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Formation of Coloured Ions
Colour of an ion can give us an idea of the oxidation number of
the ion.
Ions which have completely full or an empty 3d orbital, the ions
are colorless.
All the other ions do have a color.
Oxidation Numbers of Vanadium
Vanadium shows a sequence of distinctive color changes in its
different oxidation numbers.
Ion Oxidation number Color of solution V2+ 2+ Purple V3+ 3+
Green
VO2+ 4+ Blue VO2+ 5+ Yellow
Reactions of Copper
When exposed to air in presence of CO2 it oxidizes very slowly
to form a green film of basic copper (ii) carbonate.
CuCO3.Cu(OH)2
Copper reacts with acids only under oxidizing conditions. Copper
does not react with HCl. Copper reacts with dilute and concentrated
nitric acid to form
blue crystalline solid copper (ii) nitrate. Copper is oxidized
much more rapidly with concentrated nitric
acid.
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Copper reacts with hot, concentrated H2SO4 to give copper (ii)
sulfate which is bright blue in color.
Making CuI
2CuSO4 + 4KI ---- 2CuI + 2K2SO4 + I2
White solid
Copper (ii)
Copper (ii) ions are surrounded by 6 water molecules forming a
ligand which is:
[Cu(H2O)6]2+ Hexaaquacopper (ii)
When ammonia or NaOH is added to the ligand a precipitate of
hydrated copper(ii) hydroxide form:
[Cu(H2O)6]2+ + 2NH3 ----- [Cu(OH)2(H2O)4] + 2NH4+ Pale blue
ppt
The pale blue ppt wont react with excess NaOH. But it does react
with excess NH3.
[Cu(OH)2(H2O)4] + 4NH3 ------- [Cu(NH3)4(H2O)2] + 2OH- +
2H2O
Deep blue
When concentrated HCl is added to [Cu(H2O)6]2+: [Cu(H2O)6]2+ +
4Cl- ---- [CuCl4]2- + 6H2O Yellow-green solution
Ligands
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Ligands are compounds that have at least one lone pair of
electrons.
There are divided into groups according to the number of lone
pairs that they can donate.
For example: Monodentate ligand donates one lone pair of
electrons (eg:
aqua [hexa], amine [NH3]. Bidentate ligands donate two lone
pairs of electrons.(eg: 1,2
diethanedioate) Polydentate ligands can donate many lone pairs
of electrons.
The ligands that have no charge are called neutral ligands. (eg:
aqua and amine)
Naming Transition Metal Aqua Complex
An example of a transition aqua complex would be:
[Cr(H2O)6]3+
Cr is the transition metal of the ion. H2O is the ligand. 6 is
the number of dative covalent bonds. 3+ is the charge of the
complex. To name this type of a complex:
Check the number of ligands. Check the type of ligand. Check the
transition metal. Check the charge. If charge is positive then use
real name of metal and if
negative, use the ion name of the metal. The name of that
complex would be hexa aqua chromium (iii) ion.
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Why is the solution of an aqua complex ion acidic?
[Fe(H2O)6]3+ + H2O [Fe(OH-)(H2O)5]2+ + H3O+
Water acts as a base removing a proton from one of the
ligands.
Removal of a proton from a ligand is called deprotonation.
Since the proton is removed, the charge reduced by 1.
Colours of Ions
Cr3+ - grey green Cu2+ - pale blue
Fe2+ - pale green Co2+ - blue
Ni2+ - emerald green
Mn2+ - offwhite
Zn2+ - white
Fe3+ - red brown
Reactions of Transition Metal Aqua Complexes
With minimum NaOH/NH3
Eg 1: [Cr(H2O)6]3+ + 3OH- ---- [Cr(H2O)3(OH)3] + 3H2O
Grey green ppt
Eg 2: [Ni(H2O)6]2+ + 3OH- ----- [Ni(OH)2(H2O)4] + 2H2O
Emerald green ppt
Other colors of ppts formed:
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[Cu(OH)2(H2O)4] pale blue
[Mn(OH)2(H2O)4] off white
[Zn(OH)2(H2O)4] white
Of these precipitates, 2 are amphoteric which are:
1. [Cr(OH)3(H2O)3] 2. [Zn(OH)2(H2O)4]
They can have both acidic and basic nature.
These are the only ppts that will react with excess NaOH and
dissolve in it.
For 1. [Cr(OH)3(H2O)3] + 3OH- ----- [Cr(OH)6]3- + 3H2O
Green solution
For 2. [Zn(OH)2(H2O)4] + 2OH- ------ [Zn(OH)4(H2O)2]2- +
2H2O
White
Zinc never forms an octahedral complex. It forms a tetrahedral
one.
Even though only 2 ppts dissolve in excess of NaOH, 5 ppts
dissolve in excess of NH3.
The reactions are:
[Cr(H2O)3(OH)3] + 6NH3 ----- [Cr(NH3)6]3+ + 3H2O + 3OH-
Violet
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[Ni(H2O)4(OH)2] + 6NH3 ------- [Ni(NH3)6]2+ + 4H2O + 2OH-
Lavender blue
Zinc and copper do not undergo complete reaction.
[Cu(H2O)4(OH)2] + 6NH3 ------- [Cu(NH3)4(H2O)2]2+ + 2H2O +
2OH-
Royal blue
[Zn(H2O)4(OH)2] + 6NH3 -------- [Zn(NH3)4(H2O)2]2+ + 2H2O +
2OH-
Colourless solution
Driving force behind ligand substitution is the positive entropy
change.
Few atoms are replaced by more atoms so complexity increased
increasing randomness.
Chromium
Ion Oxidation number Colour Cr2O72-
(dichromate(iv)) +6 Orange
CrO42-(chromate(iv)) +6 Yellow Cr3+ +3 Green Cr2+ +2 Blue
When zinc and HCl are added to CrO42- , color changes from
yellow to green to blue.
This is because zinc is a reducing agent.
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REDOX TITRATIONS
KMnO4 and Fe2+
Fe2+ is oxidized to Fe3+ Fe2+ ----- Fe3+ + e-
MnO4- is reduced to Mn2+ MnO4- + 8H+ + 5e- ----- Mn2+ + 4H2O
Overall equation would be: MnO4- + 8H+ + 5Fe2+ -------- Mn2+ +
4H2O + 5Fe3+
(purple) (colourless) (colourless) A measured volume of Fe2+ is
pipetted into the conical flask. Its acidified with a small amount
of dilute sulfuric acid. KMnO4 is added slowly from the burette.
Swirl the conical flask to allow mixing. Mixture remains colorless
until all Fe2+ is oxidized to Fe3+. Then when one drop of KMnO4 is
added the mixture turns pale
pink. We dont need an indicator for this titration.
KMnO4 and Ethanedioic Acid (H
2C
2O
4)
MnO4- is reduced to Mn2+ MnO4- + 8H+ + 5e- ------ Mn2+ +
4H2O
C2O42- is oxidized to CO2 C2O42- ---- 2CO2 + 2e-
Overall equation would be: 2MnO4- + 16H+ + 5C2O42- --------
2Mn2+ + 8H2O + 10CO2
(purple) (colorless) (colorless)
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A measured volume of H2C2O4 is pipetted into a conical flask.
KMnO4 is added into the solution from a burette. Since the reaction
is very slow, H2C2O4 need to be heated to 60oC
before starting the reaction. Mn2+ is the catalyst for this
reaction. At the beginning there is none of Mn2+ and hence the
heating is
required. As Mn2+ increases the rate of reaction increases. This
is called autocatalysis. After all C2O42- is oxidized to CO2 and
one drop of KMnO4 is
added a pale pink color will be seen.
Na2S
2O
3 and I
2
S2O32- gets oxidized to S4O62- 2S2O32- ---- S4O62- + 2e-
I2 gets reduced to I- I2 + 2e- ---- 2I-
Overall equation would be: 2S2O32- + I2 ------- S4O62- + 2I-
Iodine does not well dissolve in water and hence its dissolved
in potassium iodide.
A measured volume of iodine solution is pipetted into a conical
flask.
Sodium thiosulfate is added into the solution from the burette.
Iodine solution is yellow-brown in color. It becomes colorless at
end point. But it is hard to identify when it is fully colorless.
So starch solution is added when iodine is very pale yellow in
color.
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Starch reacts with iodine to form a blue-black color. As iodine
is used up, the solution becomes colourless.
Starch should not be added at the beginning since an insoluble
starch-iodine complex form reducing the accuracy of final
reading.
ClO- and I-
ClO- reduces to form Cl- ClO- + 2H+ + 2e- ------ Cl- + H2O
I- is oxidized to form I2
2I- ----- I2 + 2e-
Overall equation would be:
ClO- + 2H+ + 2I- ------ Cl- + H2O + I2
ClO- is reacted with excess I- ions in potassium iodide. The
iodine formed makes the resulting solution have a yellow
brown color. We can estimate the amount of I2 formed by
titrating the
solution against a standardized solution of Na2S2O3. Then we can
determine the amount of ClO- present in a sample.
Cu2+ and I-
Excess of potassium iodide is reacted with a known volume of
solution containing Cu2+ ions.
CuI forms a white precipitate and the I2 formed remains in the
solution.
2Cu2+ + 4I- ---- 2CuI + I2
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Determine the amount of I2 produced by titrating the resulting
solution with a standardized solution of Na2S2O3.
Now calculate the amount of Cu2+ present in the sample.
Finding the amount of Copper in an Alloy
Dissolve a weighed amount of the alloy in concentrated nitric
acid. The nitric acid will oxidize I- ions to I2 so some sodium
carbonate
is added to remove acid. But this will cause precipitation of
copper ions. So minimal amount of ethanoic acid is added which is
enough to
keep the copper ions and to prevent oxidation of I- to I2.
Excess potassium iodide is added into the mixture. The amount of
iodine formed can be found by back-titrating with
standardized solution of Na2S2O3. Then the mass of Cu in the
alloy can be found.
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ELECTROCHEMICAL CELLS
o Zinc is the more reactive element so electrons zinc loses
electrons and copper gains electrons.
o Therefore zinc electrode is the anode and copper electrode is
the cathode.
o As a result of electrons flowing from zinc to copper, there is
a potential difference.
o The maximum potential difference (emf) can be measured using a
high resistance voltmeter.
o Written below the diagram is the half-cell notation. o The
rule in writing such a notation is:
Left-hand electrode| electrolyte on left || right-hand electrode
| electrolyte on right
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Standard Electrode Potentials
Since a voltmeter can only be connected when 2 half cells are
combined, it is impossible to measure the emf of one half-
cell.
We usually measure emf of a cell and attribute part of the emf
to each electrode reaction.
To measure the emf of a half-cell, we can use a reference-half
cell.
This half-cell has 0 potential. It is under standard conditions
of 1 atm, 298 K and in solution
concentrations of 1moldm-3. Reference-half cell is connected to
another half- cell with
unknown emf and the emf recorded will be the emf of the half
cell.
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Benzene
Benzene is represented by:
Benzene burns with a yellow smoky flame. Its the test for
benzene.
Properties
Benzene has a planar hexagonal structure. Benzene has a pi
bonding electron cloud.
Pi bonds are weaker than sigma bonds therefore benzene must be
assumed to be very reactive.
But actually benzene is inert due to delocalized pi bonding
which distributes electrons throughout the structure making it
stable.
Benzene favors electrophilic substitution.
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Benzene doesnt favor addition because this requires breaking pi
bonds.
Breaking pi bonds in benzene requires high energy.
Reactions of Benzene
1. Combustion. In a full supply of oxygen, benzene burns to give
CO2 and H2O.
C6H6 + O2 ------ CO2 + H2O 2. Reaction with hydrogen. Reagent:
hydrogen Conditions: 150oC, Raney Nickel catalyst
Cyclohexane 3. Reaction with halogens Benzene undergoes addition
reactions with bromine in the
presence of UV light. Light is needed to break the bonds within
a halogen to make free
radicals. Reagent: Bromine Condition: UV light
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1, 2, 3, 4, 5, 6 hexabromocyclohexane
4. Reaction with fuming sulfuric acid Reagent: concentrated
sulfuric acid Concentrated sulfuric acid contains sulfur trioxide.
Benzene reacts with sulfur trioxide since sulfur trioxide is an
effective electrophile.
Product is benzene sulfonic acid. 5. Reaction with halogens (in
the dark) Benzene undergoes substitution reactions with halogens in
the
dark. Benzene does not react with halogens unless there is a
catalyst
of iron (iii) bromide (halogen carrier). Benzene, bromine and
FeBr3 are refluxed together to form
bromobenzene.
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Instead of FeBr3, only Fe can also be added since it can react
with Br2 to form FeBr3
2Fe + 3Br2 ------- 2FeBr3 6. Nitration of Benzene Benzene does
not react with concentrated nitric acid alone. Benzene can react
with a mixture of concentrated HNO3 and
concentrated H2SO4. The reaction mixture is in a round bottom
flask and the flask
is kept in a cold water bath. This is because the reaction is
very exothermic. The reaction mixture should be kept below
500C.
If the temperature rises above 50oC, other substitution
reactions occur.
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7. Alkylation This involves adding an alkyl group in a benzene
ring. Reagent: halogenoalkane Catalyst: halogen carrier (AlCl3)
Benzene is fluxed with the reagent and the catalyst. CH3+ is the
electrophile.
8. Acylation This is the reaction of benzene with an acyl
chloride. Reagent: acyl chloride. Catalyst: AlCl3 Condition:
heat.
Product can be reduced to get secondary alcohol.
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Acylation and Alkylation reactions are called Friedel-Crafts
reaction.
Reactions of Phenols
1. With bromine water Phenol in water must be added to bromine
water. Multi- electrophilic substitutions takes place. No need of
heating or halogen carrier. Bromine is decolourised. Product:
2,4,6- tribromophenol (white ppt with an antiseptic
smell)
2. With dilute nitric acid Unlike benzene, phenol doesnt need
H2SO4 for it to react with
nitric acid. This is a multi-substitution reaction. Product: 2,
4 ,6- trinitrophenol.
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2, 4, 6- trinitrophenol (picric acid)
(White ppt)
Synthesis of Phenylamine
For this we need nitrobenzene and hydrogen. Nitrobenzene is
reduced by a mixture of tin and conc.HCl. Mixture of nitrobenzene
and tin is heated under reflux. When bubbles stop forming (no more
H2 forms), start adding HCl
from the top. Tin (ii) ions are oxidized to tin (iv) ions.
Sn2+ ------ Sn4+ + 2e- -NO2 group is reduced to NH2 group.
C6H5NO2 + 6H+ + 6e- ------- C6H5NH2 + 2H2O The flask is then
cooled. NaOH is added because: Phenyl amine reacts with HCl to form
phenyl ammonium chloride. NaOH is added to move equilibrium
backwards. Water is then added and steam distillation is used to
separate
the mixture of phenyl amine and water. The distillate collected
is cloudy because its an emulsion of
phenyl amine and water.
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Phenyl amine is very soluble in water. To make it less soluble,
NaCl is added. This is called salting out. The mixture is
transferred to a separating funnel and
ethoxyethane is added. The mixture is shaken. Pressure is
released occasionally by opening the tap. The lower aqueous layer
is run off into a small beaker. Ethoxyethane is run off into a
conical flask. Potassium hydroxide is added to dry the ethoxyethane
extract. KOH is better than CaCl2 because:
KOH removes traces of HCl. Phenyl amine and ethoxyethane are
separated by distillation.
Manufacture of Paracetamol
Phenol is nitrated using H2SO4 and sodium nitrate. A mixture of
2-nitrophenol and 4-nitrophenol forms. The 2 isomers are separated
by fractional distillation. 4-nitrophenol has the higher b.p
because more effective H-
bonding occurs between 4-nitrophenol molecules. 4-nitrophenol is
reduced to 4-aminophenol using
tetrahydridoborate(iii).
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4-aminophenol is reacted with ethanoic anhydride to give
paracetamol.
When 4-aminophenol reacts with ethanoyl chloride, paracetamol
forms along with HCl.
Making Azo dye
Make nitrous oxide by mixing sodium nitrite and dilute HCl. The
mixture should be kept in an ice cold water bath at below
5oC. Nitrous oxide reacts with amines above 5oC to give the
respective
alcohol, nitrogen gas and H2O. But below 5oC, phenyl amine
reacts with nitrous oxide to give
benzenediazonium chloride and water.
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Above 10oC benzenediazonium chloride decomposes so temperature
must be below that.
Reactions of Azo dye
1. With Phenol
2. With phenylamine
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For both reactions, alkaline conditions are required. Especially
for reaction with phenol since if not alkaline, OH in
phenol reacts interacts with N2+Cl-.