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Transition Elements

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					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
     elements.

Starter – What are the transition elements?
A transition element is a d-block element
that forms an ion with an incomplete d sub-
shell.

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
    d-orbital.
    Zn only forms the ion Zn2+ with a full d-
    orbital.
Electrons occupy orbitals in order of energy
level.

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
sub-shell.
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
orbitals.
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
   filled.

Cu – the 3d orbitals are full, but there is only
   one electron in the 4s orbital.
The transition metals lose electrons to form
positive ions.

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
     its compounds.

Starter – What properties would you expect the
  transition elements to have?
They display normal metallic properties.
   Shiny in appearance
   High densities
   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
absorbed.
L.O. Describe the catalytic behaviour of the
     transition elements.
     Describe simple precipitation reactions of
     transition metal ions with sodium
     hydroxide.

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
to take.

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
intermediates.
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
catalyst.
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
and oxygen.

       2H2O2(aq) à 2H2O(l) + O2(g)

The catalyst used is manganese(IV) oxide,
MnO2, in which manganese has a 4+
oxidation state.

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,
NaOH(aq).
  Ion              NaOH                        Equation
Cu2+(aq) Pale blue solution        Cu2+(aq) + 2OH-(aq) à Cu(OH)2(s)
         turns into a pale blue
         precipitate
Co2+(aq) Pink solution turns       Co2+(aq) + 2OH-(aq) à Co(OH)2(s)
         into a blue precipitate
         – that turns beige in
         air.
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
           precipitate.
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
negatively charged.
If a ligand donates 1 pair of electrons to form
a coordinate bond its called a monodentate
ligand.
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
the plane.

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
metal ion.
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.
   cis-trans isomerism
   optical isomerism
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
isomers.
Cis-trans isomers are also seen in four-
coordinate complexes with square-planar
shape.

The complex must contain two different
ligands with two of one ligand and two of
another.
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
       (en)                     (acac)




 phenanthroline              oxalate ion
     (phen)                     (ox)
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
of water.

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
enantiomers).
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
   ligand,

   1 hexadentate ligand.
L.O. Describe the process of ligand
     substitution.

Starter – What do you understand by the term
     ‘Ligand Substitution’?
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
  ammonia ligands.

  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
 equilibrium equation,

[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
 repulsion.

 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
 molecules.

 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
     constant, Kstab.
     Deduce expressions for Kstab of ligand
     substitutions and understand that a large
     Kstab results in formation of a stable
     complex ion.

Starter – What is haemoglobin?
A complex protein composed of four
polypeptide chains.

Each protein chain contains four non-protein
parts, called haem groups.

Each haem group has an Fe2+ ion at its
centre.

Oxygen binds to the Fe2+ allowing
haemoglobin to transport oxygen around the
body.
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
globin.

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
replace oxygen.

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
 reactions,
   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
substitution.
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
to form.
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
increases entropy.

[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
Fe3+.

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
Mn2+.

MnO4-(aq) + 8H+(aq) + 5e- à Mn2+(aq) + 4H2O(l)
 MnO4- is commonly used to oxidise solutions
 containing Fe(II).

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
 sample.
Redox titrations are similar to acid-base
titrations.

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
decolourises.
The end-point is the presence of permanent
pink colour.
L.O. Understand how to carry out redox
     calculations.

Starter – How do you calculate,
     Moles?
     Concentration?
     Molar Mass?
     % yield?
 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
  titration?
(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
       = 0.000509,

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
titration),
    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-
 ,and iodine.
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
solution.
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
     experimental data.

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
 concentration.

 From the results you can calculate the
 amount of Iodine (moles) and therefore the
 concentration of the oxidising agent.
Calculations
 First you calculate the number of moles of thiosulfate
 that were used in the titration,
                        n=cxV

 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,
                        c=n/V

				
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