crystal field theory, electronic spectra and mo of coordination complexes or why i decided to become...
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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
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
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]
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