CHEM 303: Periodic Table

Survey
CHEMISTRY OF TRANSITION ELEMENTS

Instructor: Dr. Cecil Kingondu

Email: kindonduc@biust.ac.bw

Tel: 493 1880

 Course Description/Content

• General properties of transition metals.

• General group trends with special reference to
electronic configuration, colour, variable
valency, magnetic and catalytic properties,
ability to form complexes.
• Stability of various oxidation states.
• Difference between the first, second and third
transition series.
• Chemistry of V, Cr, Mn, Fe and Co in various
oxidation states.
 General properties of transition metals.
• IUPAC defines transition elements as those elements whose
atoms have partially filled d orbitals in their neutral or
cationic state.
• Using this definition, the elements in Group 2B—zinc,
cadmium, and mercury —are not considered transition
metals.
• The chemical and physical properties of the transition metals
are closely linked to the presence of ns electrons outside the
(n − 1)d orbitals.
• Although transition metals have different numbers of d
electrons, they all have either 1 or 2 ns electrons in the
valence shell.
• As a result, transition metals show great similarity in
physical and chemical properties.
• Unlike the main-group elements, the properties do not
differ greatly when moving between groups.
• In general, most transition metals:
 are found in multiple oxidation states;
 have a silvery, metallic appearance (with the exception of
copper and gold);
 have high melting points (with the exception of mercury, a
liquid at room temperature);
 are highly conductive;
 form highly colored compounds; and
 are ductile (able to be drawn into a wire) and malleable (able
to be hammered into thin sheets).

Remember:
“A d-block/transition element is an element which has at least
one s-electron and at least one d-electron but no p-electrons in
its outer shell.”
 Electronic Configuration
 The relationship between the electron configuration of transition
metal elements and their ions is more complex. Consider cobalt
which forms complexes of either Co2+ and Co3+ ions.

 Before we can predict the electron configuration of the ions, we have

to know which valence electrons are removed from a cobalt atom
to form Co2+ and Co3+ .

 The electron configuration of a neutral cobalt atom is:

Co [Ar] 4s2 3d7.

 We might expect cobalt to lose electrons from only the 3d orbitals,

but this is not the case. Co2+ and Co3+ Ions have the following
electron configurations:

Co2+ [Ar] 3d7. Co3+ [Ar] 3d6.

 This shows that electrons are removed from the valence shell s
orbital before they are removed from the valence d orbitals when
metals are ionized.

 Why are electrons removed from 4s orbitals before 3d orbitals if

they are placed in 4s orbitals before 3d orbitals.

 This is due to the small difference in energy between 3d and 4s

orbitals. This explains why chromium and copper have electron
configurations different from what we expect.
 EC continued…

• The outer electronic configuration of elements in the first

row of the d-block is as follows:

4s 3d

Attention: Chromium and copper have unusual electronic structures.

Why?
 EC continued…
• The 4s and 3d subshells are very similar in energy and therefore
it is easy to promote electrons from the 4s into the 3d orbitals.

• In chromium the 4s13d5 structure is adopted because the

repulsion between two paired electrons in the 4s orbital is more
than the energy difference between the 4s and 3d subshells.

• It is thus more stable to have unpaired electrons in the higher

energy 3d orbital than paired electrons in the lower energy 4s
orbital.

• In copper, the 3d subshell is actually lower in energy than the 4s

subshell. The 3d orbitals are thus filled before the 4s orbital.
Thus copper adopts a 4s13d10 configuration.
 EC continued…
• In all ions of d-block elements, the 3d subshell is lower in energy
than the 4s subshell so the 4s electrons are always removed first. 3d
electrons are only removed after all 4s electrons have been
removed.
• Some examples of electronic configurations of transition metal ions
are shown below:

4s 3d
 Homework on EC

TM Ion Noble 4s 3d
Gas
Config.
 Atomic Size
• Effective nuclear charge and the size of the highest-energy
occupied atomic orbitals both influence the size of atoms.
 Atomic Size continued….

• In general, the atomic size of the transition metals decreases the

farther right the element appears in a period, which matches the
trend we saw for the main-group elements. However, there are
some peculiarities:
 Atomic size decreases as you move across a period, but the
change is not as significant as was observed for the main-
group elements.
 Atomic size decreases from Sc to Mn but then levels out.
 Atomic size increases slightly at the right side of the d-block.
 Within groups, the period 5 transition metals are larger than
the period 4 transition metals but are the same size or only
slightly smaller than the period 6 transition metals.
 Atomic Size continued….
• The size of a transition metal is determined by the size of the ns
orbital, the highest energy filled atomic orbital for a transition
metal. As you move across a period, both the number of protons in
the nucleus and the number of electrons are increasing, but
electrons are being added to (n − 1)d orbitals.

• These electrons screen the outer ns electrons from the nuclear

charge, effectively offsetting the effect of an increase in nuclear
charge. As a result, the atomic size of transition metals generally
decreases across a period, but not by a significant amount.

• At the end of the transition metal group, atomic radius increases

slightly due to a number of factors that include electron–electron
repulsion.
 Atomic Size continued….
• The general trend in size across a period is similar for all three
groups of transition metals, but the trend in size moving down a
group does not follow the trend of the main group elements.
• Although La is larger than Y, Hf and Zr are very similar in size, a
trend that continues across the period. This phenomenon is called
the lanthanide contraction, which is due to the inefficient shielding
of nuclear charge by the 4f electrons.
• Because the 4f electrons do not effectively shield outer (valence)
electrons from the full nuclear charge, the valence electrons are
more strongly attracted to the nucleus than predicted.

• As a result, the atomic radius for elements with a filled 4f subshell

is smaller than we would predict based on periodic trends. The
lanthanide contraction effectively offsets the increase in size
expected when moving down a group in the periodic table.
 Electronegativity
• Electronegativity continues the trend of minimal changes
moving across the transition metal series. The similarity in
the electronegativity values for the transition metals is, as
described earlier, related to the similarity in the electron
configurations of these elements, all of which have one or
two (n − 1)s electrons in their valence shell.
 Electronegativity continued…..
• Consider the increase in electronegativity within a period for
main-group elements (Li–F) when compared with the increase
within the first row of transition metals (Sc–Zn).

• In the case of the main-group elements, electronegativity

increases from 1.0 to 4.0 within the second period.

• In the first-row transition metals, the range of values is much

smaller (1.3–1.8). Notice that the electronegativity of transition
metals is higher than that of most main-group metals.
• This suggests that they are more likely to form covalent
compounds than the main-group metals, particularly those in
Groups 1A and
 Ionization Energy and Oxidation States
• One of the most interesting properties of transition metals is the
occurrence of multiple oxidation states for each element, a
property that results in extensive redox chemistry.

• The reactivity of these elements is related to the generally low

ionization energy values.

Ionization Energy
• Ionization energy, the energy required to remove an electron from
a species in the gas phase, generally increases as you move from
left to right across the periodic table, because of increasing
effective nuclear charge, and decreases moving down a group,
because of higher principal quantum number.
 IE & OS continued…..
• The first ionization energies increase
slightly across the period, following
the trend expected for a slight
increase in effective nuclear charge
across the period.
• Recall that each first-row transition
metal has at least one 4s electron and
one or more 3d electrons.

• When these elements form cations,

4s electrons are removed before 3d
electrons, so the first ionization
energy is the energy required to remove a 4s electron.
 IE & OS continued…..
• Just as with atomic radii, effective nuclear charge does not
increase significantly across the period in the transition metals,
so the first ionization energies of these elements also does not
increase very much.
• The second ionization energy is greater than the first ionization
energy, as expected, but the trend across the transition metals
still shows only a slight increase.
• Notice the higher values for Cr and Cu; both of these elements
have a single 4s electron, so the second electron is removed from
a lower-energy 3d orbital.
• The third ionization energy of the transition metals is a measure
of the energy required to remove a 3d electron (both 4s
electrons have been removed).
• In this case the anticipated result of increasing effective nuclear
charge as you move left to right across a period is observed.
 IE & OS continued…..

Oxidation states
• The oxidation state formed by an element in its compounds is
determined by the maximum number of electrons it can lose
without requiring so much energy to remove the electrons that
the energy cannot be recovered in bonding.

• s-block elements only form one stable oxidation state in their

compounds. They lose all their valence electrons easily but
cannot lose any more electrons since there is a large amount of
energy required to remove the electrons from the inner shell.

• This jump in energy is best shown graphically as follows

• Na always adopts the +1 oxidation state in its compounds because
there is a large jump between the first and the second ionization
energies.

• Mg always adopts the +2 oxidation state in its compounds because

there is a small jump between the first and the second ionization
energies but a very large jump between the second and third
ionization energies.
 IE & OS continued…..
• In the d-block elements, however, there are often a large
number of valence electrons and removing them all would
require so much energy that it would be unfeasible.
• It is usually only possible to remove some of the valence
electrons. All d-block elements can give up their 4s electrons
fairly easily but the d-electrons are harder to remove.
• Moreover, since the successive ionization energies of d-electrons
increase steadily, it is difficult to predict how many can be lost.
This effect can be shown graphically by considering the
successive ionization energies of an element such as manganese:
 IE & OS continued…..
• The ionization energies increase steadily after the removal of the
4s electrons. It turns out that the energy required to remove the
3d electrons is sometimes recovered in bonding, but not always.
The number of 3d electrons removed thus varies from compound
to compound. d-block metals are thus able to adopt a variety of
oxidation states.
• The oxidation states most commonly formed by the first-row d-
block elements are as follows:
Sc: +3 only (d0)
Ti: +3 (d1), +4 (d0)
V: +2 (d3), +3 (d2), +4 (d1), +5 (d0)
Cr: +3 (d3), +6 (d0)
Mn: +2 (d5), +3 (d4), +4 (d3), +6 (d1), +7 (d0)
Fe: +2 (d6), +3 (d5)
Co: +2 (d7), +3 (d6)
Ni: +2 (d8)
Cu: +1 (d10), +2 (d9)
Zn: +2 only (d10)
 IE & OS continued…..
• It is sometimes possible to remove all 3d electrons to form a
do configuration. However for elements beyond manganese
it is unusual to find an oxidation state containing fewer than
5, 3d electrons.
• The d5 configuration is quite stable because paired d-
electrons can be removed to attain d5 but unpaired electrons
must be removed to attain d4.
• Unpaired electrons are energetically harder to remove
because there is less repulsion between the electrons.
• Note that scandium only forms +3 ions (d0). This is because
the low effective nuclear charge on scandium enables all
three valence electrons to be removed fairly easily.
• Note also that zinc only forms +2 ions (d10). This is because
the high effective nuclear charge on zinc prevents any 3d
electrons from being removed.
 IE & OS continued…..
• All the other elements form at least one stable ion with
partially filled d-orbitals, and it is this property that defines a
transition metal.

“A transition metal is a metal which forms at least one

stable ion with partially filled d-orbitals.”

• Since zinc and scandium do not share this property, they are
not transition metals, and indeed do not show many of the
properties generally attributed to transition metals. They are
however still classified as d-block elements.

• The first row d-block elements which are also transition

metals are therefore titanium, vanadium, chromium,
manganese, iron, cobalt, nickel and copper.
 IE & OS continued…..
High Oxidation and their Stabilization
• Most transition metals exhibit more than one oxidation
state.

• Manganese, for example, forms compounds in every

oxidation state from _1 to _7. Some oxidation states,
however, are more common than others.

• Some of the oxidation states are common because they are

relatively stable.

• Others describe compounds that aren’t necessarily stable

but which react slowly. Still others are common only from a
historic perspective.
 IE & OS continued…..
 Transition metal ions with charges larger than +3 cannot exist in aqueous
solution. Consider the reaction below in which Mn is oxidized from +2 to the +7
oxidation state.

Mn+2 (aq) + 4 H2O (l) MnO4- (aq) + 8H+ (aq) + 5e-

 When Mn atom is oxidized, it becomes more electronegative. In +7 oxidation

state, the atom is electronegative enough to react with water to form a covalent
oxide, MnO4-.

 A simiar phenommenon, (stabilization of high oxidation states) is seen in the

chemistry of vanadium and chromium.

 Vanadium exists in aqueous solution as V+2. But once it is oxidized to +4 or +5

oxidation state, it reacts with water to form VO+2 or VO2+ ions.

 Likewise Cr3+ can exist in aqueous solutions, but once oxidized to +6 oxidation
state, it reacts with water to form CrO42- and Cr2O72-.
 Transition metal Complexes
• Transition metal form colored complexes with ligands.

• A complex ion is an ion comprising one or more ligands attached to a central

metal cation by means of dative covalent bonds.

• A ligand is a species, which can use its lone pair of electrons to form a dative
covalent bond with a transition metal. Examples of ligands are H2O, NH3, Cl-,
OH-, CN- …….etc

• Cations which form complex ions must have two features:

 they must have a high charge density, and thus be able to attract
electrons from ligands.
 they must have empty orbitals of low energy, so that they can accept the
lone pair of electrons from the ligands.

• Cations of d-block metals are small, have a high charge and have available
empty 3d and 4s orbitals of low energy. They thus form complex ions readily.
 Complexes cont…d
• The number of lone pairs of electrons which a cation can accept
is known as the coordination number of the cation.
• It depends on the size and electronic configuration of that
cation, and also on the size and charge of the ligand. 6 is the
most common coordination number, although 4 and 2 are also
known.
• Examples of complex ions are [Fe(H2O)6]2+, [CoCl4]2-,
[Cu(NH3)4(H2O)2]2+. Note that the formula of the ion is always
written inside square brackets with the overall charge written
outside the brackets.
Geometry of complex ions
• 6-coordinate complexes are all octahedral, and are formed with
small ligands such as H2O and NH3. Examples are:

[Fe(H2O)6]2+ [Cr(NH3)6]3+ [Cu(NH3)4(H2O)2]2+

 Complexes cont..d
• 4-coordinate complexes are either tetrahedral or square planar.
Are formed with larger ligands such as Cl-. Square planar are
common with d8 cations eg Ni, Pt, etc
[CoCl4]2- (Tetrahedral) [PtCl2(NH3)2] (square planar)
• 2-coordinate complexes are in general linear, and are formed
with Ag+ ions
[Ag(NH3)2]+
• Some rules for the likely coordination number of transition
metal complexes are:
a) Silver ions form linear complexes with a coordination
number of 2;
b) Chloride ions and other large ions form tetrahedral
complexes with a coordination number of 4;
c) Most other transition metal complexes are octahedral
with a coordination number of 6.
 Complexes cont…d
Metal salts in Solution
• When metal ions are in solution, they are usually represented as
the simple ion, such as Fe2+(aq), Co2+(aq), Cr3+(aq) or Fe2+(aq).
• This is, however, a simplified representation as all d-block
cations and many other cations with high polarizing power exist
as the hexaaqua complex, e.g Fe2+ exist [Fe(H2O)6]2+. Examples:

FeSO4(aq) consists of [Fe(H2O)6]2+ and SO42- ions

Fe2(SO4)3(aq) consists of 2[Fe(H2O)6]3+ and 3SO42- ions
CuCl2(aq) consists of [Cu(H2O)6]2+ and 2Cl- ions
• Many complex ions exist in the solid state. In these cases the
ligands are often written after the rest of the compound,
separated by a dot: Example
[Fe(H2O)6]SO4(s) is generally written FeSO4.6H2O(s)
[Cr(H2O)6]Cl3(s) is generally written CrCl3.6H2O(s)
 Chemical properties of complex ions
a) Polarising power
• The effect of ligands in a complex ion is to significantly stabilise
the metal cation by increasing its size and hence reducing its
polarizing power. Many compounds which are covalent in the
anhydrous state are actually ionic in the hydrated state, e.g

FeCl3(s) is covalent but FeCl3.6H2O(s) is ionic.

AlCl3(s) is covalent but AlCl3.6H2O(s) is ionic.

b) Precipitation reactions
• The properties of anions are also changed if they are behaving as
ligands in a complex ion. If behaving as ligands, they are much less
readily precipitated by cations.
 Examples:
1. AgNO3(aq) will form a precipitate of AgCl if added to a
solution of copper (II) chloride, [Cu(H2O)4]Cl2 but not from
sodium tetrachlorocobaltate (II), Na2[CoCl4].

2. BaCl2(aq) will form a precipitate of BaSO4 if added to a solution

of [Cr(H2O)5Cl]SO4 but not with [Cr(H2O)5SO4]Cl.
(this is due to inner-outer sphere concept to be cover in coordination chem)

c) Heating
• The properties of molecules also changes if they are behaving as
ligands. In particular, they are much less readily removed by
heating.
Examples:

1. Na2CO3.10H2O(s) can be dehydrated to Na2CO3(s) by gentle

heating at 50oC.
2. FeSO4.6H2O(s), however, can only be dehydrated to FeSO4(s) by
strong heating at 150oC, since the water is strongly bonded to
the Fe2+ as a ligand.
3. CuSO4.5H2O(s) can be dehydrated to CuSO4.4H2O(s) by gentle
heating at 50oC, and to CuSO4 by strong heating at 100oC. One
of the water molecules is a molecule of crystallization but the
others are ligands.
 Types of ligands

• Some ligands are capable of forming more than one dative

covalent bond per ligand.
Examples: Ethanedioate (C2O42-) and 1,2-diaminoethane
(H2NCH2CH2NH2), both of which donate 2 lone pairs per ligand
and are said to be bidentate.

• One unusual ligand, known as edta4-, can form 6 dative covalent

bonds per ligand. It is thus said to be hexadentate.

• Ligands such as H2O, Cl- and CN- form only one dative covalent
bond per ligand and are said to be unidentate.
[Fe(C2O4)3]3- [Cr(H2NCH2CH2NH2)3]3+ [Cu(edta)]2-

Bidentate Bidentate Hexadentate

 Color of Transition Metal Complexes
a) Formation of colored ions

• When a cation forms a complex ion, the incoming ligands

repel the d electrons in the atom, and thus they are raised in
energy.
• Some of the d-orbitals, however, are repelled more than
others and the result is that the d-orbitals are split into 2
groups of orbitals, with three orbitals being slightly lower in
energy than the other two:

The difference in energy between these groups of

orbitals is similar to the energy of visible light.
 If these d-orbitals are partially filled, some of the electrons in the
lower energy orbitals are excited into the higher energy orbitals,
and in doing so absorb the light that corresponds to that
frequency.

 The resultant light is deficient in the light of that frequency and

thus the complex appears colored

Lower/Ground state Excited state

 ‘’NB: Transition metal ions are colored because d-electrons can
absorb light and get excited into higher energy d-orbitals (d-d
transitions)’’
 Note that two criteria must be satisfied if the ion is to be colored:
1. There must be a splitting of the d-orbitals. This only happens
in the presence of ligands and thus only complex ions are
colored. Anhydrous ions do not have split d-orbitals and so
cannot absorb light in the visible spectrum and are thus white.
For instance, anydrous CuSO4 (d9) is white but hydrated
CuSO4.5H2O is blue.
2. The d-orbitals must be partially filled. If the d-orbitals are
empty (Eg Sc3+, Al3+) then there are no electrons which can be
excited into the higher energy d-orbitals and the ions will be
colorless. If the d-orbitals are full (Eg Cu+, Zn2+) then there are
no empty orbitals into which the electrons can be excited and
the ions will be colorless.
 Other Sources of Color
 Metal to Ligand and Ligand to Metal Charge Transfer (LMCT and
MLCT) transitions.

 CT absorption bands and hence colors are intense compared to

the pale/faint color due to d-d transitions.

 This is because CT transition are both spin and Laporte allowed

while d-d transitions are spin allowed and Laporte forbidden.

 LMCT arise from transfer of electrons from the ligand to the

metal. This type of transfer is predominant if complexes have
ligands with relatively high-energy lone pairs or if the metal has
low-lying empty orbitals.
Examples: MnO4− : The permanganate ion having tetrahedral geometry is
intensely purple due to strong absorption involving charge transfer from
primarily from filled oxygen p orbitals to empty manganese(VII) orbitals.
 MLCT arise from transfer of electrons from the metal to the
ligand.

 This is most commonly observed in complexes with ligands

having low-lying π* orbitals, especially aromatic ligands.

 The transition will occur at low energy if the metal ion has a
low oxidation number, for its d orbitals will be relatively high in
energy.

Examples of ligands suitable for MLCT are:

2,2'-bipyridine (bipy),
1,10-phenanthroline (phen),
CO,
CN− and
SCN−
Factors affecting Color and Use of color in analysis of
complex ions
• The color of a complex ion depends on:
the ligand
the coordination number
the oxidation state of the metal
the identity of the metal

• As transition metal ions can show a huge variety of colors, it is often

possible to identify a complex ion simply by its color.

• It is also possible to determine the concentration of a solution

containing a colored ion using ultraviolet and visible
spectrophotometry.

• As the absorbance of a solution is proportional to its concentration,

the concentration of any solution can be determined by comparing its
absorbance to the absorbance of a solution of known concentration.
• The analysis by UV-vis spectroscopy can be summarized as
follows:
a) A solution containing a known concentration of the ion is
prepared, and a suitable ligand is added in excess in order to
intensify the color (thiocyanate ions, SCN-, are often used for
this purpose).
b)A sample of the solution is placed into a cuvette in a
colorimeter and the filter which gives the maximum
absorbance is selected.
c) The absorbance of the solution is recorded.
d)This process is repeated using a range of other known
concentrations, and a graph is plotted of absorbance against
concentration (Standard curve
d)The solution of unknown concentration is mixed with an excess
of the same ligand to intensify the color.
e) The absorbance of the solution of unknown concentration is
then recorded using the same filter.
f) The concentration of this solution can then be determined
from the concentration-absorbance graph.

Homework
• Show how iron in soil sample can be analyzed using UV-VIS.
Give the name of the ligand used. Write the structure of the
Fe-ligand complex.
Magnetic properties of Transition Metals
 Transition metals form complexes which show strong magnetic
properties.

 TM compounds that are repelled by magnet are referred to as

diamagnetic. This compounds have no unpaired electrons. Those
that have unpaired electrons are said to be paramagnetic.

 Magnetic measurements can therefore be used to determine the

number of unpaired spins in a complex.

 The magnitude of paramagnetism is given in terms of dipole

moment.
 In a free atom or ion, both the orbital and spin angular momenta
give rise to dipole moment and hence contribute to paramagnetism.

 In transition metal ions, the unpaired electrons are in the outer valence shell and
due to strong ligand field, the orbital angular momentum of the transition metal
ions in the complex is quenched and hence only the spin contribution to
paramagnetism is significant.

 This is commonly referred to as spin-only paramagnetism.

 As a result, the magnetic moment, μ, is directly related to the number of

unpaired electrons, n, by the following expression:

μeff = {[n(n+2)] ½} μB

 The magnetic moment is expressed in Bohr magnetons (μB). A Bohr magneton is

given by the expression:
μB = eh/4 πmc

where h is the planck’s constant, e is the electronic charge, c is the velocity of

light and m is the mass of electron.
 Alloy and Nonstoichiometry of Transition Metals
Alloy formation
 The transition metals readily form alloys with each other because of similar
radii. The alloys so formed are hard and have high melting points. The best
known are ferrous alloys where: vanadium, chromium, molybdenum, tungsten
and manganese are used for the production of variety of steels and stainless
steel.
 Industrially important alloys of transition metals with non-transition metals are
brass (copper-zinc), bronze (copper-tin) etc.

Interstitial compound formation

 Many transition metals form interstitial compounds particularly with small non-
metal atoms such as hydrogen, boron, carbon and nitrogen.
 These small atoms enter into the voids sites between the packed atoms of the
crystalline metal. These interstitial compounds are nonstoichiometric,
e.g. TiH1.7, ZrH1.9, VH0.56, TiC, Fe3C, Fe0.94O etc.
 They are hard and have higher melting points than those of pure metals.
 They retain metallic conductivity and are chemically inert.
Other applications of transition metal complexes
(i) Haemoglobin
Haem is a complex ion consisting Fe2+ and a complex tetradentate
ligand called porphyrin. The complex is generally found with a
protein called globin, which provides the fifth coordinate bond, and
a molecule of oxygen which forms the sixth bond. The complete six
coordinate complex is called haemoglobin. This structure is
responsible for carrying oxygen in the blood throughout the human
Body. Fe2+ + porphyrin  haem
haem + globin + O2  haemoglobin

Porphryn Haemoglobin
NB: Carbon monoxide is a similar size and shape to oxygen and
forms a much stronger bond with the iron. It thus displaces the
oxygen from the complex and reduces the blood’s ability to carry
oxygen. It is thus a very poisonous gas.

(ii) Cisplatin
• The square planar complex cisplatin has the following structure:

• It kills cancerous cells and widely used in cancer treatment.

(iii) Tests for aldehydes and halide ions
• Diammine silver (I) has the formula [Ag(NH3)2]+ and is the active
ion in Tollen’s reagent.

• It is reduced to silver by reducing sugars and aldehydes (but not

ketones). It produces a characteristic silver mirror on the side of
the test-tube and thus used to test for aldehydes.

• It is also used in the test for halide ions in solution. The silver
halides are insoluble in water but AgCl and AgBr will dissolve in
ammonia due to the formation of the diammine silver complex:
AgCl(s) + 2NH3(dilute)  [Ag(NH3)2]+(aq) + Cl-(aq)
AgBr(s) + 2NH3(conc)  [Ag(NH3)2]+(aq) + Br-(aq)
 (iv) Photography
• Silver bromide, AgBr, is the substance on photographic film. In the
presence of light, it decomposes into silver and bromine:
2AgBr  2Ag(s) + Br2(l).
• The unreacted AgBr is removed when sodium thiosulpate is
added. This forms a complex with the AgBr and washes it off the
film, leaving only the silver metal on the film. The result is the
negative image.
AgBr(s) + 2S2O32-(aq)  [Ag(S2O3)2]3-(aq) + Br-(aq)
(v) Mining and Electroplating
• Metals such as silver and gold generally occur native, but in very
impure form. They can be extracted using cyanide ions which form
stable complexes with silver and gold. The Ag is oxidized to Ag+
and then complexed as [Ag(CN)2]-.
• This complex is widely used in electroplating. To coat another
object with silver, place the metal object to be coated at the
cathode and use [Ag(CN)2]- as the electrolyte. The complex breaks
up, Ag+ ions move to the cathode and the object is coated with a
layer of silver.
Catalytic properties of TM
• The ability of transition metals to form more than one stable
oxidation state means that they can accept and lose electrons
easily.
• This enables them to catalyze certain redox reactions. They can be
readily oxidized and reduced again, or reduced and then oxidized
again, as a consequence of having a number of different oxidation
states of similar stability.
• They can thus behave either as homogeneous catalysts or as
heterogeneous catalysts.
 Types of catalysts
(i) Homogeneous catalysis
• A homogeneous catalyst is a catalyst in the same phase as the
reactants.
• Homogeneous catalysis involves aqueous transition metal ions
catalyzing reactions, often between two anions. The cations
reacts with each anion in turn, thus avoiding the need for a direct
collision between two anions (this is difficult since they repel
each other).
Case 1: S2O82-(aq) + 2I-(aq)  2SO42-(aq) + I2(aq)

This can be catalyzed by Fe2+ or Fe3+ ions:

With Fe2+: S2O82-(aq) + 2Fe2+(aq)  2SO42-(aq) + 2Fe3+(aq)

2Fe3+(aq) + 2I-(aq)  2Fe2+(aq) + I2(aq)
With Fe3+: 2Fe3+(aq) + 2I-(aq)  2Fe2+(aq) + I2(aq)
2Fe2+(aq) + S2O82-(aq)  2Fe3+(aq) + 2SO42-(aq)
Case 2: 2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq)  2Mn2+(aq) + 10CO2(g)
+ 8H2O(l)
• This is a good example of auto-catalysis.
• One of the products in this reaction, Mn2+(aq), behaves as a
catalyst and thus the reaction is slow at first but is much faster
after a little of the products are formed.

(ii) Heterogeneous catalysis

• Heterogeneous catalysts react by allowing the reactant molecules
to bond to the surface of the metal, usually by attracting the
surface electrons. The reaction takes place at the surface, and the
transition metal regains electrons as the products leave.
• The reaction is catalyzed because the reactant molecules spend
more time in contact with each other than they would in the
absence of a catalyst.
Example: hydrogenation of ethene using nickel metal catalyst
 Key processes and requirements in heterogeneous catalysis
Physical adsorption and chemisorption
• The molecules usually bond to the surface of the metal using
intermolecular forces such as Van der Waal’s forces. This process is
known as physical adsorption. Sometimes they actually form a
covalent bond with the metal. This process is known as
chemisorption.
• An ideal heterogeneous catalyst adsorbs fairly strongly but not too
strongly. If it adsorbs too weakly (like silver) the reactant
molecules do not spend enough time in contact with each other. If
they adsorb too strongly (like tungsten) the products do not leave
the metal surface quickly enough and this slows the reaction
down.
• The most useful catalysts are those which adsorb moderately. As
the strength of absorption tends to decrease from left to right
along the Periodic Table, the most effective catalysts tend to be
the transition metals in the middle, like V, Fe and Ni.
 Examples of Industrial Catalysts
• Examples of heterogeneous catalysts are; Fe used in the
production of ammonia; V2O5 used in the contact process making
sulphuric acid; Pt or Ni used in the hydrogenation of ethene; and
Rh used in catalytic converter for emission control.
Reaction Catalyst
i) N2(g) + 3H2(g)  2NH3(g) Fe

ii) 2SO2(g) + O2(g)  2SO3(g) V2O5

iii) C2H4(g) + H2(g)  C2H6(g) Pt/Ni

iv) 2CO(g) + 2NO(g)  N2(g) + 2CO2(g) Rh (catalytic converter)

• The catalytic converter converts toxic carbon monoxide and

nitrogen monoxide to non-toxic carbon dioxide and nitrogen.
 Catalyst support and poisoning
(i) Catalyst support
• The catalyst needs to have a large surface area to be effective.
This means it needs to be very thinly spread out, and might need
a special support. The support disperse the catalyst and thus
prevents agglomeration/aggregation. It also cuts cost in terms of
amount of catalyst used.
• Catalytic converters in cars use rhodium (Rh) as catalyst supported
on a special ceramic.

(ii) Catalyst poisoning

• Heterogeneous catalysts can be poisoned by impurities. These
impurities bond very strongly to the catalyst surface and block the
active sites.
• The activity of iron in the Haber process is reduced by sulfur
present in iron as an impurity. It is IMPORTANT that as much
sulfur as possible is removed during the Basic Oxygen Process.

• The efficiency of catalytic converters in cars is reduced by lead,

which bonds permanently to the rhodium surface. This explains
the SWITCH to unleaded petrol.
 Redox Chemistry of TM
• Since transition metals show a variety of transition states in their
compounds, much of their chemistry is dominated by shift
between these transition states.
• When the oxidation state changes, the colour changes as the
electron distribution in the d-orbitals changes.
• The colour is also affected by the ligands involved, as the type of
ligand affects the amount of splitting in the d-orbitals.
1. Iron
• Iron exists in two common oxidation states, +2 (Fe2+) and +3
(Fe3+).
• In aqueous solution, the Fe is readily oxidized from Fe2+ to Fe3+:
Fe2+(aq)  Fe3+(aq) + e ………. oxidation

• The Fe2+ ion is thus a reducing agent. Concentrations of Fe3+ in

solution can be determined by titration with oxidizing agents.
2. Manganese
• Manganese can exist in a number of oxidation states, but is most
stable in an oxidation state of +2, +4 or +7

• In the +7 oxidation state, it exists as the intense purple ion MnO4-.

This can be reduced to pale pink Mn2+ by Fe2+ in acidic solution:
MnO4-(aq) + 8H-(aq) + 5e  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + e
Overall: MnO4-(aq) + 8H-(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

3. Vanadium
• Vanadium forms stable compounds in 4 different oxidation states,
+2, +3, +4 and +5.
• In aqueous solution, the ions formed are:

• All vanadium (V) compounds can be reduced to the +4, +3 and then
+2 oxidation state by strong reducing agents such as zinc in acid
solution:
VO2+(aq) + 4H+(aq) + 3e  V2+(aq) + 2H2O(l)
Zn(s)  Zn2+(aq) + 2e

Overall: 2VO2+(aq) + 4H+(aq) + 3Zn(s)  2V2+(aq) + 3Zn2+(aq) + 2H2O(l)

Yellow violet

• The reaction proceeds via the +4 and +3 oxidation states, so the

colour change observed is yellow (VO2+)  green (VO2+ and VO2+)
 blue (VO2+)  green (V3+)  violet (V2+).
• Other, milder reducing agents will reduce the +5 oxidation state to
+4 but no further.

• If exposed to air, the violet +2 complex will slowly oxidize to the

green +3 complex.

• The precise colors of the solutions will depend on the acids used. Cl-
ions can behave as ligands and this will affect the color.

For instance,
• if HCl is the acid, the +3 complex is [V(H2O)4Cl2], which is green.
• if H2SO4 is used, the +3 complex is [V(H2O)6]3+, which is grey-blue.

• So the colour of the complexes depends both on the oxidation

state and the nature of the different ligands in solution.
4. Chromium
• Chromium forms stable ions in three different oxidation states, +2,
+3 and +6.
• In aqueous solutions, the ions formed are:

• The actual formula of the +2 and +3 complex ions depends on the

acid used (Cl- and SO42- will also behave as ligands).

• In aqueous alkaline solutions, the ions formed are:

• The +6 chromium complexes can be readily interconverted using
acid and alkali:
2CrO42-(aq) + 2H+(aq)  Cr2O72-(aq) + H2O(l)
Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l)

a) In acidic solution
• All chromium (VI) compounds can be reduced to the +3 and then
the +2 oxidation state by strong reducing agents such as zinc in acid
solution.

• The chromium is first reduced to the +3 oxidation state:

Cr2O72-(aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7H2O(l)
Zn(s)  Zn2+(aq) + 2e

Overall: Cr2O72-(aq) + 14H+(aq) + 3Zn(s)  2Cr3+(aq) + 7H2O(l) + 3Zn2+(aq)

• It is then further reduced:
Cr3+(aq) + e  Cr2+(aq)
Zn(s)  Zn2+(aq) + 2e
Overall: Zn(s) + 2Cr3+(aq)  Zn2+(aq) + 2Cr2+(aq)

• The colour change observed on adding zinc in acid solution to

dichromate ions is orange (Cr2O72-) to green (Cr3+) to blue (Cr2+).

• Other, milder reducing agents will reduce the +6 oxidation state to

+3 but no further. Fe2+ is a good example:

Fe2+(aq)  Fe3+(aq) + e
Overall: Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq)  2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

• This reaction can be used in titrations to determine the

concentration of Fe2+ ions in a sample.
b) In alkaline solution
• In alkaline solution, it is possible to oxidize the +3 oxidation state to
the +6 oxidation state.

• The chromium ion forms the [Cr(OH)6]3- complex in alkaline

conditions.
• This complex can be oxidized to the +6 oxidation state by adding
H2O2. Hydrogen peroxide, which is a reducing agent in acidic
solution, is an oxidizing agent in alkaline solution.
[Cr(OH)6]3-(aq) + 2OH-(aq)  CrO42-(aq) + 4H2O(l) + 3e
H2O2(aq) + 2e  2OH-(aq)
2[Cr(OH)6]3-(aq) + 3H2O2(aq)  2CrO42-(aq) + 8H2O(l) + 2OH-
(aq)

• The green [Cr(OH)6]3- ion is oxidized to the yellow CrO42- ion in

alkaline solution.
• The relative tendency of species to undergo oxidation and reduction
often depends dramatically on the pH of the solution.

• The interconversion of the chromium (III) and chromium (VI)

oxidation states is an important example.

• In general, oxidation is favored by alkaline conditions and reduction

is favored by acidic conditions.

5. Cobalt
• Cobalt exists in two stable oxidation states, +2 and +3
• In acidic or neutral solution:
• In alkaline solution:

• In ammonia solution:

• The colours of these complexes are not easy to identify as they

change colour readily depending on the concentration of the
various species in solution.
• The ease of oxidation of cobalt from +2 to +3 depends on the ligand
and the pH.
• It is extremely difficult to oxidize cobalt from +2 to +3 in acidic
solution.
• It is easier to oxidize Co2+ in alkaline solution using hydrogen
peroxide. The blue Co(OH)2 is oxidized to brown Co(OH)3:
Co(OH)2(s) + OH-(aq)  Co(OH)3(s) + e
H2O2(aq) + 2e  2OH-(aq)
Overall: 2Co(OH)2(s) + H2O2(aq)  2Co(OH)3(s)

• In ammoniacal solution, it is very easy to oxidize Co2+, and it is

oxidized when left to stand in air. The straw Co(NH3)62+ is oxidized to
the brown Co(NH3)63+ (the solution gets darker):
Co(NH3)62+(aq)  Co(NH3)63+(aq) + e
O2(g) + 2H2O(l) + 4e  4OH-(aq)
Overall:4Co(NH3)62+(aq) + O2(g) + 2H2O(l)  4Co(NH3)63+(aq) + 4OH-(aq)
 Redox Titrations involving TM
• Oxidizing agents in aqueous solution can be determined by titrating
against standard solutions of reducing agents.
• Reducing agents in aqueous solution can be determined by titrating
against standard solutions of oxidizing agents.

• In the laboratory, there are two important reagents involving

transition metals that are used in these titrations: potassium
manganate (VII), KMnO4 and potassium dichromate, K2Cr2O7.

(i) Potassium manganate (VII) (KMnO4) titrations.

• The MnO4- ion is a powerful oxidizing agent:

MnO4-(aq) + 8H+(aq) + 5e  Mn2+(aq) + 4H2O(l)

• It can therefore be used to determine reducing agents

• The most important principles in titrations involving KMnO4 are:
 The color change associated with the above reduction:
MnO4- (intense purple)  Mn2+ (colorless)
The color change is so intense that no indicator is required for
the reaction.

 The KMnO4 solution is generally placed in the burette. This

causes two problems:
– the intense color of KMnO4 means that it is very difficult
to see the graduation marks on the burette, and so it is
difficult to read accurately.
– The KMnO4 reacts slightly with the glass, causing a slight
stain if the burette is used too often.
• The reason the KMnO4 is placed in the burette is as follows:
– As the KMnO4 is added to the solution under investigation, it is
immediately decolorized by the reducing agent. The end-point is
detected by the failure of the purple colour to disappear - a pink
color persistent in the conical flask indicates that the MnO4- ions
are no longer being reduced. This permanent pink color is easily
detected.

– If the KMnO4 were in the conical flask, it would slowly decolorize

until the solution became completely colorless. The gradual
disappearance of the pink color is much harder to detect than
the sudden appearance of the pink color, and for this reason the
KMnO4 is always placed in the burette, despite the difficulties it
presents in reading the burette.
  Since the MnO4- ion is only an effective oxidizing agent in acidic
medium, it is necessary to add excess acid to the conical flask
before carrying out the titration.
NB: Sulfuric acid is generally used for this purpose. Hydrochloric
acid should not be used since it is a reducing agent and will react
with the MnO4- before the MnO4- can react with the reducing
agent under investigation:

MnO4- (aq) + 8H+(aq) + 5Cl-  Mn2+(aq) + 4H2O(l) + 2.5Cl2(g)

• Common examples of reducing agents which are determined using

KMnO4 are Fe2+(aq) and C2O42-(aq)

Fe2+(aq)  Fe3+(aq) + e

C2O42-(aq)  2CO2(g) + 2e
MnO4- (aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

2MnO4- (aq) + 16H+(aq) + 5C2O42-(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

(ii) Potassium dichromate (VI) (K2Cr2O7) titrations

• Potassium dichromate (VI) is also a good oxidizing agent in acidic
medium:
Cr2O72-(aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7H2O(l)

• It can therefore be used to determine reducing agents.

• The most important principles of potassium dichromate titrations
are:
– The colour change associated with this titration is orange 
green. The colors are not, however, as intense as with the
manganate titration and an indicator is needed.
Diphenylaminesulphonate is used. It turns from colorless to
purple at the end-point.
– The dichromate is always placed in the burette as it is easier to
see the appearance of the purple colour than its disappearance.
– As with manganate, care must be taken as to which acid to use.
Sulfuric acid should be used as chloride ions could interfere with
the titration.
– Fe2+(aq) is the reducing agent most commonly determined by
titration with potassium dichromate (VI).
Fe2+(aq)  Fe3+(aq) + e
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq)  2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)