6 + MIIIY6

Self-exchange Reactions

In a self-exchange reaction, the left- and right-hand sides of theequation are identical. Only electron transfer, and no net chemicalreaction, takes place.

Example:

Fe(bpy)32+ + Fe(bpy)3

3+ Fe(bpy)33+ + Fe(bpy)3

2+

Gibbs energy ∆Go ~ 0, but activation energy needed

[Fe(H2O)6]3+ + [Fe*(H2O)6]

2+ [Fe(H2O)6]2+ + [Fe*(H2O)6]

3+

Gibbs energy of activation G‡ = 33 kJ mol-1

Gibbs Energy of Activation for Outer-sphere Electron-transfer Reactions

energy associated with bringing reductant and oxidanttogether (electrostatic repulsion!)

rearrangements within the solvent spheres

energy associated with changes in bond distances

loss of translational and rotational energy on formationof the encounter complex

Contributions to Gibbs Energy of Activation

energy associated with changes in bond distances

Usually, M-L bond lengths in MIII complexes are shorter thanthose in corresponding MII complexes. Oxidation / reduction ofMII / MIII complex is accompanied by change in bond length!

Franck-Condon Approximation: A molecular electronictransition is much faster than nuclear motions. Electron transfer faster than change of bond length

Let’s imagine that an electron is transferred from LxMII to LxM

III.As electron transfer is faster than change of bond length, thiswould result in excited states of LxM

II and LxMIII where the MIII-L

bond lengths are longer than typical MIII-L bonds and the MII-Lbonds are shorter than typical MII-L bonds. When bothcomplexes return to their ground states with “normal” bondlenghts, energy would be released. This would violate the firstlaw of thermodynamics, as a reaction with ∆Go = 0 cannotrelease energy.

Therefore the Frank-Condon restriction must apply: Theelectron transfer can only take place, when M-L bond distancesin the MII and MIII are the same; i.e. the bonds in LxM

III must beelongated and those in LxM

II must be compressed beforeelectron transfer takes place. The energy required forcompression / elongation of bond lengths contributes to theactivation energy.

Activation energy required varies depending onthe differences in bond lengths.

Variation of activation energies rates of outer-sphereself-exchange reactions vary considerably:

Rate constants for self-exchange reactions:

k [M-1 s-1], 25 °C

Cr(H2O)62+/3+ 2 x 10-5

Fe(H2O)62+/3+ 4.2

Co(H2O)62+/3+ 5

Co(NH3)62+/3+ 8 x 10-6

Co(en)32+/3+ 7.7 x 10-5

Fe(phen)32+/3+ 1.3 x 107

Co(phen)32+/3+ 12

Ru(bipy)32+/3+ 4.2 x 108

ML62+ + *ML6

3+ ML63+ + *ML6

2+

Examples:

Fe(bpy)32+ Fe-N = 1.97 Å

Fe(bpy)33+ Fe-N = 1.96 Å

k >106 dm3 mol-1 s-1

Ru(NH3)62+ Ru-N = 2.14 Å

Ru(NH3)63+ Ru-N = 2.10 Å

k = 104 dm3 mol-1 s-1

Co(NH3)62+ Co-N = 2.11 Å

Co(NH3)63+ Co-N = 1.96 Å

k = 10-6 dm3 mol-1 s-1

Marcus-Hush TheorySelf-exchange reaction (1):

ML62+ + ML6

3+ ML63+ + ML6

2+ rate constant k11

Self-exchange reaction (2):

M’L62+ + M’L6

3+ M’L63+ + M’L6

2+ rate constant k22

Cross-reaction:

ML62+ + M’L6

3+ ML63+ + M’L6

2+ rate constant k12

Marcus-Hush equation k12 = (k11k22K12f12)1/2

k = rate constantsK12 = equilibrium constant

for cross-reactionf ~ 1

log f =(logK12)

2

4 log(k11k22/Z2)

Z = effective collision frequency in solution

If the value of k12 calculated from the Marcus-Hush

equation agrees with the experimental value, this

provides strong evidence that the cross-reaction

proceeds by an outer-sphere mechanism.

If the Marcus-Hush equation is not fullfilled, this

indicates that another mechanism (e.g. inner-sphere

mechanism) is probably operative.

Example: Calculate the rate constant for the reaction

[Fe(CN)6]4- + [Mo(CN)8]

3- [Fe(CN)6]3- + [Mo(CN)8]

4-

from the following data:

1.) [Fe(CN)6]4- + [FeCN)6]

3- [Fe(CN)6]3- + [Fe(CN)6]

4-

k11 = 7.4 x 102 M-1 s-1

2.) [Mo(CN)8]3- + [MoCN)8]

4- [Mo(CN)8]4- + [Mo(CN)8]

3-

k22 = 2.5 x 104 M-1 s-1

3.) equilibrium constant: K12 = 1.0 x 102

4.) f12 = 0.85

Answer:

k12 = (k11k22K12f12)1/2 = (7.4 x 102 x 2.5 x 104 x 1.0 x 102 x 0.85)1/2

= 4 x 104 M-1 s-1 (experimentally found: 3 x 104 M-1 s-1)