Ocr Chemistry Unit 2: Module 3 - Transition Metals

A d-block element that forms an ion with an incomplete d-sub shell

They do not form ions with incomplete d - sub shells.

Sc only forms Sc 3+ in which all d-orbitals are empty.

Zn only forms Zn 2+ in which all d-orbitals are full.

By losing electrons to form positive ions

Electrons are lost from the 4s orbital first because when occupies, it has a higher energy level than the 3d orbitals.

- Shiny

- Dense

- High melting and boiling points

- When solid, transition metals exist as giant metallic lattices in which electrons are delocalised and move to conduct electricity

1) Nickel is alloyed with Copper for 'silver' coins

2) Titanium is a component for joint replacement parts - for example, the ball and socket joint for hips

3) Iron is cast to make telephone boxes and post boxes

4) Iron is alloyed for use in the construction of buildings and bridges

Compounds in which the transition metal has different oxidation states

Coloured solutions when dissolved in water that frequently catalyse chemical reactions.

Because they occupy the highest energy level when filled, but because the energy levels are so close, 3d electrons can also be lost when an atom forms a stable ion.

A transition element in its highest oxidation state

1) Manganese forms potassium permanganate - KMnO4 - a purple solid used as an oxidising agent in redox titrations 7+

2) Chromium forms potassium dichromate - Kr2Cr2O7, an orange crystalline solid that acts as an oxidising agent in the preparation of aldehydes and ketones from alcohols. 6+

A substance that increases the rate of a chemical reaction by providing an alternative route for the reaction with a lower activation energy

1) By providing a surface for the reaction to occur - Reactants adsorb to the surface of a metal and are held in place while the reaction occurs. After the reaction has occurred, the products are desorbed and the metal remains unchanged.

2) Intermediates - transition metal ions have the ability to change oxidation state by gaining or losing electrons. They then bind to reactants, forming intermediates as part of a chemical pathway with lower activation energy.

1) Haber Process

2) Contact Process

3) Hydrogenation of Alkenes

4) Decomposition of Hydrogen Peroxide, H2O2

-Vanadium (V) oxide catalyst - Vanadium is in it's +5 oxidation state

- SO3 is used in the manufacture of H2SO4 which is then used for fertilisers, detergents, adhesives and explosives.

A reaction in which soluble ions in separate solutions are mixed together to form an insoluble compound that settles out of the mixture as a solid.

Transition metal ions (aq) + OH- (aq) ---> coloured ppt.

Cu 2+

Co 2+

Fe 2+

Fe 3+

Cu2+ (aq) +2OH- (aq) -> Cu(OH)2 (s)

Pale blue solution -> Pale blue ppt.

Co2+ (aq) +2OH- (aq) -> Co(OH)2 (s)

Pink solution -> blue ppt -> beige in air

Fe2+ (aq) +2OH- (aq) -> Fe(OH)2 (s)

Pale green solution -> Green ppt -> Rusty brown on surface in the presence of air

Fe3+ (aq) +3OH- (aq) -> Fe(OH)3 (s)

Pale yellow solution -> Rusty brown ppt.

A transition metal bonded to one or more ligands by coordinate bonds

A molecule or ion that donates a pair of electrons to the central metal ion to form a coordinate bond

A bond in which one of the bonding atoms provides both the electrons in the shared pair

Complex ions.

Water :OH2 0

Ammonia :NH3 0

Chloride :Cl- -1

Cyanide :CN- -1

Thiocyanate :SCN- -1

Hydroxide :OH- -1

A ligand that donates one pair of electrons to the central metal ion to form one coordinate bond.

Species with the same structural formula but with a different arrangement of atoms in space

Cis-trans

Optical

Cis trans isomerism occurs in octahedral complex ions with two types of ligands: four of one and 2 of the other.

A ligand that can donate two pairs of electrons to the central metal ion to form two coordinate bonds

Ethane -1,2-diamine, NH2CH2CH2NH2, en.

Each nitrogen can donate an electron pair.

A ligand that has 6 lone pairs of electrons, each forming a coordinate bond to the central metal ion.

ETDA - ethylenediaminetetraaceticacid

It is used to bind metal ions.

It is known as a chelating agent - it reduces the concentration of metal ions in solution by binding them in a complex.

1) A complex with three molecules or ions of a bidentate ligand - [Ni(en)3]2+

2) A compelx with two molecules or ions of a bidentate ligand and two of a monodentate ligand - [Co(en)2Cl2]

3) A complex with one molecule or ion of a hexadentate ligand - [Cu(ETDA)]2+

Non-superimposable mirror images of one another

They rotate plane-polarised light in opposite directions.

A reaction in which one ligand in a complex ion is replaced with another ligand

1) Cu 2+ ions (aq) + NH3

2) Cu 2+ ions (aq) + HCl (conc)

3) Co 2+ ions (aq) + HCl (conc)

a) [Cu(H2O)6] 2+ (aq) + 4NH3 (aq) -> [Cu(NH3)4 (H2O)2 ]2+ + 4H2O(l)

b) [Cu(H2O)6] 2+ (aq) + 4Cl¯(aq) ——> [CuCl4] 2- (aq) + 6H2O(l)

c) [Co(H2O)6] 2+ (aq) + 4Cl¯(aq) ——> [CoCl4] 2- (aq) + 6H2O(l)

a) (Cu 2+ + NH3)

At the start: Pale blue solution

After a small amount of ammonia: Pale blue ppt as the intermediate Cu(OH)2 forms

After excess of ammonia: Ppt dissolves, deep blue solution forms

Product ( [Cu(NH3)4(H2O)2]2+ ) is DISTORTED OCTAHEDRAL due to the differences in bond lengths: C-O is longer than C-N

b) (Cu 2+ + HCl)

At the start: Pale blue solution

Initially forms: Green solution

Finally forms: Yellow solution

Chloride ligands are larger than H2O ligands so less can fit around the central ion: it forms a tetrahedral compound.

c) (Co2+ + HCl)

At the start: pink solution

End: blue solution

Tetrahedral.

A protein with four polypeptide chains and a haem group with an Fe 2+ ion at the centre, which oxygen can bind to.

When oxygen is bound, the haem group appears red in colour.

As haemoglobin travels round the body it picks up O2 from the lungs and transports it to cells where it is released.

Red blood cells can also pick up CO2 as a waste product from cells, which is then transported to the lungs where it is released.

Carbon monoxide can bind to haemoglobin at the same binding site as Oxygen. But, carbon monoxide will bind more strongly to haemoglobin than oxygen, so if both are present, carbon monoxide will bind irreversibly and less haemoglobin will be left to pick up oxygen. This means tissues will be starved of oxygen and is an example of ligand substitution.

Carbon monoxide is formed by incomplete combustion of carbon-containing fuels.

It is an odourless, colourless gas which is hard to detect (so we have sensors).

At low levels, it can cause headaches, nausea and potential suffocation.

Tobacco can also release carbon monoxide.

An equilibrium constant for the equilibrium that exists between a transition metal ion surrounded by water ligands and the complex that is formed when the same ion has undergone ligand substitution.

Kstab = [Products] / [Reactants] ( raised to the power of their balancing numbers)

Water is left out because it remains virtually constant.

The equilibrium lies to the right. The complex ion is more stable.

The loss of electrons or increase in oxidation number

The gain of electrons or the decrease in oxidation number

Fe 2+ (aq) ----> Fe 3+ (aq) + e-

Fe 2+ ions are easily oxidised in the presence of air or when in contact with an oxidising agent.

MnO4 - is a strong oxidising agent. In acidic solution MnO4- is reduced to Mn 2+:

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

MnO4- is commonly used to oxidise solutions containing Fe 2+ ions.

MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) ----> 5Fe3+ (aq) + 4H2O(l) + Mn2+ (aq)

The transfer of protons from the acid to the base, forming a salt and water. An indicator is used to establish the end-point.

The transfer of electrons from on species to another. An oxidising agent can be titrated against a reducing agent.

An indicator can be used, but sometimes there is already a colour change.

MnO4- (aq) ------------------> Mn2+ (aq)

Deep purple colour of MnO4- masks any other colours present.

Mn2+ appears almost colourless: these ions are normally pale pink but the solution is very dilute and the ions are in such a small concentration, that the solution appears colourless.

- MnO4- can be used to oxidise Fe 2+ ions.

- H2SO4 is used to acidify the solution - HCl is NOT used because it will react with the MnO4- and release chlorine gas.

- Potassium manganate (VII) is put in the burrette and the iron (II) solution in the conical flask.

- As manganate is added to the iron (II) solution, it is decolourised

- The end point is established upon the first permanent pink colour - showing an excess of MnO4- ions.

2S2O3 2- (aq) + I2 (aq) ----> 2I- (aq) + S4O62- (aq)

I2 is reduced to I- ions by aqueous sodium thiosulfate.

Aqueous sodium thiosulfate forms the tetrathionate ion, S4O6 2-.

By titration with a solution of sodium thiosulfate of known concentration.

This titration can be used to determine the concentration of a solution of an oxidising agent (Ex. ClO-, Cu 2+, and Cr2O7 2-) that reacts with iodide ions, I- to produce iodine, I2.

Iodide ions, I-, are added to the oxidising agent.

A redox reaction occurs, producing Iodine, I2.

This Iodine is titrated against a sodium thiosulfate solution of known concentration.

From the results, you can calculate the amount of idoine and thus the concentration of the oxidising agent.

2Cu 2+ (aq) + 4I- (aq) ----> 2CuI (s) + I2 (aq)

Iodide ions are oxidised to iodine: 2I- ---> I2 + 2e-

Cu 2+ ions are reduced to Cu +

Cu2+ + e- -----> Cu +

This reaction produces a light brown/yellow solution and a white precipitate of CuI (s). The precipitate, however, appears light brown due to the iodine in the solution.

The mixture is then titrated against sodium thiosulfate of known concentration. As the iodine reacts, the iodine colour gets paler during the titration. When the colour is a pale straw, a small amount of starch is added to help with the identification of the end point.

A blue/black colour forms, and disappears sharply at the end point because all the iodine has reacted.

The results can then be used to determine the concentration of iodine in the titration mixture and then the concentration of Cu 2+ ions in the original solution.

Estimate the concentration of Cu 2+ ions in an unknown solution or the percentage of copper in alloys such as brass and bronze.

The alloy is first reacted with Nitric acid.

Brass and bronze can both be reacted to produce a solution containing Cu 2+ ions. This then can be reacted with potassium iodide solution. The idoine that forms is then titrated against sodium thiosulfate.