coordination chemistry structures and isomers. electronic configuration ground state: progressive...

COORDINATION CHEMISTRY STRUCTURES AND ISOMERS

ELECTRONIC CONFIGURATIONGround State:Progressive filling of the 3d, 4d, and 5d orbitals

Exceptions: ns1 (n-1)d5 rather than ns2 (n-1)d4

ns1 (n-1)d10 rather than ns2 (n-1)d9

Transition metal ions: First in first out

TRENDS - IONIC Radii

COORDINATION COMPOUNDS Coordination compounds –

compounds composed of a metal atom or ion and one or more ligands. [Co(Co(NH3)4(OH2)3]Br6

Ligands usually donate electrons to the metal

Includes organometallic compounds

Werner’s totally inorganic optically active compound.

WERNER’S COORDINATION CHEMISTRY• Performed systematic studies to understand bonding in coordination

compounds.– Organic bonding theory and simple ideas of ionic charges were not

sufficient.• Two types of bonding

– Primary – positive charge of the metal ion is balanced by negative ions in the compound.

– Secondary – molecules or ion (ligands) are attached directly to the metal ion.

• Coordination sphere or complex ion.• Look at complex on previous slide (primary and secondary)

WERNER’S COORDINATION CHEMISTRY

• He largely studied compounds with four or six ligands.– Octahedral and square-planar complexes.

• It was illustrated that a theory needed to account for bonds between ligands and the metal.– The number of bonds was commonly more than accepted

at that time.• 18-electron rule.

• New theories arose to describe bonding.– Valence bond, crystal field, and ligand field.

COORDINATION COMPOUNDS

LIGANDS

CHELATING LIGANDS

Chelating ligands (chelates) – ligands that have two or more points of attachment to the metal atom or ion.Bidentate, tridentate,

tetra.., penta…, hexa… (EDTA).

trisoxalatochromate(III) ion or just [Cr(ox)3]3-

A HEXADENTATE LIGAND, EDTA

• There are six points of attachment to the calcium metal.– Octahedral-type geometryethylene diamine tetraacetic

acid (EDTA)

ethylenediaminetetraacetatocalcium ion or just [Ca(EDTA)]2-

LIGANDS

NOMENCLATURE

• Coordination compounds that are ionic, the cation is named first and separated by a space from the anion, as is the case for all ionic compounds. The names of neutral coordination complexes are written without spaces.

Na[PtCl3(NH3)]

Sodium amminetrichloroplatinate(II)K2[CuBr4]

Potassium tetrabromocuprate(II)

NOMENCLATURE

trans-[Co(en)2I(H2O)](NO3)2

trans-aquabis(ethylenediamine)iodocobalt(III) nitrate

mer-[Ru(PPh3)3Cl3]

mer-trichlorotris(triphenylphosphine)ruthenium(III)

NOMENCLATURE

• The name of the coordination compound (neutral, cationic or anionic) begins with the names of the ligands. The metal is listed next, following in parentheses by the oxidation state of the metal.

NOMENCLATURE

When more than one of a given ligand is bound to the same metal atom or ion, the number of such ligands is designated by the following prefixes:2 di 6 hexa 10 deca3 tri 7 hepta 11 undeca4 tetra 8 octa 12 dodeca5 penta 9 nona

NOMENCLATURE

However, when the name of the ligand in question already contains one of these prefixes or ligands with complicated names (generally ligand names that are three syllables or longer), then a prefix from the following list is used instead:

2 bis 6 hexakis3 tris 7 heptakis4 tetrakis 8 octakis5 pentakis 9 ennea

NOMENCLATURENeutral ligands are given the same name as the uncoordinated molecule, but with spaces omitted. Some examples are:

(CH3)3SO dimethylsulfoxide (DMSO)

(NH2)2CO urea

C5H5N pyridine

terpy terpyridinebpy 2,2’-bipyridineen ethylenediaminePCl3 trichlorophosphine

PPh3 triphenylphopshine

NOMENCLATURE

EXCEPTIONS: Some neutral molecules, when serving as ligands are given special names. These are:

NH3 ammine

H2O aqua

NO nitrosylCO carbonylCS thiocarbonyl

NOMENCLATURE• Anionic ligands are given names that end in the letter “o”.

When the name of the free, uncoordinated anion ends in “ate”, the ligand name is changed to end in “ato”. Some examples are :CH3CO2

- (acetate) acetato

SO42- (sulfate) sulfato

CO32- (carbonate) carbonato

acac acetylacetonato

NOMENCLATURE

When the name of the free, uncoordinated anion ends in “ide”, the ligand name is changed to end in “ido”. Some examples are:

N3- (nitride) nitrido

N3- (azide) azido

NH2- (amide) amido

NOMENCLATURE

When the name of the free, uncoordinated anion ends in “ite”, the ligand name is changed to end in “ito”. Some examples are:

NO2- (nitrite) nitrito

SO32- (sulfite) sulfito

ClO3- (chlorite) chlorito

NOMENCLATURECertain anionic ligands are given special names, all ending in “o”:

CN- cyano F- fluoroCl- chloro Br- bromoI- iodo O2- oxoO2

- superoxo OH- hydroxo

H- hydrido CH3O- methoxo

NOMENCLATURE

The ligands are named alphabetically, ignoring the prefixes bis, tris, etc…

NOMENCLATUREWhen the coordination entity is either neutral or cationic, the usual name of the metal is used, followed in parentheses by the oxidation state of the metal. However, when the coordination entity is an anion, the name of the metal is altered to end in “ate”. This is done for some metals by simply changing the ending “ium” to “ate”:

Scandium scandateTitanium titanateChromium chromate

NOMENCLATURE

Geometrical isomers are designated by cis- or trans- and mer- or fac-, the latter two standing for meridional or facial, respectively.

NOMENCLATURE

Bridging ligands are designated with the prefix -. When there are two bridging ligands of the same kind, the prefix di-- is used. Bridging ligands are listed in order with other ligands, and set off between hypens. An important exception arises when the molecule is symmetrical, and a more compact name can be given by listing the bridging ligand first.

NOMENCLATURE

Example:

NOMENCLATURE

NOMENCLATURE

Ligands that are capable of linkage isomerism are given specific names for each mode of attachment.

-SCN- thiocyanato (S-thiocyanato)-NCS- isothiocyanto (N-thiocyanto)-NCSe- isoselenocyanato (N-selenocyanato)-NO2

- nitro

-ONO- nitrito

EXAMPLES

1. [Co(NH3)5CO3]Cl

2. Potassium pentachloronitridoosmate(VI)3. [Cr(H2O)4Cl2]Cl

4. Potassium pentacyanonitrosylferrate(II)5. K4[Mn(CN)6]

6. [Ni(bipy)3(NO3)2]

7. [Co(N3)(NH3)5]SO4

NOMENCLATURE• Bridging ligands between two metal ions have the prefix ‘ ’.

– -amido--hydroxobis(tetraamminecobalt)(IV)

ISOMERISM

ISOMERISM• Four-coordinate complexes

– Square-planar complexes may have cis and trans isomers. No chiral isomers (enantiomers) are possible when the molecule has a mirror plane.

– cis- and trans-diamminedichloroplatinum(II)

– How about tetrahedral complexes?– Chelate rings commonly impose a

‘cis’ structure. Why

ISOMERISM

CHIRALITY

• Mirror images are nonsuperimposable.• A molecule can be chiral if it has no rotation-reflection axes (Sn)• Chiral molecules have no symmetry elements or only have an

axes of proper rotation (Cn).– CBrClFI, Tetrahedral molecule (different ligands)– Octahedral molecules with bidentate or higher chelating

ligands– Octahedral species with [Ma2b2c2], [Mabc2d2], [Mabcd3],

[Mabcde2], or [Mabcdef]

CHIRALITY

SIX-COORDINATE OCTAHEDRAL COMPLEXES

ML3L3’Fac isomers have three

identical ligands on the same face.

Mer isomers have three identical ligands in a plane bisecting the molecule.

ISOMERISM

SIX-COORDINATE OCTAHEDRAL COMPLEXES

The maximum number of isomers can be difficult to calculate (repeats).

Placing a pair of ligands in the notation <ab> indicates that a and b are trans to each other.[M<ab><cd><ef>], [Pt<pyNH3><NO2Cl><BrI>]

How many diastereoisomers in the above platinum compound (not mirror images)?

Identify all isomers belonging to Ma3bcd.

COMBINATIONS OF CHELATE RINGS

• Propellers and helices– Left- and right-handed propellers

• Examine the movement of a propeller required to move it in a certain direction.– For a left-handed propeller, rotating it ccw would cause it

to move away ().– For a right-handed propeller, rotating it cw would cause it

to move away ().This is called ‘handedness’. Many molecules possess it.

Tris(ethylenediamine)cobalt(III)

• this molecule can be treated like a three-bladed propeller.• look down a three fold axis to determine the ‘handedness’

of this complex ion.– the direction of rotation required to pull the

molecule away from you determines the handedness ( or ).

• do this with you molecule set and rubber bands.

DETERMINING HANDEDNESS FOR CHIRAL MOLECULES

• Complexes with two or more nonadjacent chelate rings may have chiral character.– Any two noncoplanar and nonadjacent chelate rings can be

used.– Look at Figure 9-14 (Miessler and Tarr).

• Molecules with more than one pair of rings may require more than one label.– Ca(EDTA)2+

• Three labels would be required.• Remember that the chelate rings must be noncoplanar,

nonadjacent, and not connected at the same atom.

LINKAGE (AMBIDENTATE) ISOMERISM

A few ligands may bond to the metal through different atoms. SCN- and NO2

-

How would you expect hard acids to bond to the thiocyanate ligand?

Solvents can also influence bonding.High and low dielectric constants.

Steric effects of linkage isomerism Intramolecular conversion between linkages.

[Co(NH3)5NO2]+2, Figure 9-19.

COORDINATION NUMBERS AND STRUCTURES

• Factors considered when determining structures.– The number of bonds. Bond formation is exothermic; the

more the better.– VSEPR arguments– Occupancy of d orbitals.– Steric interference by large ligands.– Crystal packing effect.It may be difficult to predict shapes.

LOW COORDINATION NUMBERS (C.N.)• CN 1 is rare except in ion pairs in the gas phase.• CN 2 is also rare.

– [Ag(NH3)2]+, Ag is d10 (how?)

– VSEPR predicts a linear structure.– Large ligands help force a linear or near-linear arrangment.

• [Mn(N[SiMePh2]2)2] in Figure 9-22.

• C.N. 3 is more likely with d10 ions.– Trigonal-planar structure is the most common.– [Cu(SPPh3)3]+, adopts a low C.N. due to ligand crowding.

COORDINATION NUMBER 4

• Tetrahedral and square planar complexes are the most common.– Small ions and/or large ligands prevent high coordination

numbers (Mn(VII) or Cr(VI)).• Many d0 or d10 complexes have tetrahedral structures (only

consider bonds).– MnO4

- and [Ni(CO)4]

– Jahn-Teller distortion (Chapter 10)


• Square-planar geometry– d8 ions (Ni(II), Pd(II), and Pt(III))

• [Pt(NH3)2Cl2]

– The energy difference between square-planar and tetrahedral structures can be quite small.

• Can depend on both the ligand and counterion.• More in chapter 10.


• Common structures are trigonal bipyramid and square pyramid.– The energy difference between the two is small. In many

measurements, the five ligands appear identical due to fluxional behavior.

– How would you modify the experiment to differentiate between the two structures?

• Five-coordinate compounds are known for the full range of transition metals.– Figure 9-27.


• This is the most common C.N. with the most common structure being octahedral.– If the d electrons are ignored, this is the predicted shape.

• [Co(en)3]3+

• This C.N. exists for all transition metals (d0 to d10).

DISTORTIONS OF COMPLEXES CONTAINING C.N. 6

• Elongation and compression (Fig. 9-29).– Produces a trigonal antiprism structure when the angle

between the top and bottom triangular faces is 60.– Trigonal prism structures are produced when the faces are

eclipsed.• Most trigonal prismatic complexes have three bidentate

ligands (Figure 9-30).• interactions may stabilize some of these structures.

The Jahn-Teller effect is useful in predicting observed distortions.

HIGHER COORDINATION NUMBERS

• C.N. 7 is not common• C.N. 8

– There are many 8-coordinate complexes for large transition elements.

• Square antiprism and dodecahedron• C.N.’s up to 16 have been observed.

MAGNETIC SUSCEPTIBILITY

• Diamagnetic versus paramagnetic complexes.• Commonly provides mass susceptibility per gram.• Magnetic moment

litysusceptibimagneticT 2

1

)(828.2

CONTRIBUTIONS TO THE MAGNETIC MOMENT

• Spin magnetic moment– S = maximum total spin in the complex

• O atom

• Orbital angular momentum– Characterized by the quantum number L which is equal the

maximum possible sum of ml values.

• O atom

)]1(41[)]1([ LLSSgLS

CONTRIBUTIONS TO THE MAGNETIC MOMENT

• Usually, the spin-only moment is sufficient to calculate the magnetic moment.– Especially for the first transition series

where g (gyromagnetic ratio) is approximated to be 2 and n is the number of unpaired electrons.

– Determine the spin-only and complete magnetic moment for Fe.

)2n(nor)1S(SgS

Calculate the spin-only magnetic moment

For the following atoms/ions:

Fe+2 (observed: 5.1), Fe, Cr, Cr+3 (observed = 3.8)

ELECTRONIC SPECTRA

• Orbital energy levels can be obtained directly from electron spectra (will be covered later).

• This chapter illustrates simple energy level diagrams that are commonly more complex.

• Based upon subtle differences in electronic spectra, the structure may be predicted with some success.

THEORIES OF ELECTRONIC STRUCTURE• Valence Bond Theory – Not commonly used, but the hybrid notation is

still common.

• Crystal Field Theory – An electrostatic approach used to describe the splitting in metal d-orbital energies. Does not describe bonding.

• Ligand Field Theory – A more complete description of bonding in terms of the electronic energy levels of the frontier orbitals. Commonly does not include energy of the bonding orbitals.

• Angular Overlap Method – Used to estimate the relative magnitude of the orbital energies in a MO calculation.

VALENCE BOND THEORY (HYBRIDIZATION)

• A set of hybrid orbitals is produced to explain the bonding.– Octahedral – d2sp3 (6 hybrid orbitals of equal energy)– Tetrahedral - ??

• Uses ‘inner’ and ‘outer’ orbitals to explain the experimentally determined unpaired electrons. – The magnetic behavior determines which d orbitals (e.g. 3d or 4d) are used

for bonding (Figure 10-2).

VALENCE BOND DESCRIPTION• Two configurations are possible for d4-d7 ions.• Fe(III) has 5 electrons in the d-orbitals.

– One unpaired electron, the ligands are ‘strong’ and force the metal d electrons to pair up.

• Strong-field (bind strongly) low spin complex• The hybridization orginates from the 3d inner orbitals

(d2sp3).

VALENCE BOND DESCRIPTION– Five unpaired electrons, the ligands are ‘weak’ and cannot

force the metal d electrons to pair up.

• Weak-field (bind weakly) high spin• The hybridization originates from the 4d outer orbitals

(sp3d2).

VALENCE BOND THEORY

Structure, hybridization, and magnetism1)[Co(NH3)6]3+, diamagnetic, octahedral

2)[CoF6]3-, paramagnetic, octahedral

3)[PtCl4]2-, diamagnetic, sq. planar

4)[NiCl4]2-, pamagnetic, tetrahedral

SAMPLE PROBLEM:

• The complexes [Mn(H2O)6]2+, [Fe(H2O)6]3+,

[MnCl4]2-, and [FeCl4]- have all magnetic moments. What does this tell about the geometric and electronic structures of these complexes?

CRYSTAL FIELD THEORY

• Focus: energies of the d orbitals

• Assumptions1. Ligands: negative point charges2. Metal-ligand bonding: entirely ionic

strong-field (low-spin): large splitting of d orbitalsweak-field (high-spin): small splitting of d orbitals

20_454

eg(dz2, dx2 – y2)

t2g(dxz, dyz, dxy)

Free metal ion3d orbitalenergies

E

= crystal field splitting

High spin Low spin

CRYSTAL FIELD THEORY• The average energy of the d-orbitals in the presence of the

octahedral field is greater than than of the free ion.

• Energy difference between the two sets is equal to O.– The t2g set is lowered by 0.4 O and the eg set is raised by 0.6 O.

• Crystal field stabilization energy (CFSE) – The energy difference between the actual distribution of electrons and that for all electrons in the uniform field.– Equal to LFSE (later)

• Drawbacks

LIGAND FIELD THEORY – OCTAHEDRAL COMPLEXES

• Consider -type bonding between the ligands and the metal atom/ion.

• Construct LGOs (performed previously).– What is the reducible representation?– Construct the LGOs (pictures).

• Construct the molecular orbitals with the metal orbitals.– Same symmetry types.

• A group of metal orbitals do not have the appropriate symmetry?– Which orbitals are these? Symmetry type? Bonding?

• Look at Figure 10-5.

SF6

= A1g + T1u + Eg


• The six bonding orbitals are largely filled by the electrons from the ligands.

• The higher MOs (e.g. t2g and eg) are largely filled by the electrons on the metal atom/ion.– The ligand field treatment largely focuses on the t2g and

higher orbitals.• The split between the two sets of orbitals, t2g and eg, is

called O.


• Ligands whose orbitals interact strongly with the metal orbitals are called strong-field ligands.– Strong-field large O low spin (why?)

• Ligands with small interactions are called weak-field ligands.– Weak-field small O high spin (why?)

• For d0 – d3 and d8-d10 only one electron configuration is possible (no difference in net spin).

• For d4 – d7 there is a difference between strong- and weak-field cases.

LOW SPIN VERSUS HIGH SPIN

• Energy of pairing electrons

– c is the Coulombic energy of repulsion (always positive when pairing) and e is the quantum mechanical exchange energy (always negative).

• e relates to the number of exchangeable pairs in a particular electron configuration. This term is negative and depends on the number of possible states.

Determine c and e for a d5 metal complex (low and high spin).

ec


• The relationship between O, c, and e determines the orbital configuration.

• is largely independent on the ligands while O is strongly dependent.

• Look at Table 10-6 which gives these parameters for aqueous (aqua) ions.– O for 3+ ions is larger than O for 2+ ions.– O values for d5 are smaller than d4 and d6.


• If O>, there is a lower energy upon pairing in the lower levels (low spin).

• If O<, there is a lower energy with unpaired electrons in the lower levels (high spin).

• In Table 10-6, [Co(H2O)6]3+ is probably the only complex that could be low spin.

Ligand Field Stabilization Energies (LFSE)

• The difference (1) the total energy of a coordination complex with the electron configuration resulting from ligand field splitting of the orbitals and (2) the total energy for the same complex with all the orbitals equally populated is the LFSE.

• -2/5O + 3/5O (d4 to d7 complexes)

• Table 10-7

BONDING IN OCTAHEDRAL COMPLEXES

• The x and z axes must be taken as a single set producing a combined LGO set. Why?

• Be able to derive the reducible representation.– = T1g + T2g + T1u + T2u

• How will the LGOs combine with orbitals from the metal atom/ion?

• Discuss the overlap between the -bonding LGOs and the p-orbitals of T1u symmetry.

PI BONDING IN OCTAHEDRAL COMPLEXES

• The main addition to the interaction diagram is between the t2g orbitals of the metal and LGOs.– These were nonbonding when only considering -

type bonding (look at Figure 10-5).

• Pi bonding may occur when the ligands have available p or * molecular orbitals.

LIGANDS WITH EMPTY * ORBITALS

• Examine the example for the CN- ligand in the book (Figure 10-9).

• The HOMO forms the LGOs from -type bonding (already discussed previously).

• The LUMO, 1*, also forms a reducible set of LGOs (T1g + T2g + T1u + T2u).– Examine Figure 10-10 to illustrate effectiveness of

overlap.

LIGANDS WITH EMPTY * ORBITALS

• The resulting t2g LGOs are generally higher in energy than the initial t2g orbitals on he metal.– Bonding/antibonding t2g orbitals will result.– What will this do to O and the bond strength?

• Figure 10-11.

• This is termed as metal-to-ligand bonding or back-bonding.– Some of the electron density in the d orbitals on the metal

is donated back to the ligands.– The ligands are termed as -acceptor ligands.

LIGANDS WITH FILLED -TYPE ORBITALS

• Ligands such as F- or Cl- will possess molecular orbitals that possess electrons.

• This set of ‘t2g’ orbitals are generally lower in energy than the t2g orbitals on the metal.

• What are the consequences? – Examine Figure 10-11.

• Ligand-to-metal bonding (-donor ligands).– This bonding is generally less favorable.

SQUARE-PLANAR COMPLEXES

• The y-axis is pointed toward the center atom.– LGOs for sigma-type bonding.

• The -bonding orbitals on the x- and z-axes have to be considered separately? Why?– These are termed as (px) and (pz)

• Examine Table 10-9.– What is the symmetry of a square-planar complex?

SQUARE-PLANAR COMPLEXESSIGMA-TYPE BONDING ONLY

• Finding the LGOs.– red = A1g + B1g + Eu

• What are the orbitals on the central metal atom that can interact with these LGOs?

• Inspecting the character table reveals that the metal d-orbitals are split into three representations. Why?

• Examine Figure 10-13.– The energy difference between the eg/b2g nonbonding

orbitals and the a1g antibonding is .

SQUARE-PLANAR COMPLEXESINCLUDING PI-BONDING

• px = A2g + B2g + Eu ()– What are the interacting orbitals on the metal?

• pz = A2u + B2u + Eg ()– What are the interacting orbitals on the metal?

• The effective overlap of the p orbitals on the metal to form bonds is small.

• Examine Figure 10-15.

THE ‘SETS’ OF ORBITALS IN FIGURE 10-15

• The 1st set contains bonding orbitals (mostly sigma).– 8 electrons from the ligands largely fill these orbitals.

• The 2nd set contains 8 -donor orbitals of the ligands. – This interaction is small and decreases the energy differences in orbitals

the next higher set.• The 3rd set is primarily metal d-orbitals with some modifications

due to interactions with the ligands.– 3, 2, and 1 are in this set.

• The 4th set largely originates from the * orbitals of the ligands (if present).– One of the main effects of these orbitals is the increase in the gap energy

labeled 1.

ANGULAR OVERLAP (CRYSTAL FIELD)

• Estimates the strength of interaction between individual ligand orbitals and d-orbitals based on the overlap between them. These values are then combined for all ligands and d-orbitals.

• The value for a given d-orbital is the sum of the numbers for the appropriate ligands in a column.– This number can be positive or negative depending on location of the

ligand and d-orbitals.• The value for a given ligand is the sum of the numbers for all d-

orbitals in the row.– This number can also be positive or negative depending on location of the

ligand and d-orbitals.

ANGULAR OVERLAP

• Sigma-donor interaction (no pi-orbitals are available).– [M(NH3)6]n+

• The strongest interaction is between the metal dz2 orbital and a ligand p-orbital (or appropriate MO).

• Describe the interaction based on this method.– Table 10-11 and Figure 10-20.

ANGULAR OVERLAP

• Pi-acceptor ligands (available -type orbitals).• Strongest interaction is between dxz and * on the ligand.• The * orbitals are almost always higher in energy.

– Reverse the signs.• Figure 10-22 and Table 10-12

– There is a lowering of 4e due to this interaction.• Why is magnitude e always smaller than that of e?

• Understand -donor interactions.

SAMPLE PROBLEM

Using the angular overlap model, determine the splitting pattern of the d orbitals for a tetrahedral complex of formula ML4.where L is a capable of interactions only.

SAMPLE PROBLEM

Determine the energies of the d orbitals predicted by the angular overlap model for square planar complexes

a) considering interactions onlyb) considering both -donor and -

acceptor interactions

THE SPECTROCHEMICAL SERIES

• depends on the relative energies and the degree of overlap.

• How ligands effect – -donor ligands– -donating – -accepting (or back bonding)

• Understand the spectrochemical series (page 368)

MAGNITUDE OF E, E, AND

• Changing the metal and/or ligand effects the magnitudes of e and e, thereby changing the value of .– Aqua species of Co2+ and Co3+

– [Fe(H2O)6]2+ versus [Fe(H2O)6]3+

• Tables 10-13 and 10-14 (Angular Overlap)– e > e (always)– Values decrease with increasing size and decreasing

electronegativity– Negative values for e. Why?

THE JAHN TELLER EFFECT

• There cannot be unequal occupation of orbitals with identical energies. The molecule will distort so that these orbitals are no longer degenerate.– Cu(II) d9 ion, The complex will distort. How?– The low-spin Cr(II) complex is octahedral with

tetragonal distortion (Oh D4h)• Two absorption bands are observed instead of one.

DETERMINING FOUR- AND SIX-COORDINATE PREFERENCES

• General angular overlap calculations of the energies expected for different number of d electrons and different geometries can give us some indication of relative stabilities.– Larger number of bonds usually make the octahedral

complexes more stable. Why are the energies equal in the d5, d6, and d7 cases?

– Figure 10-27.

DETERMINING FOUR- AND SIX-COORDINATE PREFERENCES

• The success of these simplistic calculations is variable.– The s- and p-orbitals of the metal are not included.– No -type interactions are included in Figure 10-27.– The orbital potential energies for the metals change

with increasing atomic number (more negative).• Can add –0.3e (increase in Z) as a rough correction to

the total enthalpy.

THE PROCESS FOR A COMPLEX OF D3h SYMMETRY

• Construct the sigma-type bonding LGOs for the complex.

• Determine the interacting orbitals on the center atom.• Construct a table to determine e (and e if

appropriate).• Construct the MO diagram and overlap energy figure.Homework: Determine the e contribution.

Symmetry and Group Theory

The symmetry properties of molecules and how they can be used to predict vibrational spectra, hybridization, optical activity, etc.

POINT GROUPS

Molecules are classified and grouped based on their symmetry. Molecules with similar symmetry are but into the same point group.

A point group contains all objects that have the same symmetry elements.

SYMMETRY ELEMENTS

Symmetry elements are mirror planes, axis of rotation, centers of inversion, etc.

A molecule has a given symmetry element if the operation leaves the molecule appearing as if nothing has changed (even though atoms and bonds may have been moved.)

Symmetry ElementsElement Symmetry Operation Symbol

Identity En-fold axis Rotation by 2π/n Cn

Mirror plane Reflectionσ

Center of inversion Inversion in-fold axis of Rotation by 2π/n Sn

improper rotation followed by reflectionperpendicular to the axis of rotation

C3 or three-fold rotational axis of theammonia molecule. If we rotate the ammoniamolecule by 360/3 or 120º about thisaxis, its appearance is unchanged.

C3

Rotational axes of BF3

three-fold axis three-fold axis two-fold axis two-fold axisviewed from viewed from viewed from viewed from above the side the side above

Note: there are 3 C2 axes

C3 C3 C2 C2

principal axis(highest value of Cn)

.

SAMPLE PROBLEM

• How many axes of rotation does borazine possess?

• Ethane in the eclipsed conformation?N

B

N

B

N

B

H

H

H

H

H

H

Mirror planes (σ) of BF3:

Mirror planes can contain the principal axis (σv) or be at right angles to it (σh). BF3 has one σh and three σv planes: (v = vertical, h = horizontal)

σv

mirror plane C3

principal axis

σh

mirror plane C3

principal axis

σv mirror plane

contains the C3 axis

σh mirror plane

is at right angles to the C3 axis

SAMPLE PROBLEM

• Mirror planes of symmetry for Borazine, naphtlalene, diborane, dxy orbital?

(Note: The center of symmetry is important in deciding whether orbitals are g or u (lecture 2.))

center of symmetrycenter of symmetry

SAMPLE PROBLEM

• Which of the following flourine compounds has center of inversion? BF3, SiF4, PF5, XeF5

-, SF6, C2F4,

rotateby 360o/4

The S4 improper rotation axis here is also a C2 axis

Rotational axes and mirror planes of the water molecule:

C2

principal axis C2C2 σv

mirror plane

σv

mirror plane

The water molecule has only one rotational axis, its C2 axis,which is also its principal axis. It has two mirror planes thatcontain the principal axis, which are therefore σv planes. It has no σh mirror plane, and no center of symmetry.

C6

principal axisC2 C2

C2C6C2

σvσv

Rotational axes and mirror planes of benzene

σh

C6

principal axis

C6

principal axis

Rotational axes and mirror planes of boron trifluoride

C3

principal axis

C3principal axis

σh

σh

σv σv

C2

C2 C2

boron trifluoride has a C3 principal axis and three C2 axes, a σh mirror planethree σv mirror planes, but no center of inversion

Identity, E

All molecules have Identity. This operation leaves the entire molecule unchanged. A highly asymmetric molecule such as a tetrahedral carbon with 4 different groups attached has only identity, and no other symmetry elements.

Improper Rotation

An improper rotation is rotation, followed by reflection in the plane perpendicular to the axis of rotation.

Improper Rotation

The staggered conformation of ethane has an S6 axis that goes through both carbon atoms.

Improper Rotation

Note that an S1 axis doesn’t exist; it is same as a mirror plane.

Improper Rotation

Likewise, an S2 axis is a center of inversion.

Sample problem

Draw the structure for the following showing the correct geometry and identify all the symmetry elements present in each:

a) SCN- b) S2O32-, c) IF4

- d) 1,8-dichloronaphthalene e) formaldehyde

Point Groups

Molecules with the same symmetry elements are placed into point groups.

Group theory, the mathematical treatment of the properties of groups can be used to determine the molecular orbitals, vibrations, and other properties of the molecule.

∞ ∞

Point Groups

In general, you will not need to assign a molecule to its point group. Recognition of the features of some common point groups is useful.

Point GroupsWater and

ammonia both belong to the Cnv class of molecules. These have vertical planes of reflection, but no horizontal planes.

Point GroupsThe Dnh groups

have a horizontal plane in addition to vertical planes. Many inorganic complexes belong to these symmetry groups.

XX

XX

Y

Y

POINT GROUPS

Highly symmetrical molecules, such as identically substituted tetrahedrons or octahedrons belong to their own point groups (Td or Oh respectively).

Point Groups

In assigning a point group, we typically ignore the fine detail, such as conformation isomers, of the ligands.

In working problems using group theory, the point group of the molecule will usually be provided to you.

Example:

• PF5, SF6, IOF3, XeF4, ethane (eclipsed and staggered), ethylene and chloroethylene. Ferrocene (eclipsed and staggered)

COORDINATION CHEMISTRY III:REACTIONS OF METAL COMPLEXES

The ability to predict products and chooseappropriate reaction condition to obtainthe desired products is still a matter of artas well as science.

substitution reactions

kinetic consequences of reaction pathways

experimental evidence in octahedral substitution

substitution reactions of square-planar complexes

the trans effect

oxidation-reduction reactions

reactions of coordinated ligand

SUBSTITUTION REACTIONS

SUBSTITUTION REACTIONS

MLn + L' MLn-1L' + L

Labile complexes <==> Fast substitution reactions (< few min)Inert complexes <==> Slow substitution reactions (>h)

a kinetic concept

Not to be confused withstable and unstable (a thermodynamic concept Gf <0)

Inert Intermediate Labile

d3, low spin d4-d6& d8 d8 (high spin) d1, d2, low spin d4-d6& d7-d10

SUBSTITUTION REACTIONS – INERT AND LABILE

INERT, LABILE vs STABLE, UNSTABLE

kinetic terms thermodynamic terms

Stable but labile

unstable but inert

MLnX + Y MLnY + X

Mechanisms of ligand EXCHANGE REACTIONSIN OCTAHEDRAL COMPLEXES

MLnX MLnY

MLnX Y

Dissociative (D)MLnX MLnY

MLnXYY X

Associative (A)

MLnX MLnY

[MLn]°Y X

X Y

Interchange (I)

Ia if associationis more important

Id if dissociationis more important

KINETICSOF DISSOCIATIVE

REACTIONS

Kineticsof interchange

reactions

Fast equilibriumK1 = k1/k-1

k2 << k-1

For [Y] >> [ML5X]

Kinetics of associative reactions

Principal mechanisms of ligand exchange in octahedral complexes

ML5Xk1

slowML5 + X

k2

fast

+YML5Y

r = k1 [ML5X]

ML5X + Yk1

slowML5XY

k2

fast

-XML5Y

r = k1 [ML5X][Y]

Dissociative

Associative

Dissociative pathway(5-coordinated intermediate)

Associative pathway(7-coordinated intermediate)

MOST COMMON

Experimental evidence for dissociative mechanisms

Rate is independent of the nature of L

Experimental evidence for dissociative mechanisms

Rate is dependent on the nature of L

Inert and labile complexesSome common thermodynamic and kinetic profiles

Exothermic(favored, large K)

Large Ea, slow reaction

Exothermic(favored, large K)

Large Ea, slow reactionStable intermediate

Endothermic(disfavored, small K)Small Ea, fast reaction

LM

L L

L

L

X

LM

L L

L

L

X

LM

L L

L

L

G

Ea

Labile or inert?

LFAE = LFSE(sq pyr) - LFSE(oct)

Why are some configurations inert and some are labile?

Inert !

Other metal on factors that affect reaction rates

Oxidation state of the central atom: Central atom with higher oxidation states have slower ligand exchange rates

[AlF6]- > [SiF6]- > [PF6]- > SF6

Ionic radius. Smaller ions have slower exchange rates [Sr(H2O)6]2+ > [Ca(H2O)6]2+ > [Mg(H2O)6]2+ 112 pm 99 pm 66 pm

Both effects due to higher electrostatic attraction between central atom and attached ligands.

Substitution reactions in square-planar complexesthe trans effect

T

M

L X

L T

M

L Y

L

+X, -Y

(the ability of T to labilize X)

Synthetic applicationsof the trans effect

Mechanisms of ligand exchange reactions in square planar complexes

-d[ML3X]/dt = (ks + ky [Y]) [ML3X]

LM

L L

X

LM

L L

Y

LM

L L

X

LM

L L

X

LM

L L

S

LM

L L

S

S

Y

Y

+Y

+S

-X

+Y

-S

-X

THE trans EFFECT

SIGMA-BONDING EFFECTS

Sigma-Bonding Effect. A strong bond between Pt and T weakens the Pt-X bond.

H- > PR3 > SCN- ~ CH3- ~ CO ~ CN- > Br- > Cl- > NH3 > OH-

PI-BONDING EFFECTS

If back donation occurs to a ligand, the flow of electron density from the metal leaves less electron density to be donated in the opposite direction.

C2H4 ~ CO > CN- > NO2- > SCN- > I- > Br- > Cl- > NH3 > OH-

Overall trans effect:

CO ~ CN- ~ C2H4 > PR3 ~ H- > CH3- ~ SC(NH2)2 > C6H5

- >NO2-

~ SCN ~ I- >Br- > Cl- > py , NH3 ~ OH- ~H2O

SAMPLE PROBLEM:

Predict the products of the reactions (there may be one product when there are conflicting preferences)

[PtCl4-] + NO2

- → (a) (a) + NH3 → (b)

[PtCl3NH3]- + O2- → (c) (c) + NO2

- → (d)

SAMPLE PROBLEM:

Is it possible to prepare different isomers of Pt(II) complexes with 4 different ligands?

Predict the products expected if 1 mole of [PtCl4]- is reacted successively with the following reagents: (the product of reaction a is used in reaction b)

a)2 moles NH3

b)2 moles pyc)2 moles Cl-

d)1 mole NO2-

Electron transfer (redox) reactions

M1(x+)Ln + M2

(y+)L’n M1(x +1)+Ln + M2

(y-1)+L’n

-1e (oxidation)

+1e (reduction)

Very fast reactions (much faster than ligand exchange)

May involve ligand exchange or not

Very important in biological processes (metalloenzymes)

REDOX MECHANISMS:

Inner sphere mechanism: When two molecules are connected by a common ligand which the electron is transferred, in which case the reaction is called bridging or innersphere reaction.

Outer sphere mechanism:Exchange may occur between two separate coordination sphere in outersphere reaction.

Outer sphere mechanism

[Fe(CN)6]4- + [IrCl6]2- [Fe(CN)6]3- + [IrCl6]3-

[Co(NH3)5Cl]2+ + [Ru(NH3)6]2+ [Co(NH3)5Cl]+ + [Ru(NH3)6]3+

Reactions ca. 100 times fasterthan ligand exchange(coordination spheres remain the same)

r = k [A][B]

Ea

A B+

A B

A' B'+

G

"solvent cage"

Tunnelingmechanism

Inner sphere mechanism

[Co(NH3)5Cl)]2+ + [Çr(H2O)6]2+ [Co(NH3)5Cl)]2+:::[Çr(H2O)6]2+

[Co(NH3)5Cl)]2+:::[Çr(H2O)6]2+ [CoIII(NH3)5(-Cl)ÇrII(H2O)6]4+

[CoIII(NH3)5(-Cl)ÇrII(H2O)6]4+ [CoII(NH3)5(-Cl)ÇrIII(H2O)6]4+

[CoII(NH3)5(-Cl)ÇrIII(H2O)6]4+ [CoII(NH3)5(H2O)]2+ + [ÇrIII(H2O)5Cl]2+

[CoII(NH3)5(H2O)]2+ [Ço(H2O)6]2+ + 5NH4+

Inner sphere mechanism

Reactions much faster than outer sphere electron transfer(bridging ligand often exchanged)

r = k’ [Ox-X][Red] k’ = (k1k3/k2 + k3)

Ox-X + Red Ox-X-Redk1

k2

k3

k4Ox(H2O)- + Red-X+

Ea

Ox-X Red+

Ox-X-Red

G

Ox(H2O)- + Red-X+

Tunnelingthrough bridgemechanism

coordination chemistry structures and isomers. electronic configuration ground state: progressive...

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