Transcript Coordination Chem

COORDINATION CHEMISTRY
STRUCTURES AND ISOMERS
ELECTRONIC CONFIGURATION
Ground 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
COORDINATION COMPOUNDS
LIGANDS
LIGANDS
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 geometry
ethylene 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 deca
3 tri
7 hepta
11 undeca
4 tetra
8 octa
12 dodeca
5 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 hexakis
3 tris
7 heptakis
4 tetrakis
8 octakis
5 pentakis
9 ennea
NOMENCLATURE
Neutral 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
terpyridine
bpy
2,2’-bipyridine
en
ethylenediamine
PCl3
trichlorophosphine
PPh3
triphenylphopshine
NOMENCLATURE
EXCEPTIONS: Some neutral molecules, when serving as ligands
are given special names. These are:
NH3 ammine
H2O aqua
NO
nitrosyl
CO
carbonyl
CS
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
NOMENCLATURE
Certain anionic ligands are given special names, all ending in “o”:
CNcyano
Ffluoro
Clchloro
Brbromo
Iiodo
O2oxo
O2superoxo
OHhydroxo
Hhydrido
CH3O- methoxo
NOMENCLATURE
The ligands are named alphabetically, ignoring the prefixes bis,
tris, etc…
NOMENCLATURE
When 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
scandate
Titanium
titanate
Chromium
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
NOMENCLATURE
NOMENCLATURE
Ligands that are capable of linkage isomerism are given specific
names for each mode of attachment.
-SCNthiocyanato (S-thiocyanato)
-NCSisothiocyanto (N-thiocyanto)
-NCSeisoselenocyanato (N-selenocyanato)
-NO2nitro
-ONOnitrito
EXAMPLES
1.
2.
3.
4.
5.
6.
7.
[Co(NH3)5CO3]Cl
Potassium pentachloronitridoosmate(VI)
[Cr(H2O)4Cl2]Cl
Potassium pentacyanonitrosylferrate(II)
K4[Mn(CN)6]
[Ni(bipy)3(NO3)2]
[Co(N3)(NH3)5]SO4
NOMENCLATURE
• Bridging ligands between two metal ions have the prefix ‘ ’.
– -amido--hydroxobis(tetraamminecobalt)(IV)
ISOMERISM
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 transdiamminedichloroplatinum(II)
– How about tetrahedral complexes?
– Chelate rings commonly impose a
‘cis’ structure. Why
ISOMERISM
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)
COORDINATION NUMBER 4
• 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.
COORDINATION NUMBER 5
• 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.
COORDINATION NUMBER 6
• 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 
  2.828( T )
1
2
  magnetic susceptibility
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
 S  L  g [ S ( S  1)]  [ 1 4 L( L  1)]
CONTRIBUTIONS TO THE MAGNETIC MOMENT
• Usually, the spin-only moment is sufficient to
calculate the magnetic moment.
– Especially for the first transition series
S  g S(S  1) or
n(n  2)
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.
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 magnetism
1) [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
• Assumptions
1. Ligands:
negative point
charges
2. Metal-ligand bonding:
entirely ionic
strong-field (low-spin): large splitting of d orbitals
weak-field (high-spin): small splitting of d orbitals
20_454
eg(d z2, d x 2 – y 2)

t2g (d xz, d yz, d xy)
E
 = crystal field splitting
Free metal ion
3d orbital
energies
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
LIGAND FIELD THEORY – OCTAHEDRAL
COMPLEXES
• 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.
LIGAND FIELD THEORY – OCTAHEDRAL
COMPLEXES
• 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 weakfield cases.
LOW SPIN VERSUS HIGH SPIN
• Energy of pairing electrons
  c  e
– 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).
LOW SPIN VERSUS HIGH SPIN
• 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.
LOW SPIN VERSUS HIGH SPIN
• 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.
• -2/5O + 3/5O (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 COMPLEXES
SIGMA-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 COMPLEXES
INCLUDING 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 dorbitals 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.
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 only
b) 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 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.)
Element
n-fold axis
Mirror plane
σ
Center of inversion
n-fold axis of
improper rotation
Symmetry Operation
Symmetry
Elements
Identity
Rotation by 2π/n
Reflection
Inversion
Rotation by 2π/n
followed by reflection
perpendicular to the
axis of rotation
Symbol
E
Cn
i
Sn
C3
C3 or three-fold rotational axis of the
ammonia molecule. If we rotate the ammonia
molecule by 360/3 or 120º about this
axis, its appearance is unchanged.
Rotational axes of BF3
principal axis
(highest value of Cn)
C3
C3
C2
C2
.
three-fold axis
viewed from
above
three-fold axis
viewed from
the side
two-fold axis
viewed from
the side
two-fold axis
viewed from
above
Note: there are 3 C2 axes
SAMPLE PROBLEM
• How many axes of rotation does borazine
possess?
• Ethane in the eclipsed conformation?
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
σv mirror plane
contains the C3 axis
C3
principal axis
σh
mirror plane
C3
principal axis
σh mirror plane
is at right angles to the C3 axis
SAMPLE PROBLEM
• Mirror planes of symmetry for Borazine,
naphtlalene, diborane, dxy orbital?
center of symmetry
center of symmetry
(Note: The center of symmetry is important in deciding whether orbitals
are g or u (lecture 2.))
SAMPLE PROBLEM
• Which of the following flourine compounds
has center of inversion? BF3, SiF4, PF5, XeF5-,
SF6, C2F4,
rotate
by 360o/4
The S improper rotation axis here is also a C axis
Rotational axes and mirror planes of the water molecule:
C2
principal axis
C2
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 that
contain the principal axis, which are therefore σv planes. It
has no σh mirror plane, and no center of symmetry.
C2
σv
mirror plane
Rotational axes and mirror planes of benzene
C6
C2
principal axis
C2
C6
σh
C2
C2
σv
C6
principal axis
σv
C6
principal axis
Rotational axes and mirror planes of boron trifluoride
C2
C3
principal axis
C2
C2
σh
σh
boron trifluoride has a C3 principal
axis and three C2 axes, a σh mirror plane
three σv mirror planes, but no center of inversion
σv
σv
C3
principal axis
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 Groups
Water and
ammonia both
belong to the Cnv
class of molecules.
These have vertical
planes of reflection,
but no horizontal
planes.
Point Groups
The Dnh groups
have a horizontal
plane in addition to
vertical planes.
Many inorganic
complexes belong to
these symmetry
groups.
Y
X
X
X
X
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 choose
appropriate reaction condition to obtain
the desired products is still a matter of art
as 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-1L' + L
MLn + L'
Labile complexes <==> Fast substitution reactions (< few min)
Inert complexes <==> Slow substitution reactions (>h)
a kinetic concept
Not to be confused with
stable and unstable (a thermodynamic concept Gf <0)
Inert
Intermediate
d3, low spin d4-d6& d8
d8 (high spin)
Labile
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
MECHANISMS OF LIGAND EXCHANGE REACTIONS
IN OCTAHEDRAL COMPLEXES
ML n Y + X
ML n X + Y
Dissociative (D)
MLn X
X
Associative (A)
MLn Y
MLn
Y
Y
MLn X
Interchange (I)
Y
Ia if association
is more important
MLn Y
MLn X
[ML n]°
X
Y
MLn XY
MLn Y
X
Id if dissociation
is more important
X
KINETICS
OF DISSOCIATIVE
REACTIONS
Kinetics
of interchange
reactions
Fast equilibrium
K1 = k1/k-1
k2 << k-1
For [Y] >> [ML5X]
Kinetics of associative reactions
Principal mechanisms of ligand exchange in octahedral complexes
Dissociative
Associative
Dissociative pathway
(5-coordinated intermediate)
MOST COMMON
Associative pathway
(7-coordinated intermediate)
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 complexes
Some common thermodynamic and kinetic profiles
Exothermic
(favored, large K)
Large Ea, slow reaction
Exothermic
(favored, large K)
Large Ea, slow reaction
Stable intermediate
Endothermic
(disfavored, small K)
Small Ea, fast reaction
Labile or inert?
L
L
L
M
L
L
Ea
L
L
L
L
M
L
L
M
L
L
L
X
L
X
G
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 complexes
the trans effect
L
X
M
T
L
+X, -Y
L
Y
M
T
(the ability of T to labilize X)
L
Synthetic applications
of the trans effect
Mechanisms of ligand exchange reactions in square planar
complexes
L
L
X
L
S
+S
M
L
L
M
X
L
+Y
-X
Y
L
L
L
-d[ML3X]/dt = (ks + ky [Y]) [ML3X]
M
X
L
L
M
S
L
+Y
Y
L
-X
L
L
L
L
M
Y
-S
L
M
S
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)
[PtCl3NH3]- + O2- → (c)
(a) + NH3 → (b)
(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)
b)
c)
d)
2 moles NH3
2 moles py
2 moles Cl1 mole NO2-
Electron transfer (redox) reactions
-1e (oxidation)
M1(x+)Ln + M2(y+)L’n
M1(x +1)+Ln + M2(y-1)+L’n
+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]3- + [IrCl6]3-
[Fe(CN)6]4- + [IrCl6]2-
[Co(NH3)5Cl]+ + [Ru(NH3)6]3+
[Co(NH3)5Cl]2+ + [Ru(NH3)6]2+
Reactions ca. 100 times faster
than ligand exchange
(coordination spheres remain the same)
A
B
"solvent cage"
r = k [A][B]
Ea
Tunneling
mechanism
A
+
B
A'
G
+
B'
Inner sphere mechanism
[Co(NH3)5Cl)]2+ + [Çr(H2O)6]2+
[Co(NH3)5Cl)]2+:::[Çr(H2O)6]2+
[CoIII(NH3)5(-Cl)ÇrII(H2O)6]4+
[CoII(NH3)5(-Cl)ÇrIII(H2O)6]4+
[CoII(NH3)5(H2O)]2+
[Co(NH3)5Cl)]2+:::[Çr(H2O)6]2+
[CoIII(NH3)5(-Cl)ÇrII(H2O)6]4+
[CoII(NH3)5(-Cl)ÇrIII(H2O)6]4+
[CoII(NH3)5(H2O)]2+ + [ÇrIII(H2O)5Cl]2+
[Ço(H2O)6]2+ + 5NH4+
Inner sphere mechanism
Ox-X + Red
k1
Ox-X-Red
k2
Reactions much faster
than outer sphere electron transfer
(bridging ligand often exchanged)
k3
k4
Ox(H2O)- + Red-X+
Ox-X-Red
Tunneling
through bridge
mechanism
r = k’ [Ox-X][Red] k’ = (k1k3/k2 + k3)
Ea
Ox-X
+
Red
Ox(H 2O) - + Red-X +
G