Bonding in Coordination Compounds. Crystal Field Theory. Bonding in Transition Metal Complexes
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1 Bonding in Transition Metal Complexes 1) Crystal Field Theory (ligand field theory) Crystal Field Theory Treat igands as negative charges (they repel the e- in the d orbitals deals only with d orbitals 2) Molecular orbital theory M n+ The six negative charges are equally distributed in a sphere around the metal Bonding in Coordination Compounds Isolated transition metal atom Bonded transition metal atom Crystal field splitting ( Δ) is the energy difference between two sets of d orbitals in a metal atom when ligands are present S=5/2 S=5/2 S=1/2
2 S = 2 S = 1 S = 5/2 S = 1/2 S = 1 S = 2 S = 0 S = 2 S = 3/2 S = 1/2 Magnetic Susceptibility Bonding in Coordination Compounds Unpaired electrons: Paramagnetic compounds attracted to magnetic field weigh more Δ o Paired electrons: Diamagnetic compounds repelled by magnetic field weigh less Δ ο = hν Bonding in Coordination Compounds Spectrochemical Series Spectrochemical Series I - < Br - < - < OH - < F - < H 2 O < H 3 < en < C - < CO Weak field ligands Small Δ Strong field ligands arge Δ
3 Quiz: Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Valence electron count for neutral metal How many d electrons in? Fe(H 2 O) 6 +2 d electrons =? Quiz: Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Valence electron count for neutral metal How many d electrons in? Ti(H 2 O) 6 +4 d electrons =? A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 G. 7 H. 8 A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 G. 7 H. 8 Quiz: Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Valence electron count for neutral metal How many d electrons in? Co(H 3 ) d electrons =? Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Quiz: What is spin state? S =? Fe(H 2 O) 6 +2 A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 G. 7 H. 8 I - < Br - < - < OH - < F - < H 2 O < H 3 < en < C - < CO A. 1/2 B. 1 C. 3/2 D. 2 E. 5/2 F. 3 Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Quiz: What is spin state? S =? Sc Ti V Cr Mn Fe Co i Cu Zn Y Zr b Mo Tc Ru Rh Pd Ag Cd a Hf Ta W Re Os Ir Pt Au Hg Quiz: What is spin state? S =? Fe(C) 6-3 Co(H 3 ) 6 +3 low spin. I - < Br - < - < OH - < F - < H 2 O < H 3 < en < C - < CO A. 1/2 B. 1 C. 3/2 D. 2 E. 5/2 F. 3 I - < Br - < - < OH - < F - < H 2 O < H 3 < en < C - < CO A. 1/2 B. 1 C. 3/2 D. 2 E. 5/2 F. 3
4 Tetrahedral Coordination Δ t = 4/9Δ o All tetrahedral compounds are High Spin Why do d 8 metal compounds often form square planar compounds z M y x Thought experiment: Make a square planar compound by removing two ligands from an octahedral compound M Square Planar Complexes trans Pt(H 3 ) 2 2 d x 2 -y 2 d x 2 -y 2 d z 2 d z 2 [Pt(H 3 ) 4 ] 2+ Octahedral Square Planar Pt 2+ d 8 cis Pt(H 3 ) 2 2 H 2 O H 2 O OH 2 2 OH 2 i OH 2 H 2 O C C 2- Octahedral Coordination number =6 i(ii) d 8 S = 1 C i C 2- Square Planar (C=4) Co(I) Rh(I) Ir(I) i(ii) Pd(II) Pt(II) i Tetrahedral (C=4) i(ii) d 8 S =1 i(ii) d 8 S = 0
5 d x 2 -y 2 d z 2 Octahedral d x d 2 -y 2 z 2 Tetrahedral d z 2 d x 2 -y 2 Square Planar Molecular Orbital Theory for M 6 M n+ OH 2 2 H 2 O OH 2 i H 2 O OH 2 H 2 O Octahedral Coordination number =6 i(ii) d 8 S = 1 i 2- Tetrahedral (C=4) i(ii) d 8 S =1 2- C C i C C Square Planar (C=4) i(ii) d 8 S = 0 Metal ligands σ antibonding The an orbitals is much more diffuse in space than are the orbitals. They have a higher quantum number n and the have fewer nodes (0 or 1) versus 2 for the d orbitals. This means that when a transition metal atom bonds to other atoms, the largest interactions are with the s and p orbitals, not the d orbitals. σ bonding
6 d xy, d xz, d xy can t over lap with ligand orbitals eg* σ Antibonding y y t2g nonbonding x x d xy d xy σ bonding eg nonbonding d xy, d xz, d xy become nonbonding σ bonding between the d orbitals and the ligand orbitals is less than that between the ligand orbitals and the s and p orbitals on the metals * [Co( H 3 ) 6 ] 3+ Co 3+ d 6 2 x 6 = 12e- Total 18e- Δ o 6 M- σ bonding MO s (12 e-) + pi bonding d xy pi* - pi antibonding pi Other combinations are non bonding. odes don t match
7 pi pi pi* pi* Spectrochemical Series Quiz. For each of the following show the d orbital electron occupancies. Count d electrons? What is the spin, S? pi no pi pi antibonding Fe() -3 Mn(C) Cu() -2 4 Pd(C) -2 4 A. 0 B. 1/2 C. 1 D. 3/2 E. 2 F. 5/2 G. 3
8 The following complexes are not colored? Why not? Ti(H 2 O) 6 +4 Zn(H 2 O) 6 +2 Mn(H 2 O) 6 +2
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