1

Topic 4:

d-Block Chemistry

Required Background

• Ground state electronic configuration

• Trends in first ionization energy

• Metallic radii

• Aquated cations: formation and acidic properties

• Solubility of ionic salts

• Stability constants for metal complexes

• Selected ligand structures and abbreviations

• Redox chemistry in aqueous solution (potential diagrams)

• Geometrical isomerism

• Chiral molecules

• Binary metal hydrides

4-1

2

4-2

Electronic Configurations

heavier d-block metals

a triad

platinum-group metals

copper triangle

Trends in Metallic Radii4-3

–> little variations across a given row of the d-block–> greater for 2nd and 3rd row metals than for first row metals–> similar for the 2nd and 3rd row metals in a given a triad

3

Oxidation States4-4

Sc Ti V Cr Mn Fe Co Ni Cu Zn

0 0 0 0 0 0 0 [0]

1 1 1 1 1 1 1 2

2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3

4 4 4 4 4 4 4 [4]

5 5 5

6 6 6 0 0 0

7

4-5

Problem 4-1: For each of the following complexes, give the oxidation state of themetal and its dn configuration:

a) [Mn(CN)6]4–

b) [CoCl3(py)3]

c) [Ru(bipy)3](PF6)2

d) [Ni(en)3]2+

e) [Cr(acac)3]

f) [Ti(H2O)6]3+

4

Pauling’s Electroneutrality Principle4-6

The distribution of charge in a molecule or ion is such that the charge on asingle atom is within the range +1 to –1 (ideally close to zero)

–> the electroneutrality principle results in a bonding description for the [Co(NH3)6]3+

ion which is 50% ionic and 50% covalent.

Coordination Numbers4-7

Note: if the energy difference between different structure is small, fluxional behaviorin solution may occur

Coordination Number

Common Geometry Less Common Geometry

2345

67

8

9

lineartrigonal Planartetrahedral, square planartrigonal bipyramidal,square based pyramidaloctahedralpentagonal pyramidal

dodecahedralsquare antiprismatichexagonal bipyramidaltricapped trogonal prismatic

trigonal pyramidal

trigonal prismaticmonocapped trigonal prismaticmonocapped ocathedralcube, bicapped trigonal prismatic

5

VSEPR Theory and Kepert Model4-8

• VSEPR theory successfully predicts the shapes of molecular species of themain group elements,but fails for d-block elements:

[V(H2O)6]3+, [Mn(H2O)6]2+, [Co(H2O)6]2+, [Ni(H2O)6]2+, [Zn(H2O)6]2+ havedifferent electron configurations, but identical shape (octahedral)–> the Kepert model rationalizes the shapes of d-block metal complexes by

considering only the repulsions between the ligands:

23456

lineartrigonaltetrahedraltrigonal bipyramidaloctahedral

• The Kepert model fails for square planar complexes (due to electroniceffects) or inherent constraints of a ligand (e.g. tripodal ligand)

4-9

Problem 4-2: Within the Kepert model, what geometries do you predict for thefollowing coordination complexes?

a) K4[Mn(CN)6]

b) [Cu(SPMe3)3][BF4]

c) [Ag(NH3)2]+

6

Coordination Number Two4-10

• Uncommon, generally restricted to Cu(I), Ag(I), Au(I) and Hg(II) (d10 ions)

• Examples:[CuCl2]–, [Ag(NH3)2]+

[Au(CN)2]–, Hg(CN)2

[Au{P(cyclo-C6H11)3}22]+

Note: Bulky ligands generally stabilize coodination numbers

Coordination Number Three4-11

• Usually trigonal planar structures observed, typical for d10 metal centers”

• Examples:Cu(I) in [Cu(CN)3]2–, [Cu(SPMe3)3]+

Ag(I) in [AgTe7]3–, [Ag(PPh3)3]+

Au(I) in [Au{PPh(C6H11)2}3]+

Hg(II) in [Hg(SPh3)3]–

Pt(0) in (Pt(PPh3)3]

[Fe{N(SiMe3)2}33][AgTe7]3–

7

Coordination Number Four4-12

• Very common, most frequently with tetrahedral geometry–> distortions often due to steric or crystal packing effects

• Examples:

• Transition metals with d8 configuration strongly favor a square planargeometry!

d0: [VO4]3–, [CrO4]2–

d1: [MnO4]2–, [ReO4]2–, [RuO4]–

d2: [FeO4]2–, [RuO4]2–

d5: [FeCl4]–, [MnCl4]2–

d6: [FeCl4]2–

d7: [CoCl4]2–

d8: [NiCl4]2–, [NiBr4]2–

d9: [CuCl4]2–

d10: [ZnCl4]2–, [CdCl4]2–, [Cu(CN)4]3–, [Ni(CO)4]

[Rh(PMe2Ph)4]+

Coordination Number Five4-13

• Limiting case: Trigonal pipyramid, square-based pyramidBut: many structures lie between these two extremes (often small energy

difference)

• Examples:

d0: [NbCl4(O)]–

d1: [V(acac)2(O], [WCl4(O)]–

d2: [TcCl4(O)]–

[Cu(bpy){NH(CH2CO2)2}4]

Conformational constraint ofligand gives preference for asquare-based pyramidal complex

[Zn{N(CH2CH2NH2)3}Cl]+

• Found for manypolydentate amine or

phosphine ligands:

8

4-14

Problem 4-3: Suggest a structure for [CuCl(L)]+, assuming that all donor atomsare coordinated to the Cu(II) center.

N

N

NN

L =

Coordination Number Six4-15

• Vast majority show octahedral geometry, trigonal prismatic complexes arevery rare

• Examples: aqua complexes

[Ti(H2O)6]3+, [V(H2O)6]3+, [Cr(H2O)6]3+, [Mn(H2O)6]2+, [Fe(H2O)6]2+,[Zn(H2O)6]2+, etc.

[Re(S2C2Ph2)3]

Example for a trigonal prismaticstructure:

Difference between octahedral andtrigonal prismatic geometry:

9

Coordination Number Seven4-16

• Coordination number 7 or greater are observed most frequently for ions ofthe early, second, and third row d-block metals, and for lanthanoids andactinoids –> rcation must be large to accommodate seven or more ligands

[TaCl4(PMe3)3] [ScCl2(15-crown-5)]+[ZrF7]3–

Coordination Number Eight4-17

• With increasing coordination number also increasing number of possiblestructures:

[Nb(ox)4]4– [CdBr2(18-crown-6)][Y(H2O)8]3+

10

Isomerism in Coordination Complexes4-18

4-19

• Ionization isomers: result from the interchange of an anionic ligand withinthe first coordination sphere with an anion outside the coordination sphere:

[Co(NH3)5Br][SO4] (violet), and [Co(NH3)5(SO4)]Br (red)

• Hydration isomers: result from interchange of H2O and another ligandbetween the first coordination sphere and the ligands outside it:CrCl3•6H2O: [Cr(H2O)4Cl2]Cl•2H2O, [Cr(H2O)5Cl]Cl2•H2O, [Cr(H2O)6]Cl3

• Coordination isomers: only possible for salts in which boh cation and anionare complex ions –> the isomers arise from interchange of liagnds between

the two metal centers:[Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]

• Linkage isomers: arise when one or more of the ligands can coordinate tothe metal in more than one way, e.g. [SCN]–:[Co(NH3)5(NCS–N)]2+ and [Co(NH3)5(NCS–S)]2+

Structural Isomerism

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4-20

• Cis- and trans-isomers of a square planar complex can be distinguished by

IR spectroscopy or NMR spectroscopy:

Geometrical Isomerism

31P NMR of [PtCl2(PBu3)2:

195Pt(I=1/2) constitutes 33.8%–> gives rise to “satellite” peaks(doublet coupling pattern)–> JPPt (cis) = 3508 hz and JPPt(trans) = 2380 HzNote: for an IR active vibration a change in

the molecular dipole moment is required

4-21

Problem 4-4: In the solid state Fe(CO)5 posses a trigonal bipyramidal structure.

a) How many carbon environments are there?

b) Explain why only one signal is observed in the 13C NMR of Fe(CO)5, even atlow temperature.

12

4-22

Problem 4-5: Using spectroscopic methods, how would you distinguish betweenfac- and mer-[RhCl3(PMe3)3]?

4-23

• Simplest case: metal cation surrounded by three identical bidentate ligands,e.g. [Co(en)3]3+:

Optical Isomerism

–> the two isomers are non-superimposable images of each other andtherefore chiral molecules

and prefixes: labels to identify the two enantiomers.The octahedron is viewed down a 3-fold axis, and the chelate definesthen a right (∆) or left (Λ) handed helix.

Λ ∆

13

4-24

Problem 4-6: Comment on the possibility of isomer formation for each of thefollowing complexes:

a) [Ru(py)3Cl3]

b) [Ru(bipy)2Cl2]+

c) [Ru(tpy)Cl3]

N

N

N

tpy

4-25

Problem 4-7: State the types of isomerism that may be exhibited by the followingcomplexes, and draw structures of the isomers.

a) [Co(en)2(ox)]+

b) [Co(en)(NH3)2Cl2]2+

14

– The atomic orbitals required for an octahedral complex are the 3dz2,3dx2–y2, 4s, 4px, 4py, and 4pz

– These orbitals must be unoccupied so as to be available to accept sixpairs of electrons from the ligands:

4-26

• Valence bond theory (Pauling, 1930) not much used today, but the conceptis still useful for a qualitative understanding–> hybridization scheme used for main group elements can be also appliedto d-block elements: an empty hybrid orbital on the metal center can accepta pair of electrons from a ligand to form a σ-bond.

Bonding in Metal Complexes

Example: octahedral complex of Cr(III)

– The atomic orbitals required for an octahedral complex are the 3dz2,3dx2–y2, 4s, 4px, 4py, and 4pz

– These orbitals must be unoccupied so as to be available to accept sixpairs of electrons from the ligands:

3d 4s 4p

vacant orbitals acceptligand electrons

3d d2sp3

hybridization

4-27

CoordinationNumber

23454566

Geometry

LinearTrigonal planarTetrahedralSquare planarTrigonal bipyramidalSquare-based pyramidalOctahedralTrigonal prismatic

Orbitals

s, pzs, px, pys, px, py, pzs, px, py, dx2-y2

s, px, py, pz, dz2

s, px, py, pz, dx2-y2

s, px, py, pz, dz2, dx2-y2

s, dxy, dyz, dxz, dz2, dx2-y2

Description

sp sp2

sp3

sp2d sp3d sp3d sp3d2

sd5

Example

[Ag(NH3)2]+

[HgI3]–

[FeBr4]2–

[Ni(CN)4]2–

[CuCl5]3–

[Ni(CN)5]3–

[Co(NH3)6]3+

[Mo(S2C2Ph2)3]

15

4-28

Problem 4-8: Describe the hybridization scheme and electron configuration of thefollowing complexes:

a) low spin [Fe(CN)6]3–

b) high spin [FeF6]3–

Splitting of d-Orbital Energy in Octahedral Complexes

z

y

x

How does the ligand field influencethe energy of the five metal d-orbitals?

dz2 d

x 2 − y 2

dxy d xz d yz

Orientation along ligand-metal bond axis=> raised to higher energy

Orientation between ligand-metal bond axis=> raised to less higher energy

4-29

16

4-30

Ligand Field Theory

The ligand field is an electrostatic model which predicts the energy splittingof of the d orbitals in a complex. Ligands are considered point charges, andthere are NO metal-ligand covalent interactions

Metal ion ingas phase

Spherical chargedistribution aroundmetal ion

charge distributionwith octahedralsymmetry aroundmetal ion

eg

t2g

Octahedral Complexes

orbitalenergy

dz2 d

x 2 − y 2

dxy d xz d yz

∆oct = 10 Dq3/5 ∆oct

2/5 ∆oct

∆oct (weak field) < ∆oct (strong field)

4-31

eg

t2g

17

4-32

Field Strength

• The splitting energy ∆oct can be determined from spectroscopic data:

Example: [Ti(H2O)6]3+:

ener

gy

eg

t2g

λmax = 20,300 cm–1

Influence of Ligand

Br–

Cl–

(H2N)2C=O

NCS–

F–

H2O

CN–

11,400

13,000

17,550

18,400

18,900

20,100

22,300

∆O/cm–1Ligand

• The absorption maximum in Ti(III) complexes is sensitive to the ligandenvironment (= field strength).Since the absorption is due to transition of an electron from the t2g orbitals to theeg orbitals, the absorption energy corresponds to ∆O.

4-33

18

Magnitude of Splitting Energy (∆O)

• Charge on the metal (oxidation state)For first row transition elements ∆O varies from about 7,500 cm–1 to 12,500 cm–1 fordivalent ions, and 14,000 cm–1 to 25,000 cm –1 for trivalent ions.

• Position in a groupvalues for analogous complexes of metal ions in a group increases by 25% to 50%going from one transition series to the next:[M(NH3)6]3+ with M=Co 23,000cm–1, M=Rh 34,000cm–1, M=Ir 41,000 cm–1

• Geometry and coordination numberFor identical (or nearly identical) ligands ∆t (tetrahedral) is 4/9 of ∆O. This is a resultof the reduced number of ligands and their orientation relative to the d orbitals.Note: The energy ordering of the orbitals is reversed in tetrahedral complexesrelative to that in octahedral complexes.

• Identity of ligandThe dependence of ∆ on the nature of the ligand follows an empirical order, knownas the spectrochemical series, for all metals in all oxidation states and geometries.

4-34

Spectrochemical Series

• Order of ligand field strength with decreasing Dq:

CN– > phen ~ NO2– > en > NH3 ~ py > H3O > C2O4

2– > OH– > F– > S2– > Cl–

> Br– > I–

• Order of metals with increasing Dq:

Mn(II) < Ni(II) < Co(II) < Fe(III) < Cr(III) < Co(III) < Ru(III) < Mo(III) < Rh(III) < Pd(II) < Ir(III) < Pt(IV)

Note: H2O > OH–: this cannot be explained in terms of the electrostatic model

4-35

19

Estimation of Splitting Energies

10 Dq ~ f (ligand) • g (metal)

Ligand

Br–

SCN–

Cl–

N3–

F–

C2O4–

H2ONCS–

gly–

pyNH3enbpyCN–

f factor

0.720.730.780.830.900.991.01.021.181.231.251.281.331.70

Metal ion

Mn(II)Ni(II)Co(II)V(II)Fe(III)Cr(III)Co(III)Ru(II)Mn(IV)Mo(III)Rh(III)Tc(IV)Ir(III)Pt(IV)

g factor [1000 cm–1]

8.08.79.0121417.418.2202324.627.0303236

4-36

4-37

Problem 4-9: Estimate the splitting energy ∆oct for the following octahedralcomplexes:

a) [Cr(NH3)6]3+

b) [Co(en)2Cl2]+

20

Ligand Field Stabilization Energy (LFSE)

energy

dz2 d

x 2 − y 2

dxy d xz d yz

∆O = 10 Dq

4 Dq

6 Dq

Example: [Cr(NH3)6]3+ => octahedral complex, d3 electron configuration

LFSE = 3 • (–4) Dq = – 12 Dq

Note: The LFSE for any octahedral complex with d3 electron configuration is –12 Dq,but the absolute value of Dq varies with ligand and metal!

4-38

4-39

Problem 4-10: Calculate the crystal field stabilization energy (in cm–1) for[Co(NH3)6]2+. (∆oct = 10,200 cm–1)

21

Influence of Splitting Energy

energy

[CrCl6]4–

dz2 d

x 2 − y 2

d xy d xz d yz

4 Dq

6 Dq

LFSE = 6 Dq + 3(–4 Dq) = – 6 Dq

dz2 d

x 2 − y 2

d xy d xz d yz

4 Dq

6 Dq

stronger

field

LFSE = 4(–4) Dq = – 16 Dq

[Cr(CN)6]4–

4-40

∆oct < SPE ∆oct > SPE

Spin State and Ligand Field Strength4-41

22

Ligand Field Stabilization Energies4-42

Splitting Energies for Aqueous Complexes

Π = Spin pairing energy

Note: Only Co3+ has a splitting energy similar to the spin pairing energy –> It is the only low-spin aqua complex of the listed examples!

4-43

23

Low-Spin vs. High-Spin Complexes

• Strong-field ligands = low-spin complexesStrong field ligands have pi-acceptor orbitals or low-lying d-orbitals:π* as in CO or CN–, π* as in CH2=CH2, low lying d as in PR3, PF3

• Weak field ligands = high-spin complexesWeak field ligands have pi-donor orbitals:Usually multiple p-orbitals as in X–

• Intermediate field ligands = usually high-spin for +2 ions, low-spin for+3 ionsIntermediate field ligands have few, or no pi-donor or acceptor orbitals, orthere is a poor match in energy of available pi-orbitals: NH3, H2O, pyridine

• Note:Complex stability and reactivity do not necessarily correlate with ligand fieldstrength–> Thermodynamic stability refers to the energetics of a given reaction–> Kinetic stability (=reactivity) refers to the rate with which a given reactionoccurs (activation energy)

4-44

4-45

Molecular Orbital Theory: Octahedral Complexes

Complexes with s-bonding only (e.g. [Co(NH3)6]3+:

Note: in a complex without p-bonding, the differencebetween eg* and t2g corresponds to the splitting energy ∆oct

24

Effect of pi-donor/acceptor interactions

sigma bondingonly

intermediate fieldligands

energy

∆O

∆O

∆O

sigma +pidonor

low field ligands

sigma + piacceptor

high field ligands

eg

t2g

eg eg

t2g

t2g

4-46

Cyanide is a Pi-Acceptor

energy

MO-Diagram of CN–

Overlap of d, π*, and p-orbitals withmetal d orbitals:–>Overlap is good with ligand d andp π*-orbitals, poorer with ligand p-orbitals

4-47

25

π-donor ligands π-acceptor ligands

4-48

Effect of pi-Bonding

energy

4-49

26

4-50

• Octahedral complexes of d9 and high-spin d4 ions are often distorted (axial

bonds are different in lengths compared to equatorial bonds)–> system of lower symmetry with non-degenerate orbitals is more stable

Jahn-Teller Distortion

dz2 d

x 2 − y 2

d xy d xz d yz

4 Dq

6 Dq elongation

dz2

dx 2− y 2

d xy

d xzd yz

4 Dq

6 Dq

+1/2 β

–1/2 β

ML

L L

L

L

L

ML

L L

L

L

L

+2/3 β

–1/3 β

4-51

Problem 4-11: Which of the following complexes exhibit a distorted octahedralgeometry. Determine, whether the distortion is due to elongation or contractionalong the axial molecule axis.

a) [Cu(NH3)6]2+

b) [Cr(NH3)6]3+

c) [Ti(H2O)6]3+

27

Tetrahedral Complexes

orbitalenergy

dz2 d

x 2 − y 2

dxy d xz d yz

∆tet = 10 Dq3/5 ∆tet

2/5 ∆tett

4-52

∆tet = 4/9 ∆oct

e

t2

Tetrahedral Complexes

• least ligand-ligand repulsion

• Small ligand field splitting energy (only four ligands, and they are not aligned alongthe orbital axis)–> the ligand field splitting energy is in most cases too small to overcome the spinpairing energy, therefore tetrahedral low spin complexes are very rare!

• Ideal geometry for metal cations with no LFSE (d0, d5 and d10), or only littleLFSE (d2, d7)

Examples: MnO4– (d0), FeO4

2– (d2), FeCl4– (d5, high spin)

CoCl42– (d7, high spin), ZnCl42– (d10)

NO

Cr

LL

LN

Si

Si

L =Cr[N(SiMe3)2]3NO

A rare example for a low spin complexwith tetrahedral coordination geometry

4-53

28

energy

dz2 d

x 2 − y 2

d xy d xz d yz

Removal of z ligands

d xy

dx 2 − y 2

dx 2 − y 2

dz2

dz2

d xy

d xz d yz

dxz dyz

4-54

Square Planar Complexes4-55

orbitalenergy

dz2 d

x 2 − y2

d xy dxz dyzd xy

dx 2 − y2

dz2

dxzdyz

Tetrahedral[NiCl4]2–

Square planar[Ni(CN)4]2–

paramagnetic diamagnetic

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Square Planar Complexes

• Square planar complexes are expected for all metal cationswith d8 electron configuration–> this geometry offers the greatest stabilization accordingto the ligand field theory, since the highest energy orbitaldx2-y2 remains unoccupied

Examples: Ni(II), Pt(II), Pd(II), Au(III)

• Square planar complexes with d8 configuration are alwaysdiamagnetic

Examples: Ni(H2O)62+: octahedral –> paramagnetic

dx 2 − y 2

dz2

dxy

d xz d yz

energy

Ni(CN)42–: square planar –> diamagnetic

Ni(Cl)42–: tetrahedral –> paramagnetic

4-56

4-57

Problem 4-12:

a) The anion [Ni(SPh)4]2– is tetrahedral. Explain why this complex isparamagnetic.

b) The complexes NiCl2(PPh3)2 and PdCl2(PPh3)2 are paramagnetic anddiamagnetic. What dies this tell you about their structures.

30

Energy Diagram Overview4-58

4-59

• Studies of electronic spectra of metal complexes provide information aboutstructure and bonding. Absorptions arise from transitions between electronic

energy levels:

– Transitions between metal-centered orbitals possessing d-character =d- d transitions (MC) (weak intensity, Laporte-forbidden)

– Transitions between metal- and ligand-centered orbitals = metal-to-ligand or liagnd-to-metal charge transfer transitions (MLCT, LMCT)(strong intensity)

• Absorption bands are usually broad

due to vibrational and rotationalsublevels

Electronic Spectra

Amax = max ⋅ c ⋅ l

31

4-60

• Spin Selection Rule: ∆S = 0

–> Transitions may occur from singlet to singlet, or triplet to triplet states

etc., but a change in spin multiplicity is forbidden.

• Laporte Selection Rule: ∆l = ± 1

–> There must be a change in parity: allowed transitions: g ↔ u forbidden transitions: g ↔ g and u ↔ u

Therefore, allowed transitions are s → p, p → d, d → f,forbidden transitions are s → s, p → p, d → d, s → d, p → f

Selection Rules

• Spin-forbidden transitions can be still observed due to spin-orbit couplingand vibronic coupling

• Charge transfer transitions are allowed by the selection rules and havetherefore greater intensity than d-d transitions

4-61

Problem 4-13:

Explain why [Mn(H2O)6]2+ is very pale in color, but [MnO4]– is intensively purplecolored.

32

4-62

• For multielectron atoms with open shells, there are several (energeticallynon-equivalent) possibilities to place the electrons in the orbitals, e.g.:

Term Symbols

etc.

• The angular momentum and spin multiplicity of an atom is specified by asymbolic representation:

aT

multiplicity a = 2S+1 S, P, D, F, … (L=0, 1, 2, 3…)

• Atoms in S, P, D, F … state have the same angular momentum L, just as for

the H-atom with a single electron in s, p, d, f…

• The quantum number S is the summation of all electron spins in the atom

Hunds Rule: Ground state has maximum spin multiplicity. If there are two states with

identical multiplicity, the ground state has the highest value of L

4-63

Example: Ground State of Carbon (for details see Box 20.5)

1s2 2s2 2p2

filled shell => S = 0 (all spins are paired) L = 0 (all momentum cancel each other out)=> only 2p2 must be considered:

n = 2, l = 1

ml = +1, 0, –1 (py, pz, px)ms = +1/2,–1/2 (spin state)

ML = Σ ml (= total angular momentum)MS = Σ ms (= total spin)

ML = 0, ±1, ±2,… ±L

MS = S, S–1, S–2,… –S

ml +1 0 –1 +1 0 –1 +1 0 –1

ML +1 0 –1

Therefore: L = 1, S = 1 => 3P is ground state

33

4-64

Problem 4-14: Determine the ground state term for Ti3+ and Co2+.

4-65

• The energy states of a metal cation in a octahedral environment are furthersplitted by the ligand field:

Electronic Spectra of Octahedral Complexes

Term

S

P

D

F

G

H

I

Degeneracy

1

3

5

7

9

11

13

States in octahedral ligand field:

A1g

T1g

Eg + T2g

A2g + T1g + T2g

A1g + Eg + T1g + T2g

Eg + T1g + T1g + T2g

A1g + A2g + Eg + T1g + T2g + T2g

34

4-66

• Example: [Ti(H2O)6]3+:

Ground state: 2D => splitting to T2g and Eg in octahedral ligand field:

Electronic Spectra of d1 Complexes

Note: For a d9 ion in octahedral field, the splitting diagram is an inversion of thatfor the octahedral d1 ion (d9 configuration contains one ‘hole’)=> the ground state is 2Eg

d z2 d x2 − y 2

dxy d xz dyz

energy

dz2 d

x2 − y 2

d xy d xz d yz

hν

4-67

• The relative splitting of the d-orbital energy isqualitatively identical for octahedralcomplexes with d1 or d6 electron configuration

and for d4 or d9 tetrahedral complexes

• Similarly, the splitting for d1, d6 tetrahedralcomplexes is the same as for d4 and d9

octahedral complexes

Note: the magnitude of splitting is alwayssmaller for the tetrahedral geometry: ∆tet = 4/9 ∆oct

Orgel Diagram for d1, d4, d6 and d9 ions

Only one electronic transition is possible from a ground state to the excited state:

Eg ← T2g: octahedral d1 and d6 ion

T2g ← Eg: octahedral d4 and d9 ion

T2 ← E: tetrahedral d1 and d6 ion

E ← T2: tetrahedral d4 and d9 ion

35

4-68

Orgel Diagram for d2, d3, d7 and d8 ions

There are two terms arising from a d2 ion:3F (ground state) and 3P (excited state)

• 3P state does not split in Oh (T2g symmetry)• 3F splits into T1g, T2g, and A2g

Non-crossing rule:states of same symmetry are not allowedto cross –> bending

Note: all transitions from the ground stateare allowed –> three absorption bands

8 Dq

18 Dq

10 Dq

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Problem 4-15: a) Assign the absorption bands in the electronic spectra of (a)[Ni(H2O)6]2+ and (b) [Ni(NH3)6]2+ to the corresponding transitions in the Orgeldiagram on previous page.

b) Estimate the splitting energy ∆oct (=10 Dq) for [Ni(NH3)6]2+ based on abovespectrum

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More Advanced Term Splitting Diagrams...

ligand field d-d interactionfree ion

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• More advanced treatment of the energies ofelectronic states

• The energy of the ground state is taken zero forall field strengths, an dthe energy of all otherterms are plotted with respect to the ground

state energy

• The energy and field strength are expressed interms of the Racah parameter B.

Tanabe Sugano Diagram

Tanabe-sugano diagram for the d2

configuration in an octahedral field

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Problem 4-16: An aqueous solution of [V(H2O)6]3+ shows absorptions at 17200and 25600 cm–1.

a) Assign the absorption band to the corresponding transitions using the Tanabe-Sugano diagram on previous page.

b) Estimate the values for the Racah parameter B and the splitting energy ∆oct.

Electronic Spectra for d5 Complexes: Mn(H2O)62+

• Mn(II) has a d5 high spin electron configuration –>none of the possible (d-d) transitions is spin allowed,since for any transition the spin of the electron must bereversed (both higher energy eg orbitals containalready one electron, according to the Pauli principlethe spin of the second electron must be reversed)Therefore: all possible transitions are very weak, andMn(H2O)6

2+ is very pale in color

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38

• There is evidence for sharing electrons between metal and ligands

– Pairing energies are lower in complexes than in gaseous Mn+ ions,indicating less interelectronic repulsion–> effective size of the metal orbital has increased

• Nephelauxetic = “cloud expanding”

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Nephelauxetic Effect

metal ion

Co(III)Rh(III)Co(II)Fe(III)Cr(III)Ni(II)Mn(II)

k

0.350.280.240.240.210.120.07

ligands

6 Br–

6 Cl–

6 [CN]–

3 en6 NH36 H2O6 F–

h

2.32.02.01.51.41.00.8

B0 − B

B0

≈ hligands × kmetal ion

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Problem 4-17: Using the data in the table on previous page, estimate thereduction in the interelectronic repulsion in going from the gaseous Fe3+ ion to[FeF6]3–.

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• The ligand field stabilization energy has important consequences for thethermodynamic properties of the transition metal complexes

Thermodynamic Aspects of the LFSE

Hydration enthalpies of M2+ and M3+

LFSE as a function of d-electron configuration

Lattice energies for MCl2

Irving-Williams Series

• The stability for many transition metal complexes follows the followingorder:

Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II)

• Based on the increasing electron affinity (increasing effective nuclearcharge!) an increase of log K (or ∆H) would be naturally expected

But: The complex stability is also influenced by d-electron-electronrepulsion–> complexes with greater ligand field stabilization energy are expected tobe thermodynamically more stable–> therefore, copper(II) is expected to form the most stable (high-spin)complexes

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40

Effect of LFSE

[M(OH2)6]2+ + 3 en [M(en)3]2+ + 6 H2O

log β3

∆H

∆LFSE

Mn2+

5.6

–46

0

Fe2+

9.7

–67

4.8

Co2+

13.9

–93

19

Ni2+

18.4

–120

44

Cu2+

20.0

–105

Zn2+

12.9

–87

0

• The overall complex stability parallels the ligand field stabilization energycontribution and is greatest for Cu(II)

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