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Ligand Field Theory. Ligand Field Thoery (aka Crystal Field Theory) assumes that the only interaction between the metal and the ligands is purely ionic. Imagine the five d-orbitals in an isolated metal ion; energetically degenerate.

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ligand field theory
Ligand Field Theory
  • Ligand Field Thoery (aka Crystal Field Theory) assumes that the only interaction between the metal and the ligands is purely ionic.
  • Imagine the five d-orbitals in an isolated metal ion; energetically degenerate.
  • Next, imagine that the metal ion is surrounded by s spherically symmetrical field of negative ions and that that sphere is contracting towards the metal. As the sphere contracts, the repulsion between the negative field and the negative electrons in the metal will cause the energy of all of the d-orbitals to rise, but rise equally. That is, they will remain degenerate.
  • In “real life” can’t have an infinite number of ligands, so symmetry will be less.

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spherical field

M

M

E

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no ligand field

octahedral ligand field
Octahedral Ligand Field
  • Imagine six ligands approaching the free metal ion to form an Oh ligand field. Assume that the six approach the metal along the x, y, and z axes.
  • Under these conditions, which d-orbitals will interact most strongly with the ligands? dx2-y2 and dz2.
  • These two orbitals (designated eg) will increase in energy to the same extent. The remaining orbitals (designated t2g) will also be repelled by the incoming ligands, but to a much lesser extent.
  • The difference in energy between the eg and t2g orbitals is defined as “10Dq”.
  • Relative to a barycenter of the spherical field, the eg orbitals are seen as being repelled by 6Dq, while the t2g orbitals are stabilized by 4Dq.
ligand field stabilization energy lfse
Ligand Field Stabilization Energy (LFSE)
  • What happened when 1 electron is added to the Oh ligand field?
    • it goes in a t2g orbital and, therefore, d1 (t2g1eg0) has an LFSE of -4Dq.
  • A second electron will also go into a t2g orbital (won’t pair).
    • thus, d2 (t2g2eg0) has an LFSE of -8Dq.
  • A third electron will also go into a t2g orbital.
    • thus, d2 (t2g3eg0) has an LFSE of -12Dq.
  • Note now that the t2g orbitals are half filled. If we add a fourth electron there are two places it can go.
  • Which is more stable?

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ligand field stabilization energy
Ligand Field Stabilization Energy
  • The difference in energy bewteen the t2g and eg orbitals is relatively small. Also, pairing “costs” energy (has to, otherwise wouldn’t need Hund’s Rule).
  • So, need to determine which “costs” less, putting the electron in the eg orbital or pairing with an electron already in the t2g orbital.
  • Weak Field Case (a.k.a High Spin).
    • if the splitting of the orbitals (10Dq) is small with respect to the pairing energy (P), the fourth electron will enter the eg orbital.
    • hs d4 (t2g3eg1) has an LFSE of (3 x -4Dq) + (1 x 6Dq) = -6Dq.

S = (4 x ½) = 2

    • hs d5(t2g3eg2) has an LFSE of (3 x -4Dq) + (2 x 6Dq) = 0Dq.

so, no stabilization by Oh field

S = (5 x ½) = 5/2

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ligand field stabilization energy5
Ligand Field Stabilization Energy
  • Weak Field Case (a.k.a High Spin).
    • hs d6 (t2g4eg2) has an LFSE of (4 x -4Dq) + (2 x 6Dq) = -4Dq.

has to pair; better to pair in lower t2g orbital.

S = (4 x ½) = 2

    • hs d7(t2g5eg2) has an LFSE of (5 x -4Dq) + (2 x 6Dq) = -8Dq.

so, no stabilization by Oh field

S = (3 x ½) = 3/2

  • d8 to d10 neither hs nor ls.
    • hs d8 (t2g6eg2) has an LFSE of (6 x -4Dq) + (2 x 6Dq) = -12Dq.
    • hs d9(t2g6eg3) has an LFSE of (6 x -4Dq) + (3 x 6Dq) = -6Dq.
    • hs d10(t2g6eg4) has an LFSE of (6 x -4Dq) + (4 x 6Dq) = -8Dq.

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ligand field stabilization energy6
Ligand Field Stabilization Energy
  • Strong Field Case (a.k.a Low Spin).
    • if the splitting of the orbitals (10Dq) is large with respect to the pairing energy (P), it is energetically more favorable for the electrons to pair than to go into the upper orbital.
    • ls d4 (t2g4eg0) has an LFSE of (4 x -4Dq) + P = -16Dq + P.

S = (2 x ½) = 1

    • ls d5(t2g5eg0) has an LFSE of (5 x -4Dq) + 2P = -20Dq + 2P.

S = (1 x ½) = 1/2

    • Pairing energies based on change in pairing on going from hs to ls and are NOT calculated from absolute number of pairs.
    • ls d6(t2g6eg0) has an LFSE of (6 x -4Dq) + 2P = -24Dq + 2P.

S = (0 x ½) = 0

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a comparison of the weak field and strong field cases
A Comparison of the Weak Field and Strong Field Cases

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10Dq > P

low spin case

10Dq < P

high spin case

evidence for ligand field stabilization energy
Evidence for Ligand Field Stabilization Energy
  • First evidence that LFSE might be important in transition metal complexes came from an examination of lattice energies (E α 1/r).
    • Since ionic radii decrease gradually across 1st transition row, would expect gradual increase in energy.
    • However, curve is not smooth and only Ca+2, Mn+2 and Zn+2 fall on theoretical straight line.
    • What can you say about the LFSE of these 3 ions (assume high spin case)? d0 d5 d10 LFSE for these? = 0
    • Why are the other ions above this line?

d1 and d6 LFSE = -4Dq

d2 and d7 LFSE = -8Dq

d3 and d8 LFSE = -12Dq

d4 and d9 LFSE = -6Dq

measuring 10dq
Measuring 10Dq

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  • How is 10Dq measured? Look at, for example, Ti(H2O)6+3.
    • Ti+3 is a d1 ion.
    • Say this ion absorbs a photon of appropriate energy to allow transition to the eg level. What can you say about the energy of this photon? Ephoton = 10Dq.
    • The wavelengths of these photons are mostly in the UV and visible ranges. Therefore, can use spectrophotometer to determine λmax, and from that the energy.
    • For λmax = 500nm (20,000 cm-1 wavenumbers).
    • This corresponds to 240 KJ/mole = 10Dq.
    • Note: the d1 case is quite simple; more involved for other situations (to be discussed later.)

10Dq = 240 KJ/mole

λmax

calculating 10dq from max
Calculating 10Dq from λmax

hc (6.62 x 10-34 J.s) (3.0 x 108 m/s)

λ (500 x 10-9 m)

ΔE = hν = =

= (3.972 x 10-19 J) (6.022 x 1023 mole-1)

= (2.39 x 105 J/mole) (1KJ/103J)

= 239 KJ/mole

factors affecting pairing energies and 10dq
Factors Affecting Pairing Energies and 10Dq.
  • Metals in Oh coordination have the potential to be either high or low spin.
  • So, it is important to be aware of factors affecting P and 10Dq.
  • Pairing Energy a result of:
    • inherent repulsion of “forcing” two electrons to occupy the same orbital; this energy more or less constant from one metal to another.
    • energy which is required as electrons with parallel spins are forced to “flip” in order to pair (a.k.a. exchange energy).
    • Pairing energies tend to fall in a fairly narrow range (~230 – 280 KJ/mole).

if 10Dq > P then low spin

10Dq < P then high spin

factors affecting pairing energies and 10dq12
Factors Affecting Pairing Energies and 10Dq
  • 10Dq affected by:
    • Oxidation state of the metal; as the oxidation number increases, so does 10Dq. Can justify this in terms of the ligand field model. The increased positive charge on the metal will bring the ligands in towards it, thus destabilizing the eg orbitals.
factors affecting pairing energies and 10dq13
Factors Affecting Pairing Energies and 10Dq
  • 10Dq affected by:
    • Number and geometry of the ligands; the splitting in Oh field ~9/4 times that in Td field. Due to 1/3 fewer ligands in Td case (will decrease the field by 33.3%) and the fact that the ligands are not overlapping with orbitals as much as in the Oh case.
factors affecting pairing energies and 10dq14
Factors Affecting Pairing Energies and 10Dq
  • 10Dq affected by:
    • Nature of the ligands. Table 16.2 shows 10Dq’s for various Cr(III) complexes. Note that as the ligands change, so does 10Dq. Table 9.7 shows 10Dq values for a number of metals with a number of ligands. Notice that although the absolute values of the 10Dq’s are different, the trend of the effect that a ligand has on a metal ion is the same from metal to metal.

Cr(III): Cl- < dtp < F- < H2O < NH3 < en < CN

Co(III): dtp < H2O < NH3 < en < CN

Rh(III): Cl- < dtp < H2O < NH3 < en < CN

Ir(III): Cl- < dtp < NH3 < en

factors affecting pairing and 10dq
Factors Affecting Pairing and 10Dq
  • 10Dq affected by:
    • Nature of the ligands.
    • If the ligands are listed in order of increasing field strength, a spectrochemical series is obtained.
    • Although no one metal could be used to derive this series, overlapping sequences can be used.
    • You should know the following spectrochemical series:
    • Nature of the metal.
    • across the periodic table, no large change, but down the table 10Dq increases significantly (3d < 4d < 5d).
    • result of this is that most complexes of the 2nd and 3rd transition series are low spin.

I- < Br- < Cl- < dtp < F- < OH- < H2O < NH3 < en < CN < CO

weak field strong field

predicting 10dq values
Predicting 10Dq values
  • C.K. Jørgensen (no relation) has attempted to predict values of 10Dq based on the spectrochemical series and various metals.

e.g. [CrCl6]-3

10Dq = fligand x gion x 103

= (0.78) x (17.4) x 103

= 13,572 cm-1 (736 nm)

(compare to 13,200 on Table 9.7)

10Dq = fligand x gion x 103cm-1

note that increasing f values correspond to increasing field strength in the spectrochemical series.

jahn teller theorem
Jahn-Teller Theorem
  • For a non-linear molecule in an electronically degenerate state, distortion must occur to lower the symmetry, remove the degeneracy, and lower the energy. This is an empirical observation.
  • What is meant by degeneracy?

e.g. d1 could be: or or triply degenerate.

while d3 can only be: singularly degenerate.

  • How can this degeneracy be removed?
  • Through what is called “tetragonal distortion.”

The ligands along the z-axis moved out (z-out) or moved in (z-in).

As a result, these ligands interact less with those orbitals having a d-component (dz2, dxz and dyz) with a resultant stabilization of these orbitals; concomittantly, the dx2-y2 and dxy orbitals will increase in energy.

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jahn teller theorem18
Jahn-Teller Theorem
  • The splitting by the eg orbitals (δ1) will be larger than splitting by the t2g orbitals (δ2). Both δ1 and δ2 are minute compared to 10Dq.
  • Although there are two possible tetragonal distortions, the Jahn-Teller theorem does not predict which will occur, nor does it address the extent of distortion.
  • Experimentally, z-out is seen most often.

z-out Oh z-in

square planar coordination
Square Planar Coordination
  • If we start with Oh coordination and then introduce tetragonal distortion (z-out) until those ligands along the z-axis are removed to infinity, a square planar coordination is achieved.
  • Metal ions which have a d8 configuration and ligands high in the spectrochemical series (low spin) favor square planar coordination. For example, [Ni(CN)4]-2 , [PdCl4]-2 and [Pt(NH3)4]+2 are all d8 square planar complexes.
  • Why only low spin square planar complexes?
  • The weak field/high spin case doesn’t result in any stabilization, whereas the low spin situation does result in stabilization.
  • Note: xy → x2-y2 = 10Dq.

Sq Pl

Oh

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tetrahedral coordination
Tetrahedral Coordination
  • Again, imagine the set of five d-orbitals in spherical symmetry. Next imagine eight ligands approaching the central metal from along the corners of a cube.
  • In this case the ligands don’t approach directly any of the d-orbitals, but do come closest to the t2g orbitals, thus destabilizing them. This would represent cubic symmetry.
  • The energy level diagram for tetrahedral symmetry is qualitatively similar to cubic symmetry, but 10Dq is only half as large.
  • Because tetrahedral systems have no center of symmetry, can’t have t2g and eg, just t2 and e.
  • Because 10Dq for tetrahedral complexes is 4/9 of the octahedral case, only weak field (high spin) species are observed.
molecular orbital theory
Molecular Orbital Theory
  • Although LFT is useful and can be used to account for a large number of observations, it does have its drawbacks.
    • It assumes that both metal and ligands are point charges (with no volume).
    • It assumes that all bonding is strictly ionic.
    • Anomoly: the ligand with the strongest field is CO, which has no ionic charge.
  • Is there covalent bonding?
    • Indirect evidence: electron-electron repulsions in metal complexes are somewhat less than in free ions. This may be due to the formation of molecular orbitals, which are larger than atomic orbitals and would reduce repulsions. This is known as the nephelauxetic effect (cloud expanding).
    • Direct evidence: in electron spin resonance (esr) unpaired electrons can align parallel or antiparallel to an applied magnetic field and we can get a transition by applying radiation (like nmr). An isolated metal ion shows a “singlet”. However, many complexes show a more complicated spectra indicating that the electron is spending some of its time “away” from the metal. That is, it is behaving covalently.
comparison of lft and mot approaches to bonding
Comparison of LFT and MOT Approaches to Bonding
  • Consider a simple system: a Lewis base (B) having one orbital with an electron pair to be donated, and a Lewis acid (A+), with two sp orbitals available for bonding and a single odd electron.
  • The Ligand Field Theory explanation:
    • In the isolated A+ ion the two sp orbitals (A1 and A2) are degenerate.
    • When the ligand approaches, the two degenrate orbitals will be split into higher and lower energies; the odd electron will occupy the lower energy orbital.
    • A2* → A1* represents 10Dq.

B :+ A ● + → [AB]+

A1 A2

comparison of lft and mot approaches to bonding23
Comparison of LFT and MOT Approaches to Bonding
  • The Molecular Orbital Theory explanation:
    • As B approaches, it will overlap and mix with A1 forming bonding (Ψb) and antibonding (Ψa) orbitals.
    • The second metal orbital, A2, faces away from B and won’t overlap; it is a non-bonding orbital (Ψn).
    • Ψn → Ψa represents 10Dq.
    • Note similar results.

B :+ A ● + → [AB]+

A1 A2

using symmetry to produce a mo correlation diagram
Using Symmetry to Produce a MO Correlation Diagram

Co+3 6NH3

  • Assume that the compound [Co(NH3)6]+2 has pure Oh symmetry.
  • From the Oh Character Table, determine the irreducible representations to which the 3d, 4s and 4p orbitals of cobalt belong.
  • Next, use the σ-bonds as a basis to find the reducible representation for the ligands.
  • Reduce this: Γ = A1g + Eg + T1u

t1u

a1g

eg + t2g

4p

4s

3d

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a1g + eg + t1u

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6L

Oh E 8C3 6C2 6C4 3C2’ i 6S4 8S6 3σh 6σd

Γ 6 0 0 2 2 0 0 0 4 2

using symmetry to produce a mo correlation diagram25
Using Symmetry to Produce a MO Correlation Diagram

Co+3 6NH3

t1u*

a1g*

eg*

t2g

eg

a1g

t1u

  • Both the maetal and the ligands have orbitals with t1u symmetry, so there will be overlap and the formation of triply degenerate MO’s.
  • The a1g atomic orbitals will also overlap.
  • Similarly the doubly degenerate eg orbitals.
  • There are no t2g ligand orbitals, so those metal orbitals are non-bonding.
  • What does the center of this diagram look like?

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t1u

a1g

eg + t2g

4p

4s

3d

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a1g + eg + t1u

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6L

pi bonding in metal complexes
Pi-Bonding in Metal Complexes
  • Can only be accomodated through MO Theory.
  • Π-bonding only possible via t2g orbitals as the eg’s are involved in σ bonding.
  • Three types of metal-ligand π-bonding.
  • Simple p-π type orbitals.
    • e.g. Cl-.
  • Simple d-π type orbitals.
    • e.g. phosphenes R3P:
  • Molecular orbitals.
    • π* orbitals of polyatomic ligands.
    • e.g. CO, CN, pyridine.
pi bonding in the octahedral case
Pi-Bonding in the Octahedral Case.
  • As a basis, assume each ligand has a possibility of a pair of π-orbitals perpendicular to the σ-orbitals.
  • This reduces to: t1u + t2u + t2g + t1g. Only the t2g can overlap with the metal ion.
  • Case 1: empty π-orbitals on ligand that are higher in energy than the metal t2g orbitals (e.g. phosphenes, CO, CN).
    • in this case π-bonding increases 10Dq.
  • Case 2: ligand has filled π-bonding orbitals which are lower in energy than metal t2g orbitals (e.g. F-, O2-).
    • in this case π-bonding decreases 10Dq.
  • These two cases explain the discrepencies in the spectrochemical series!

Oh E 8C3 6C2 6C4 3C2’ i 6S4 8S6 3σh 6σd

Γ 12 0 0 0 -4 0 0 0 0 0