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.

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

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.

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:

[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

[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.

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)

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.

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

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.

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

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.

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.

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

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.

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.

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.

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.

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.

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).

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.

Above 10oC benzenediazonium chloride decomposes so temperature must be below that.

Reactions of Azo dye

1. With Phenol

2. With phenylamine

For both reactions, alkaline conditions are required. Especially for reaction with phenol since if not alkaline, OH in

phenol reacts interacts with N2+Cl-.