Crystal Field Theory It is not a bonding theory Method of - - PDF document

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Crystal Field Theory

 It is not a bonding theory

• Method of explaining some physical properties that occur in

transition metal complexes.

• Involves

a simple electrostatic argument which can yield reasonable results and predictions about the d orbital interactions

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in metal complexes.

d orbitals

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Consider metal ion, Mm+, lying at the centre of an

• ctahedral set of point charges.

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 Suppose the metal atom has a single d electron outside of the closed shells (Ti3+ or V4+).  In the free ion, the electron can be in any one of the 5 orbitals, since all are equivalent. Is this True??? q

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 Recall the shapes of the d orbitals

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2 groups of orbitals

t2g

dxy , dyz , dzx

eg ∆ 0.4 ∆o eg barycentre

dz

2, dx 2 ‐ y 2

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∆o 0.6 ∆o t2g

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Δo is the difference in energy between eg and t2g. The net energy of a t2g

x eg yconfiguration relative to the barycentre is called

the ligand field stabilization energy (LFSE).

LFSE = (0.4x – 0.6y)Δo

• Let us see what happens when we withdraw the 2 trans ligands in an
• ctahedral complex (let it be the z ligands)
• When this happens, we have a tetragonally distorted octahedral complex.
• As soon as the distance from Mm+ to these 2 ligands becomes greater than

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the other 4 ligands, new energy differences are established.

• z2 orbital becomes more stable than x2‐y2orbital.
• yz and xz are equivalent more stable than xy

eg dx

2

• y

2

t2g E Δo dxy dz

2

dzy , dzx

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Whether this happens depends on the metal ion and the ligands concerned. Square complexes of CoII, NiII and CuII lead to energy level diagrams shown as follows:

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dx

2

• y

2

M = CoII, NiII and CuII

eg t2g Δo Δo exactly 2/5 Δo 1/12 Δo dz

2

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• ctahedral

square MX6 MX4

dyz , dzx

High- Spin vs Low- Spin in Octahedral complexes

d 1, d 2, d 3 - simple

eg

d 4

t2g e

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high-spin low-spin t2g eg

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High- spin d 4 t2g

3 eg 1

Low- spin d 4

x = 3 y = 1 x = 4 y = 0

t2g

4 eg

x = 3 , y = 1 x = 4 , y = 0 E = (0.4x – 0.6y)Δo E = (0.4x – 0.6y)Δo

= 0.6 Δo = 1.6 Δo + P

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What is the LFSE for octahedral ions of the following configurations? (a) d 3 (b) high‐spin d 5 (a) electronic configuration : t2g

3eg 0, x = 3, y = 0

Therefore, LFSE = (0.4x – 0.6y)Δo = [(0.4)(3) – (0.6)(0)]Δo = 1.2 Δo (b) electronic configuration : t2g

3eg 2, x = 3, y = 2

Therefore LFSE = (0 4x 0 6y)Δ = [(0 4)(3) (0 6)(2)]Δ = 0

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Therefore, LFSE = (0.4x – 0.6y)Δo = [(0.4)(3) – (0.6)(2)]Δo = 0

What is LFSE for both high‐ and low‐spin d 6 configuration?

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Octahedral Geometry

Energy

eg t2g Δo

dxy

dxz dyz

dx

2

• y

2

dz

2

LFSE = (0.4x – 0.6y)Δo

Tetrahedral Geometry

Energy

eg t2g Δt

dxy

dxz dyz

dx

2

• y

2

dz

2 4 9

Δt = Δo

x y z

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Square-Planar Geometry

dx

2

• y

2

Δo

Energy

Δo

dxy

dxz dyz

dz

2 2 3

LFSE = (dx

2

• y

2 -dz 2) Δo + (dz 2 – dxy)2/3 Δo

1) What is the LFSE for high-spin d 5 of the following geometries? (a) Square planer (b) Tetrahedral (a) Square-planer (b) Tetrahedral 2) What is LFSE for both geometries (Q1) for a low spin d 5 configuration?

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The spectrochemical series

 The splitting of d orbitals in the CF model depends on a number of factors. E.g.  geometry of the complex  nature of the metal ion  charge on the metal ion  ligands that surround the metal ion Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

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• When the geometry and the metal are held constant, the

splitting of the d- orbitals increases in the following order: I- < Br- < [NCS]- < Cl-< F- < OH- < H2O < NH3 < en < CN- < CO weak-field strong-field

• The ligand‐field splitting parameter, Δo varies with the identity of the

ligand.

• In the series of complexes [CoX(NH3)5]n+ with X = I‐, Br‐, Cl‐, H2O and

NH th l f l (f X I ) th h i k (X Cl ) NH3, the colours range from purple (for X = I‐) through pink (X = Cl‐) to yellow (with NH3).

• This observation indicates that energy of the lowest electronic

transition increases as the ligands are varied along the series.

• Ligands that give rise to high energy transition (such as CO) are

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• Ligands that give rise to high energy transition (such as CO) are

referred to as a strong‐field ligand.

• Ligands that give rise to low energy transitions (such as Br‐) are

referred to as weak‐field ligand.

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Magnetic measurements

• Used to determine the number of unpaired spins in a complex, hence

identify its ground‐state configuration.

• Compounds are classified as diamagnetic if they are repelled by a

Compounds are classified as diamagnetic if they are repelled by a magnetic field and paramagnetic if they are accepted by a magnetic field.

• The spin‐only magnetic moment, μ, of a complex with total spin

quantum number is given by:

μ = 2 {S (S + 1)}½ μ

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μ = 2 {S (S + 1)}½ μB

μB = Bohr magneton

Calculated spin-only magnetic moments Ion N S μ / μB μ μB

• Calc. Expt.

Ti3+ 1 ½ 1.73 1.7 ‐ 1.8 V3+ 2 1 2.83 2.7 ‐ 2.9 Cr3+ 3 1½ 3.87 3.8

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Mn3+ 4 2 4.90 4.8 ‐ 4.9 Fe3+ 5 2½ 5.92 5.9

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The magnetic moment of a certain octahedral Co(II) complex is 4.0 μB . What is its d- electron configuration? A Co(II) complex is d 7 Two possible configurations: t2g

5eg 2 (high-spin, S = 1½) with 3 unpaired

electrons or t2g

6eg 1 (Low-spin, S = ½) with 1 unpaired electron.

μ = 2 {S (S + 1)}½ μB

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High-spin Low-spin

μ = 2 {1½ (1½ + 1)}½ μB μ = 2 {1½ (1½ + 1)}½ μB μ = 3.87μB μ = 1.73μB

• Therefore, the only consistent assignment is the

5 2

• The spin‐only magnetic moments are 3.87 μB and 1.73 μB.

The magnetic moment of an octahedral complex [Mn(NCS)6]4‐ is 6.06 μB. high-spin configuration t2g

5eg 2.

What is its electron configuration?