Crystal Field Theory, Crystal Field Theory, Electronic Spectra and MO of Electronic Spectra and MO of

Coordination ComplexesCoordination Complexes

Or why I decided to become an Or why I decided to become an inorganic chemistinorganic chemist

ororOhhh!!! The Colors!!!Ohhh!!! The Colors!!!

• Corundum mineral, Al2O3: Colorless

• Cr Al : Ruby• Mn Mn Al: Al: Amethyst• Fe Fe Al: Al: Topaz• Ti &Co Ti &Co Al: Al: SapphireSapphire

• Beryl mineral, BeBeryl mineral, Be3 3 AlAl 2 2SiSi66OO1818: Colorless: Colorless

• Cr Cr Al : Al : EmeraldEmerald• Fe Al : Aquamarine

• Corundum mineral, Al2O3: Colorless

• Cr Al : Ruby• Mn Mn Al: Al: Amethyst• Fe Fe Al: Al: Topaz• Ti &Co Ti &Co Al: Al: SapphireSapphire

• Beryl mineral, BeBeryl mineral, Be3 3 AlAl 2 2SiSi66OO1818: Colorless: Colorless

• Cr Cr Al : Al : EmeraldEmerald• Fe Al : Aquamarine

Gemstone owe their color from trace transition-metal ions

o

or10 Dq

dxy dyz dxz

dz2x2-y2

d

dxy dyz dxz

dz2x2-y2

d

o

or+ 6 Dq

o

or- 4 Dq

Let’s Look at 4 Co 3+ complexes:

Config. Color of Complex Absorbs

[Co(NH3)6]3+ d6

[Co(NH3)5(OH2)]3+ d6

[Co(NH3)5Br]2+ d6

[Co(NH3)5Cl]2+ d6

350-400 600-700

600-650

570-600520-570

400-500

Values are in nm

Greater

Splitting

So there are two ways to put the electrons

Low Spin High Spin

Which form for our 4 cobalt(III) complexes?

And why the difference between Cl- and Br-?

OTHER QUESTIONS

R. Tsuchida (1938) noticed a trend in while looking at a series of Cobalt(III) Complexes.

With the general formula : [Co(NH3)5X]

look at that! The same ones we just looked at….

He arrived a series which illustrates the effect of ligands on o (10Dq)

He called it:

The Spectrochemical Series

Tsuchida, R. Bull. Chem. Soc. Jpn. 1938, 13, 388

Ligand effect on o :

Small o

I- < Br- < S2- < Cl- < NO3- < F- < OH- < H2O < CH3CN < NH3 < en < bpy < phen < NO2- < PPh3 < CN- < CO

Large o

Or more simply :X < O < N < C

Metals also effect o :Mn2+ < Ni2+ < Co2+ < Fe2+ < V2+ < Fe3+ < Co3+ < Mn4+ < Mo3+ <

Rh3+ < Ru3+ < Pd2+ < Ir3+ < Pt2+

Fe3+ << Ru3+Ni2+ << Pd2+

Important consequences result!!!

The Spectrochemical Series

I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO

Spectrochemical Series

Strong field ligandsLarge

Weak field ligandsSmall

[Fe(H2O)6]3+

[Co(H2O)6]2+[Ni(H2O)6]2+

[Cu(H2O)6]2+[Zn(H2O)6]2+

S=5/2S=5/2 S=1/2

S = 2

S = 1

Spectrochemical Series

Another important question arises:

How does filling electrons into orbitals effect the stability (energy) of the d-orbitals relative to a spherical environment

where they are degenerate?

We use something called Crystal Field Stabilization Energy (CFSE) to answer these questions

For a t2gx eg

y configuration : CFSE = (-0.4 · x + 0.6 · y)o

d1 config. [t2g1]: S=1/2

CFSE = –0.4 o

d2 config. [t2g2]:

S=1 CFSE = –0.8 o

d3 config. [t2g3]: S=3/2

CFSE = -1.2

So Lets take walk along the d-block…….and calculate the CFSE

BUT WHEN YOU GET TO:d4

THERE ARE TWO OPTIONS!!!!!

CFSE = -1.6 o + CFSE = -0.6 o

When is one preferred over the other ?????

It depends. ( 14,900 cm-1 / e- pair)

= o > o < o

both are equally stabilized high spin (weak field) stabilized low spin (weak field)

stabilized NOTE: the text uses the symbol P, for spin pairing energy

Low Spin High Spin

, Spin Pairing Energy is composed of two terms

(a)The coulombic repulsion –

This repulsion must be overcome when forcing electrons to occupy the same orbital. As 5-d orbitals are more diffuse than

4-d orbitals which are more diffuse than 3-d orbitals, the pairing energy becomes smaller as you go down a period. As a

rule 4d and 5d transition metal complexes are generally low spin!

(b) The loss of exchange energy –

The exchange energy (Hünd’s Rule) is proportional to the number of electrons having parallel spins. The greater this number,

the more difficult it becomes to pair electrons. Therefore, d5 (Fe3+ , Mn2+) configurations are most likely to form high spin

complexes.

Pairing energy for gaseous 3d metal ions

M2+ (cm-1) M3+ (cm-1)

d4 Cr2+ 23,500 Mn3+ 28,000

d5 Mn2+ 25,500 Fe3+ 30,000

d6 Fe2+ 17,600 Co3+ 21,000

d7 Co2+ 22,500 Ni3+ 27,000

Pairing energies in complexes are likely to be 15-30% lower, due to covalency in the metal-ligand bond.

These values are on average 22% too high.

C. K. Jørgensen’s f and g factors

o = f (ligand) · g (metal)o in 1000 cm-1 (Kkiesers)

g factors f factors

3d5 Mn(II) 8.0Br -

0.72

3d8 Ni (II) 8.7SCN -

0.73

3d7 Co(II) 9.0Cl -

0.78

3d3 V(II) 12.0N3

- 0.83

3d5 Fe(III) 14.0F -

0.90

3d3 Cr(III) 17.4oxalate2-

0.99

3d6 Co(III) 18.2H2O

1.00

3d9 Cu(II) 9.5NCS -

1.02

3d4 Cr(II) 9.5 CH3CN 1.22

4d6 Ru(II) 20.0pyridine

1.23

3d3 Mn(IV) 23.0NH3

1.25

3d3 Mo(III) 24.6en (ethylenediamine)

1.28

4d6 Rh(III) 27.0bipy (2,2’-bipyridine)

1.33

4d3 Tc(IV) 30.0Phen (1:10-phenanthroline)

1.34

5d6 Ir(III) 32.0CN -

1.70

5d6 Pt(IV) 36.0

Note: Rh3+ and Ir3+ are a lot different than Co3+

g3d < g4d ≤ g5d

EXAMPLE:

Calculate the o (10Dq) for [Rh(OH2)6]3+ in cm-1 and nm.

for [Rh(pyr)3Cl3]

Tetrahedral Coordination

t = 4/9o

All tetrahedral compounds areHigh Spin

Why do d8 metal compounds often form square planar compounds

L

M

L

L L

LLM

L L

LL

z

x

y

Thought experiment: Make a square planarcompound by removing two ligands from anoctahedral compound

dx2-y2

dz2

dxy

dxz,dyz

dx2-y2

dz2

dxydxz,dyz

Octahedral Square Planar

Octahedral

dxz,dyzdxy

dz2

dx2-y2

dxz,dyzdxy

dz2

dx2-y2

TetrahedralSquare Planar

dxz,dyz

dxy

dz2

dx2-y2

OH2

Ni

H2O

OH2

OH2

H2O

H2O

2

Octahedral Coordination number =6

Ni

Cl

ClCl

Cl2-

Tetrahedral (CN=4)

Ni(II) d8 S =1

Ni

C

C

C

C

N

N

N

N

2-

Square Planar (CN=4)

Ni(II) d8 S = 0Ni(II) d8 S = 1

The Energy Levels of d-orbitals in Crystal Fields of Different Symmetries