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

6/4/2011 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. CHEM261HC/SS2/01 d orbitals CHEM261HC/SS2/02 1

6/4/2011 Consider metal ion, M m+ , lying at the centre of an octahedral set of point charges. CHEM261HC/SS2/03  Suppose the metal atom has a single d electron outside of the closed shells (Ti 3+ or V 4+ ).  In the free ion, the electron can be in any one of the 5 orbitals, since all are equivalent. Is this True??? q  Recall the shapes of the d orbitals CHEM261HC/SS2/04 2

6/4/2011 CHEM261HC/SS2/05 2 groups of orbitals t 2g d xy , d yz , d zx e g d z 2 , d x 2 2 ‐ y e g barycentre 0.4 ∆ o ∆ ∆ o 0.6 ∆ o t 2g CHEM261HC/SS2/06 3

6/4/2011 Δ o is the difference in energy between e g and t 2g . x e g y configuration relative to the barycentre is called The net energy of a t 2g the ligand field stabilization energy (LFSE). LFSE = (0.4x – 0.6y) Δ o  Let us see what happens when we withdraw the 2 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 2 ligands becomes greater than  the other 4 ligands, new energy differences are established. z 2 orbital becomes more stable than x 2 ‐ y 2 orbital.   yz and xz are equivalent more stable than xy CHEM261HC/SS2/07 2 2 d x -y e g d xy Δ o E 2 d z t 2g 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: CHEM261HC/SS2/08 4

6/4/2011 M = Co II , Ni II and Cu II 2 2 d x -y e g Δ o exactly Δ o 2/5 Δ o d z 2 t 2g 1/12 Δ o d yz , d zx octahedral square MX 6 MX 4 CHEM261HC/SS2/09 High- Spin vs Low- Spin in O ctahedral complexes d 1 , d 2 , d 3 - simple e g t 2g d 4 e e g t 2g high-spin low-spin CHEM261HC/SS2/10 5

6/4/2011 High- spin d 4 Low- spin d 4 3 e g 1 t 2g 4 e g 0 t 2g x = 3 , y = 1 x = 3 y = 1 x = 4 , y = 0 x = 4 y = 0 E = (0.4 x – 0.6 y ) Δ o E = (0.4 x – 0.6 y ) Δ o = 0.6 Δ o = 1.6 Δ o + P CHEM261HC/SS2/11 What is the LFSE for octahedral ions of the following configurations? (a) d 3 (b) high ‐ spin d 5 3 e g 0 , x = 3, y = 0 (a) electronic configuration : t 2g Therefore, LFSE = (0.4 x – 0.6 y ) Δ o = [(0.4)(3) – (0.6)(0)] Δ o = 1.2 Δ o 3 e g 2 , x = 3, y = 2 (b) electronic configuration : t 2g Therefore, LFSE = (0.4 x – 0.6 y ) Δ o = [(0.4)(3) – (0.6)(2)] Δ o = 0 Therefore LFSE = (0 4 x 0 6 y ) Δ = [(0 4)(3) (0 6)(2)] Δ = 0 What is LFSE for both high ‐ and low ‐ spin d 6 configuration? CHEM261HC/SS2/12 6

6/4/2011 Octahedral Geometry e g 2 2 2 d x d z -y Δ o Energy t 2g d yz d xz d xy LFSE = (0.4x – 0.6y) Δ o Tetrahedral Geometry Energy t 2g d xz d yz d xy Δ t e g 2 2 2 d x d z x -y y z 4 Δ o Δ t = 9 7

6/4/2011 Square-Planar Geometry 2 2 d x -y Δ o 2 d z 2 Δ o d xy Energy 3 d xz d yz 2 - d z 2 – d xy )2/3 Δ o 2 2 ) Δ o + ( d z LFSE = ( d x -y 1) What is the LFSE for high-spin d 5 of the following geometries? (a) Square planer (a) Square-planer (b) Tetrahedral (b) Tetrahedral 2) What is LFSE for both geometries (Q1) for a low spin d 5 configuration? CHEM261HC/SS2/12 8

6/4/2011 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 2+ > Co 2+ > Ni 2+ > Mn 2+  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 - < H 2 O < NH 3 < en < CN - < CO weak-field strong-field CHEM261HC/SS2/13  The ligand ‐ field splitting parameter, Δ o varies with the identity of the ligand. In the series of complexes [CoX(NH 3 ) 5 ] n+ with X = I ‐ , Br ‐ , Cl ‐ , H 2 O and  NH 3 , the colours range from purple (for X = I ‐ ) through pink (X = Cl ‐ ) NH th l f l (f X I ) th h i k (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 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. CHEM261HC/SS2/14 9

6/4/2011 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 μ = 2 {S (S + 1)} ½ μ B μ B = Bohr magneton CHEM261HC/SS2/15 Calculated spin-only magnetic moments Ion N S μ / μ B μ μ B Calc. Expt. Ti 3+ 1 ½ 1.73 1.7 ‐ 1.8 V 3+ 2 1 2.83 2.7 ‐ 2.9 Cr 3+ 3 1½ 3.87 3.8 Mn 3+ 4 2 4.90 4.8 ‐ 4.9 Fe 3+ 5 2½ 5.92 5.9 CHEM261HC/SS2/16 10

6/4/2011 The magnetic moment of a certain octahedral Co(II) complex is 4.0 μ B . What is its d - electron configuration? A Co(II) complex is d 7 2 (high-spin, S = 1½) with 3 unpaired Two possible configurations: t 2g 5 e g 1 (Low-spin, S = ½) with 1 unpaired electron. electrons or t 2g 6 e g μ = 2 {S (S + 1)} ½ μ B High-spin Low-spin μ = 2 {1½ (1½ + 1)} ½ μ B μ = 2 {1½ (1½ + 1)} ½ μ B μ = 3.87 μ B μ = 1.73 μ B CHEM261HC/SS2/17  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 2g 5 e g 5 2 2 . The magnetic moment of an octahedral complex [Mn(NCS) 6 ] 4 ‐ is 6.06 μ B. What is its electron configuration? 11