LATTICE ENERGY

Lattice energy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions. Lattice energy is always exothermic, the more exothermic, the stronger the lattices ionic bond. The standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions. The first electron affinity is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous 1- ions under standard conditions. (exothermic) The second electron affinity is the enthalpy change when 1 mole of electrons is added to one mole of gaseous 1- ions to form 1 mole of gaseous 2- ions under standard conditions. (2nd and 3rd electron affinities are always endothermic because there must be an input of energy to overcome the repulsive forces between the electron and the negative ion) Born-Haber Cycle

1. Write its elements in standard conditions on the left-hand side in the amount as needed for the lattice (e.g. O2 (g) is allowed).

2. Point downwards to the bottom for Hf of the lattice.

3. Continue the steps as needed to form the ions in the gaseous state.

Point the arrows accordingly whether the reaction is endothermic or exothermic.

Pay attention to the ionisation energy and electron affinity. (e.g. to make Mg2+, the 1st and 2nd ionisation energies must be taken into account; to make 2Cl(g) from Cl2 (g), calculate 2Hat; 2Cl(g) + 2e- 2Cl-(g) requires 2Hea1)

4. Once the elements are in its ionic gaseous state and with the right amount needed to form the lattice, point downwards to the bottom to form Hlatt.

Factors affecting the value of lattice energy: Lattice energy arises from the electrostatic force of attraction of oppositely charged ions when the crystalline is formed.

More exothermic as the size of the ion decreases.

Ions with the same charge have a lower charge density if their radius is larger. A higher charge density results in stronger electrostatic forces of attraction in the ionic lattice, so more exothermic.

More exothermic as the charge on the ion increases.

Ion polarisation The positive charge on the cation in an ionic lattice may attract the electrons in the anion towards it resulting in ion polarisation. Polarising power: The ability of a cation to attract electrons and distort an anion. Factors affecting ion polarisation: The degree of polarisation of an anion depends on:

The charge density of the cation. The ease with which the anion can be

polarised its polarisibility. An anion is more likely to be polarised if:

The cation is small The anion is large The cation has a charge of 2+ or 3+ The anion has a charge of 2- or 3-

Many ionic compounds have some covalent character due to ion polarisation. Many covalent compounds have some degree of charge separation, i.e. they are polar, due to bond polarisation. Soluble and insoluble are only relative terms. No metallic salts are absolutely insoluble in water, but dissolves to a very small extent. If salts were completely insoluble they could not have a value for Hsol.

The thermal stability of Group II carbonates and nitrates The Group II carbonates decompose to their oxides and CO2 on heating. The more positive the enthalpy change, the more stable is the carbonate relative to its oxide and CO2. The relative stabilities of these carbonates increases down the group:

MgCO3 < CaCO3 < SrCO3 < BaCO3

(least positive Hr) (most positive Hr)

The carbonate ion has a relatively large ionic radius so it is easily polarised by a small highly charged cation.

The ionic radius of the Group II cations increase down the group (the smaller the better it is at polarising)

The greater the polarisation, the easier it is to weaken a C-O bond in the carbonate and form CO2 and the oxide on heating.

The same goes for Group II nitrates: these decompose to form the oxide, NO2 and O2. The order of stability is:

Mg(NO3)2 < Ca(NO3)2 < Sr(NO3)2 < Ba(NO3)2 (most polarised) (least polarised)

Enthalpy changes in solution The enthalpy change of solution is the energy absorbed or released when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution. (can be exothermic or endothermic) A compound is likely to be soluble in water only if Hsol is negative or has a small positive value. Substances with large positive values of Hsol are relatively insoluble. The enthalpy change of hydration is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution. (always exothermic because it forms bonds with water) ions (g) ions (aq) When an ionic solid dissolves in water, bonds are formed between water molecules and the ions (ion-dipole bonds).

The - oxygen atoms in water molecules are attracted to the positive ions in the ionic compound. The + hydrogen atoms in water molecules are attracted to the negative ions in the ionic compound. The energy released in forming ion-dipole bonds is sufficient to compensate for the energy that must be put in to separate the ionic bonds in a lattice. The value of Hhyd is more exothermic for ions with higher charge density. Hlatt + Hsol = Hhyd (add the Hhyd values for both cations and anions) Energy level diagram to calculate for the enthalpy change in solution

1. Write the gaseous ions. 2. Point downwards for Hlatt to form the ionic

solid, then downwards again for Hsol to form the ions (aq).

3. Point downwards from the gaseous ions to the bottom for Hhyd to form the ions (aq).

The solubility of Group II sulfates The solubility decreases as the radius of the metal ion decreases. The order of the solubility is:

MgSO4 > CaSO4 > SrSO4 > BaSO4

Change is hydration enthalpy down the group Smaller ions with the same charge have

greater enthalpy changes of hydration. So the enthalpy change of hydration gets

less exothermic down Group II. This decrease is relatively large down the

group and it depends entirely on the increase in the size of the cation since the anion is unchanged.

Change in lattice energy down the group

Lattice energy is greater if the ions with the same charge forming the lattice are small.

The lattice energy decreases down Group II. The lattice energy is inversely proportional

to the sum of radii of the anion and cation. The decrease in lattice energy is relatively

smaller down the group and it is determined more by the size of the much larger sulfate ion than the size of the cations.

Difference in enthalpy change of solution of Group II sulfates

The lattice energy of the sulfates decreases by relatively smaller values down the group.

The enthalpy change of hydration decreases by relatively larger values down the group.

So applying Hesss law, the value of Hsol gets more endothermic down the group.

So the solubility of the Group II sulfates decreases down the group.

ELECTRODE POTENTIALS

The standard hydrogen electrode consists of: H2 gas at 101kPa pressure (1 atm) H+ ions of concentration 1.00 mol dm-3, in

equilibrium with H2 gas A platinum electrode

The more positive the electrode potential, the easier it is to reduce the ions. So the metal on the right is relatively unreactive and is a relatively poor reducing agent. The more negative the electrode potential, the more difficult it is to reduce the ions. So the metal on the right is relatively reactive and is a relatively good reducing agent. Combining half-cells Standard electrode potential (E):

Concentration of ions at 1.00 mol dm-3 A temperature of 25C (298 K) Pressure of 1 atm (101 kPa)

The standard electrode potential for a half-cell is the voltage measured under standard conditions with a standard hydrogen electrode as the other half-cell. Half-cells are connected together using:

Wires connecting the metal rods in each half-cell to a high-resistance voltmeter; the electrons flow round this external circuit not through the electrolyte solution from the more negative electrode potential to the more positive electrode.

A salt bridge to complete the chemical circuit allowing the movement of ions between the two half-cells so that ionic balance is maintained; a salt bridge does not allow the movement of electrons.

A salt bridge can be made from a strip of filter paper (or other inert porous material) soaked in a saturated solution of KNO3. Metal/metal ion half-cell

Reduction takes place at the positive terminal of the cell (the half-cell with the more positive E value)

Oxidation takes place at the negative terminal of the cell (the half-cell with the more negative E value)

Non-metal/non-metal ion half-cell Electrical contact with the solution is made by using platinum wire or platinum foil as an electrode. The redox equilibrium is established at the surface of the platinum. The platinum electrode is inert so plays no part in the reaction. The platinum must be in contact with both the element and the aqueous solution of its ions. Ion-ion of the same element in different oxidation states half-cell The concentration of each ion present is 1 mol dm-3. Using E values Standard cell potential: the difference in standard electrode potential between two half-cells. In order to calculate the cell voltage, always subtract the less positive E value from the more positive E- value. The more positive the value of E, the greater is the tendency for the half-equation to proceed in the forward direction. The more positive the value of E, the easier it is to reduce the species on the left of the half-equation.

A reaction is said to be feasible if it is likely to occur. Will ____ oxidise __ ions to __ ions?

1. Compare the E values for both half-cell reactions.

2. The reaction with the more positive E values will go to the forward reaction, while the other reaction will go in the reverse direction.

3. Check whether the combined equation satisfies the question.

4. If so, then the reaction is said to be feasible. A reaction will occur if the E value is positive. How does E vary with ion concentration?

If the E values of the two half reactions involved under non-standard conditions differ by more than 0.30 V, then the reaction predicted by the E values is highly likely to occur. The feasibility of a reaction based on E values is no guarantee that a reaction will proceed quickly. It only tells us that a reaction is possible, and that the reverse reaction does not occur. Some reactions are feasible, but they proceed so slowly that they do not seem to be taking place. It is the rate of reaction rather than then value of E which is determining the lack of reactivity. Cells and batteries Primary cells use redox reactions until the reactants reach a low concentration and the voltage of the cell declines. Secondary cells can be recharged by passing an electric current through them. The products are then changed back to reactants so the cell can function again.

Advantages of fuel cells: Water is the only product made. They produce more energy. They are very efficient. There are no moving

parts where energy is wasted as heat. Limitations to hydrogen-oxygen fuel cells:

High cost: the materials used to make the electrodes and membrane are expensive.

Manufacturing of fuel cells involves the production of toxic by-products

Storage of hydrogen: high-pressure tanks are needed in order to store a sufficient amount of fuel. At present, refueling has to be done more often compared to petrol.

Manufacturing hydrogen: the hydrogen needed for fuel cells can only be produced cheaply by using fossil fuels.

Fuel cells do not work well at low temperatures.

Electrolysis of molten electrolytes When pure molten ionic compounds containing 2 simple ions are electrolysed, a metal is formed at the cathode and a non-metal at the anode. Electrolysis of aqueous solution Aqueous solution of electrolytes contain more than one cation and more than one anion. The cation in the half-equation with the most positive E value will be discharged. At the anode, the ease of discharge of anions follows the order: SO42- (aq) NO3- (aq) Cl- (aq) OH- (aq) Br- (aq) I- (aq) --------------------------------------------------------

increasing ease of discharge/oxidation

Electrolysis products and solution concentration An ion, Z, higher in the discharge series (more difficult to discharge) may be discharges in preference to one below it if Z is present at a relatively high concentration than normal. For this to be possible, the E values of the competing ions are usually less than 0.30V different from each other.

Electrode potential

Concen

tration Reactant Product

+ +

Quantitative electrolysis Faradays first law: the mass of substance produced at an electrode is proportional to the quantity of electricity passed Faradays second law: the number of Faradays required to discharge 1 mole of an ion at an electrode is equal to the charge on the ion. (the number of moles of the electrons needed to reduce/oxidise 1 mole of a substance) 1 F = 96 500 C mol-1

Finding for Avogadros constant: L = charge on 1 mole of electronscharge on 1 electron (1.60 10!!" C)

How to find the charge on 1 mole of e- 1. Weigh the pure copper anode and pure

copper cathode separately. 2. Do an electrolysis experiment. 3. Remove the cathode and anode and wash

and dry them with distilled water and propanone.

4. Reweigh the cathode and anode. 5. Measure the decrease in mass of the anode. 6. Use proportions to find the charge required

for 1 mole of the element. 7. Calculate for the charge on 1 mole of

electrons by using proportions again.

IONIC EQUILIBRIA

Strong acids ionise completely in water. Weak acids only ionise to a small extent in water. The ionic product of water, Kw

H2O (l) H+ (aq) + OH- (aq) Kw at 298 K = 1.00 10-14 mol2 dm-6 pH = log10[H+] [H+] = 10-pH log10 [H+] log10 [OH-] = 14 The pH of strong acids The concentration of hydrogen ions in solution is approximately the same as the concentration of the acid. Diluting the acid 10 times reduces the value of the H+ ion concentration by one-tenth and increases the pH by a value of one. The pH of strong bases The concentration of hydroxide ions in a solution is approximately the same as the concentration of the base. H! = K![OH!] Weak acids Ka is the acid dissociation constant.

HA (aq) H+(aq) + A- (aq) K! = H! [A!][HA] = H! ![HA]

A high value of Ka indicates that the position of equilibrium lies to the right. The acid is almost completely ionised. A low value for Ka indicates that the position of equilibrium lies to the left. The acid is only slightly ionised and exists mainly as HA molecules and comparatively few H+ and A- ions.

pKa = log10 Ka

The less positive the value of pKa, the more acidic is the acid. In order to calculate the value of Ka, we make 2 assumptions:

We ignore the concentration of [H+] produced by the ionisation of the water molecules present in the solution. This is reasonable because the ionic product of water (Kw) is negligible compared with the values for most weak acids.

We assume that the ionisation of the weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as that of the original acid.

Acid-base indicators The acid and conjugate base have different colours. HIn H+ + In-

un-ionised indicator conjugate base

colour A colour B

Adding an acid to this indicator solution shifts the position of equilibrium to the left.

Adding an alkali shifts the position of equilibrium to the right. There are now more ions of colour B.

The colour of the indicator depends on the relative concentrations of HIn and In-. The colour of the indicator during a titration depends on the concentration of H+ ions present.

The midpoint of the sharp fall in the titration graph corresponds to the point at which the H+ ions in the acid have exactly reacted with the OH- ions in the alkali; this is the end-point of the titration. For weak acids with weak bases: There is no sharp fall in the graph line. No acid-base indicator is suitable to determine the end-point of this reaction. Buffer solution A buffer solution is a solution in which the pH does not change significantly when small amounts of acids or alkalis are added. A buffer solution is used to keep pH (almost) constant. One type of buffer solution is a mixture of a weak acid and one of its salts. The weak acid stays mostly in the unionised form and only gives rise to a low concentration of ions in solution. The pH of a buffer solution depends on the ratio of the concentration of the acid and the concentration of its conjugate base. If this does not change very much, the pH changes very little. CH3COOH (aq) H+ (aq) + CH3COO- (aq) (high conc. of ethanoic acid and ethanoate ion) When H+ ions are added to the buffer solution:

P.O.E. shifts to the left The large reserve supply of ethanoate ion

and ethanoic acid ensures that the conc. in solution does not change significantly.

So the pH does not change significantly.

When OH- ions are added to the buffer solution: The added OH- combine with H+ to form H2O P.O.E. shifts to the right The large reserve supply of ethanoate ion

and ethanoic acid ensures that the conc. in solution does not change significantly.

So the pH does not change significantly. If very large amounts of acid or alkali are added, the pH will change significantly. Buffer solutions which resist changes in pH in alkaline regions are usually a mixture of a weak base and its conjugate acid. pH of buffer solution: pH = pK! + log!" [salt][acid] Uses of buffer solutions:

Industrial processes including electroplating Manufacture of dyes Treatment of leather To calibrate pH metres

In humans, blood pH is kept constant between 7.35 7.45 by a number of buffers:

HCO3- Haemoglobin and plasma proteins H2PO4- and HPO42-

CO2 + H2O H+ + HCO3- (catalysed by carbonic anhydrase)

If H+ ion concentration increases: P.O.E. shifts to the left. Reduces the concentration of H+ in the

blood and keeps the pH constant. If H+ ion concentration decreases:

P.O.E. shifts to the right. Increases the concentration of H+ in the

blood and keeps the pH constant. Equilibrium and solubility A solution is saturated when no more solute dissolves in it. Solubility product Ksp is the product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K, raised to the power of their relative concentrations.

The idea of solubility product only applies to ionic compounds which are only slightly soluble. The smaller the value of Ksp, the lower is the solubility of the salt. If Ion product > Ksp, precipitation occurs. The common ion effect The common ion effect is the reduction in the solubility of a dissolved salt achieved by adding a solution of a compound which has an ion in common with the dissolved salt. This often results in precipitation.

For example, AgCl (s) Ag+ (aq) + Cl- (aq)

Then add NaCl:

The Cl- is common to both NaCl and AgCl. P.O.E. shifts to the left AgCl is precipitated because [Ag+][Cl-] > Ksp

The solubility of an ionic compound in aqueous solution containing a common ion is less than its solubility in water.

REACTION KINETICS

Methods for following the course of a reaction Sampling This method involves taking small samples of the reaction mixture at various times and then carrying out a chemical analysis on each sample. Samples are removed at various times and quenched to stop or slow down the reaction. E.g. by cooling the sample in ice. Continuous A physical property of the reaction mixture is monitored over a period of time. The rate of reaction Rate of reaction = k [A]m[B]n

k = rate constant m and n = orders of the reaction (usually 0, 1, 2, 3) The order of reaction with respect to a particular reactant is the power to which the concentration of reactant is raised in the rate equation. How to identify the order of a reaction:

Plot a graph of reaction rate against concentration of reactant

Plot a graph of concentration of reactant against time

Deduce successive half lives from graphs concentration against time

Graphs of reaction rate against concentration It is very rare to obtain an order with respect to a particular reagent higher than second order.

Zero-order The graph is a horizontal straight line. The rate does not change with concentration. First-order The graph is an inclined straight line going through the origin. The rate is directly proportional to the concentration. Second-order The graph is an upwardly curved line. The rate is directly proportional to the square of the concentration. Graphs of concentration of reactant against time Zero-order The graph is a descending straight line. The rate of reaction is the slope of the graph. First-order Declines in a shallow curve. Second-order Declines in a deeper curve which then levels out.

Half-life and reaction rates Half-life, t1/2, is the time taken for the concentration of a reactant to fall to half of its original value. Zero-order Has successive half-lives which decrease with time. First-order Has a half-life which is constant. Second-order Has successive half-lives which increase with time. Calculating k from half-life For a first-order reaction, half-life is related to the rate constant by the expression k = 0.693t!/! Kinetics and reaction mechanisms

A reactant that appears in the chemical equation may have no effect on the reaction rate.

A substance which is not a reactant in the chemical equation can affect reaction rate.

Rate-determining step: the slowest step in a reaction mechanism. If a substance does not appear in the overall rate equation, it does not take part in the rate-determining step. (and vice versa) The mechanism is not deduced from the kinetic data. The kinetic data simply shows us that a proposed reaction mechanism is possible. The slow step may not involve molecules A and B but the intermediate which is involved in the slow step which is derived from A and B, then A and B will appear in the rate reaction. If a molecule is not involved until after the rate-determining step, then the rate reaction does not depend on the concentration of that molecule.

Catalysis Homogeneous catalysis occurs when the catalyst is in the same phase as the reaction mixture. Homogeneous catalysis often involves changes in oxidation number of the ions involved in the catalysis. Ions of transition elements are often good catalysts because of their ability to change oxidation number. Heterogeneous catalysis occurs when the catalyst is in a different phase to the reaction mixture. A redox reaction occurs in heterogeneous catalysis. Energy must be supplied to overcome the negative charges of the two reactants by using a catalyst with a positive charge. Heterogeneous catalysis often involves gaseous molecules bonding to the surface of a solid catalyst in a process called adsorption. Transition elements such as nickels are good at chemisorbing hydrogen gas. The weak van der Waals forces link the hydrogen molecule to the nickel, causes the weakening of the H-H covalent bond. Heterogeneous catalysis in the Haber process The iron catalyst works by allowing H2 and N2 molecules to come close together on the surface of the iron. They are then more likely to react.

1. Diffusion of H2 and N2 to the surface of Fe. 2. Adsorption. The bonds formed between the

reactants and Fe are strong enough to weaken the covalent bonds within H2 and N2 so they can react with each other but weak enough to break and allow the products to leave the surface.

3. Reaction of H2 with N2 4. Desorption. The bonds between ammonia

and the surface of Fe weaken and broken. 5. Diffusion. Ammonia diffuses away from the

surface of Fe.

Transition elements in catalytic convertersThe honeycomb structure inside the catalytic converter contains small beads coated with platinum, palladium or rhodium which act as heterogeneous catalysts.

The steps are: Adsorption of NO and CO onto the catalyst surface Weakening of the covalent bonds within NO & CO Formation of new bonds to form N2 and CO2 Desorption of N2 and CO2 from the catalyst surface.

GROUP IV

The metallic character of the elements increases down the group. Metalloids differ from metals in two ways:

Metalloids have very small electrical conductivities at room temperature; metals conduct electricity very well.

The electrical conductivity of metalloids increases with increase in temperature; the conductivity of metals decreases with increase in temperature.

Tin and lead are less reactive than metals from Group I and II. Melting points C, Si, Ge have a similar structure to diamond (giant molecular), the difference in melting points can be related to their bond energies. Sn and Pb have a metallic structure at room temperature. The ions are held together by the electrostatic attraction between their positive charges and the delocalised electrons. Sn and Pb have relatively large ions, so the metallic bonding is relatively weak and the melting points are relatively low. Electrical conductivity Carbon (diamond) does not conduct electricity because of the strong covalent bonds so no delocalised electrons free to move around. Carbon (graphite) does conduct electricity. Si and Ge conduct to a small extent. The weaker bonding allows some electrons to move out of position especially if there are traces of other contaminating atoms in the lattice, but the electrons are not delocalised. Sn and Pb are conductors because of the free delocalised electrons.

The tetrachlorides The tetrachlorides all have:

The general formula XCl4 Simple covalent molecules A tetrahedral structure They are all volatile liquids at room temp. Low boiling points

*arrow head always pointing to the largest/the most ___

Thermal stability

CCl4, SiCl4, GeCl4 are stable at high temp. They do not decompose on heating.

SnCl4 decomposes readily on heating. SnCl! l SnCl! s + Cl!(g) PbCl4 decomposes explosively.

Reaction with water

All the tetrachlorides except CCl4 are hydrolysed by water. SiCl! l + 2H!O l SiO! s + 4HCl (g)

The ease of hydrolysis increases from SiCl4 to PbCl4 as the metallic nature increases.

The oxides

Oxidation state +2 Oxidation state +4 CO CO2 SiO SiO2 GeO GeO2 SnO SnO2 PbO PbO2

Most of the Group IV oxides do not decompose on heating because of the strong bonding in their structure.

Element Symbol Metal or non-metal? MP /C Electrical Conductivity Bond

Bond energy/ kJ mol-1

Carbon C Non-metal 3550 Non-conductor Strong covalent 350 Silicon Si Metalloid 1410 Semi-conductor Strong covalent 222 Germanium Ge Metalloid 937 Semi-conductor Strong covalent 188 Tin Sn Metal 232 Conductor Metallic bonding Lead Pb Metal 327 Conductor Metallic bonding

Tetrachloride BP /C Thermal Stability Can be

hydrolysed? Ease

CCl4 76

SiCl4 57

GeCl4 87 SnCl4 114 PbCl4 105

The +2 oxides Oxidation state +2 Structure, bond

Thermal stability Acid-base character

Strength as a reducing agent

CO SM, Strong Covalent

Very weakly acidic

SiO GM, Weak Covalent GeO GM, Weak Covalent Amphoteric SnO Ionic Amphoteric PbO Ionic Amphoteric (B) Bonding and thermal stability

CO has a simple molecular structure. The strong triple covalent bond makes the molecules thermally stable.

SiO and GeO have giant molecular structure. They have weak covalent bonds and they decompose on heating in a disproportionation reaction. 2GeO s GeO! s + Ge(s)

SnO and PbO do not decompose on heating. When PbO is heated, it forms a bright red

solid called triplumbic tetroxide Pb3O4, also known as red lead oxide. This compound behaves chemically as PbO2.2PbO

Acid-base properties

CO is very slightly soluble in water. It forms a neutral solution. It reacts with hot conc. NaOH solution. So it is an acidic oxide.

GeO, SnO, PbO all react with HCl. SnO s + 2HCl (aq) SnCl! aq + H!O(l) GeO, SnO, PbO all react with NaOH. PbO s + 2OH! (aq) PbO!!! aq + H!O(l)

The +4 oxides Oxidation state +4 Structure, bond

Thermal stability Acid-base character

Strength as an oxidising agent

Ease of reaction with alkali

CO2 SM, Strong Covalent

Acidic

SiO2 GM Very weakly acidic GeO2 GM Amphoteric (A) SnO2 GC (~ionic) Amphoteric PbO2 GC (~ionic) Amphoteric Bonding and thermal stability

PbO2 is the only one that decomposes readily on heating. 2PbO! s 2PbO s + O!(g)

Acid-base properties

CO2 is slightly soluble in water. CO! aq + H!O l HCO!! aq + H! aq CO!!! aq + 2H! (aq) The P.O.E. lies well over to the left. Most of the dissolved CO2 is in the form of CO2 (aq), so carbon dioxide solution is only weakly acidic.

CO2 also reacts with alkalis CO! g + 2OH! (aq) CO!!! aq + H!O(l) SiO2 does not react with acids (except

hydrofluoric acid). It reacts with hot concentrated alkalis to form silicate (IV) ions and water, so it is an acidic oxide.

SiO2 + 2OH- SiO32- + H2O GeO2, SnO2, PbO2 are all amphoteric. They

react with concentrated HCl to form the tetrachlorides. SnO2 (s) + 4HCl (aq) SnCl4 (l) + 2H2O (l)

GeO2, SnO2, PbO2 all react with hot concentrated alkalis.

SnO2 (s) + 2OH- SnO32- (aq) + H2O (l) GeO2 and PbO2 react in a similar manner to

form germinate (IV) and plumbate (IV) ions (needs molten NaOH to make plumbate (IV))

The reason why +4 oxidation state is less stable at the bottom of the group can only be explained using complex Born-Haber cycles:

Compounds low in the group having the +2 ox. state have greater ionic character than those in the +4 ox. state.

+4 compounds low in the group have more weak covalent character. The energy released on forming these bonds is low.

It takes less energy to oxidise an element to a low oxidation state than to a high ox. state

At the bottom of the group, the energy released on forming covalent bonds or ions in the +4 ox. state is not enough to compensate for the extra energy required to form the +4 ox. state.

At the bottom of the group, the +4 ox. state becomes less energetically stable with respect to the +2 ox. state.

Towards the bottom, the oxidized form (e.g. Ge4+ or Pb4+) is more readily reduced to the +2 state.

The value of E gets more positive down the group.

Ceramics from silicon (IV) oxide Also known as silicon dioxide. It is used to make a variety of ceramics. It is either used on its own or mixed with clay.

Ceramics containing silicon (IV) oxide are used:

For furnace linings As abrasives In the manufacture of glass and porcelain

The properties of silicon (IV) oxide which make it an ideal ceramic include:

Very high melting and boiling point. It needs a high temperature to break the strong covalent bonds in the giant molecular structure.

It is an electrical and thermal insulator (no free electrons)

It is hard. It is difficult to break the 3D network of strong covalent bonds.

It is generally chemically unreactive. It can be moulded at a very high

temperature into a variety of shapes without affecting its strength.

TRANSITION ELEMENTS

A transition element is a d-block element which forms one or more stable ions with an incomplete d sub-shell. Sc and Zn are not transition elements.

Sc forms only one ion Sc3+ and this has no electrons in its 3d sub-shell. (Ar) 3d0 4s0

Zn forms only one ion Zn2+ and this has a complete 3d sub-shell. (Ar) 3d10 4s0

SPECIAL EXCEPTIONS Cr: 3d5 4s1 Cu: 3d10 4s1

The transition elements are all metals. Their atoms tend to lose electrons so they

form positively charged ion. Transition metals have variable oxidation

states and the resulting ions often have different colours.

The existence of variable oxidation states means that the names of compounds containing transition elements must have their oxidation number included. E.g. manganese (IV) oxide.

When transition elements form ions, their atoms lose electrons from the 4s sub-shell first, followed by 3d electrons.

The most common ox. state is +2, formed when the two 4s electrons are lost.

The max. ox. number of the transition elements at the start of the row involves all the 4s and 3d electrons in the atoms.

At the end of the row, (from iron onwards) the +2 ox. state dominates as 3d electrons become increasingly harder to remove as the nuclear charge increases across the period.

The higher ox. states are found in complex ions or in compounds formed with O2 or F. e.g. chromate (VI) ion & manganate (VII) ion.

PHYSICAL PROPERTIES OF TRANSITION ELEMENTS

High melting point High density Hard and rigid, useful for construction Good conductors of electricity & heat The 1st ionisation energy, the atomic radius

& the ionic radius do not vary much as we go across the first row.

Comparing the transition elements with an s-block element (Ca)

The melting point of Ca < TE The density of Ca < TE The atomic radius of Ca > TE The ionic radius of Ca > TE The 1st ionisation energy of Ca > TE The electrical conductivity of Ca > TE

(except copper) Redox Reactions When a compound of a transition element is treated with a suitable reagent, the oxidation state of the transition element can change. Balance a redox reaction by making the number of electrons equal to both half-equation (does not effect the value of E) The redox reaction equation can be used to calculate the amount of a reactant by carrying out a titration. Use Cr2O72- rather than MnO4- to obtain a more accurate result by titration because compounds such as K2Cr2O7 can be prepared to a higher degree of purity than KMnO4. Ligands and complex formation A ligand is a species that contains one or more lone pairs of electrons that forms a dative bond to a metal ion. A complex is a compound formed by a central metal atom surrounded by one or more ligands. Co-ordination number is the number of dative bonds formed by ligands to the central transition metal ion in a complex. Monodendate are ligands (e.g. H2O & NH3) which can form only one dative bond from each ion/molecule to the central transition metal ion. Bidentate are ligands which can form 2 dative bonds from each ion/molecule to the central transition metal ion. The shape of a complex with 6 ligands is octahedral. charge of a complex = charge on (CMI + ligands)

Tetrahedral complex e.g. [CuCl4]2- and [CoCl4]2-. The copper(II) and cobalt(II) ions have four chloride ions bonded to them rather than six, because the chloride ions are too big to fit any more around the central metal ion. Square planar complex e.g. [Ni(CN)4]2- and Pt(NH3)2Cl2 Pt(NH3)2Cl2 is neutral because the 2+ charge of the original platinum(II) ion is exactly cancelled by the two negative charges supplied by the chloride ions. Substitution of ligands The ligands in a complex can be exchanged, wholly or partially, for other ligands. This is a type of substitution reaction. It happens if the new complex formed is more stable than the original complex. The colour of complexes The colour of complexes containing transition metal ions arises because part of the visible spectrum is absorbed by transition metal ions. The 5 d orbitals in an isolated transition metal atom/ion are described as degenerate (the orbitals are at the same energy level) In the presence of ligands, a transition metal ion is not isolated. The co-ordinate bonding from the ligands causes the 5 d orbitals in the transition metal ion to split into 2 states of non-degenerate orbitals at slightly different energy levels.

The lone pairs donated by the ligands into the transition metal ion repel electrons in the two d!!!!! and d!! orbitals more than those in the other three d orbitals. This is because these d orbitals line up with the co-ordinate bonds in the complexs octahedral shape and so they are closer to the bonding electrons in the octahedral arrangement, increasing repulsion between electrons. The difference between the non-degenerate orbitals is E. When light shines on the solution, an electron absorbs E and jumps into the higher energy level among the 2 non-degenerate orbitals. E is affected by many factors: The identity of the ligand that surrounds the transition metal ion. If E is different, then the amount of energy being absorbed by electrons will be different. Therefore, a different colour is seen.

APPLICATIONS OF ANALYTICAL CHEMISTRY

Electrophoresis: The separation of charged particles by their different rates of movement in an electric field. The sample is placed on absorbent paper or on a gel supported on a solid base such as a glass plate. A buffer solution carries the solution along. The rate at which the ions move towards the oppositely charged electrode depends on the size and the charge on the ions. (smaller and higher charge is faster) You get a series of lines or bands on the paper/gel one a chemical is applied. Sometimes UV light is

used to show the bands up. The series of bands is called an electropherogram. Electrolysis is used in biochemical analysis. It can be used to separate, identify and purify proteins & nucleic acids. Buffer solution is needed to maintain the pH because pH will affect the movement of ions during electrophoresis. When separating a mixture of proteins, they are usually first treated with a chemical that makes them negatively charged. A dye can also be added.

All the proteins move towards the positive electrode but larger proteins move more slowly. Uses of DNA fingerprinting/profiling Forensic Science Genetic fingerprinting: a technique based on matching the minisatellite regions of a persons DNA to a database of reference samples. We inherit DNA half from our mother and half from our father. The process:

1. Restriction enzymes are used to cut the DNA molecule at specific places where the same sequences occur, making smaller fragments for analysis.

2. DNA fragments are all negatively charged because of the P groups present, so they all move towards the + electrode. The larger ones find it harder to get through the matrix formed by the gel so they do not travel as far in a given time.

3. The bands are made visible by radioactive labeling of the bands with the P-32 isotope, which causes the photographic film to fog. Alternatively, use a probe that makes the bands fluoresce in UV light.

Short-tandem repeat (STR) analysis

1. Short sequences of bases that make up genes are multiplied using DNA polymerase to copy selected sequences.

2. People differ in the number of these short sequences that the DNA contains. This affects the distances that the bands travel.

How an NMR works The nucleus of each hydrogen atom (proton) in an organic molecule behaves like a tiny magnet. This proton can spin. This movement of the (+) charged proton causes a very small magnetic field to be set up. In NMR, we put the sample to be analysed in the magnetic field. The proton, by spinning, either line up with or against the field. There is a tiny difference in energy between the oppositely spinning nuclei. This difference corresponds to the energy carried by waves in the radio wave range of the E.M. radiation spectrum.

The nuclei flip between the 2 energy levels. They absorb energy in the range of frequencies that are analysed. The size of the gap between nuclear energy levels varies depending on the other atoms in the molecule (the molecular environment) In NMR spectroscopy, we vary the magnetic field rather than the wavelength of the radiowaves. As the magnetic field is varied, the H nuclei in different molecular environment flip at different field strengths. The different field strengths are measured relative to tetramethylsilane (TMS SiCl4 an inert, volatile liquid which mixes well with most organic compounds and its H atoms are equivalent) which is zero. Spin-spin coupling: Peaks are made up of a series of closely grouped peaks because the magnetic fields generated by spinning nuclei interfere slightly with those of neighbouring nuclei. The number of signals a peak splits into equals n+1 where n is the number of H atoms on the adjacent C atom. Steps to interpret high-resolution NMR spectrum

1. Use values to identify the environment of the protons present at each peak.

2. Look at the relative areas under each peak to determine how many of each type of non-equivalent protons are present.

3. Apply the n + 1 rule to see which protons are on adjacent C atoms in the unknown molecule.

4. Put all this information together to identify the unknown molecule.

For ethanol: The peak of the OH group is not split by the H on the neighbouring CH2 group because the OH proton exchanges very rapidly with protons in any traces of water (or acid) present. (labile proton). This also occurs in amines and amides which contain the NH group.

Identifying the OH or NH signal in an NMR spectrum Add a small amount of D2O (deuterium oxide) to the sample then their peaks disappear from the spectra. The deuterium (2H) atoms exchange reversibly with the protons in the OH or NH groups. The deuterium atoms do not absorb in the same region of the E.M. spectrum as protons, so the OH or NH signal disappears. X-ray crystallography To get the best results, we need to have a very pure crystal of the sample. Many large biological molecules can be crystallised from solution. This technique relies on the diffraction of X-rays as they pass into a crystal which is caused by the electrons in the atoms present. The larger the atom, the more electrons it contains and the more intense the spot it produces in its diffraction pattern.