Coordination Chemistry Bonding in transitionmetal complexes Summary of

Coordination Chemistry Bonding in transition-metal complexes

Summary of key points on isomerism

Crystal field theory: an electrostatic model The metal ion will be positive and therefore attract the negatively charged ligands But there are electrons in the metal orbitals, which will experience repulsions with the negatively charged ligands

Ligand/d orbital interactions Orbitals point at ligands (maximum repulsion) Orbitals point between ligands (less repulsion)

The two effects of the crystal field

Splitting of d orbitals in an octahedral field eg 3/5 Do Do 2/5 Do t 2 g Do is the crystal field splitting E(t 2 g) = -0. 4 Do x 3 = -1. 2 Do E(eg) = +0. 6 Do x 2 = +1. 2 Do

Ligand effect of splitting Strong field Weak field The spectrochemical series, D 0 depends on ligand CO, CN- > phen > NO 2 - > en > NH 3 > NCS- > H 2 O > F- > RCO 2 - > OH- > Cl- > Br- > I-

Effect of metal ion on splitting Strong field Weak field D increases with increasing formal charge on the metal ion, ligands closer D increases on going down the periodic table, more diffuse orbitals

The splitting constant must depend on both the ligand the metal. Do ≈ M ∑ nl. Ll x 103 cm-1 Predicts value of D (cm-1) nl is # of ligands Ll Observe that ML 4 expected to have smaller splitting than ML 6

Placing electrons in d orbitals (strong vs weak field) Strong field Weak field d 1 d 2 Strong field Weak field d 4 d 3 So, what is going on here!!

When the 4 th electron is assigned it will either go into the higher energy eg orbital at an energy cost of D 0 or be paired at an energy cost of P, the pairing energy. Strong field Weak field d 4 Strong field = Low spin (2 unpaired) Pairing Energy!!. P < Do Weak field = High spin (4 unpaired) D 0, P > Do

Pairing Energy, P The pairing energy, P, is made up of two parts. 1) Pc: Coulombic repulsion energy caused by having two electrons in same orbital. Destabilizing energy contribution of Pc for each doubly occupied orbital. 2) Pe: Exchange stabilizing energy for each pair of electrons having the same spin and same energy. Stabilizing contribution of Pe for each pair having same spin and same energy 3) 1. = sum of all Pc and Pe interactions 2. How do we get these interactions?

Placing electrons in d orbitals High Low d 5 1 u. e. 5 u. e. 0 u. e. 2 u. e. Low d 6 4 u. e. d 8 2 u. e. High d 7 1 u. e. 3 u. e. d 9 1 u. e. d 10 0 u. e.

Detail working out…. High Field (Low Spin) Low Field (High Spin) d 5 1 u. e. Low Field 5 u. e. Coulombic Part = 0 High Field Exchange part = for Coulombic Part = 2 Pc Exchange part = for For 1 Pe What are the energy terms for both high spin and low spin? 3 Pe + Pe P = 4 Pe 3 Pe P = 2 Pc + 4 Pe LFSE = 5 * (-2/5 D 0) = -2 D 0 LFSE = 3*(-2/5 D 0) + 2 (3/5 D 0) = 0 High Field – Low Field = -2 D 0 +2 Pe When D 0 is larger than Pe the high field, the result is negative and high field (low spin) is favored.

Positive favors high spin. Neg favors low spin.

Interpretation of Enthalpy of Hydration of hexahydrate using LFSE d 0 d 1 LFSE (in D 0). 0 . 4 d 2 d 3 d 4. 8 1. 2. 6 d 5 d 6 d 7. 0. 4 . 8 d 9 d 10 1. 2. 6 . 0

Splitting of d orbitals in a tetrahedral field t 2 Dt e Dt = 4/9 Do Always weak field (high spin)

Magnetic properties of metal complexes Diamagnetic complexes very small repulsive interaction with external magnetic field no unpaired electrons Paramagnetic complexes attractive interaction with external magnetic field some unpaired electrons

Measured magnetic moments include contributions from both spin and orbital spin. In the first transition series complexes the orbital contribution is small and usually ignored.

Coordination Chemistry: Molecular orbitals for metal complexes

The symmetry of metal orbitals in an octahedral environment A 1 g T 1 u

The symmetry of metal orbitals in an octahedral environment T 2 g Eg

The symmetry of metal orbitals in an octahedral environment s

Metal-ligand s interactions in an octahedral environment Six ligand orbitals of s symmetry approaching the metal ion along the x, y, z axes We can build 6 group orbitals of s symmetry as before and work out the reducible representation

s If you are given G, you know by now how to get the irreducible representations G = A 1 g + T 1 u + Eg

Now we just match the orbital symmetries s

“d 0 -d 10 electrons” anti bonding “metal character” non bonding 6 s ligands x 2 e each 12 s bonding e “ligand character”

Introducing π-bonding 2 orbitals of π-symmetry on each ligand We can build 12 group orbitals of π-symmetry

Gπ = T 1 g + T 2 g + T 1 u + T 2 u

Anti-bonding LUMO(π) The CN- ligand

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

ML 6 s-only bonding “d 0 -d 10 electrons” anti bonding “metal character” non bonding 6 s ligands x 2 e each The bonding orbitals, essentially the ligand lone pairs, 12 s bondingwill e not be worked with further. “ligand character”

π-bonding may be introduced as a perturbation of the t 2 g/eg set: Case 1 (CN-, CO, C 2 H 4) empty π-orbitals on the ligands M L π-bonding (π-back bonding) t 2 g (π*) t 2 g eg eg Do D’o t 2 g Do has increased Stabilization t 2 g (π) ML 6 s-only ML 6 s+π (empty π-orbitals on ligands)

π-bonding may be introduced as a perturbation of the t 2 g/eg set. Case 2 (Cl-, F-) filled π-orbitals on the ligands L M π-bonding eg Do has decreased eg D’o t 2 g (π*) Do Destabilization t 2 g Stabilization t 2 g (π) ML 6 s-only ML 6 s+π (filled π-orbitals)

Putting it all on one diagram. Strong field / low spin Weak field / high spin

Spectrochemical Series Purely s ligands: D: en > NH 3 (order of proton basicity) p donating which decreases splitting and causes high spin: D: H 2 O > F > RCO 2 > OH > Cl > Br > I (also proton basicity) Adding in water, hydroxide and carboxylate D: H 2 O > F > RCO 2 > OH > Cl > Br > I p accepting ligands increase splitting and may be low spin D: CO, CN-, > phenanthroline > NO 2 - > NCS-

Merging to get spectrochemical series CO, CN- > phen > NH 3 > NCS- > H 2 O > F- > RCO 2 - > OH- > Cl- > Br- > I- Strong field, p acceptors large D low spin s only Weak field, p donors small D high spin

Turning to Square Planar Complexes Most convenient to use a local coordinate system on each ligand with y pointing in towards the metal. py to be used for s bonding. z being perpendicular to the molecular plane. pz to be used for p bonding perpendicular to the plane, p^. x lying in the molecular plane. px to be used for p bonding in the molecular plane, p|.

ML 4 square planar complexes ligand group orbitals and matching metal orbitals

ML 4 square planar complexes MO diagram s-only bonding π - bonding

A crystal-field aproach: from octahedral to tetrahedral Less repulsions along the axes where ligands are missing

A crystal-field aproach: from octahedral to tetrahedral A correction to preserve center of gravity

The Jahn-Teller effect Jahn-Teller theorem: “there cannot be unequal occupation of orbitals with identical energy” Molecules will distort to eliminate the degeneracy

Angular Overlap Method An attempt to systematize the interactions for all geometries. The various complexes may be fashioned out of the ligands above Linear: 1, 6 Tetrahedral: 7, 8, 9, 10 Trigonal: 2, 11, 12 Square planar: 2, 3, 4, 5 T-shape: 1, 3, 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 s 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 dz 2 dx 2 -y 2 dxy dxz dyz 1 1 es 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 7 0 0 1/3 1/3 8 0 0 1/3 1/3 9 0 0 1/3 1/3 10 0 0 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 dx 2 -y 2 orbital is destabilized by (3/4 +6/16) es = 18/16 es in a trigonal bipyramid complex due to s interaction. The dxy, equivalent by symmetry, is destabilized by the same amount. The dz 2 is destabililzed by 11/4 es.