High Spin Ground States: d2, d3, d6, and d7

We have taken care of the d0, d1, d4, d5, d6, d9, and d10 configurations. Now have to do d2, d3, d6, and d7 configurations. It turns out that all we have to do is solve d2.

We saw earlier that F in octahedral environment splits to A2g + T1g + T2g; in tetrahedral we would get A2 + T1 + T2. Our problem is the energy ordering. Which is GS?

Thus the 3F GS for d2 splits into 3A2g + 3T1g + 3T2g. The 4F GS for d3 splits into 4A2g + 4T1g + 4T2g. Where did the spin multiplicities come from??

But it is not so easy. Here is our approach:

We know the symmetry of the GS of the free d2 ion. How? We can get the terms for d2 using the methods applied earlier to p2, etc. They are 3F, 1D, 3P, 1G, 1S.

We identify the GS as 3F. How?

But how do we decide on what becomes the GS after the splitting due to the ligands?

We use a correlation diagram. It shows the affect of increasing the ligand field strength from zero (free ion) to very high where energy ordering is determined solely by the occupancy of the t2g and the eg orbitals.

d2

1

3

electrons

d7 = d5 + d2

2 d3=d5-hole2

3

1

2

d8=d10-hole2

holesd2 and d7 both are electrons on top of a spherical shell yielding a splitting pattern: 1, 2, 3

d3 and d8 are both two d-holes in a spherical shell, yielding reversed splitting: 3, 2, 1

d2

Real complexes

We have two electrons in the eg orbitals. It can be shown that these give rise to 1A1g,

1Eg, and 3A2g which have same energy in strong ligand field.

Connect the terms of the same symmetry without crossing.

Splitting of free ion terms.

Similarly, splitting occurs for these occupancies.

We have included the 3T1g originating from the 3P. We will need it immediately. Same symmetry as lower energy 3T1g from the 3F.

Free Ion terms

Configurations based on splitting of d electrons. Dominant in very strong fields.

F

P

Ligand field strength (Dq)

Energy

Orgel diagram for d2, d3, d7, d8 ions

d2, d7 tetrahedral d2, d7 octahedral

d3, d8 octahedral d3, d8 tetrahedral

0

A2 or A2g

T1 or T1g

T2 or T2g

A2 or A2g

T2 or T2g

T1 or T1g

T1 or T1g

T1 or T1g

And now

d2 and d7 in tetrahedral (reversed due to tetrahedral field)

and

d3 and d8 in octahedral (reversed due to d-holes).

Note the reversed ordering of the splitting coming from F (T1/T2/A2). The lower T1(g) now aims up and should cross the upper T1(g) but does not due to interaction with the upper T1(g). Now have strong curvature to avoid crossing.

Same symmetry; crossing forbidden

T1 or T1g

T1 or T1g

First look at

d2 and d7 in octahedral (2 elecs on a spherical cloud)

and

d3 and d8 in tetrahedral (double reversal: d-holes and tetrahedral)

All states shown are of the same spin. Transitions occur between them but weakly.

Note the weak interaction of the two T1, the curvature.

This curvature will complicate interpretation of spectra.

Move to Tanabe-Sugano diagrams. d1 – d3 and d8 – d9 which have only high spin GS are easier. Here is d2.

Correlation diagram for d2.

Convert to Tanabe-Sugano.

Tanabe-Sugano

Electronic transitions and spectra

d2 Tanabe-Sugano diagram

V(H2O)63+, a d2 complex

Configurations having only high spin GS

d1 d9

d3

d2

d8

Note the two lines curving away from each other.

Note the two lines curving away from each other.

Slight curving.

Configurations having either high or low spin GS

The limit betweenhigh spin and low spin

Determining o from spectra

d1d9

One transition allowed of energy o

Exciting electron from t2g to eg

Exciting d-hole from eg to t2g

Exciting d-hole from eg to t2g

Exciting electron from t2g to eg

Lowest energy transition = o

mixing

mixing

Determining o from spectra

Here the mixing is not a problem since the “mixed” state is not involved in the excitation.

d3

d8

Ground state and excited state mixing which we saw earlier.

E (T1gA2g) - E (T1gT2g) = o

For d2 and d7 (=d5+d2) which involves mixing of the two T1g states, unavoidable problem.

Make sure you can identify the transitions!!

But note that the difference in energies of two excitations is o.

d2

d7

Can use T-S to calculate Ligand Field Splitting. Ex: d2, V(H2O)63+

Again, the root, basic problem is that the two T1 s have affected each other via mixing. The energy gap depends to some extent on the mixing!

E/B

O/B

Technique: Fit the observed energies to the diagram.

We must find a value of the splitting parameter, o/B, which provides two excitations with the ratio of 25,700/17,800 = 1.44

Observed spectrum

17,800 cm-1

2: 25,700 cm-1

First, clearly 1 should correspond to 3T1 3T2 But note that the 2 could correspond to either 3T1 3A2 or 3T1 3T1. The ratio of 2/1 = 1.44 is obtained at o / B= 31

Now can use excitation energies

For1: E/B = 17,800 cm-1 /B = 29 yielding B = 610 cm-1

By using 31 = o/B = o/610 obtain o = 19,000 cm-1

The d5 case

All possible transitions forbiddenVery weak signals, faint color

Jahn-Teller Effect found if there is an asymmetrically occupied e set.

octahedral d9 complex

z2 x2-y2

xy xz yz

effect of octahedral field elongation along thex axis

x2-y2

z2

xy

xz yz

b1g

a1g

b2g

eg

Can produce two transitions.

This picture is in terms of the orbitals. Now for one derived from the terms.

Continue with d9

2D

Free ion termfor d9

2T2g

2Eg

effect ofoctahedral field

Eg

B2g

A1g

B1g

effect of elongation along z

GS will have d-hole in either of the two eg orbitals. ES puts d-hole in either of the three t2g orbitals.

For example, the GS will have the d-hole in the x2-y2 orbital which is closer to the ligands.

Some examples of spectra

Charge transfer spectra

LMCT

MLCT

Ligand character

Metal character

Metal character

Ligand character

Much more intense bands

Coordination ChemistryReactions of Metal Complexes

Substitution reactions

MLn + L' MLn-1L' + L

Labile complexes <==> Fast substitution reactions (< few min)Inert complexes <==> Slow substitution reactions (>h)

a kinetic concept

Not to be confused withstable and unstable (a thermodynamic concept; Gf <0)

Inert Intermediate Labile

d3, low spin d4-d6& d8 d8 (high spin) d1, d2, low spin d4-d6& d7-d10

MLnX + Y MLnY + X

Mechanisms of ligand exchange reactionsin octahedral complexes

Ia if associationis more important

Id if dissociationis more important

Dissociative (D)

MLnX-x

MLn

YMLnY

Associative (A)

MLnXY

YMLnX-X

YMLn

Interchange (I)

MLnXY

Y- -MLn- -X-X

YMLn

Association or Dissociation step may be more important and the process classifiedas such.

Kineticsof dissociative reactions

Using Steady State Approximation, concentration of ML5 is always very low; rate of creation = rate of consumption

Kineticsof interchange reactions

Fast equilibriumK1 = k1/k-1

k-1 >> k2

Again, apply Steady State.

For [Y] >> [ML5X]common experimental

condition!

Kinetics of associative reactions

Principal mechanisms of ligand exchange in octahedral complexes

ML5Xk1

slowML5 + X

k2

fast

+YML5Y

r = k1 [ML5X]

ML5X + Yk1

slowML5XY

k2

fast

-XML5Y

r = k1 [ML5X][Y]

Dissociative

Associative

Dissociative pathway(5-coordinated intermediate)

Associative pathway(7-coordinated intermediate)

MOST COMMON

Experimental evidence for dissociative mechanisms

Rate is independent of the nature of L

Experimental evidence for dissociative mechanisms

Rate is dependent on the nature of L

Inert and labile complexesSome common thermodynamic and kinetic profiles

Exothermic(favored, large K)

Large Ea, slow reaction

Exothermic(favored, large K)

Large Ea, slow reactionStable intermediate

Endothermic(disfavored, small K)Small Ea, fast reaction

LM

L L

L

L

X

L

ML L

L

L

X

L

ML L

L

L

G

Ea

Labile or inert?

LFAE = LFSE(sq pyr) - LFSE(oct)

Why are some configurations inert and some are labile?

Inert !

Substitution reactions in square-planar complexesthe trans effect

T

M

L X

L T

M

L Y

L

+X, -Y

(the ability of T to labilize X)

Synthetic applicationsof the trans effect

Cl- > NH3, py

Mechanisms of ligand exchange reactions in square planar complexes

-d[ML3X]/dt = (ks + ky [Y]) [ML3X]

LM

L L

X

LM

L L

Y

LM

L L

X

LM

L L

X

LM

L L

S

LM

L L

S

S

Y

Y

+Y

+S

-X

+Y

-S

-X

Electron transfer (redox) reactions

M1(x+)Ln + M2

(y+)L’n M1(x +1)+Ln + M2

(y-1)+L’n

-1e (oxidation)

+1e (reduction)

Very fast reactions (much faster than ligand exchange)

May involve ligand exchange or not

Very important in biological processes (metalloenzymes)

Outer sphere mechanism

[Fe(CN)6]4- + [IrCl6]2- [Fe(CN)6]3- + [IrCl6]3-

[Co(NH3)5Cl]2+ + [Ru(NH3)6]2+ [Co(NH3)5Cl]+ + [Ru(NH3)6]3+

Reactions ca. 100 times fasterthan ligand exchange(coordination spheres remain the same)

r = k [A][B]

Ea

A B+

A B

A' B'+

G

"solvent cage"

Tunnelingmechanism

Inner sphere mechanism

[Co(NH3)5Cl)]2+ + [Cr(H2O)6]2+ [Co(NH3)5Cl)]2+:::[Cr(H2O)6]2+

[Co(NH3)5Cl)]2+:::[Cr(H2O)6]2+ [CoIII(NH3)5(-Cl)CrII(H2O)6]4+

[CoIII(NH3)5(-Cl)CrII(H2O)6]4+ [CoII(NH3)5(-Cl)CrIII(H2O)6]4+

[CoII(NH3)5(-Cl)CrIII(H2O)6]4+ [CoII(NH3)5(H2O)]2+ + [CrIII(H2O)5Cl]2+

[CoII(NH3)5(H2O)]2+ [Co(H2O)6]2+ + 5NH4+

Inner sphere mechanism

Reactions much faster than outer sphere electron transfer(bridging ligand often exchanged)

r = k’ [Ox-X][Red] k’ = (k1k3/k2 + k3)

Ox-X + Red Ox-X-Redk1

k2

k3

k4Ox(H2O)- + Red-X+

Ea

Ox-X Red+

Ox-X-Red

G

Ox(H2O)- + Red-X+

Tunnelingthrough bridgemechanism