D1d1 d2d2 d3d3 d4d4 Placing electrons in d orbitals (strong vs weak field) Strong field Weak field So, what is going on here!!

d1 d2

d3 d4

Placing electrons in d orbitals (strong vs weak field)

Strong field Weak field Strong field Weak field

Strong field Weak field Strong field Weak field

So, what is going on here!!

d4

Strong field =Low spin

(2 unpaired)

Weak field =High spin

(4 unpaired)

< o > o

When the 4th electron is assigned it will either go into the higher energy eg orbital at an energy cost of 0 or be paired at an energy cost of , the pairing energy.

0,

Pairing Energy!!.

Strong field Weak field

Pairing Energy,

The pairing energy, , is made up of two parts.

1) c: Coulombic repulsion energy caused by having two electrons in same orbital. Destabilizing energy contribution of c for each doubly occupied orbital.

2) e: Exchange stabilizing energy for each pair of electrons having the same spin and same energy. Stabilizing contribution of e for each pair having same spin and same energy

= sum of all c and e interactions

How do we get these interactions?

Placing electrons in d orbitals

1 u.e. 5 u.e.

d5

0 u.e. 4 u.e.

d6

1 u.e. 3 u.e.

d7

2 u.e. 2 u.e.

d8

1 u.e. 1 u.e.

d9

0 u.e. 0 u.e.

d10

High Low High Low High Low

Detail working out….

1 u.e. 5 u.e.

d5

High Field Low Field (Low Spin) (High Spin)

What are the energy terms for both high spin and low spin?

High Field

Coulombic Part = 2c

Exchange part = for 3e For 1e

= 2c + 4e

Low Field

Coulombic Part = 0

Exchange part = for 3e + e

= 4e

High Field – Low Field = -20 +2e

LFSE = 5 * (-2/50) = -20

LFSE = 3*(-2/50) + 2 (3/50) = 0

When 0 is larger than e the high field arrangement (low spin) is favored.

Positive favors high spin. Neg favors low spin.

Interpretation of Enthalpy of Hydration of hexahydrate using LFSE

d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 d10

LFSE (in 0) .0 .4 .8 1.2 .6 .0 .4 .8 1.2 .6 .0

Splitting of d orbitals in a tetrahedral field

t2

e

t

t = 4/9o

Always weak field (high spin)

Extreme elongation: from octahedral to square planar

LM

L L

L

L

L

LM

L L

L

Less repulsions along the axeswhere ligands are missing

A crystal-field aproach: from octahedral to square planar

A correction to preservecenter of gravity

Magnetic properties of metal complexes

Diamagnetic complexesvery small repulsive

interaction with external magnetic field

no unpaired electrons

Paramagnetic complexesattractive interaction with

external magnetic fieldsome unpaired electrons

)2( nns

High spin Low spin M dn # u.e. (expt)

BM (calc) BM

# u.e. (expt) BM

(calc) BM

Ti3+ 1 1 1.73 1.73 V4+ 1 1 1.68-1.78 1.73 V3+ 2 2 2.75-2.85 2.83 V2+ 3 3 3.80-3.90 3.88 Cr3+ 3 3 3.70-3.90 3.88 Mn4+ 3 3 3.8-4.0 3.88 Cr2+ 4 4 4.75-4.90 4.90 2 3.20-3.30 2.83 Mn3+ 4 4 4.90-5.00 4.90 2 3.18 2.83 Mn2+ 5 5 5.65-6.10 5.92 1 1.80-2.10 1.73 Fe3+ 5 5 5.70-6.0 5.92 1 2.0-2.5 1.73 Fe2+ 6 4 5.10-5.70 4.90 0 Co3+ 6 4 5.4 4.90/5.48* 0 Co2+ 7 3 4.30-5.20 3.88/5.20* 1 1.8 1.73 Ni3+ 7 3 3.88 1 1.8-2.0 1.73 Ni2+ 8 2 2.80-3.50 2.83 Cu2+ 9 1 1.70-2.20 1.73

*total magnetic moment (S+L)

Values of magnetic moment

Coordination Chemistry:Molecular orbitals for metal complexes

The symmetry of metal orbitals in an octahedral environment

A1g

T1u

T2g

Eg

The symmetry of metal orbitals in an octahedral environment

The symmetry of metal orbitals in an octahedral environment

s

M

z

Metal-ligand interactions in an octahedral environment

Six ligand orbitals of symmetry approaching the metal ion along the x,y,z axes

We can build 6 group orbitals of symmetry as beforeand work out the reducible representation

s

If you are given , you know by now how to get the irreducible representations

= A1g + T1u + Eg

s

Now we just match the orbital symmetries

6 ligands x 2e each

12 bonding e“ligand character”

“d0-d10 electrons”

non bonding

anti bonding

“metal character”

Introducing π-bonding

2 orbitals of π-symmetryon each ligand

We can build 12 group orbitalsof π-symmetry

π = T1g + T2g + T1u + T2u

The T2g will interact with the metal d t2g orbitals. The ligand pi orbitals do not interact with the metal eg orbitals.

We now look at things more closely.

Anti-bonding LUMO(π)

First, the CN- ligand

Some schematic diagrams showing how p bonding occurs with a ligand having a d orbital (such as in P), or a * orbital, or a vacant p orbital.

6 ligands x 2e each

12 bonding e“ligand character”

“d0-d10 electrons”

non bonding

anti bonding

“metal character”

ML6 -only bonding

The bonding orbitals, essentially the ligand lone pairs, will not be worked with further.

t2g

eg

t2g

ML6

-onlyML6

+ π

Stabilization

(empty π-orbitals on ligands)

o

’oo has increased

π-bonding may be introducedas a perturbation of the t2g/eg set:

Case 1 (CN-, CO, C2H4)empty π-orbitals on the ligands

ML π-bonding (π-back bonding)

t2g (π)

t2g (π*)

eg

These are the SALC formed from the p

orbitals of the ligands that can interac with the d on the metal.

t2g

eg

t2g

ML6

-onlyML6

+ π

π-bonding may be introducedas a perturbation of the t2g/eg set.

Case 2 (Cl-, F-) filled π-orbitals on the ligands

LM π-bonding

(filled π-orbitals)

Stabilization

Destabilization

t2g (π)

t2g (π*)

eg’o

o

o has decreased

Strong field / low spin Weak field / high spin

Putting it all on one diagram.

Spectrochemical Series

Purely ligands:

en > NH3 (order of proton basicity)

donating which decreases splitting and causes high spin:: H2O > F > RCO2 > OH > Cl > Br > I (also proton basicity)

accepting ligands increase splitting and may be low spin

: CO, CN-, > phenanthroline > NO2- > NCS-

Merging to get spectrochemical series

CO, CN- > phen > en > NH3 > NCS- > H2O > F- > RCO2- > OH- > Cl- > Br- > I-

Strong field, acceptors large low spin

onlyWeak field, donors small high spin

Turning to Square Planar Complexes

y

x

zMost convenient to use a local coordinate system on each ligand with

y pointing in towards the metal. py to be used for bonding.

z being perpendicular to the molecular plane. pz to be used for bonding perpendicular to the plane, .

x lying in the molecular plane. px to be used for bonding in the molecular plane, |.

ML4 square planar complexesligand group orbitals and matching metal orbitals

bonding

bonding (in)

bonding (perp)

ML4 square planar complexesMO diagram

-only bonding Sample- bonding

eg

Angular Overlap Method

An attempt to systematize the interactions for all geometries.

M

1

65

4 2

3

M

109

78

M 2

6

1

12

11

The various complexes may be fashioned out of the ligands above

Linear: 1,6

Trigonal: 2,11,12

T-shape: 1,3,5

Tetrahedral: 7,8,9,10

Square planar: 2,3,4,5

Trigonal bipyramid: 1,2,6,11,12

Square pyramid: 1,2,3,4,5

Octahedral: 1,2,3,4,5,6

Cont’d

All interactions with the ligands are stabilizing to the ligands and destabilizing to the d orbitals. The interaction of a ligand with a d orbital depends on their orientation with respect to each other, estimated by their overlap which can be calculated.

The total destabilization of a d orbital comes from all the interactions with the set of ligands.

For any particular complex geometry we can obtain the overlaps of a particular d orbital with all the various ligands and thus the destabilization.

ligand dz2 dx2-y2dxy dxz dyz

1 1 e 0 0 0 0

2 ¼ ¾ 0 0 0

3 ¼ ¾ 0 0 0

4 ¼ ¾ 0 0 0

5 ¼ ¾ 0 0 0

6 1 0 0 0 0

7 0 0 1/3 1/3 1/3

8 0 0 1/3 1/3 1/3

9 0 0 1/3 1/3 1/3

10 0 0 1/3 1/3 1/3

11 ¼ 3/16 9/16 0 0

12 1/4 3/16 9/16 0 0

Thus, for example a dx2-y2 orbital is destabilized by (3/4 +6/16) e

= 18/16 e in a trigonal bipyramid complex due to interaction. The dxy, equivalent by symmetry, is destabilized by the same

amount. The dz2 is destabililzed by 11/4 e.