Crystal Field Theory, Electronic Spectra and MO of Coordination Complexes Or why I decided to become an inorganic chemist or Ohhh!!! The Colors!!!

Corundum mineral, Al 2 O 3 : Colorless Cr  Al : Ruby Mn  Al:Mn  Al:Amethyst Fe  Al:Fe  Al:Topaz Ti &Co  Al:SapphireTi &Co  Al:Sapphire Beryl mineral, Be 3 Al 2 Si 6 O 18 : ColorlessBeryl mineral, Be 3 Al 2 Si 6 O 18 : Colorless Cr  Al : EmeraldCr  Al : Emerald Fe  Al : Aquamarine Corundum mineral, Al 2 O 3 : Colorless Cr  Al : Ruby Mn  Al:Mn  Al:Amethyst Fe  Al:Fe  Al:Topaz Ti &Co  Al:SapphireTi &Co  Al:Sapphire Beryl mineral, Be 3 Al 2 Si 6 O 18 : ColorlessBeryl mineral, Be 3 Al 2 Si 6 O 18 : Colorless Cr  Al : EmeraldCr  Al : Emerald Fe  Al : Aquamarine Gemstone owe their color from trace transition-metal ions

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 o or 10 Dq d xy d yz d xz d z2z2 x 2 -y 2 d

d xy d yz d xz d z2z2 x 2 -y 2 d  o or + 6 Dq  o or - 4 Dq

Let’s Look at 4 Co 3+ complexes: Config. Color of Complex Absorbs [Co(NH 3 ) 6 ] 3+ d 6 [Co(NH 3 ) 5 (OH 2 )] 3+ d 6 [Co(NH 3 ) 5 Br] 2+ d 6 [Co(NH 3 ) 5 Cl] 2+ d Values are in nm Greater Splitting

So there are two ways to put the electrons Low SpinHigh 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(NH 3 ) 5 X] 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 - < H 2 O < NH 3 < en < CN - < CO Spectrochemical Series Strong field ligands Large  Weak field ligands Small 

[Fe(H 2 O) 6 ] 3+ [Co(H 2 O) 6 ] 2+ [Ni(H 2 O) 6 ] 2+ [Cu(H 2 O) 6 ] 2+ [Zn(H 2 O) 6 ] 2+

S=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 t 2g x e g y configuration : CFSE = (-0.4 · x · y)  o

d 1 config. [t 2 g 1 ]: S=1/2 CFSE = –0.4  o d 2 config. [t 2 g 2 ]: S=1 CFSE = –0.8  o d 3 config. [t 2 g 3 ]: S=3/2 CFSE = -1.2   So Lets take walk along the d-block…….and calculate the CFSE

BUT WHEN YOU GET TO: d 4 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 stabilizedhigh spin (weak field) stabilized low spin (weak field) stabilized NOTE: the text uses the symbol P, for spin pairing energy Low SpinHigh 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, d 5 (Fe 3+, Mn 2+ ) configurations are most likely to form high spin complexes.

Pairing energy for gaseous 3d metal ions M 2+  (cm -1 ) M 3+  (cm -1 ) d4d4 Cr 2+ 23,500Mn 3+ 28,000 d5d5 Mn 2+ 25,500Fe 3+ 30,000 d6d6 Fe 2+ 17,600Co 3+ 21,000 d7d7 Co 2+ 22,500Ni 3+ 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 factorsf factors 3d 5 Mn(II) 8.0Br d 8 Ni (II) 8.7SCN d 7 Co(II) 9.0Cl d 3 V(II) 12.0N d 5 Fe(III) 14.0F d 3 Cr(III) 17.4oxalate d 6 Co(III) 18.2H 2 O d 9 Cu(II) 9.5NCS d 4 Cr(II) 9.5CH 3 CN d 6 Ru(II) 20.0pyridine d 3 Mn(IV) 23.0NH d 3 Mo(III) 24.6en (ethylenediamine) d 6 Rh(III) 27.0bipy (2,2’-bipyridine) d 3 Tc(IV) 30.0Phen (1:10-phenanthroline) d 6 Ir(III) 32.0CN d 6 Pt(IV) 36.0

Note: Rh 3+ and Ir 3+ are a lot different than Co 3+ g 3d < g 4d ≤ g 5d EXAMPLE: Calculate the  o (10Dq) for [Rh(OH 2 ) 6 ] 3+ in cm -1 and nm. for [Rh(pyr) 3 Cl 3 ]

Tetrahedral Coordination

 t = 4/9  o All tetrahedral compounds are High Spin

Why do d 8 metal compounds often form square planar compounds Thought experiment: Make a square planar compound by removing two ligands from an octahedral compound

Ni(II) d 8 S =1 Ni(II) d 8 S = 0 Ni(II) d 8 S = 1

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