Lecture note CHM62-221 Part II

PRINCIPLES OF INORGANIC CHEMISTRY I
(CHM62-221)

PART II

By
Assoc. Prof. Dr. Phimphaka Harding

2

PRINCIPLES OF INORGANIC CHEMISTRY I

PART II Page
1
LIST OF CONTENTS 1
3
The transition elements 6
The d orbitals 7
The first transition series 9
The properties of transition metal complexes 11
Complexes and ligands 12
Coordination vs organometallic chemistry 13
Chelate complexes 15
Oxidation state and coordination number 18
Nomenclature for complexes 20
Isomerism in complexes 21
Magnetism 23
Crystal field theory 24
Crystal field splitting for an octahedral complex 27
Crystal field splitting for a tetrahedral complex 30
Crystal field splitting for a square planar complex 34
Crystal field theory and magnetism 36
Factors affecting the size of crystals field theory 37
Explanation of anomalous magnetic behaivour 37
Octahedral vs tetrahedral coordination 40
Square planar vs tetrahedral coordination 43
Thermodynamic effects of crystal field splitting 46
The colour of transition metal complexes 52
Jahn Teller effects
Molecular orbital theory
Ligand field theory

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THE TRANSITION ELEMENTS
Transition elements have partly filled d or f shells (including complexes of the various oxidation
states). There are 68 elements. The order of filling the orbitals is generally given by:

The rules for determining the ground states (lowest energy, greatest stability) of atoms are as
follows:

❖ Aufbau Principle: Electrons are added in sequence into the lowest orbitals available.
❖ Pauli Exclusion Principle: No two electron can have the same set of quantum

numbers e.g. if two electrons have the same values of n, l and ml, one must have ms
= +1/2 and the other must have ms = -1/2. So two electrons in an orbital must have
paired (opposite) spins. It follows that you can’t have more than two electrons per
orbital.
❖ Hund’s First Rule: Degenerate orbital (e.g. 2px, 2py, 2pz) are filled singly for as long
as possible with parallel spins (same ms).
THE d ORBITALS
The main block elements have their chemistry dictated by their valence s and p orbitals. For
transition metal atoms the valence orbitals will be s and d orbitals.
It is essential that you are able to draw and label the d orbitals correctly because the
later section on Crystal Field Theory rely heavily on an appreciation of the directional
properties and names of these orbitals.
❖ Particularly note that two orbitals are axial whist three lie between the axes.
❖ The names of the d orbitals give valuable information on how to draw them. For
example 3dz2 will have positive lobes along both +z and –z (z2 always +).
Similarly, 3dxy is a function of x and y and lie between the x and y axes. The
product xy (++, +-, -+ and --) gives the correct sign for the lobes.

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n Principal quantum number n = 1, 2, 3, …

Essentially defines orbital size. Small n implies low

energy

l Azimuthal quantum number l = (n-1),…, 1, 0

Defines orbital type: l = 0 is s, l = 1 is p, l = 2 is d and

l = 3 is f

ml Magnetic quantum number ml = l, (l-1), …, 0, …, - l

Defines orientation in space

ms Spin quantum number ms = +1/2 or -1/2

nl ml number of orbitals (n2)
1 0(1s) 0 1
2 0(2s) 0
+1, 0, -1 1+3=4
1(2p) 0 1 + 3+ 5 = 9
3 0(3s) +1, 0, -1
+2, +1, 0, -1, -2 1 + 3+ 5 + 7 = 16
1(3p) 0
2(3d) +1, 0, -1
4 0(4s) +2, +1, 0, -1, -2
1(4p) +3, +2, +1, 0, -1, -2, -3
2(4d)
3(4f)

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THE FIRST TRANSITION SERIES

Electronic configuration and ionization energies for transition metal atoms (kJ mol-1)

Sc Ti V Cr Mn Fe Co Ni Cu Zn

4s23d1 s2d2 s2d3 s2d4 s2d5 s2d6 s2d7 s2d8 s1d10 s2d10

1st 631 656 650 653 717 762 758 737 745 909

2nd 1235 1309 1414 1592 1509 1561 1644 1752 1958 1739

3rd 2389 2650 2828 3056 3251 2956 3231 3489 3545 3843

4th 7130 4173 4600 4900 5020 5510 5114 5404 5683

7000

6000

5000

IP kJ/mol 4000 3rd IP
3000 2nd IP
1st IP

2000

1000

0

Sc Ti V Cr Mn Fe Co Ni Cu Zn
element

Notes
1. Half-filled shell stability for Cr (s1d5) and filled shell stability for Cu (s2d10) neutral atoms.

This arise because of spin exchange effects:

cf.

For Cr [Ar]4s13d5 For Cr [Ar]4s23d4

ab, ac, ad, ae ab, ac, ad

bc, bd, be bc, bd

cd, ce cd

de

Spin exchange: 10 x 100 kJ mol-1 Spin exchange: 6 x 100 kJ mol-1

So, 3d5 is 400 kJ mol-1 more stable due to spin exchange but less stable by E. If E is < 400

kJ mol-1 4s13d5 will be preferred.

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2. N.B. Transition metal Ions (as found in complexes) always have ground states with a dn

configuration.

Cr is 4s13d5, therefore Cr3+ is 3d3 Fe is 4s23d6, therefore Fe3+ is 3d5

Co is 4s23d7, therefore Co+ is 3d8

For a Co atom [Ar]4s23d7 N.B. The presence of Ligands will split
the degeneracy of the d orbitals – see
later
For a Co(I) ion [Ar]3d8

The relative energies of atomic orbitals depend on the nuclear charge; in the ions, where the
nuclear charge dominates, the 3d orbitals are of lower energy than the 4s. This occurs
because the s and p orbitals penetrate well and tend to see a fairly constant effective nuclear
charge which doesn’t change much between the neutral atom and its cation. In contrast, the
d orbitals penetrate poorly and are much sensitive to the loss of valence electrons on
ionization. The d orbital energy therefore drops sharply as valence electrons are removed
because there is less competition for the effective nuclear charge.
For metal complexes in zero oxidation state the starting d electron counts to use are:

4 5 6 7 8 9 10
Ti V Cr Mn Fe Co Ni
Zr Nb Mo Tc Ru Rh Pd
Hf Ta W Re Os Ir Pt

3. There is a gradual increase in ionization energy along the series as the effective nuclear
charge increase and atomic and ionic sizes decrease. There are nasty peaks and troughs
in all the series. These are not always easy to explain for the 1st and second IP’s because
it is not always clear whether we are removing and s or a d electron on ionization. For the
3rd ionization potentials we can be sure that all the electrons are now occupy d orbitals e.g.
Sc Ti V Cr Mn Fe Co Ni Cu Zn
M2+ 3d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
M3+ 3d0 d1 d2 d3 d4 d5 d6 d7 d8 d9

It follows that spin exchange effects nicely explain why the 3rd I.P. for Mn is greater than that
of Fe:

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ab, ac, ad, ae ab, ac, ad

bc, bd, be bc, bd

cd, ce cd

de

Spin exchange: 10 x 100 kJ mol-1 Spin exchange: 6 x 100 kJ mol-1

So ionization is unfavorable because 4 exchange pairs are lost.

ab, ac, ad, ae ab, ac, ad, ae

bc, bd, be bc, bd, be

cd, ce cd, ce

de de

Spin exchange: 10 x 100 kJ mol-1 Spin exchange: 10 x 100 kJ mol-1

No loss of spin exchange and unfavorable spin pair in Fe2+ is removed.

4. The increase in a particular ionization energy along the series is relatively small, ca. 50%.

This result in generally similar chemistry cf. elements with more substantial variations such

as the First Short Period where there is a four-fold increase in the first ionization energy:

Li Be B C N O F Ne

518 899 798 1087 1404 1313 1680 2077 kJ mol-1

The chemistry of Fe and Co is, therefore, much more alike than that of C and N.

5. The higher ionization energies of transition metals are still quite low and Mn+ ions in

complexes are therefore formed readily since they can be substantially stablillised by

electron donation from ligands.

6. Some chemistry can be predicted from ionization energies.

e.g. Sc3+ should be formed easily, but Sc4+ is unlikely.

Ni3+, Cu3+ and Zn3+ are unlikely.

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But note that the stability of oxidation states in metal complexes is influenced by many factors
other than ionization energy. In particular as we shell different ligands are good at stabilizing
either high or low oxidation states.

THE PROPERTIES OF TRANSITION METAL COMPLEXES

❖ Colour: Arises from electronic transitions between d-orbitals, resulting the absorption

of visible light.

❖ Magnetism: Arises from the presence of unpaired electrons (up to five) and provides

a very useful insight into the chemistry and structures of transition metal compounds.

❖ Variable Oxidation state:

e.g. for iron: Fe (-II, -I, 0, I, II, III, IV, V)

The stability of an oxidation state is strongly influence by the nature of the ligands

attached to the metal centre.
❖ Variable Co-ordination Number and Geometry:

Metal complexes show a wide range of coordination numbers and have geometries

which do not necessarily follow the sterically based ideas of VSEPR theory which work

well for the main group elements.

e.g. [Ni(CN)4]2- four coordinate square planar (not possible for

main group elements)

[CoCl4]- four coordinate tetrahedral
[Co(NH3)6]2+ six coordinate regular octahedral

Five coordinate complexes may have either trigonal bipyramidal or square pyramidal

geometry:

Not found for main group
elements

N.B. In general, VSEPR rules do not apply to the compounds of transition elements.

The Importance of Transition Elements
❖ Metallurgy

❖ Catalysis Home work: Find an example
❖ Biochemistry for each application and explain.
❖ Medicine

❖ Cancer

❖ Magnetic Resonance Imaging (MRI)

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COMPLEXES AND LIGANDS

Transition metals from many compounds which are described as co-ordination compounds

or complexes.

Complexes contain a central metal atom or ion surrounded by an array of ligands which bond

to the metal. Depending on the charges of the ligands which bond to the metal. Depending

on the charges of the metal and the ligands, the complex ca be neutral, cationic or anionic.

e.g.

neutral: [PtF6], [NiCl2(PR3)2]

cationic: [Co(NH3)6]3+, [Ni(H2O)6]2+

anionic: [CoCl4]2-, [Fe(CN)6]3-

Ligands are molecules or ions capable of bonding directly to a metal atom or ion.

e.g.

neutral molecules: NH3, H2O, pyridine, CO, PR3
anions: F-, Cl-, CN-, OH-, SCN-

-donor ligands

All the above ligands are formally electron pair -donor, i.e. a pair of electrons in a  ligand

orbital is formally donated to an empty metal valence orbital.

This is a stabilizing interaction for M2+, M3+, etc., but would destabilize metals in low oxidation

states such as M(0) or M(-I) which are electron rich.

-donor ligands

Ligands such as F-, Cl-, OH-, H2O use one lone pair for -bonding but still have one (H2O) or
two (F-, Cl-, OH-) additional lone pairs of electrons in p-orbital which may also contribute

electron density to the metal via a -bonding interaction with an empty metal d orbital.

-donation -donation

unused ligand empty metal d
lone pair in a orbital

p orbital

Again this is a stabilizing interaction for M2+, M3+, etc., but would destabilize metals in low
oxidation states such as M(0) or M(-I) which are electron rich. Because the degree of  overlap
is less than that in a -bonding this interaction is energetically less important than the primary
-bond.

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-acceptor ligands

In order to form stable complexes with metals in low oxidation states special ligands are

necessary i.e. -acceptor ligands. These ligands not only donate electrons to a metal in a -

fashion, but also remove electron density from the metal d-orbitals through -backbonding.

-forward donation -backbonding donation

lone pair on ligand empty orbital(s) on empty * antibonding filled metal d orbital

metal (s, p or d) orbital

An example of a good -donor and a good -acceptor ligand is CO. This type of synergic

metal-ligand bonding will be discussed in Organometallic compounds.

Bridging Ligands

Some ligands, when bound to a metal, still electron pairs available for donation to a second

metal atom; they are thus able to bridge between two metal atoms. Typically (though not

always since CO is a notable exception) these ligands fall in the -donor category mentioned

above but here an unused electron pair is donated to a second metal centre.

e.g.

Bridging is also possible for OH-, NR2-, PR2- and CO. Notice that two of these complexes have
metal-metal single bonds. Complexes are also known with M=M double (triple or quadruple)
bonds and these will be discussed later.
Unsaturated Hydrocarbon Ligands
These very important ligands are -acceptors, i.e. they can stabilize low oxidation states of
transition metals.

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In these complexes the hydrocarbon donates electron density to the metal from a -bonding
molecular orbital and receives electron density back from the metal into an unfilled *-
antibonding molecular orbital. This results in a substantial change in the normal reactivity of
the hydrocarbon. The modification of reactivity upon co-ordination to a transition metal can be
used in organic synthesis, and is an important aspect of the industrial catalysis of organic
reactions by transition metals.

COORDINATION VS ORGANOMETALLIC CHEMISTRY

Coordination chemistry is often associated with metals in higher oxidation states. To stabilize
the metal in high oxidation states we need good -donor ligands such as amines and ligands
which are also -donors (Cl-, OH- etc.) will be even better.
Organometallic compounds have metal-carbon bonds. As we have seen above many
important carbon ligands (e.g. CO, alkenes, etc.) are modest -donor and good -acceptors.
A great many organometallic compounds will therefore exist with electron rich metals in low
oxidation states which will stabilized by the good -acceptor characteristics of the carbon
based ligands.
There are of course many grey areas. Zeise’s salt is organometallics because it has metal to
alkene bonds but it also has many features of a typical coordination complex e.g. a fairly high
oxidation sate [Pt(II)] and Cl- ligands which are good - and -donor ligands. The alkene is
however a -acceptor.

Chelating Ligands
A chelate is a ligand containing two or more donor atoms which can simultaneously bond to
a transition metal ion.

e.g. 5- and 6-membered rings are
most stable; there are strain
problems in smaller rings.

A ligand with one donor atom is said to be monodentate (not a chelate),

Two donor atoms bidentate,

Three donor atoms tridentate,

Four donor atoms tetradentate,

Five donor atoms pentadentate,

Six donor atoms hexadentate,
“dentate ….from teeth”

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Name of ligand Donor atom Example
Monodentate
H- H [ReH9]2-
F-, Cl-, Br- and I- F, Cl, Br and I [NiCl4]2-, [CrF6]3-
O2- O [MnO4]-
HO- O [Zn(OH)4]2-
RS- S [Fe(SPh)4]-
NCS- N or S [Cr(NH3)5(NCS)]2+
[Cr(NH3)5(SCN)]2+
NO2- N or O [Co(NH3)5(NO2)]2+
[Co(NH3)5(ONO)]2+
CN- C [Fe(CN)6]3-
CO C [Mo(CO)6]
NR3, amines N [Co(NH3)6]2+
N [Ni(py)6]2+
, pyridine, py
PR3, phosphines P [Pt(PMe3)4]
Bidentate O, O [Fe(ox)3]3-

, oxalato, ox O, O [Cr(acac)3]

, acac

S, S [Re(S2C2Ph2)3]

, bpy N, N [Ru(bpy)3]2+
H2NCH2CH2NH2, en
Ph2PCH2CH2PPh2, dppe N, N [Co(en)3]3+
Tridentate P, P [Fe(dppe)(CO)]

, dien N, N, N [Co(dien)2]3+

Polydentate Ligands can be designed to co-ordinate specific metals, for example to bind Na+
or K+ selectively (antibiotics) or to separate Rh in ores from other platinum group metals.

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Note that chelating ligands may be used to change the nature of the complex and mixed
complexes containing both monodentate and chelating ligands are possible:

also
[Co(en)2(NH3)2]3+
[CoCl4(ox)]3-

CHELATE COMPLEXES
The chelate Effect
Chelate complexes are considerably more thermodynamically than complexes with
comparable monodentate ligands. For example [Ni(NH3)6]2+ is 1010 more stable than
[Ni(en)3]2+.

G = H -TS
In the above reactions there is little difference in the strength of 6 metal-N bonds. Therefore
H for the reactions should be comparable.
In reaction (i) we start and finish with 7 species (ignoring salvation effects). There is no
increase in disorder (entropy S). In reaction (ii) 4 species become 7. There is an increase in
entropy. Overall G is more negative for reaction (ii) because of the increase of entropy which
results from the use of a chelating ligands. Hence the chelate complex is more stable.
Applications
The stability of chelate complexes is made use of:

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❖ In analysis, e.g. in gravimetric analysis where it is essential to precipitate all of a
complex.

❖ In removal of unwanted metal ions, e.g. Ca2+ and Mg2+ from water in “softening”,
detoxification.

❖ In ensuring that MRI imaging agents such as Gd3+ are not allowed into the blood
stream which would kill the patient!

Note also that haemoglobin and other biologically important molecules contain chelated
metals; it is vital that these complexes are very stable so that the metal is not readily lost.

N.B. Denoting a chelate ring

for M(glycinate) It matters that the two
ends of the ligand are
different.

OXIDATION STATE AND CO-ORDINATION NUMBER
Oxidation state

The formal charge remaining on a metal when all the ligands are given their usual charges.

[MnO4]- O2- Mn(VII) d0
F- Pt(VI) d4
[PtF6] NH3 Cr(III) d3
[Cr(NH3)6]3+ CN- Fe(III) d6
[Fe(CN)63- CO Ni(0) d10
CO Mn(-III) d10
[Ni(CO)4]
[Mn(CO)4]3-

Note

❖ Effect of ligand in stabilizing different oxidation states. Highly electronegative ligands
for high oxidation states with -donor ability; -acceptors for low oxidation states.

❖ Remember to treat all the valence electrons as d electrons!
❖ Oxidation states does not indicate the true charge on the metal. It is only a useful

formalism.
e.g. It is not Mn7+ in [MnO4]-. Written as Mn(VII)

It is not Pt6+ in [PtF6], a covalent volatile solid. Written as P(VI).

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Co-ordination Number

The number of ligand atoms (not ligands) bonded to a metal

[CoCl4]- co-ordination number 4
[Co(NH3)6]3+ co-ordination number 6
[Co(en)3]3+ co-ordination number 6, not 3
[CoCl2(en)2]+ co-ordination number 6, not 4

The most common co-ordination numbers for the first transition metal series are four
(tetrahedral and square planar) and six (octahedral). Higher co-ordination numbers are
possible for the larger second and third row transition metals. The factors controlling co-
ordination number and geometry will be discussed later.

NOMENCLATURE FOR COMPLEXES
Writing formulae
In writing the formula place the metal atom first, then add the ligands in alphabetical order
(according to the first symbol of their formula, e.g. H2O before NH3); anionic ligands are placed
before neutral ligands. Place square brackets round the formula. For ionic complexes the
cation and anion are separated, with square brackets around each: the cation is placed before
the anion
Naming compounds

❖ Name the ligands fist and then metal.
❖ Ligands are named in the order in which they appear in the formula, i.e. alphabetically.

The names of anionic ligands end in “o”; e.g. chloro, hydroxo. Those of neutral ligands
are spelled as normal, except for H2O (“aquo”), NH3 (“amine”), and carbon monoxide
(“carbonyl”).
❖ Names of complexes cations and neutral molecules terminate with the normal spelling
of the metal. Metals in complex anions end in “ate”; e.g. ferrate, chromate, cobaltate.
❖ The number of each ligand is shown by the prefixes mono, di, tri, tetra, penta, hexa,
etc. Used without hyphens. If the name of the ligand itself includes such a prefix, the
number of ligands is denoted by bis, tris, tetrakis, pentakis, hexakis, etc. And the ligand
is bracketed.
❖ A bridging ligand is separated in the name by hyphen and prefixed by --
❖ The oxidation state of the metal is usually indicated by a Roman numeral after the
metal.

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Examples: Hexaaquonickel(II) dichloride Ni(II), d8
[Ni(H2O)6]Cl2
K3[Fe(CN)6] Tripotassium hexacyanoferrate(III) Fe(III), d5
[CoCl2(en)2]Cl
Dichlorobis(ethylenediamine)cobalt(III) Co(III), d6
[PtCl2(NH3)4][PtCl4]
chloride
[Co2(H2O)8(-OH)2][SO4]2
Dichlorotetraamineplatinum(IV) Pt(IV), d6
[Ni(CO)4]
[PtCl2(PPh3)2] tetrachloroplatinate(II) Pt(II), d8

Octaaquodi--hydroxodicabalt(III) Co(III), d6

disulphate

Tetracarbonylnickel(0) Ni(0), d10

Dichlorobis(triphenylphosphine)platinum(II) Pt(II), d8

Coordination Number and Geometries

Coordination Number Geometry Examples

2 [H3N−Ag−NH3]+

Linear [CuCl2]-

3 R 2N Fe NR 2
Trigonal Planar (rare)
NR2 R = SiMe3

4 [Ni(CO)4], Ni(0)
Tetrahedral [FeCl4]2-, Fe(II)

Square Planar [NiCl2(py)2], Ni(II)
[PtCl4]2-, Pt(II)
5
Trigonal Bipyramidal [Fe(CO)5], Fe(0)
[Zn(acac)2(H2O)], Zn(II)

Square Pyramidal [VO(acac)2], V(IV)
[NiBr3(PEt3)2], Ni(III)
6
Octahedral [Cr(CO)6], Cr(0)
[Co(NH3)6]3+, Co(III)

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Coordination Number Geometry Examples
Trigonal Prismatic (rare)
[ReL3], L = -S Ph
7 -S Ph
Pentagonal Bipyramidal
[UF7]3-
[ZrF7]3-

Capped Trigonal [TaF7]2-
Prismatic [NbF7]2-

8 Archimedean Anti prismatic, [TaF8]3-
Dodecahedral, [Mo(CN)8]3-;
9
Hexagonal Bipyramidal, [VO2(O2CMe)3]-
Tricapped Trigonal Prismatic, [ReHg]2-

ISOMERISM IN COMPLEXES
Geometrical Isomerism
Different spatial arrangements of the ligands in a complex.

❖ Square Planar [MA2B2]

e.g. both cis- and trans-[PtCl2(PPh3)2] are
isolable.

❖ Octahedral [MA2B4] Octahedral [MA3B3]

cis- same ligands are on the same side
trans- same ligands are opposite
fac- same ligands on the same triangular
mer- same ligands are on the different triangular

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Optical Isomerism
Optical isomers (enantiomers) are non-superimposable on their mirror images. They rotate
the plane of a beam of polarized light in opposite directions. It is often difficult to judge whether
a molecule is superimposable on its mirror image. A better approach is to examine the
molecular symmetry. If a molecule has a mirror plane of symmetry (or a centre of symmetry)
it will be superimpose on its mirror image and will not be optically active (this is almost, but not
quite, a complete test).
N.B. A molecule with a plane of mirror symmetry (or a centre of symmetry) will not be
chiral.
In organic chemistry you are often taught that if a carbon atom has four different substituents
it will be chiral. This generally works well for carbon chemistry but is a dangerous
oversimplification as the following examples show:

Chiral but no carbon has Two chiral centres but Two chiral centres and
four different substituents overall molecule not chiral overall molecule is chiral

(meso isomer)

Four different substituents but not chiral Five different substituents
❖ Tetrahedral Metal Complexes but not chiral

[MABCD] [M(A-B)2]
Chiral as in carbon chemistry Chiral with unsymmetrical chelate ligand
❖

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❖ Octahedral Metal Complexes

Chiral [M(A-A)3] (i.e. tris chelates) Chiral [M(A-A)2B2] (i.e. bis chelates)
e.g. [Cr(en)3]3+ e.g. [CoCl2(en)2]+

NOT chiral. There is no mirror plane BUT there is a centre of symmetry!

Linkage Isomerism

Arises when a ligand can coordinate via either one of two sites. In a formula the ligand is

written with the donor atom first:

[Co(NO2)(NH3)5]2+ (nitro, bonded through N)
[Co(ONO)(NH3)5]2+ (nitrite, bonded through O)

[Mn(SCN)(CO)5] (thiocyanato, bonded through S)

[Mn(NCS)(CO)5] (isothiocyanato, bonded through N)

Ionization Isomerism

Isomers with the same overall composition, but containing different ions.

[CoCl2(NH3)4]NO2 and [CoCl(NO2)(NH3)4]Cl

Hydration Isomerism

[Cr(H2O)6]Cl3 and [CrCl(H2O)5]Cl2H2O

Ligand Isomerism

H2NCH2CH2CH2NH2 and H2NCH2CH(CH3)NH2

(1,3- and 1,2-diaminopropane)

Co-ordination Isomerism

When both the cation and the anion of a salt are complexes the distribution of ligands between

the two can vary.

[CoIII(NH3)6][CrIII(CN)6] and [CrIII(NH3)6][CoIII(CN)6]
[PtII(NH3)6][PtIVCl6] and [PtIVCl2(NH3)4][PtIICl4]

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MAGNETISM
This is a very important property of transition metal complexes and measurements of magnetic
properties can be very useful in helping us to understand transition metal chemistry.
Diamagnetism
All substances show diamagnetism which is due to the presence of paired electrons. If a
substance has no unpaired electron it will only show diamagnetism and is referred to as
diamagnetic. Diamagnetic substances are weakly repelled by a magnetic field. Diamagnetism
does not exist in the absence of an external magnetic field.
Paramagnetism
Paramagnetism is associated with the presence of on or more unpaired electrons.
Paramagnetic substances are strongly attracted to an external magnetic field. Paramagnetism
is a permanent property of compounds eith unpaired electrons and does exist in the absence
of a magnetic field.
N.B.

❖ Paramagnetism is ca. 103 greater than diamagnetism.
❖ Paramagnetic compounds will also show diamagnetism (because of their paired

electrons). Because diamagnetism is relatively weak this will only very slightly reduce
the amount by which a paramagnetic compounds is attracted to an external magnetic
file.
Origin of Diamagnetism
In an applied magnetic field (B0) electrons (paired) tend to circulate about the nucleus and
produce an induced field (Bi) which opposes the applied field at the nucleus. This is because
electrons are charged and a circulating charge produces a magnetic field (c.f. a solenoid).
Because the induced field opposes the applied field diamagnetic compounds are repelled from
a magnet.
The effective field at the nucleus will be B = B0 – Bi. In an electron rich environment the
induced field (Bi) will be bigger and the effective field B will be smaller (and vice versa). This
is origin of diamagnetic shielding and chemical shifts in NMR spectroscopy.

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Origin of Paramagnetism
Electron are charged and are spinning (spin quantum number ms = +1/2 or -1/2). Spinning
charged mean there is a magnetic moment associated with this behaviour. In diamagnetic
compounds all the electrons are paired so the spins cancel. In compounds with one or more
unpaired spins the spin magnetic moments all add together to give a permanent magnetic
moment.

❖ Electrons will also circulate in their orbitals i.e. they have orbital angular momentum.
As electron swap position between degenerate orbitals this will produce and additional
orbital contribution to the overall magnetic moment.

❖ Generally for the 1st row transition metals the spin only magnetic moment gives a good
idea of what the measured magnetic moment for the complex will be. You will
encounter more detail about measurement later.

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CRYSTAL FIELD THEORY

The Origin of Crystal Field Theory
This was discovered in the 1930’s to explain the magnetism and colours observed for ionic
crystalline transition metal salts e.g. oxides and fluorides. It was applied to transition metal
complexes in the 1950’s. It was later modified by Van Vleck to include covalence in the metal-
ligand bonding and was renamed “Ligand Field Theory”.
Crystal or ligand filed theory offer a very simple way of understanding some of the properties
of transition metal complexes but it essentially NOT a good model of what really happens in
the complex. A much better qualitative description of bonding in transition metal complexes
may be obtained from Molecular Orbital Theory and this will be briefly introduced latter.
Simple Crystal Field Theory
The theory is purely ionic and based upon the electrostatic attraction between a positively
charged metal ion and the ligands which are modeled by point charges. For an octahedral
complex ML6:

Situation for a metal ion in an octahedral site in an ionic
crystal lattice: Hence the name “crystal field”.

There will be a strong attraction between the negatively charged ligands and the positively
charged metal. This will be offset by repulsion between the metal electrons and the negatively
charged ligands. Overall there is net stabilization and a complex is formed.
Using this approach for main group compounds (which do not have partially filled valence d-
orbitals) such as [SiCl6]2- and [PF6]- quite good bond energies may be estimated. This is
perhaps a little surprising given that the bonding in these ions is certainly not purely ionic.
For transition metals the same approach sometimes gives bond energies which are too small.
To account for this we need to examine in more detail what happens to the energies of the
metal d-orbitals under the influence of the crystal field.
First Recall that the d orbitals break into two sets which point along and between the axes,
respectively:

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CRYSTAL FIELD SPLITTING FOR AN OCTAHEDRAL COMPLEX
1. First surround the metal ion by a sphere of uniform negative charge equal to 6 x 1-ve

charge and consider the energy of the d orbitals. Obviously the d orbital energies rise
because if you placed electrons in them they would be destabilized by the cloud of
negative charge. Another viewpoint would be that the effective nuclear charge for the d
orbitals would obviously decrease because of the extra negative charge.
2. Now localize the –ve charges so that there is one point negative charge (representing
each ligand) located at each end of the octahedral set of axes. Electrostatic theory says
that this cannot change the total energy of the system so the AVERAGE energy of the
system must remain the same.
3. An electron entering an axial d orbital [dz2 or d(x2-y2)] will obviously find itself right next to
strongly destabilizing –ve charges. These orbitals therefore rise in energy and for good
reasons (although they don’t look the same these orbitals are in fact completely
equivalent) both are affected equally.
4. An electron entering an interaxial orbital [dxy, dyz and dzx] will obviously be a much happier
beasty, being as far as possible away from the –ve charges on the axes. These orbitals
are clearly lower in energy than the axial orbitals and the preserve the overall energy of
the d orbital set the three interaxial orbitals must drop in energy by an amount which
matches the rise in energy of the axial orbitals.

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Notes
❖ We have just focused on the energy of the d orbitals which on average have gone
up in energy but remember overall the complex will be stable because of the big
gain in stabilization energy from the attraction of the point negative charges
(ligands) to the positively charged nucleus.
❖ o (or oct) is the crystal filed stabilization energy (CFSE) for a metal in an
octahedral crystal field of ligands.
❖ The term t2g and eg come from group theory and describe the symmetry of the
orbitals and their degeneracy.
❖ Notice that the three t2g orbitals go down in energy by 3 x (-0.4 o) = -1.2 o and
the eg orbitals rise in energy by a corresponding amount 2 x (0.6 o) = 1.2 o. So
the average energy of the d orbitals is preserved as required by electrostatic theory.
❖ The CSFE o is not a very large quantity being about the order of one bond energy.
CSFE energy is therefore not negligible but in isolation it cannot tell us whether a
complex will be stable and cannot be used to say whether it will have four, five or
six ligands. It can however provide important information about the chemistry of
transition metal complexes.
❖ The point negative charges model the electron pair -donor ability of the ligands.
Note that -donor and -acceptor properties of the ligands have been ignored in
this simple model but these turn out to be significant as we shall see.

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Example
Consider [Ti(H2O)6]3+ Ti(III) d1

The single d electron enters the lowest energy t2g set (Aufbau principle applies) and is -
0.4 o more stable than expected if the shapes of the d orbitals had been neglected. We
say that CSFE is -0.4 o.

CRYSTAL FIELD SPLITTING FOR A TETRAHEDRAL COMPLEXE

1. When dealing with awkward geometries in chemistry the rule is always to superimpose
the structure on the axes as symmetrically as possible.

2. The on-axis d orbitals now point between the ligand and are more stable than the
interaxial d orbitals which lie closer to the ligands. Exactly the reverse of the octahedral
situation.

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Notes
❖ Learn the octahedral diagram really well and simply remember to invert it for a
tetrahedral ML4 complex.
❖ Calculations reveal that t is  4/9o (a bit less than half) which is obviously sensible
since there are fewer ligands.
❖ Note t2 and e, not t2g and eg.

CRYSTAL FIELD SPLITTING FOR A SQURE PLANAR COMPLEX
This can be derived in two ways:
First principles

1. Superimpose the molecule as symmetrically as possible on the axes and as a general
rule make z the unique axis.

2. The dx2-y2 orbital will obviously be highest in energy because its lobes overlap with those
of the ligands.

3. The dxy orbital is next highest in energy because it lies in the ligand plane.
4. The dzx and dyz orbitals are equivalent and low in energy because they are well from the

ligand plane.
5. You might have guessed that dz2 would have been lowest of all in energy because it

points along the z axis and its major lobes are as far away as possible from the ligands.
However, notice that this orbital has a small “donut” lying in the xy plane which is
unfavourable and accounts for dz2 lying above the dzx and dyz orbitals.
6. The names of the orbitals also give the game away. Anything with x and y only in the
name will lie in the xy plane and be high in energy. Anything with z in it name will be
lower in energy.

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7. The dx2-y2 and dxy orbitals have exactly the same relationship to each other as found in

an octahedral complex i.e. the presence or absence of ligands on the z axis will affect
both of them equally because they lie in the xy plane. It follows that they will still be
separated by o.
Derivation of the Square Planar Crystal Field Diagram by Tetragonol Distortion of an
Octahedral

Notes
❖ Progressively removing the two ligands from the z axis of an octahedral complex
(tetragonal distortion) in the limit affords a square planar complex.
❖ Any orbital which has lobes in the z direction (i.e. has z in its name) will clearly fall in
energy. Orbitals without lobes in the z direction (names use only x and y) will
proportionately rise in energy.
❖ The dx2-y2 and dz2 orbitals both lie in the xy plane and have an identical relationship to
the z axis. As we remove the ligands from the z axis they will both equally affected.
These orbitals are separated by o and this is maintained throughout the tetragonal
distortion right through to the final square planar geometry.

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Summary of Key Crystal Field Diagrams

Other more uncommon geometries

Using the same ideas we may develop the crystal field diagrams for less common metal
geometries.

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CRYSTAL FEILD THEORY AND MAGNETISM

For transition metal ATOMS we use the following rules to predict the electronic configurations:

❖ Aufbau Principle: Electrons are added in sequence into the lowest orbitals available.
❖ Pauli Exclusion Principle: No two electron can have the same set of quantum

numbers e.g. if two electrons have the same values of n, l and ml, one must have ms
= +1/2 and the other must have ms = -1/2. So two electrons in an orbital must have
paired (opposite) spins. It follows that you can’t have more than two electrons per
orbital.
❖ Hund’s First Rule: Degenerate orbital (e.g. 2px, 2py, 2pz) are filled singly for as long
as possible with parallel spins (same ms).
Sometimes we modify these rules to take account of spin exchange effects. Thus half filled
and filled shells have particular stability (exchange energy effects). Moreover, we should
remember that a consequence of Hunds rule is that if electrons prefer to occupy degenerate
orbitals singly with parallel spins it must cost energy (P = pairing energy: energy required to
accommodate two electrons in one orbital) to have two electrons paired in the same orbital.

cf.

For Cr [Ar]4s13d5 For Cr [Ar]4s23d4

ab, ac, ad, ae ab, ac, ad

bc, bd, be bc, bd

cd, ce cd

de

Spin exchange: 10 x 100 kJ mol-1 Spin exchange: 6 x 100 kJ mol-1

For transition metal ATOMS the d orbitals are degenerate and Hunds rule requires that all d1

to d9 metal atoms should be paramagnetic because there will always be at least one unpaired

electron present is these species.

Transition metal COMPLEXES can be either paramagnetic or diamagnetic which is a direct
consequence of the fact that the crystal filed splits the degeneracy of the d orbitals.

Consider two non-degenerate orbitals and two electrons. Let the energy required to pair two
electrons be P:

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Case 1 Case 2

Energy = 2E + P Energy = E + (E+E)

= 2E + E

Case 1 E > P

The normal way we fill non-degenerate orbitals following the Aufbau principle and the Pauli
Exclusion principle.

Case 2E < P
Hund’s rule prevails even though the orbitals are not degenerate.

Magnetic Properties of Octahedral Complexes

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1. For octahedral complexes d1, d2 and d3 fill normally (Hund and Aufbau). Affording

paramagnetic complexes.
2. For d4, d5, d6 and d7 there is a choice of high spin or low spin complexes. This rise

because for octahedral complexes o and P have roughly the same magnitude. As we
shall see, the size of o depends (amongst other factors) on the type of ligand bound to
the metal. So depending on the ligand used either high spin or low spin complexes may
arise. We can easily decide which configuration is adopted by measuring the magnetic
moment and hence determining the number of unpaired electrons.
3. For d8, d9 or d10 there is no choice e.g. exited states need not concern us here.

An excited state

Magnetic Properties of Tetrahedral Complexes

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1. In principle for d3 to d6 there is a choice of low spin and high spin compounds.
2. In practice because t is so small (ca. 4/9o) t is always < P and all tetrahedral

compounds will be high spin.
3. It follows that all d1 to d9 tetrahedral complexes will be paramagnetic (just like metal

atoms with degenerate d orbitals)

Magnetic Properties of Square Planar Complexes

The gap between dx2-y2 and dxy is very large (= o) and all the lower orbitals (apart from dxy)
are strongly stabilizing. This means that electrons will enter the lower 4 orbitals first. You will
normally encounter this geometry for sterically non-hindered d8 complexes but d7 and d9 are
also possible.
FACTORS AFFECTING THE SIZE OF CRYSTALS FIELD THEORY
If the factors which affect the size of Crystal Field Splitting are known, magnetic behaviour can
be predicted or at least understood. The following factors are illustrated with values of o but
the same arguments can be applied to other geometries. Unfortunately the pairing energy P
is not constant and also varies but we can usually get away with ignoring this and assuming
(incorrectly) that it does not vary very much.
1. The ligand
For a given ligand, with ions of the same Transition Series in the same oxidation state, o lies
in a narrow range.
e.g. for [M(H2O)6]2+ (M = First Transition Series):

o (minimum) = 7,500 cm-1 for Mn2+
o (maximum) = 14,000 cm-1 for Cr2+

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Other ligands have different, but equally narrow ranges of o. It is therefore possible to place
the ligands in order of their ability to cause Crystal Field Splitting, i.e. in order of their Crystal
Field Strength. This order is called The Spectrochemical Series.

The Spectrochemical Series

good -donors -donors only good -acceptors

I- < Br- < SCN- < F- < OH- < H2O < NCS- < NH3  py < en < bipy < phen < NO2- < PR3 < CO  CN-  C2H4

Small o Large o
Tend to give high spin complexes Tend to give low spin
Weak field ligands Strong field ligands
Weak as salt water Strong as cyanide

halide- < OH- < H2O < NCS- < NH3  py < en < bipy < PR3 < CO, CN-
halide- < O-donor < N-donor < P-donor < CO, CN-

It is difficult to explain the series in terms of the purely ionic (i.e. electrostatic) Crystal Field
Theory; e.g. why F- < H2O? We need to allow for covalency and -bonding between metal and
ligand. However, in many ways the series is intuitive. Ligands such as halides which are happy
as ions and strongly electronegative will not tend to share their electron density too readily.
Their bonding will tend towards ionic character and they can (and often are) readily displaced
from the metal coordination sphere as happy ions by “better” ligands. Contrast this with CO
which by virtue of being both a good -donor and a good -acceptor has substantial covalent
multiple bond character when bound to the metal. Clearly ligands like CO which bind strongly
will give a bigger perturbation of the d orbitals and a bigger o.

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For some metal ions there is a point in the Spectrochemial Series where the electron
configuration changes from high spin to low spin, i.e. where there is a charge from o.< P to
o.> P:

[CoL6]3+ d6 L = OH- and stronger low spin o.> P (high ox state, big
o., see 2)

NB If we know that [Co(H2O)6]3+ Co(III) d6 is low spin then we can be sure that it would
be low spin for all the other ligands to the right in the Spectrochemical series.

[FeL6]2+ d6 L = bpy and stronger low spin o.> P

[MnL6]2+ d5 L = CN- and stronger low spin o.> P (low ox state,
smaller o., see 2)

2. Oxidation State of the Metal
The greater the positive charge on the metal the greater is o. The ligands are attracted closer
to the metal and interact more strongly with the d-orbitals, e.g. for the First Transition Series:

[M(H2O)6]2+ o = 10,000 cm-1

[M(H2O)6]3+ o = 20,000 cm-1

[Fe(NH3)6]2+ d6 o < P high spin  = 4.9 BM 6 unpaired e

[Co(NH3)6]3+ d6 o > P low spin  = 0.0 BM 0 unpaired e

3. Transition Series of the Metal
Within a transition metal group, o increase by 20-50% on going from the First Series metal to
the Second, and by another 20-50% on going from the Second Series metal to the Third.

The size of the metal d-orbitals increases (3d < 4d < 5d) with consequent increases in the
interaction with the ligand electrons. Note also that larger d-orbitals have lower pairing
energies, favouring low spin configurations.

[Co(NH3)6]3+ [Rh(NH3)6]3+ [Ir(NH3)6]3+

o 22,900 34,100 41,000

All have o > P M(III) d6 low spin  = 0.0 BM with no unpaired
electrons

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M(III) d6 o [CoF6]3- < o [RhF6]3- < o [IrF6]3- <

High spin Low spin Low spin

o < P o > P o > P

4. General Rules
Without a great deal of experience it will often not be possible to unambiguously guess
whether a complex will be high spin or low spin by using the above rules. However some
general would be:

❖ CN- and CO are such high field ligands that one might guess that many of their
complexes will be low spin.

❖ Similarly we expect many F- (and to a lesser extent H2O) complexes will be high
spin.

❖ A third row metal in a high oxidation state may well be low spin for most ligands.

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EXPLANATION OF ANOMALOUS MAGNETIC BEHAIVOUR

Earlier we saw that depending on the ligands chosen transition metal complexes could either
be diamagnetic or paramagnetic. We are now in a position to use this information.

Complex Metal No. of d electrons obs Predicted no. of unpaired electrons
(BM)

[CoF6]3- Co(III) d6 4.9 4
d6 0.0 0
[Co(NH3)6]3+ Co(III)

Both the Co complexes are octahedral. To account for the observed magnetic moments the
F- complex must be high spin whist the NH3 complex is evidently low spin. Notice that a free
Co3+ ion would have 4 unpaired electrons. Key factors are:

❖ A 1st row metal tends to make o small.
❖ High oxidation state (Co(III)) tend to make o big.
❖ Only the ligand changes. So F- is a very weak field and overall gives a high spin

complex whilst NH3 is low spin. In this instance the only common high spin complex is
the fluoride.
Complex Metal No. of d electrons obs Predicted no. of unpaired electrons

(BM)

[NiCl4]2- Ni(II) d8 3.0 2

[Ni(CN)4]2- Ni(II) d8 0.0 0

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Note that the measured magnetic moment tells us the geometry. But why does Cl- give a
tetrahedral complex whilst CN- give a square planar complex? Key factors:

❖ The cyanide ligand is very strong field. For d8 this makes it energetically very
favourable to populate the four lower levels found for a square planar complex.

❖ Cyanide is also small so there will be little steric hindrance even though Ni is a first row
metal with a small covalent radius.

❖ Chloride is a weak field ligand. Whilst it would still be better for the chloride complex to
be square planar on CFSE grounds the difference between the two geometries is not
so grate for a week field ligand.

❖ Chloride is much bigger than cyanide. Steric hindrance between the ligands is
minimized by the tetrahedral geometry.

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OCTAHEDRAL VS TETRAHEDRAL COORDINATION

1. Electronic Effects (i.e. Crystal Field Stabilization Energy)
e.g. consider V3+ (d2)

Thus, octahedral coordination is favoured.CFS E (oct)

The same calculation for other dn configurations gives the graph below, showing that on CFSE
grounds octahedral coordination is always preferred over tetrahedral (except for d0, high spin
d5 and d10). Note, however, that for some dn configurations (e.g. d3 and d8) it is preferred more
than others. Note also that the calculations are for high spin octahedral complexes. Low spin
octahedral complexes are even more stabilized than tetrahedral complexes with the same d-
configuration.

N.B. o is typically about the energy associated with the metal ligand enthalpy of one bond.
So CFSE alone cannot tell you whether you will get a four coordinate or a six coordinate
complex. It can only help you to understand the situations which tend to favour one geometry
over another.

1.4
1.2

1
0.8

Oct
T et
0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8 9 10
No. of d-electrons

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2. Steric Effects
The bigger the ligand the greater are the ligand-ligand repulsions and the more likely it is that
a four-coordinate complex will be formed rather than a six-coordinate complex. Large,
negatively charged ligands such as Cl-, Br- and I- are therefore found in tetrahedral complexes
such as [CoCl4]2- (Co(II) d7) and [NiCl4]2- (Ni(II) d8). These are also ligands which cause low
Crystal Field splitting. Therefore, the octahedral geometry is not so favoured on CFSE
grounds.

3. Conclusion
Octahedral complexes will generally be favoured over tetrahedral complexes on CFSE
grounds and having six ligands may well be better in terms of the overall heat of formation.

For d0, high spin d5 and d10 there is no CFSE for either geometry. Tetrahedral complexes are
therefore quite possible particularly if the ligands are demanding. The product of a reaction
may depend on the amounts of the reagents used. For example both TiCl4 and [TiCl6]2- are
known. These are both Ti(IV) d0 complexes and therefore there is no CFSE preference for
either geometry.

SQUARE PLANAR VS TETRAHEDRAL COORDINATION

See previous Ni example. For d8 complexes there will be a strong preference for square planar
geometry on CFSE grounds. For First row transition metals (small covalent radius) sterically
demanding ligands may favour tetrahedral geometry. Steric effects will be less important for
the larger second and third row transition metal complexes.

THERMODYNAMIC EFFECTS OF CRYSTAL FIELD SPLITTING

Hydration Energy

-3.0

Hhyd / MJ mol-1
140 kJ /mol
-2.5

Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

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Hydration energies of the M2+ ions of the first row transition metals
The enthalpy change for the reaction:

1. Can measure H by calorimetry for most of the 1st row metals.
2. If there were no crystal field effects we would expect H to increase gradually because

the effective nuclear charge increases and the size decreases as we move from left to
right.
3. There is an increase, but some ions seem much more stable (bigger H) than expected.
This is due to additional CFSE.
4. We have a series of Octahedral complexes. All the metals are in the same oxidation
state (II), with the same ligands (H2O), and all are 1st row metals. Therefore we expect
them to have the same Hoct. They will all be high spin because H2O is a weak field
ligand, we have a first row metal and the oxidation sate is not too high. We can ignore
spin paring effects because the electrons in the five degenerate d orbitals of the free ions
will have the same number of paired electrons as found in the complexes.

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CFS E (oct) 39

1.4
1.2

1
0.8

Oct
T et
0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8 9 10
No. of d-electrons

The compound CFSE’s give us the “double hump” curve which we saw in a previous section.
The Hydration energy graph has the same form except that it slops upwards because there is
a gradual increase in H superimposed on top of the CFSE graph.
For d0 Ca(II), d5 Mn(II) and d10 Zn(II) the CFSE is zero and this is shown as a smooth curve
on the hydration energy graph. The deviations for the other M(II) ions are due to the additional
CFSE. If we knew H we could correct all the values by subtracting the relevant CFSE’s.
Measurement of o (H2O)
We can use the graph of H to estimate o. Consider [Ti(H2O)6]2+.
This ion is ca. 140 kJ/mol more stable than expected.
Ti(II) is d2 and the CFSE is -0.8o = 140 kJ and o.(H2O) = -175 kJ.
This corresponds to a weak bond energy. The same approach could be used for other ligands.
Lattice Energies
For MCl2 complexes of the first row transition metals we have a related situation. In the crystal
lattice the metal(II) ions pack such that they are octahedrally coordinated by six chloride ions
(i.e. each metal shares its chlorides to achieve this). The lattice energy is the energy released
when free MCl2 forms the crystal.
So we have octahedral metal ions which are again high spin because Cl- is a weak field ligand.
Again some crystal lattice energies seem too high due to additional CFSE.

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Measurement of o (Cl-)
We can use the graph of lattice energy to estimate o. Consider [Ti(Cl)2].
This ion is ca. 120 kJ/mol more stable than expected.
Ti(II) is d2 and the CFSE is -0.8o = 120 kJ and o.(Cl-) = -150 kJ.
N.B. o.(Cl-) = -150 kJ < o.(H2O) = -175 kJ. This is what expect from the Spectrochemical
Series and this is one of the methods which can be used to derive the series in the first place.

-3.0

Lattice E nergy / MJ mol-1 -2.5
120 kJ /mol

-2.0 Sc Ti V Cr Mn Fe Co Ni Cu Zn
Ca

Lattice energies of the dichlorides of the first row transition metals

THE COLOUR OF TRANSITION METAL COMPLEXES

The colours of transition metal complexes may be explained by Crystal Field Theory. For
Ocatahedral complexes typical values of o are roughly the same energy as the visible region
of the spectrum.
A simple one electron system
For [Ti(H2O)6]3+ the metal is Ti(III) d1 and octahedral.

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Absorption of a photon of light of energy E = h can promote an electron from the ground state
to an exited state orbital. The UV-Visible spectrum should therefore have a single absorption
corresponding to an energy o. The actual spectrum is shown below.

The absorption is not sharp because there are actually two overlapping bands and not one as
predicted. This arise because the symmetry of the molecule in the exited state is lower than
octahedral. There is also broadening because of vibrational coupling. Despite this we can
assign a frequency of ca. 20,000 cm-1 to the absorption and then: E = o = h = h x 20,000 =
244 kJ/mol.
c.f. [Ti(H2O)6]3+ o = 244 kJ/mol (from spectroscopy)

[Ti(H2O)6]2+ o = 175 kJ/mol (obtained from heats of hydration)
Notice that the value for the higher oxidation state is greater which agrees with our general
rules for the factors which affect the value of o.
The colour of the complex is red/purple. This is due to the remaining light which is transmitted
through a solution of the sample after the compound has absorbed photons in the blue/green
region of the spectrum.
The colour is NOT particularly intense for a given amount of the complex. This is because
strictly speaking electrons are not allowed to undergo transitions between two d orbitals (d-d).
This arises because of the selection rules of quantum mechanics. The fact that this “forbidden”
d-d transition is observed arises because we are not dealing with pure d orbitals i.e. as we
shall see later a more sophisticated treatment of bonding reveals that the orbitals actually have
some ligand character.
If we were to examine [TiCl6]2- Ti(IV) d0 there are no d electrons and the complex is colourless
because there are no absorptions in the visible region of the spectrum.

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Similarly a d10 system will be colourless because all the d orbitals are filled and you cannot
promote an electron to an empty eg orbital.
Multi Electron System
Here the same ideas apply but the situation is MUCH more complex. There may be more
electronic transitions possible and electron-electron interactions become important and these
effectively reduce the degeneracy of the d orbitals leading to even more possibilities. You
should not attempt to predict the number of bands for these systems by drawing simple
diagrams of the type we used for d1, it won’t work and you will only get frustrated. To
understand this area you need to know all about spectroscopic term symbols.
Example are shown below for the octahedral complexes [Ni(H2O)6]2+ Ni(II) d8 and [Ni(en)3]2+
Ni(II) d8. Notice that there are more bands and that they extend into the UV and IR regions of
the spectrum.
The [Ni(H2O)6]2+ complex absorbs photon in both the purple and red regions of the visible
spectrum allowing only green light to pass thourgh- it is a green compound.
The [Ni(en)3]2+ absorbs in the green (UV and IR not relevant) and therefore transmits
red/purple and has the latter red/purple colour.
Both complexes are relatively pale in colour because the transitions are formally d-d (Laporte)
forbidden. Notice as you go from en (moderately strong field ligand) to H2O (weak field) the
spectrum shifts to lower frequency. This reflects the fact that o will be smaller for the aquo
complex and with a full understanding of the spectrum o can be estimated from these complex
spectra.

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JAHN TELLER EFFECTS
The Jahn-Teller Theorem (1937)
“Any non-linear molecular system in a degenerate electronic state will lower its symmetry to
split that state.”
In English this means: “If the electron density is non spherical the molecule will distort its
geometry to avoid having a regular stereochemistry.”
Classic Example is Cu2+ (d9) octahedral
The ion is degenerate because two alternative electronic configurations are possible:

Ligands along z axis are
shielded by an electron pair
from the full nuclear charge.
Therefore tend to be longer
axial bonds because only a
single electron shields the
ligands in the xy plane.

Ligands xy plane are
shielded by an electron pair
from the full nuclear charge.
Therefore tend to be longer
equatorial bonds in xy plane.

A tetragonally distorted geometry is produced in each case but we cannot predict which will
occur. X-ray diffraction shows that the observed distortion involves lengthening of the axial
bonds.

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Crystal Field Theory for Jahn Teller Distortions
We saw earlier how to derive a square planar CFSE diagram by tetragonal distortion of an
octahedron.

Examples of ground state Jahn Teller distortions from X-ray diffraction studies
[CuF2] Cu(II) d9, octahedral is solid state, shared halides

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Other Systems

e.g. Cr(II) high spin d4 in CrCl2 has 4 x Cl at 2.39 Å and 2 x CL at 2.90 Å.

N.B. ions like those shown below which have a symmetrical electron density do not show Jahn
Teller distortions because all the ligands are equally shielded from the nuclear charge.

In principle distortions should also be seen for unsymmetrical t2g distributions and for
tetrahedral complexes. But in the latter cases the orbitals are interaxial and the effects are
expected to be small and are not observed.

Jahn Teller Distortion of Exited States

The absorption band in the visible region of the [Ti(H2O)6]3+ is asymmetric and in fact is made
up of two transitions.

The exited state suffers Jahn-Teller distortion because of the asymmetric electron density
distribution. The exited state has a short life time but the electrons move faster than the
molecule can adjust its geometry and the in effectively becomes permanently distorted.

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MOLECULAR ORBITAL THEORY
The Molecular Orbital Theory, developed by Robert S. Mullikan, incorporates the wave like
characteristics of electrons in describing bonding behavior. In Molecular Orbital Theory, the
bonding between atoms is described as a combination of their atomic orbitals. While the
Valence Bond Theory and Lewis Structures sufficiently explain simple models, the Molecular
Orbital Theory provides answers to more complex questions. In the Molecular Orbital Theory,
the electrons are delocalized. Electrons are considered delocalized when they are not
assigned to a particular atom or bond (as in the case with Lewis Structures). Instead, the
electrons are “smeared out” across the molecule. The Molecular Orbital Theory allows one to
predict the distribution of electrons in a molecule which in turn can help predict
Introduction
Atoms form bonds by sharing electrons. Atoms can share two, four, or six electrons, forming
single, double, and triple bonds respectively. Although it is impossible to determine the exact
position of an electron, it is possible to calculate the probability that one will find the electron
at any point around the nucleus using the Schrodinger Equation. This equation can help
predict and determine the energy and spatial distribution of the electron, as well as the shape
of each orbital. The figure below shows the first five solutions to the equation in a three
dimensional space. The colors show the phase of the function. In this diagram, blue stands
for negative and red stands for positive. Note, however, that the 2s orbital has 2 phases, one
of which is not visible because it is inside the other.

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Principles of Molecular Orbital Theory

Principle Details/Examples

1 total number of molecular The molecule H2 is composed of two H atoms. Both
orbits is equal to the total H atoms have a 1s orbital, so when bonded together, there are

number of atomic orbitals therefore two molecular orbitals.

from combining atoms

2 Bonding molecular orbitals Bonding molecular orbitals help stabilize a system of atoms
have less energy than the since less energy is associated with bonded atoms as opposed
constituent atomic orbitals to a system of unbound atoms.
before bonding Likewise, antibonding molecular orbitals cause a system to be
Antibonding molecular unstabilized since more energy is associated with bonded
orbitals have more energy atoms than that of a system of unbound atoms.
than the constituent atomic
orbitals before bonding.

3 Following both the Pauli Electrons fill orbitals with the lowest energy first. No more than
exclusion principle and 2 electrons can occupy 1 molecular orbital at a time.
Hund's rule, electrons fill in Furthermore, all orbitals at an energy level must be filled with
orbitals of increasing energy. one electron before they can be paired. (see second diagram
below)
4 Atomic orbitals are best Molecular Orbital Configuration of Li2:
formed when composed of (1s)2(*1s)2(2s)2
atomic orbitals of The bonding (1s)2 and antibonding (*1s)2 cancel each other
like energies. out, leaving (2s)2 as the valence electrons involved in the
atoms' bonding.

Bonding vs. Antibonding Molecular Orbitals
In molecules, atomic orbitals combine to form molecular orbitals which surround the molecule.
Similar to atomic orbitals, molecular orbitals are wave functions giving the probability of finding
an electron in certain regions of a molecule. Each molecular orbital can only have 2 electrons,
each with an opposite spin. Take a hydrogen molecule (H2) for example. It has two molecular
orbitals, an antibonding orbital and a bonding orbital. Compared to the original atomic orbitals,
a bonding molecular orbital has lower energy and is therefore more stable. Where the atomic
orbitals overlap, there is an increase in electron density and therefore an increase in the
intensity of the negative charge. This increase in negative charge causes the nuclei to be
drawn closer together. Due to the lower potential energy in molecular bonds than in separate
atomic orbitals, it is more energy efficient for the electrons to stay in a molecular bond rather

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than be pushed back into the 1s orbitals of separate atoms. This is what keeps bonds from
breaking apart. A bonding orbital can only be formed if the orbitals of the constituent atoms
have the same phase (here represented by colors). The wave functions of electrons of the
same phase interfere constructively which leads to bonding.

If the atomic orbitals have the different
phases, they interfere destructively and
an antibonding molecular orbital is
formed (see the top part of the figure
below). Antibonding molecular orbitals
have a higher energy than the atomic
orbitals of their constituent atoms. When
antibonds are formed, the interaction
creates a decrease in the intensity of the
negative charge, which causes a
decrease in the plus minus attraction in the molecular bond. This smaller attraction leads to
the higher potential energy. This type of bond destabilizes the attraction between atoms, so
the number of antibonding orbitals in a molecule must be less than the number of bonding
orbitals.

Sigma Bonds
Molecular orbitals that are symmetrical about the axis of the bond are called sigma molecular
orbitals, often abbreviated by the Greek letter σ. The diagram to the left shows the 1s orbitals
of 2 Hydrogen atoms forming a sigma orbital. There are two types of sigma orbitals formed
(abbreviated σ), and antibonding sigma orbitals (abbreviated σ*). In sigma bonding orbitals,
the in phase atomic orbitals overlap end to end causing an increase in electron density along
the bond axis. Where the atomic orbitals overlap, there is an increase in electron density and
therefore an increase in the intensity of the negative charge. This increase in negative charge
causes the nuclei to be drawn closer together. In sigma antibonding orbitals (σ*), the out of
phase 1s orbitals interfere destructively which results in a low electron density between the
nuclei as seen on the top of the diagram.
The diagram below is a representation of the energy levels of the bonding and antibonding
orbitals formed in the hydrogen molecule. Two molecular orbitals were formed, one
antibonding (σ*) and one bonding (σ).The two electrons in the hydrogen molecule have
antiparallel spins. Notice that the σ* orbital is empty and has a higher energy than the σ
orbital.

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