Crystal Field Theory

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Lenard Moody
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1 6/4/011 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 in metal complexes. CHEM61HC/SS/01 d orbitals CHEM61HC/SS/0 1

6/4/011 Consider metal ion, M m+, lying at

a single d electron outside of the closed

2 6/4/011 Consider metal ion, M m+, lying at the centre of an octahedral set of point charges. CHEM61HC/SS/03 Suppose the metal atom has a single d electron outside of the closed shells (Ti 3+ or V 4+ ). Inthefreeion,theelectroncanbeinanyoneofthe5orbitals, sinceallareequivalent. Is this True??? Recall the shapes of the d orbitals CHEM61HC/SS/04

3 6/4/011 CHEM61HC/SS/05 groups of orbitals d xy, d yz, d zx t g d z, d x y barycentre 0.4 o o 0.6 o t g CHEM61HC/SS/06 3

4 6/4/011 is the difference in energy between and t g. The net energy of a t gx y configuration relative to the barycentre is called the ligand field stabilization energy (LFSE). LFSE = (0.4x 0.6y) Let us see what happens when we withdraw the trans ligands in an octahedral complex (let it be the z ligands) When this happens, we have a tetragonally distorted octahedral complex. As soon as the distance from M m+ to these ligands becomes greater than the other 4 ligands, new energy differences are established. z orbital becomes more stable than x y orbital. yz and xz are equivalent more stable than xy CHEM61HC/SS/07 d x -y E d xy d z t g d zy, d zx Whether this happens depends on the metal ion and the ligands concerned. Square complexes of Co II, Ni II and Cu II lead to energy level diagrams shown as follows: CHEM61HC/SS/08 4

5 6/4/011 M = Co II, Ni II and Cu II d x -y exactly t g /5 d z 1/1 Δo octahedral square d yz, d zx MX 6 MX 4 CHEM61HC/SS/09 High- Spin vs Low- Spin in Octahedral complexes d 1, d, d 3 - simple t g d 4 t g high-spin low-spin CHEM61HC/SS/10 5

6 6/4/011 High- spin d 4 t g 3 1 Low- spin d 4 t g4 0 x =3, y =1 x =4, y =0 E = (0.4x 0.6y) E = (0.4x 0.6y) = 0.6 = P CHEM61HC/SS/11 What is the LFSE for octahedral ions of the following configurations? (a) d 3 (b) high spin d 5 (a) electronic configuration : t g3 0, x = 3, y = 0 Therefore, LFSE = (0.4x 0.6y) = [(0.4)(3) (0.6)(0)] = 1. (b) electronic configuration : t g3, x = 3, y = Therefore, LFSE = (0.4x 06y)Δ 0.6y) = [(0.4)(3) (0.6)()] = 0 What is LFSE for both high and low spin d 6 configuration? CHEM61HC/SS/1 6

7 6/4/011 Octahedral Geometry Energy d x -y d z d xy d xz d yz t g LFSE = (0.4x 0.6y) Tetrahedral Geometry Energy d xy d xz d yz d x -y d z x y z t g Δ t Δ t = 4 9 7

8 6/4/011 Square-Planar Geometry d x -y Energy d z d xy 3 d xz d yz LFSE = (d x -y -d z ) + (d z d xy )/3 1) What is the LFSE for high-spin d 5 of the following geometries? (a) Square-planer (b) Tetrahedral ) What is LFSE for both geometries (Q1) for a low spin d 5 configuration? CHEM61HC/SS/1 8

9 6/4/011 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 Pt 4+ > Ir 3+ > Rh 3+ > Co 3+ > Cr 3+ > Fe 3+ > Fe + > Co + > Ni + > Mn + When theometry and the metal are held constant, the splitting of the d- orbitals increases in the following order: I - < Br - < [NCS] - < Cl - < F - < OH - <H O < NH 3 < en < CN - < CO weak-field strong-field CHEM61HC/SS/13 The ligand field splitting parameter, varies with the identity of the ligand. In the series of complexes [CoX(NH 3 ) 5 ] n+ with X = I,Br,Cl,H Oand NH 3, the colours range from purple (for X = I ) throughh pink (X = Cl ) to yellow (with NH 3 ). 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 referred to as a strong field ligand. Ligands that give rise to low energy transitions (such as Br ) are referredto asweak field ligand. CHEM61HC/SS/14 9

10 6/4/011 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 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 = {S (S + 1)} ½ μ B μ B = Bohr magneton CHEM61HC/SS/15 Calculated spin-only magnetic moments Ion N S μ / μ B Calc. Expt. Ti 3+ 1 ½ V Cr ½ Mn Fe 3+ 5 ½ CHEM61HC/SS/16 10

11 6/4/011 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: t g5 (high-spin, S = 1½) with 3 unpaired electrons or t g6 1 (Low-spin, S = ½) with 1 unpaired electron. μ = {S (S + 1)} ½ μ B High-spin Low-spin μ = {1½ (1½ + 1)} ½ μ B μ = {1½ (1½ + 1)} ½ μ B μ = 3.87 μ B μ = 1.73 μ B CHEM61HC/SS/17 The spin only magnetic moments are 3.87 μ B and 1.73 μ B. Therefore, the only consistent assignment is the high-spin configuration t 5 g5. The magnetic moment of an octahedral complex [Mn(NCS) 6 ] 4 is 6.06 μb. What is its electron configuration? 11

Crystal Field Theory

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