Bonding in Octahedral and Tetrahedral Metal Complexes. Predict how the d orbitals are affected by the Metal- Ligand Bonding

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1 Bonding in Octahedral and Tetrahedral Metal Complexes 327 Molecular Orbital Theory and Crystal Field/Ligand Field Theory Predict how the d orbitals are affected by the Metal- Ligand Bonding d z 2, d x 2-y 2 Δ o 3/5 Δ o d orbitals with same energy -2/5 Δ o d xy, d xz, d yz d orbitals in ML 6 M n+ (no ligands) (octahedral) the d z 2 and d x 2-y2 orbitals point exactly at the negative charges of the ligand electrons, so electrons in these orbitals are repelled

2 Electrons go into the and orbitals to form the d 1 -d 1 possible electronic configurations: 328 d 1, d 2, d 3, d 4, d 5, d 6, d 7, d 8, d 9, d 1 d 1 d 2 d 3 d 4 d 4 d 5 d 5 H.S. L.S. H.S. L.S. d 6 d 6 d 7 d 7 d 8 d 9 d 1 H.S. L.S. H.S. L.S. d 1 one unpaired e - ; d 2 two unpaired e - ; d 3 three unpaired e - ; d 4 H.S.(4) L.S.(2); d 5 H.S.(5) L.S.(1); d 6 H.S.(4) L.S.(); d 7 H.S.(3) L.S.(1); d 8 (2); d 9 (1); d 1 ()

3 So paramagnetic electronic configurations are: (those with unpaired electrons) 329 H.S. H.S. H.S. H.S. d 1, d 2, d 3, d 4, d 5, d 6, d 7, d 8, d 9 that is, most of them! L.S. L.S. L.S. Write the electronic configurations with the correct, filling d 1 1 d 2 2 d 3 3 d 4 t 3 1 2g and t 4 2g d 5 t 3 2g e 2 g and t 5 2g H.S. L.S. H.S. L.S. d 6 t 4 2 2g and t 6 2g d 7 t 5 2g e 2 g and t 6 1 2g H.S. L.S. H.S. L.S. d d d 1 6 4

4 Consider the series below: 33 Complex o (in cm -1 ) [Co(H 2 O) 6 ] 2+ 9, Co(II) [Co(H 2 O) 6 ] 3+ [Co(NH 3 ) 6 ] 2+ [Co(NH 3 ) 6 ] 3+ 18, Co(III) 12, Co(II) 22, Co(III) [Co III F 6 ] 3-13, weak field Co(III) [Co III (CN) 6 ] 3-32, strong field Co(III) Compare the same oxidation states and then look at the ligand type. For Co(III) F - < H 2 O < NH 3 < CN - in terms of donor strength

5 Bonding in Square Planar Complexes 331 All you have to do is begin with the octahedral d-orbital splitting and consider how the repulsions between metal electrons and ligand electrons change when you pull two trans ligands away d x 2-y 2 d z 2 d xz d yz d xy Δ o d x 2-y 2 d z 2 d xy d xz, yz Removal of z ligands d x 2-y 2 Δ o d xy These orbitals are raised in energy due to increased interactions These d z 2 orbitals drop due to decreased d xz, yz repulsion along z z character

6 Bonding in Tetrahedral Complexes 332 Q. Why? d xz d yz d xy d z 2 d x 2-y 2 Δ t t 2 e much smaller relative to the octahedral field splitting, Δ o A. Interactions are weaker in tetrahedral geometries (ligands do not point to orbitals on the metal) None of the ligands in a tetrahedron point directly at the x, y, z coordinates (and therefore they do not directly contact the d z 2 or d x 2-y 2 orbitals. They have more contact with the orbitals between the axes i.e. d xy, d xz, d yz, so these are destabilized.

7 in tetrahedral geometry, one can have d 1 - d 1 configurations but since t = 4/9 o, there are no L.S. possibilities (Pairing Energy of two electrons > 4/9 o ) 333 d 1 e 1 t 2 d 4 e 2 t 2 2 d 2 e 2 t 2 d 5 e 2 t 2 3 d 3 e 2 t 2 1 d 6 e 3 t 2 3 d 7 e 4 t 2 3 d 8 e 4 t 2 4 d 9 e 4 t 2 5 d 1 e 4 t 2 6 all have unpaired electrons except for d 1! How many unpaired electrons does each have? d 1 (1) d 2 (2) d 3 (3) d 4 (4) d 5 (5) d 6 (4) d 7 (3) d 8 (2) d 9 (1) d 1 () e.g. d 6 d 7 t 2 t 2 etc., e e

8 Ligands dictate how big o and t are - For the same metal and ligands, t = 4/9 o 334 Δ o Δ o Δ o octahedral t 2g π-acceptors σ-donors π-donors t 2 t 2 Δ t Δ t Δ t e e π-acceptors σ-donors π-donors t 2 e tetrahedral

9 Ligand Field Stabilization Energy (LFSE) 335 LFSE in Octahedral Complexes each electron is assigned the energy -2/5 o each electron is assigned the energy +3/5 o (P E pairing energy) Calculate LFSE for L.S. vs H.S. Fe III [Fe III (CN) 6 ] 3-5 L.S. LFSE = 5(-2/5 o ) + 2P E = -1/5 o + 2P E = -2 o + 2P E 2e - sets are paired [Fe III (H 2 O) 6 ] H.S. LFSE = 3(-2/5 o ) + 2(+3/5 o ) = [Fe(CN) 6 ] 3- is much more stable!!

10 336 Example: d 4 high-spin 3 1 H.S. L.S. = 3 x (-2/5 o ) + 1(3/5 o ) = -6/5 o + 3/5 o = -3/5 o stabilization d 4 low-spin t 4 2g 4 x (-2/5 o ) + P E = -8/5 o + P E In the H.S. case, obviously the pairing energy P E > o In the L.S. case, the opposite is true o > P E

11 Electronic Spectroscopy Promoting an electron from thround state (g.s.) to an excited state (e.s.) takes energy. 2. The energy of the transition depends on o, t primarily for octahedral and tetrahedral complexes respectively 3. The number of transitions that we observe depends on the number of available excited states that match with thround state in terms of selection rules. Two main selection rules (a) orbital selection or Laporte selection rule (b) spin selection rule

12 Δ o hν Δ o 338 Simplistically, you may want to view the promotion of an electron from the to the level as the basis for the hv absorption peak. In the case of 1 compounds [Ti(H 2 O) 6 ] 3+, this is sort of correct. In other words the energy, hv, will be very similar to o. Realistically, transitions occur between states and not orbitals 1 can be defined by a state symbol, called a term symbol.