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Crystal Field Theory, Electronic Spectra and MO of Coordination Complexes. Or why I decided to become an inorganic chemist or Ohhh!!! The Colors!!!. Gemstone owe their color from trace transition-metal ions. Corundum mineral, Al 2 O 3 : Colorless Cr Al : Ruby Mn Al: Amethyst
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Crystal Field Theory, Electronic Spectra and MO of Coordination Complexes Or why I decided to become an inorganic chemist or Ohhh!!! The Colors!!!
Gemstone owe their color from trace transition-metal ions • Corundum mineral, Al2O3: Colorless • Cr Al : Ruby • Mn Al: Amethyst • Fe Al: Topaz • Ti &Co Al: Sapphire • Beryl mineral, Be3 Al 2Si6O18: Colorless • Cr Al : Emerald • Fe Al : Aquamarine
3 6 6 3 6
d d x2-y2 z2 Do or 10 Dq dxy dyz dxz
d d x2-y2 z2 + 0.6 Do or + 6 Dq - 0.4 Do or - 4 Dq dxy dyz dxz
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 Greater Splitting 350-400 600-700 600-650 400-500 520-570 Values are in nm 570-600
OTHER QUESTIONS 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-?
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 Do (10Dq) He called it: The Spectrochemical Series Tsuchida, R. Bull. Chem. Soc. Jpn.1938, 13, 388
The Spectrochemical Series Ligand effect on Do : Small Do I- < Br- < S2- < Cl- < NO3- < F- < OH- < H2O < CH3CN < NH3 < en < bpy < phen < NO2- < PPh3 < CN- < CO Large Do Or more simply : X < O < N < C Metals also effect Do : Mn2+ < Ni2+ < Co2+ < Fe2+ < V2+ < Fe3+ < Co3+ < Mn4+ < Mo3+ < Rh3+ < Ru3+ < Pd2+ < Ir3+ < Pt2+ Fe3+ << Ru3+ Ni2+ << Pd2+ Important consequences result!!!
Strong field ligands Large D Weak field ligands Small D Spectrochemical Series I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
[Fe(H2O)6]3+ [Ni(H2O)6]2+ [Zn(H2O)6]2+ [Co(H2O)6]2+ [Cu(H2O)6]2+
S=5/2 S=5/2 S=1/2
S = 1 S = 2
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 egy configuration : CFSE = (-0.4 · x + 0.6 · y)Do
So Lets take walk along the d-block…….and calculate the CFSE d1config. [t2g1]: S=1/2 CFSE = –0.4 Do d2config. [t2g2]: S=1 CFSE = –0.8 Do d3 config. [t2g3]: S=3/2 CFSE = -1.2 Do
BUT WHEN YOU GET TO: d4 THERE ARE TWO OPTIONS!!!!! Low Spin High Spin CFSE = -1.6 Do + PCFSE = -0.6 Do When is one preferred over the other ????? It depends. (P 14,900 cm-1 / e- pair) P = DoP > DoP < Do both are equally stabilizedhigh spin (weak field) stabilizedlow spin (weak field) stabilized NOTE: the text uses the symbol P, for spin pairing energy
P , Spin Pairing Energy is composed of two terms • 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 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 Do = f (ligand) · g (metal) Do in 1000 cm-1 (Kkiesers)
Note: Rh3+ and Ir3+ are a lot different than Co3+ g3d < g4d ≤ g5d EXAMPLE: Calculate the Do(10Dq) for [Rh(OH2)6]3+ in cm-1 and nm. for [Rh(pyr)3Cl3]
Dt = 4/9Do All tetrahedral compounds are High Spin
Why do d8 metal compounds often form square planar compounds Thought experiment: Make a square planarcompound by removing two ligands from anoctahedral compound
Ni(II) d8 S = 0 Ni(II) d8 S = 1 Ni(II) d8 S =1
The Energy Levels of d-orbitals in Crystal Fields of Different Symmetries