UNIT 2 Module 3
L.O. Deduce the electron configurations of
atoms and ions of the d-block elements.
Describe the elements Ti-Cu as transition
Starter – What are the transition elements?
A transition element is a d-block element
that forms an ion with an incomplete d sub-
Although Scandium and Zinc are the first and
last elements in the first of the d-block they
are not transition elements,
Sc only forms the ion Sc3+ with an empty
Zn only forms the ion Zn2+ with a full d-
Electrons occupy orbitals in order of energy
Remember that the 4s sub-shell has a lower
energy than the 3d sub-shell. This means
that the 4s orbital fills before the orbitals in
the 3d sub-shell.
The d-block metals will have filled their 4s
orbital before electrons are added to the 3d
A maximum of two electrons are put into
orbitals in the order of increasing orbital
energy: the lowest-energy orbitals are filled
before electrons are placed in higher-energy
Chromium and Copper do not follow the
Aufbau principle for placing electrons.
Cr – the 3d and 4s orbitals all contain one
electron with no orbital being completely
Cu – the 3d orbitals are full, but there is only
one electron in the 4s orbital.
The transition metals lose electrons to form
They lose their 4s electrons before the 3d
electrons. This seems surprising as the 4s
orbitals are filled first.
The 4s and 3d orbitals are close in energy but
once full the 4s electrons have more energy
and so lost first.
L.O. Describe the existence of more than one
oxidation state for a transition element in
Starter – What properties would you expect the
transition elements to have?
They display normal metallic properties.
Shiny in appearance
High melting and boiling points
Have giant metallic lattices
Delocalised electrons – allows conduction
In addition the transition elements form
compounds with different oxidation states,
which give the compounds different colours.
Ti to Cu, the transition elements all form ions
with 2 or more oxidation states.
They all form an ion with a 2+ oxidation state.
When white light passes through a sample of
a compound containing a transition element,
some of the wavelengths are absorbed.
The colour we observe is a mixture of the
wavelengths of light that have not been
L.O. Describe the catalytic behaviour of the
Describe simple precipitation reactions of
transition metal ions with sodium
Starter – What is a catalyst?
A catalyst is a substance that increases the
rate of a chemical reaction by providing an
alternative lower energy route for the reaction
Transition metals can act as a surface for the
reactions to take place on.
Or the metal ions can change oxidation states
and bind to the reactants forming
The Haber process is used to make ammonia,
NH3, from nitrogen and hydrogen.
N2(g) + 3H2(g) ⇌ 2NH3(g)
The catalyst used is iron metal, it is used to
increase rate and lower the temperature
needed for the reaction.
Much of the ammonia produced in the Haber
process is used to manufacture fertilisers.
The contact process is used to convert
sulphur dioxide into sulphur trioxide in the
manufacture of sulphuric acid.
2SO2(g) + O2(g) ⇌ 2SO3(g)
Vanadium(V) oxide, V2O5, is used as the
Vanadium here has its 5+ oxidation state.
Hydrogen can be added across the C=C
double bonds in unsaturated compounds to
make saturated compounds – hydrogenation.
H2C=CH2 + H2 à H3C-CH3
The catalyst is Ni. It lowers the temperature
and pressure needed to carry out the reaction.
This process is used to make margarines.
Hydrogen peroxide decomposes slowly at
room temperature and pressure into water
2H2O2(aq) à 2H2O(l) + O2(g)
The catalyst used is manganese(IV) oxide,
MnO2, in which manganese has a 4+
This reaction is used to generate oxygen.
A precipitation reaction is one where soluble
ions in different solutions are mixed together
to form an insoluble compound, a precipitate.
Transition metal ions in aqueous solution
react with aqueous sodium hydroxide,
Ion NaOH Equation
Cu2+(aq) Pale blue solution Cu2+(aq) + 2OH-(aq) à Cu(OH)2(s)
turns into a pale blue
Co2+(aq) Pink solution turns Co2+(aq) + 2OH-(aq) à Co(OH)2(s)
into a blue precipitate
– that turns beige in
Fe2+(aq) Pale green solution Fe2+(aq) + 2OH-(aq) à Fe(OH)2(s)
that forms a green
precipitate – turns
rusty in air.
Fe3+(aq) Pale yellow solution Fe3+(aq) + 3OH-(aq) à Fe(OH)3(s)
forms a rusty-brown
L.O. Understand the terms ligand, complex
ion, and coordinate number.
Starter – What is the oxidation state of the
transition metal in the following compounds?
KMnO4, K2Cr2O7, V2O5, FeCl6.
A complex ion is a transition metal ion
bonded to one or more ligands by coordinate
bonds (dative covalent bonds).
The coordinate number is the total number of
coordinate bonds formed between a central
metal ion and its ligands.
A ligand is a molecule or ion that can donate
a pair of electrons with the transition metal
ion to form a coordinate bond.
All ligands have 1 or more lone pairs of
electrons in their outer shell.
Some ligands are neutral and others are
If a ligand donates 1 pair of electrons to form
a coordinate bond its called a monodentate
These ions have an octahedral shape. Four of
the ligands are in one plane, with the fifth
one above the plane, and the sixth one below
It doesn't matter what the ligands are. If you
have six of them, this is the shape they will
take up. Easy!
Tetrahedral ions are ions that have four
ligands bonded to them rather than six.
They adopt this shape because the ligands
are too big to fit any more around the central
Occasionally a 4-co-ordinated complex turns
out to be square planar. There's no easy way
of predicting that this is going to happen.
Cis-platin is a neutral complex, Pt(NH3)2Cl2.
The platinum, the two chlorines, and the two
nitrogens are all in the same plane.
L.O. Describe stereoisomerism in transition
metal complexes using examples of cis-
trans and optical isomerism.
Starter – What is a stereoisomer?
Stereoisomers are species with the same
structural formula but with a different
arrangement of the atoms in space.
There are 2 types of stereoisomerism in
transition element chemistry.
Octahedral complex ions often contain 2
different ligands, four of one type (in the
plane), and two of another (out of the plane).
These complex ions can exist as cis and trans
Cis-trans isomers are also seen in four-
coordinate complexes with square-planar
The complex must contain two different
ligands with two of one ligand and two of
You recognise optical isomers because they
have no plane of symmetry.
They are not easy to visualise or draw.
The examples you are most likely to need
occur in octahedral complexes which contain
bidentate ligands - ions like
[Ni(NH2CH2CH2NH2)3]2+ or [Cr(C2O4)3]3-.
L.O. Explain and use the term bidentate ligand
(e.g. NH2CH2CH2NH2, ‘en’).
Starter – What does the word bidentate mean?
Bidentate ligands are Lewis bases that donate
two pairs ("bi") of electrons to a metal atom.
Bidentate ligands are often referred to as
chelating ligands ("chelate" is derived from
the Greek word for "claw") because they can
"grab" a metal atom in two places.
A complex that contains a chelating ligand is
called a chelate .
Some Bidentate Ligands
ethylenediammine acetylacetonate ion
phenanthroline oxalate ion
A hexadentate ligand has six points at which
it can attach to the central atom. An example
of such a ligand is: ethylene diamine
tetraacetic acid (EDTA)
Each of the acetic acid groups can give up a
proton, yielding an anion with a -4 charge. In
addition, each of the two nitrogen's has a pair
of non-bonding electrons.
EDTA is used in foods to remove metal ions
that might catalyse the oxidation of the food.
It is also used in detergents to bind calcium
and magnesium ions to reduce the hardness
EDTA also has medical applications, it is
added to blood samples to prevent clotting
and treat lead and mercury poisoning.
They are two compounds that have the same
molecular and structural formula BUT can
have mirror image forms which are NOT
superimposable on each other.
The non-superimposable mirror image
isomers are called optical isomers (or
Occur in complexes with,
3 molecules or ions of a bidentate ligand,
2 molecules or ions of a bidentate ligand
and 2 molecules or ions of a monodentate
1 hexadentate ligand.
L.O. Describe the process of ligand
Starter – What do you understand by the term
A ligand substitution reaction is a reaction in
which one ligand in a complex ion is replaced
by another ligand.
Aqueous solutions of Cu(II) ions contain
[Cu(H2O)6]2+ complex ions with a
characteristic pale blue colour.
When ammonia is added a pale blue
precipitate forms, Cu(OH)2.
With the addition of excess ammonia, a deep
blue solution is formed.
Four of the water ligands are replaced by four
The product is [Cu(NH3)4(H2O)2]2+(aq)
This is a ligand substitution reaction.
[Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
When conc. HCl is added to aq Cu(II) ions the
pale blue solution forms a green solution
before forming a yellow solution.
This reaction can be reversed by adding water
to the yellow solution.
The reaction can be explained by the
[Cu(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CuCl4]2-(aq) + 6H2O(l)
The complex goes from 6 to 4 ligands
because the chlorine ligands are larger than
the water ligands and have a stronger
The complex has a tetrahedral shape.
Aqueous Co(II) ions contain [Co(H2O)6]2+ ions
which are characteristically pink.
When conc. HCl is added the solution turns a
dark blue colour.
The concentrated HCl provided a high
concentration of chlorine ions.
Four chloride ions replace the six water
This is a reversible reaction,
[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
L.O. Use and understand the term stability
Deduce expressions for Kstab of ligand
substitutions and understand that a large
Kstab results in formation of a stable
Starter – What is haemoglobin?
A complex protein composed of four
Each protein chain contains four non-protein
parts, called haem groups.
Each haem group has an Fe2+ ion at its
Oxygen binds to the Fe2+ allowing
haemoglobin to transport oxygen around the
In a haem group the central Fe2+ ion is
surrounded by four coordinate bonds to
nitrogen atoms in the haem structure.
It also has a coordinate bond to the protein
And a final coordinate bond can be formed to
an oxygen molecule.
CO can also bind to haemoglobin at the same
site that oxygen can.
Sadly CO binds more strongly than oxygen,
this leaves less haemoglobin to carry oxygen.
This is a ligand substitution as CO can
This is also an irreversible reaction.
When chlorine ions are added to an aqueous
solution of some transition metal ions, an
equilibrium is set up,
[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
An expression can be written for these
Kstab = conc. products / conc. Reactants
Kstab = [CoCl42-] / [[Co(H2O)6]2+] [Cl-]4
Water is left out as it is in excess and its
concentration is virtually constant.
The stability constant, Kstab is the equilibrium
constant for an equilibrium existing between
a transition metal ion surrounded by water
ligands and the complex formed when the
same ion has undergone a ligand
A large value of Kstab indicates that the
position of an equilibrium lies to the right.
Complex ions with high stability constants
are more stable than those with lower values.
The more stable a complex ion the easier it is
Bidentate and multidentate ligands (EDTA)
form particularly stable complex ions.
The additional stability is a result of the
increase in entropy.
Ligand substitution by EDTA results in an
increase in the number of moles, which
[Cu(H2O)6]2+ + edta4- ⇌ [Cu(edta)]2- + 6H2O
L.O. Describe redox behaviour in transition
metals using suitable examples.
Starter – What is a Redox reaction?
Reduction is the gain of electrons, or a
decrease in oxidation number.
Oxidation is the loss of electrons, or an
increase in oxidation number.
Iron has 2 common oxidation states, Fe2+ and
The Fe(II) oxidation state is less stable than
the Fe(III) oxidation state. This means Fe(II)
easily oxidises in air to Fe(III)
Fe2+(aq) à Fe3+(aq) + e-
Mn exists in a number of oxidation states,
but the most important are +2 and +7.
MnO4- is a strong oxidising agent. In acidic
solution, MnO4- can be reduced to form
MnO4-(aq) + 8H+(aq) + 5e- à Mn2+(aq) + 4H2O(l)
MnO4- is commonly used to oxidise solutions
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) à
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Redox reactions like this are used to find the
percentage composition of a metal in a
Redox titrations are similar to acid-base
An oxidising agent can be titrated against a
reducing agent, the end-point can be found
either with an indicator or using the colour
change of the metal ion.
MnO4- (purple) à Mn2+ (almost colourless)
MnO4- is typically used to oxidise solutions
containing Fe(II) ions.
KMnO4 is used in the burette and the Fe(II)
solution in the conical flask.
As the manganate solution is added it
The end-point is the presence of permanent
L.O. Understand how to carry out redox
Starter – How do you calculate,
24.3 cm3 of 0.02 mol dm-3 KMnO4 reacted
with 20.0 cm3 of an iron(II) solution.
(1) Calculate the molarity of the iron(II) ion.
(2) How do recognise the end-point in the
(1) mol MnO4- = 0.02 x 24.3 / 1000
= 0.000486 moles
mol Fe2+ = 5 x 0.000486
= 0.00243 in 20 cm3,
so scaling up to 1 dm3, the molarity of Fe 2+
= 0.00243 x 1000 / 20 = 0.122 mol dm -3.
(2) The end point is the first faint permanent
pink due to a trace excess of KMnO4.
Calculate the percentage of iron in a sample
of steel wire if 1.51 g of the wire was
dissolved in excess of dilute sulphuric acid
and the solution made up to 250 cm3 in a
standard graduated flask. 25.0 cm3 of this
solution was pipetted into a conical flask
and needed 25.45 cm3 of 0.02 mol dm-3
KMnO4 for complete oxidation.
mol MnO4- = 0.02 x 25.45 / 1000
mol Fe = 5 x 0.000509 = 0.002545,
mass Fe = 0.002545 x 55.9 = 0.1423 g,
total Fe in wire = 0.1423 x 10 = 1.423 g
(1/10th of the made up solution used in
so % Fe = 1.423/1.51 x 100 = 94.2 %
Given the following two half-reactions
(i) S4O62-(aq) + 2e- à 2S2O32-(aq)
(ii) I2(aq) + 2e- à 2I-(aq)
Construct the full ionic redox equation for
the reaction of the thiosulphate ion S2O32-
2S2O32-(aq) + I2(aq) à S4O62-(aq) + 2I-(aq)
What mass of iodine reacts with 23.5 cm3 of
0.012 mol dm-3 sodium thiosulphate
mole S2O32- = 0.012 x 23.5/1000
= 0.000282 mol
mol I2 = mol S2O32-/2 (1:2 in equation)
= 0.000141 mol
mass of iodine = 0.000141 x 126.9 x 2
= 0.0358 g
25cm3 of a solution of iodine in potassium
iodide solution required 26.5 cm3 of 0.095
mol dm-3 sodium thiosulphate solution to
titrate the iodine.
What is the molarity of the iodine solution
and the mass of iodine per dm3?
mol S2O32- = 0.095 x 26.5/1000
= 0.002518 mol
mol of iodine = mole 'thio' / 2
= 0.002518 / 2
= 0.001259 in 25 cm3
scaling to 1 dm3 gives 0.001259 x 1000/25
= 0.0504 mol dm-3 of molecular iodine I2.
mass concentration of I2 = 0.0504 x 2 x 126.9
= 12.8 g dm-3 of iodine
L.O. Carry out calculations based on
Starter – Write the half equations for the
oxidation of Fe2+ and the
reduction of MnO4-.
I2 + S 2O 3 2-
The reaction between iodine and thiosulfate
ions is a useful redox reaction.
I2(aq) + 2S2O32-(aq) à 2I-(aq) + S4O62-(aq)
This reaction is used to reduce iodine to
iodide ions using sodium thiosulfate which
forms tetrathionate ions.
Analysing Oxidising Agents
Iodide ions are added to the oxidising agent
which produces Iodine.
The Iodine is titrated against sodium
thiosulfate solution of a known
From the results you can calculate the
amount of Iodine (moles) and therefore the
concentration of the oxidising agent.
First you calculate the number of moles of thiosulfate
that were used in the titration,
Then work out number of moles of Iodine used in
titration from molar ratio.
Then the number of moles of the oxidising agent
from molar ratio (different equation).
Finally, calculate concentration of oxidising agent,