PeriodicityAll covalent oxides are acidic. WHY: MO(S) + H2O(l) MO(OH)-(aq)+ H+(aq) All ionic oxides are alkaline. WHY: O2-(aq)+ H2O(l) 2OH-(aq)Oxides of Period 3Bonding TypePropertiesEquations

Na2OIonic Bonding: large difference in electronegativity.1) Solid2) High melting point; to strong electrostatic force of attraction between negative and positive ions.3) Conducts electricity when molten.4 ) Brittle; different sized ions2Na + 1/2O2 Na2O(s)Yellow flame is seen

Very exothermic reaction (H= - ) with water:-Na2O(s) + H2O(l) 2NaOH(aq)O2-(s) + H2O(l) 2OH-The O2- gains a proton (H+) and therefore it is:Strongly alkaline (base); pH= 14.0Conducts electricity when molten as ions free to move

MgOIonic Bonding: large difference in electronegativity.1) Solid with higher melting point than Na2O.WHY: a) Mg smaller ionb) Mg has higher chargec) Stronger electrostatic force of attraction.2Mg +O2 2MgO(s)Bright white light is seen and white fumes of the oxide formed

Almost totally insoluble in water (forms suspension in water):-MgO(s) + H2O(l) Mg(OH)2(aq)O2-(s) + H2O(l) 2OH-Almost not feasible as reversible sign shows; pH= 8.0

Al2O3Ionic with Covalent Character: polarisation introduces covalence1) Solid; due to ionic bonding2) The fact that it has covalent character makes it amphoteric4Al + 3O2 2Al2O3Bright white light is seen and white fumes of the oxide formed

Totally insoluble in water or moist air as G is very positive and lattice energy is large; pH= 7.0

SiO2Giant Molecular: Lots of strong covalent bonds that require a lot of energy to break1) Very High melting point2) Very hard3) Good insulator4) High melting pointSi + O2 SiO2Bright white light is seen and white fumes of the oxide formed

Totally insoluble in water or moist airpH= 7.0

P4O10(It not P4O10 as this is its most stable form)Molecular Covalent: Phosphorus and Oxygen have similar electronegativity.1) Solid as the strongest force is Van Der Waals, this molecule is big therefore Van Der Waals is higher2) Van Der Waals increase as surface area increases3) Higher melting point than SO34P + 5O2 P4O10Bright white light is seen and white fumes of the oxide formed

Very exothermic reaction (H= - ) with water (makes phosphoric acid) :-P4O10(s) + 6H2O(l) 4H3PO4(l)This dissolves again in water:-H3PO4(l) 3H+(aq) + PO43-(aq)Strongly acidic; pH= 1-0

SO2Molecular Covalent: Sulphur and Oxygen have similar electronegativity.1) Gas; Van Der Waals forces BETWEEN MOLECULES not high enoughS + O2 SO2(g) - colourlessBlue flame is seen and pungent smell of gas is observed.

Extremely soluble in water but endothermic reaction (H= + ) reaction (dissolves in water-colourless):-SO2 SO2 H2SO3 H+ + HSO3- 2H+ + SO32-Weak acid; pH= 3-4

SO3Molecular Covalent: Sulphur and Oxygen have similar electronegativity.1) Van Der Waals forces BETWEEN MOLECULES very high as molecule is big2) Melting point not as high as P4O10 as it is not as bigS + 3/2O2 SO3 ; can also be made by reacting SO2 with O2 (Contact Process).

Very exothermic reaction (H= - ) with water:-SO3 + H2O(l) H2SO4(l) This dissolves again in water:-H2SO4(l) 2H+(aq) + SO42-(aq)Strongly acidic; pH= 1-0

Order of melting point:-MgO > Al2O3 > SiO2 > Na2O > P4O10 > SO3 > SO2Also remember that if we wanted to show that an oxide contained ions, we would heat it until it is molten, and see whether it conducts electricity.As you go down oxides of metal groups (group 1 and 2): melting point decreases as the ion gets bigger and therefore there is less electrostatic force of attraction between the positive metal ion and negative Oxygen ion. In a question such as this, never mention electro-negativity, nuclear force or shielding at all.

Melting point of MgO is higher than Na2O: Because Mg is a smaller ion with a more negative charge than Na, Therefore stronger electrostatic force of attraction with negative Oxygen ion.

Also remember, that when it says aqueous in the question, it wants ions. So if it says: aqueous Sodium Hydroxide, DO NOT write NaOH, write OH- or Na+.

Cations (+) form Acids. Anions (-) form Bases. As a general rule Acid + Base + (Water) Salt + (Water)REMEMBER: Hydroxide (OH) is not the same as a substance being soluble. Aluminium is not soluble in water; however it does form a hydroxide. Hydroxide involves the ion OH-.OxideContainsEquation as to how cation/anion formedSalt formedIt isReacts with

NaOH / Na2ONa+ (cation)Na2O(s) + 2H+(aq) 2Na+(aq) + H2O(l)Sodium saltsBASEACID

Mg(OH)2 / MgOMg2+ (cation)MgO(s) + 2H+(aq) Mg2+(aq) + H2O(l)Magnesium saltsBASEACID

Al2O3 / Al(OH)3Al3+ (cation)Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)Aluminium saltsBASEACID

Al(OH)4- (anion)Al2O3(s) + 3H2O(l) + 2OH-(aq) 2Al(OH)4-(aq)Aluminate saltsACIDBASE

SiO2SiO32- (anion)SiO2(s) + 2OH-(aq) SiO32-(aq) + H2O(l)Silicate saltsACIDBASE

P4O10 / H3PO4PO43- (anion)P4O10(s) + 12OH-(aq) 4PO43-(aq) + 6H2O(l)Phosphate (II) saltsACIDBASE

SO2 / H2SO4SO42- (anion)SO3(g) + 2OH-(aq) SO42-(aq) + H2O(l)Sulphate (VI) saltsACIDBASE

SO3 / H2SO3SO32- (anion)SO2(g) + 2OH-(aq) SO32-(aq) + H2Ol)Sulphate (IV) saltsACIDBASE

Examples:-1) NaOH + H3PO4 Na+ + PO43-3NaOH + H3PO4 Na 3PO4 + 3H2O6Na2O + P4O10 4Na3PO42) NaOH + H2SO4 Na+ + SO42-2NaOH + H2SO4 Na 2SO4 + 2H2O3) NaOH + Al2O3 Al(OH)-+ Na+3H2O + 2NaOH + Al2O3 2NaAl(OH)4The tip for this is to always draw the anion (negative ion) first such as Al(OH)4- SO42-, PO43-, then add the cation. Also when a weak alkaline is used to neutralise a strong acid, we can use excess and no problems will occur. However if a strong alkaline is used (NaOH), we must not use excess as we will create a strong alkaline solution, and this as we know could cause environmental problems in lakes.ElementEquationProperties

Na

2Na(s) + H2O(l) 2Na+(g) + 2OH-(aq) + H2(g)EXTREMELY EXOTHERMIC (H= - )1) Sodium melts, we see fizzing2) Large lump explodes, due to Hydrogen gas, which burns with a yellow flame3) Low density metal, therefore floats on water5) Sodium hydroxide forms in alkaline solution

MgMg(s) + H2O(g) MgO(s) + H2(g)(Originally Mg(OH)2 was made, but it decomposed to MgO at the temperature of the reaction.DOES NOTHING1) Mg sinks2) Very slow reaction giving small quantity of H2(g) after a week3) HOWEVER, if steam is added, a very exothermic reaction occurs because Mg melts into gas.

Also remember that Phosphorus is stored under water as it spontaneously ignites in air.If you want to know whether something basic (alkali) or acidic, add it to water, if you get H+ its acidic, if you get OH- its basic (alkali)If in the exam, the word moist is used, it is a reference to water (H2O(l)).Sometimes, to prevent corrosion in moist air, aluminium oxide is coated over a metal, this is because aluminium oxide is insoluble.ThermodynamicsBonds being broken = Endothermic (+)Bonds being made = Exothermic (-)Things you should know about S and H:H: 1) If it is (exothermic), then it is going to products, but increase the temperature and it will move to reactants due to Le Chatelliers principle.2) If it is + (endothermic), then it is going to reactants, but increase the temperature and it will move to products due to Le Chatelliers principle.S: 1) If it is + (endothermic), then it is going to products, and if you increase the temperature it will be even more powerful towards products. (La Chatelliers principle ineffective)SHWhat happens

++Reaction reversible as H is going to reactants but S is going to products. Hence if you increase the temperature H moves to products (LCP), as does S. This tells us that a reaction of this sort is only feasible (i.e. will go to products) at very high temperatures.

--Reaction reversible as H is going to products but S is going to reactants. Hence if you increase the temperature H moves to reactants (LCP), as does S. This tells us that a reaction of this sort is only feasible (i.e. will go to products) at very low temperatures.

+-This reaction is never feasible at any temperature (i.e. will never go to products) as H is going to reactants but S is also going to reactants.

-+This reaction is feasible at all temperatures (i.e. will always go to products)as H is going to products but S is also going to products.

2) If it is - (exothermic), then it is going to reactants, and if you increase the temperature it will be even more powerful towards reactants. (La Chatelliers principle ineffective)Type of EnthalpyInformation Regarding It

Enthalpy of AtomisationEndothermic (+); you make one mole of gaseous atoms. Remember you make one mole of atoms, not ions just yet. In bond enthalpy you make atoms of that element but it is not one mole, it is two. Therefore you multiply this number by two. For example: Cl2 will have a bond enthalpy twice the enthalpy of atomisation of chlorine as breaking Cl2 gives you 2Cl.

Enthalpy of IonisationEndothermic (+); each ionisation more endothermic (+) than the other. WHY?Ion is gaining more charge; therefore the outer-electron is closer to the nucleus, and the ion is smaller, therefore it becomes more difficult to remove the outer-electron. Here, we make ions.

Electron Affinity1st Electron Affinity: Exothermic (-) WHY? Electrons lose energy as they fall into orbitals.Attractive force between nucleus and external electron.Remaining are: Endothermic (+) WHY? Repulsion between negative ion and negative electron

Lattice Enthalpy of FormationExothermic (-) as you make bonds to form ionic lattice from gaseous ions.

Lattice Enthalpy/ Lattice Disassociation EnthalpyEndothermic (+) as you break an ionic lattice to form gaseous ions. Complete opposite of Lattice Enthalpy of Formation. It will usually be the same number as Lattice Enthalpy of Formation but it will have a + in front of it instead. It increases as the force between the lattice increases. Hence smaller, higher charged ions will be harder to pull apart and therefore will have a higher Lattice Enthalpy. Size decreases across a period, and increases down a group. Charge increases across a period and stays the same down a group.

Enthalpy of HydrationYou make one mole of aqueous ions from one mole of gaseous ions. Hence if you are making 2 moles of aqueous ions from 2 moles of gaseous ions, then multiply this number by 2. When the aqueous ion is in the water, e.g. Na+, it will have dipole with the O-- of H2O, or if it is F-, it will have a dipole with the H+ of H2O. Hydrogen bonding will ONLY take place between molecules.

Enthalpy of SolutionYou dissolve the ionic lattice in water and this gives you aqueous ions. It is worked by first doing the Lattice enthalpy of the ionic lattice, this will give you gaseous ions, now use Enthalpy of hydration to work out aqueous ions. Add them together to give you the Enthalpy of solution.

Bond EnthalpyYou work out enthalpy of breaking (+)/ making (-) each bond in a compound. However, all products and reactants must be in the gas state. To work it out: Bonds broken - Bonds made. Bond Enthalpy differs from Enthalpy of formation in 2 ways: 1) Enthalpy of Formation has products and reactants in standard states, bond enthalpy has them in gas state2) Bond Enthalpy considers mean values of enthalpy for that bond

Always remember, when drawing Born-Haber Cycles, is for endothermic (+) is for exothermic (-).

Examinations also tend to as to why the enthalpy of hydration or lattice disassociation is higher for a certain molecule or atom than another molecule or atom, the answer should always mention the SIZE of the ion involved and its charge and how this will affect its force of attraction.

For example, the enthalpy of lattice disassociation of Calcium Fluoride is greater than that for Calcium Chloride because Fluoride ions are smaller, therefore they form a stronger electrostatic force of attraction with positive ion (Ca2+), and therefore more energy is required to break them.

Also for example, the enthalpy of hydration of Aluminium ions is much higher than enthalpy of hydration of magnesium ions because Al3+ is more polarising (higher charge density/size to charge ratio), and therefore attracts water more strongly.

It is also important to remember rather more confusing question such as why HF(l) has a higher enthalpy of vaporisation (turn substance to gas) than NH3. Well this is because although BOTH have hydrogen bonding, the hydrogen bonding in HF(l) is stronger because H-F is more polar than H-N, therefore more energy required to separate it.

Entropy is the colleague of Enthalpy. It explains as to how endothermic reactions take place, if you think about it, in endothermic reactions (+), enthalpy is pushing to reactants, so how does the reaction even take place? Its because Entropy is pushing to products. However, entropy is very weak (compared to enthalpy), hence endothermic reactions that are feasible (i.e. go to products) are very rare).

Entropy is defined as randomness. Gases have the highest entropy. Liquids have the second highest entropy. Solids have the least entropy. At 0K, there is no entropy as there is no randomness. Then entropy increases as the temperature increases as particles vibrate more. At Tm, the melting point (solid to liquid) takes place and entropy jumps as the particles become even more random. At Tb, the boiling point (liquid to gas) takes and entropy jumps higher than it did for Tm, this is because there is more randomness in a higher volume, there bigger increase in disorder.In solids: particles vibrating in small volumeIn liquids: particles moving in small volumeIn gases: particles moving in large volume

Entropy is worked out by: Entropy products - Entropy reactantsWe can tell if the entropy is going to be positive by looking at the states of the reactants and products. Hence, if a mole of solid is converted into two moles of gas, then entropy is most definitely positive. WHY? Two big numbers (gas) one small number (solid)Entropy is measured in Joules. Hence Enthalpy is measured in Kilojoules. Therefore Entropy is divided by 1000 to make it into Kilojoules when required.

Feasible: Thermodynamically possibleSpontaneous: takes place without any energy inputAll spontaneous reactions are obviously feasible. However a feasible reaction may not be spontaneous, maybe because it has very high activation energy.

Also remember that the enthalpy of solution (dissolve in water) is at times exothermic too, but when it is endothermic (+) we should be providing energy, then the reason for the reaction taking place SPONTANEOUSLY (it happens without us giving it energy) is because entropy is highly positive, pushing the reaction to products.

Free Energy is known as G. It is measured in Kilojoules. It tells us whether a reaction is feasible or not, i.e. will it take place or not. For a reaction to take place:- G0.Hence, this is how we work it out:G = H T (S/1000)We can conclude a couple of things:1) As S gets more positive, G becomes more negative2) At equilibrium, when the products and reactants are EXACTLY the same amount, G = 0. When a liquid boils G = 0, this is because boiling is a spontaneous change and mainly because water and water vapour are in equilibrium. 3) When G = 0, the reaction is just about feasible. You can work out this temperature by re-arranging the equation to: T = H/ (S/1000). This will tell you when G = 0 (i.e. when the reaction is just about feasible)4) In reversible reactions, G = 0 when the reaction has no tendency to move spontaneously in one direction or the other. At G = 0, you can work out the temperature, if this temperature is increased/decreased the reaction will no longer remain spontaneous in that direction. Hence yield in that direction would decrease. This is why when a liquid boils (reversible reaction) G = 0, because the water and water vapour are at equilibrium and the reaction has no tendency to move spontaneously in one direction or the other.5) When moles on both sides are equal and phase is also the same, entropy is close to zero; hence as long as H is constant, G will remain constant at any temperature.

When H and S are both negative, we must use a low temperature (above), hence such a reaction if carried out will not be feasible, i.e. will not produce products in an internal combustion engine as this type of engine uses very high temperatures.Industrialists however sometimes break this feasibility rule and do the reaction at a temperature higher than the limit of the feasible temperature that we have calculated. But we dont need to know at this point much about this; all we need to know is that they do this because the reaction may have high activation energy or they need the reaction to speed up.

In the exam, a common question is one wherein two equations are given, and we must work out the temperature at which both reactions will have the SAME VALUE for G. Hence we do it like this:

G (reaction A) = G (reaction B)T = [H (reaction A) - H (reaction B)] [S (reaction A) - S (reaction B)]

Also, in the exam, when it gives you a Born-Haber cycle of a reaction, and asks you to name the enthalpies, do not just write enthalpy of atomisation or enthalpy of ionisation, ensure to write what the enthalpies are of. I.e. write enthalpy of atomisation of... (chemical) and enthalpy of ionisation of ... (chemical).

A key reaction in enthalpy is the reaction in which NO is produced. In an internal engine, it is made in the following way: N2 + O2 2NO, then the NO is turned into NO2. The temperature in an internal combustion is above 794k, therefore this reaction does not take place: NO + 1/2O2 NO2 as it is not feasible above 794K.As a general rule, whenever something does not form, which according to enthalpy and entropy, should form, it is due to high activation energy. Also remember that melting points and boiling points in molecular covalent structures have nothing to do with the breaking of bonds, it is always about the intermolecular forces, and these intermolecular forces (mainly H-Bonding) only arise between two molecules, NOT between two ions. There are no H-Bonds between H2O molecules and F- ions.Sometimes a graph that is presented in the exam is a graph which shows how the free-energy change for a reaction varies with temperature above a certain temperature, such as:Here, the value of the slope is known as the gradient which is S. The reason the graph stops to slope, is because at this point there is a phase change, hence the graph moves vertically upwards. The units of the SLOPE are KJmol-1K-1. Where the graph crosses the x-axis, G becomes 0 and hence the reaction at this temperature is now feasible.

Redox Equilibria

General Rule:Reducing agents lose electrons as they themselves are oxidised.Oxidising agents gain electrons as they themselves are reduced.

Electrochemical cells contain two electrodes (electrical conductors) immersed in an electrolyte (ionic conductor). The electrolyte can be an aqueous solution (the solution cannot be due to dissolving in water is SOME cells, because water could SOMETIMES react with very reactive metals such as Lithium) or a molten salt. An electrode with its electrolyte is called an electrode compartment. If two compartments are used, then a salt bridge is used to connect the two compartments (cells).The salt bridge allows the flow of ions (in this case: M+ and H+) and the electrode allows the flow of electrons (current).As you can see, in one cell there is an oxidation process and in the other, there is a reduction process.SALT BRIDGE (needed to join two half-cells and complete circuit whilst preventing phase change, this is because every phase change causes a change in voltage): Contains the salt KCl which is unreactive and allows ions to move.In a cell that involves a reactive metal such as Lithium, water is never used as a solvent as it will react with the reactive metal. Sometimes, in batteries a Porous Separator is used, it has the same function as a salt bridge.

ELECTRODE: Platinum as it is unreactive, it is a good conductor (can carry electrons) and allows equilibrium to be established. It is also a good catalyst and therefore Platinum Rods are made by coating ceramic material with Platinum so as to increase the surface area. The purpose of the electrode is to carry electrons.

The representation of the cell is this: Reducing agent I Oxidising agent II Oxidising agent I Reducing agentThe double line (II) separates the two half-cells. The single line (I) separates a phase change- i.e. solid, liquid and gas.

The metal electrode is the actual metal in equilibrium with a solution of its ions (left hand side). For example: Cu2+(s) I Cu(s). In this cell, the Cu(s) will be the actual electrode.The gas electrode is a platinum electrode surrounded by a gas in equilibrium with a solution of its ions, the gas and its ions are obviously in a different phase, but the platinum electrode is needed as the gas cannot CARRY ELECTRONS to the other cell as an electrode is supposed to do, also the platinum acts as a catalyst for the gas and its ions to reach equilibrium. For example: H+(aq) I H2(g). The H2(g) gas will be bubbled in.The redox electrode is the platinum electrode surrounded by a solution in which the phase (state-solid, liquid, and gas) is the same, but the oxidation number is different. For example: Fe3+(aq) , Fe2+(aq)

Every half-cell has an electrode potential (E-0; we work out this potential by joining it with the half-cell of Hydrogen as seen in the picture above. The Standard Hydrogen Electrode (SHE) is assigned the electrode potential 0; this is by definition so that we can use it as a reference for all other half-cell electrode potentials. SHE: Hydrogen gas is bubbled into a 1.00 moldm-3 solution of HCl (contains H+) at 298K, with a Platinum electrode.The positive electrode will be considered to be the one with the more positive electro potential, by convention the positive electrode is placed on the right.

N.B. Always remember we cannot have a solution of ions alone, therefore when we say a metal in equilibrium with a solution of its ions, we dont mean for example Mg and Mg2+, we mean Mg and MgCl2, however, when in solution the MgCl2 becomes Mg2+. Basically it is not possible to have a jar of Mg2+ on your shelf.A voltage (measure of flow of electrons) is set up between the two half-cells. Each half-cell has an electrode potential as discussed in the paragraph above, the difference between these two potential is called the voltage/e.m.f. It is worked out by:Electrode Potential of right hand cell Electrode potential of left hand cell The following shows the conditions needed in order to get the correct e.m.f of a cell, any changes with the below and the e.m.f will change.Standard Conditions of half-cell: Zero-current conditions (not when ions are both aqueous) Concentration must be 1.00 moldm-3 Temperature must be 298K Pressure must be 100kPa (in gases only)The most powerful reducing agent (gives away most amount of electrons) is the product (right-hand side) of the most negative/least positive electro potential half-cell.The most powerful oxidising agent (takes away from others the most amount of electrons) is the reactant (left-hand side) of the most positive/least negative electro potential half-cell.This is because the best oxidising agent has the highest E-0The worst oxidising agent has the lowest E-0Therefore as the Electro potential decreases, the reducing power of the substance on the right (product) increases and it will be able to reduce everything to the left (reactant) that is above it.

In the same way as the Electro potential increases, the oxidising power of the substance on the left (reactant) increases and it will be able to oxidise everything to the right (product) that is above it.EquationE-0

1Cl(aq) + 2e- 2Cl-+1.36

2H2O2(I) + 2H+(aq+ 2e- 2H2O(I)+1.77

3O2(g) + 4H+(aq) + 4e 2H2O(I)+1.23

4O2(g) + 2H+(aq) + 2e- H2O2(aq)+0.68

Most powerful reducing agent: H2O2(aq)Most powerful oxidising agent: Cl(aq)Hence when the most powerful reducing agent is at the bottom right and the most powerful oxidising agent is at the top left. The reactions occur as:Bottom Right + Top Left Bottom Left + Top Right(Ensuring the electrons are balanced)

For example: when 2 and 4 discharge (react):-H2O2(aq) (Bottom Right) + H2O2(I) + 2H+(aq+ 2e- (Top Left ) 2H2O(I) (Top Right) + O2(g) + 2H+(aq) + 2e- (Bottom Left)

Similarly when the most powerful reducing agent is at the top right and the most powerful oxidising agent is at the bottom left. The reactions occur as:Bottom Left + Top Right Bottom Right + Top Left (Ensuring the electrons are balanced)

When these reactions are shown, remember that a reaction in alkaline conditions will contain OH- in the equation, whereas a reaction under acidic conditions will contain H+ in the equation.

Also remember that the concentrations must be constant (no increase or decrease) for our worked out e.m.f to be correct. But if we DID change the concentration of a substance how would that affect e.m.f?For example: we have two half cells:Mg2+(aq) + 2e- Mg(s) = -2.37 Fe2+(aq) + 2e- Fe(s) = -0.44Now, the e.m.f is right left. Therefore, the e.m.f is -0.44 (-2.37) = 1.93. But if we increased the concentration of Mg2+ions, by Le Chatelliers Principle the half-cell equilibrium will reduce it by moving towards Mg, Mg is a reducing agent and as we know if were making more reducing agent, the Electro potential of that cell becomes MORE NEGATIVE. So lets just say it now goes from -2.37 all the way down to -4.37, now lets look at our e.m.f, it will be -0.44 (-4.37) = 3.93. Hence you see how our e.m.f increased. In the exam they will ask about this. Always remember, if you increase the concentration of a weak oxidising agent such as Mg2+, the half cell reaction gets MORE NEGATIVE, if you increase the concentration of a strong oxidising agent such as O2, the half cell reaction gets MORE POSITIVE.

Also remember any time when a metal is being converted into its ions in the overall reaction, then that metal electrode will be corroding away.

Batteries that are not re-chargeable only discharge in the way shown above, hence their e.m.f will eventually decrease and go down to zero, and if the voltmeter were replaced by an ammeter, the ammeter would read zero, as the ions/elements/molecules reacting are used up (H+ will not get used up as constant supply of H2 that is being oxidised, however reactants in the other half-cell will get used up) and because the reaction is not reversible, the battery cant be re-charged either.

However, in re-chargeable batteries, such as Lead-acid batteries, this does not happen, because as soon as reactants are used up, the reverse reaction begins to take place. Hence, first the discharge reaction takes place (as shown above), then when it wishes to re-charge the reaction will turn itself around, hence in the exam, just turn the discharge reaction the other way round to show that the battery is now re-charging. This is similar to a reversible reaction (). The advantage of re-chargeable batteries is that it reduces landfill.

N.B. When working out oxidation states (oxidation numbers), always remember that in H2O2, Oxygen has oxidation state -1, not -2 as it usually has.

Fuel cells convert stored chemical energy directly into chemical energy.Fuel cells have a constant e.m.f. WHY? 1) Constant supply of external fuel 2) Constant concentration of reagentsIn a hydrogen/oxygen fuel cell, H2 and O2 are added to a cell in which there is a platinum electrode with a cathode and anode. It releases H2O. It has advantages such as: 1) H2 is easy to make2) No pollutionIt is not carbon-neutral as making H2 does release CO2. How? Electrolysis of water produces H2.

Their advantage over a re-chargeable battery is that they have a constant supply of external fuel and do not need to re-charge.

You can also get an ethanol-oxygen fuel cell, wherein ethanol is oxidised to CO2 and O2 is reduced as above to H2O. However Ethanol CAN be considered a carbon-neutral fuel because the number of moles in photosynthesis and fermentation are equal (6:6) UNIT 2.

Also remember that half-cells use chemical change to produce electricity.However electrolysis uses electricity to bring about a chemical change.

Transition Metals

Transition metals have incompletely occupied d-orbitals. This means any ion/element with a d0 or d10 will not be a transition metal ion/transition metal. Look at electronic structures of the d-block elements to see which ones are transition metals and which ones are not.They have 4 main characteristics: 1) Complex formation: due to incompletely occupied d-orbital2) Coloured Ions: due to complex formation3) Catalysts: due to variable oxidation states4) Variable Oxidation numbers/statesIn transition metal elements the 4s orbital is before the 3d orbital, however in transition metal ions the 3d orbital is before the 4s orbital as the 4s orbital in a transition metal ion is empty.ScTiVCrMnFeCoNiCuZnAl

1+

2+2+2+2+2+2+2+

3+3+3+3+3+3+3+3+

4+4+4+

5+(VO2+)

6+(CrO42-, Cr2O72-)6+6+

7+(MnO4-)

So as you can see CrO42-, Cr2O72-, MnO4-, VO2+ are good oxidising agents, i.e. they gain electrons. However Zn2+, Cu+, Cu2+ are good reducing agents, i.e. they lose electrons.

1) Complex formation; this is done by the metal forming dative covalent bonds with a molecule/ion. This molecule/ion is called a ligand. Hence all the hard work is done by the ligands as they provide both the electrons in the each bond.2) Obviously, the fact that ligands form dative covalent bonds with metal ions, tell us that the ligand must have a lone pair/s.3) Depending on the size of the ligand, the metal ion can form 2, 4 or 6 coordinate bonds with a ligand. Therefore you may get a complex that is linear (2 coordinate bonds), Tetrahedral (4 coordinate bonds with a big ligand), square planar (4 coordinate bonds with big ligands) or octahedral (6 coordinate bonds with a small ligand)The coordination number is the number of coordinate bonds formed. This is not always the same number as the number of ligands as some ligands form 2 or more coordinate bonds.Unidentate Ligands form one coordinate bond a metal ion, Multidentate Ligands form 2 or more coordinate bonds with a metal ion (obviously they must have 2 lone pairs).Common ligands are: :NH3 = Ammonia (unidentate) -:OH = Hydroxide(unidentate) NC-: = Cyanide(unidentate) Cl-: = Chloride (unidentate) H2O: = Water (unidentate) :O2 = Oxygen H3N:-CH2-CH2-:NH3 = 1,2 Diaminoethane (en) (multidentate - 2 coordinate bonds) -:OOC-COO-: = Ethanedioate Ion (C2O42-) (multidentate - 2 coordinate bonds) EDTA4-Notice that Chlorine is NOT a ligand, a chloride is. Also as you can see, some ligands have charges, and some dont.The ligands form different shapes with the metal ions, the ones that are most common and ones we are required to know are: Linear, Tetrahedral, Square Planar and Octahedral.

Just as a side note, when excess Hydrochloric Acid is added to H3N:-CH2-CH2-:NH3 = 1,2 Diaminoethane (en), it becomes ClH3N-CH2-CH2-NH3Cl, the Chloride (Cl-) bonds with the Nitrogen.

UNIDENTATE LIGANDS: The ligand is forming only 1 coordinate bond with the metal ion.

Linear Complexes: Formed by Cu+ and Ag+ (remember that neither are transition metals; complete d-shell)Ligand in use: Cl-:[CuCl2]- The Cu has charge +1; the 2Cl- has charge -2, therefore -2+1 = -1It is colourless.

Ligand in use: :NH3[Ag(NH3)2]+ The Ag has charge +1; the NH3 has no charge, therefore = +1This complex is known as Tollens Reagent (Unit 2) and is reduced by Aldehydes. Also, it is colourless, but when added to Aldehyde such as Ethanal, we get a silver mirror colour, which is the colour of Ag(s). The reduction reaction is: [Ag(NH3)2]2+ + 2e- Ag + 2NH3As a side note, you can also get [AgCl2]- as you did with the Cu.

Square Planar Complexes: Formed by Pt2+

Ligand/s in use: Cl- and NH3It called Cis Platin and it forms 4 coordinate bonds only (instead of 6) as the Cl- ligand is a very big ligand big ligands are caused by the negative charge, as you should know by now, negative charges are quite big in size. Cis Platin is a cancer drug that has one risk-you could die, as it is toxic.

Ligand/s in use: :Cl- and :NH3This is called Trans Platin, it is much like Cis Platin but the Cl- and the NH3 arranged differently.

Octahedral Complexes: Formed by Fe2+, Co2+, and all other transition metals.

Ligand in use: H2O:[Fe(H2O)6]2+ The H2O: has charge 0, the Fe has charge +2, therefore = +2It is called hexa aqua Iron (II) Hexa because of the 6 ligands, Aqau because of the ligands being water and Iron (II) because of the Fe2+This is known as the Aqua Complex, we will discuss regarding this later on.

However, as mentioned above, some ligands form more than one coordinate bond with the metal ion.Obviously this is only possible if the ligand has more than 1 lone pair.

MULTIDENTATE LIGANDS: The ligand is forming 2 or more coordinate bonds with the metal ion.

Examples of BIIDENTATE LIGAND (2 coordinate bonds with metal ion) COMPLEXES:

Octahedral Complexes: Formed by Co2+, Fe2+, and couple others

Ligand in use: H3N:-CH2-CH2-:NH3 = 1, 2 Diaminoethane (en)The NH3 has no charge, the Co has charge +2, therefore = +2There is a shorter way of drawing it (easy peasy):

Ligand in use: -:OOC-COO-: = Ethanedioate Ion (C2O42-) and H2O:The O has no charge, the Fe has charge +2, therefore = +2

Example of QUADRIDENTATE LIGANDS (4 coordinate bonds with metal ion) COMPLEXES:Octahedral: Formed by Fe2+ (Haemoglobin), Mg2+ (Chloroplast), Co2+ (Vitamin B12)Ligand in use: Porphyrin Ring (PR)The N has no charge, the Fe has charge +2, therefore = +2If youre wondering why we have said octahedral whereas you can only see 4 coordinate bonds of N to Fe, it is because there are two more ligands above and below Fe that are difficult to draw.

The object of this structure is to carry oxygen. (Biology)

Example of HEXADENTATE LIGANDS (6 coordinate bonds with metal ion) COMPLEXES:

Octahedral:Formed by EDTA4-. However, it is important to understand that although the shape is octahedral, the ligand is only one; this is because EDTA4- looks like this:

Fear not, they will not ask you to draw this ligand in a complex, it would be very difficult. However, they could ask you about it, especially in Ligand Substitution.N.B. The picture is not a complex, its just a ligand. A mighty bigun.Ligand Substitution

Ligand Substitution: A ligand is attached to a metal ion, it is then chucked out and replaced by another stronger ligandEXAMPLE: [Co(H2O)6]2+ + EDTA4- [CoEDTA]2- + 6H2OCan you see how the EDTA4- chucked out the 6H2O ligands and joined the Co2+? This is why we see 6H2O in the products. However [CoEDTA]2- is still octahedral because although now only 1 ligand is attached to the Co2+, this ligand is a Hexadentate ligand, which means that it is forming 6 coordinate bonds with the Co2+.The [CoEDTA]2- is a very stable complex as there has been an increase in Entropy, from 2 moles you created 7, and you replaced 6 ligands of water with 1 ligand of EDTA4-. G for this reaction is very negative as Enthalpy is close to zero. WHY? Same number and same type of bonds broken and made, but Entropy is very positive making G very negative.Its also important that you notice how ligand substitution will always involve the sign.Ligand Substitution can be very dangerous too. This is because Haemoglobin (red blood cells) contains Fe2+ with :O2 ligands, however CO: is a stronger ligand, so ligand substitution would occur and the O2 would be chucked out and replaced by CO:, which would obviously kill you. Thats why you shouldnt breathe in Carbon Monoxide (CO:), ever. It will kill you.However, Oxygen being a weak ligand is actually rather important as it allows Oxygen to be carried by the Haem and then removed from the Haem so that it may be used inside the body. If it was a strong ligand, it would forever remain attached to the Haem and the body would be deprived of Oxygen.

A common ligand substitution question involves HCl, where 4Cl- ions replace the water molecules in a complex.FOR EXAMPLE:

[Cu(H2O)6]2+ + 4Cl- [CuCl4]2- + 6H2OThe product [CuCl4]2- is a yellow-green colour, It will be asked regarding the colour too in the exam.Also ensure that you do not write HCl as a reactant, when added to water HCl becomes Cl- and H+, by writing HCl, you are implying that Cl- is using up its lone pair to bond with the H+ and make HCl hence no longer remaining a ligand. (The [CuCl4]2- can become [CuCl4]3-, wherein the oxidation state of Cu is +1, and Cu metal acts as a reducing agent

In Vitamin B12, Co2+ is co-ordinately bonded to a porphyrin ring (above) and 2 other ligands, one above Co2+ and one below that are difficult to draw. One of these ligands is CN:- (Cyanide if it is on its own in the body, it is toxic), but yet Vitamin B12 is not toxic, this is because CN:- (Cyanide) is strongly bonded to Co, and therefore does not separate from it.

Different types of Ligand Substitution:

Type of Ligand SubstitutionCHANGE IN

Co-ordination NumberCharge

[Co(H2O)6]2+ + 6NH3 [Co(NH3)6]2+ + 6H2O

[Co(H2O)6]2+ + 4Cl- [CoCl4]2- + 6H2O

[Co(H2O)6]2+ + 3C2O42- [Co(C2O4)3]4- + 6H2O

Notice how [CoCl4]2- is formed, the Cl- is a big ligand, therefore it has coordination number 4 (tetrahedral).We can ascertain an important fact here; this is that H2O and NH3 are similar in size.

Colourimetry

Ligands are only one of many ways a substance receives colour, for example pure ice is pale blue when illuminated by white light because ice absorbs frequencies from light spectrum and transmits the colour it has not absorbed which is blue. But in transition metals, colour is transmitted due to the ligands. A metal that is not transition metal and is in solution may form a ligand, but not a colour = [Al(OH4)]- is colourless.

How Ligands cause colour in TRANSITION METALS (must be transition metal):

White light is directed at a transition metal complex, the d-orbital electrons absorb certain frequencies from the visible light spectrum that are equal to hf and move from a ground state to an excited state (i.e. the d-orbital splits into 2 levels, an excited state and a ground state). Hence the change in energy of the electrons is equal to: E = hf (sometimes they write it as E = hv, theres no difference, v is frequency (f)). Frequency of the wave of light that it absorbs = f, Planks Constant = h. The colour that we see is the colour of that frequency of light that the electrons did not absorb. Hence its like this:Although it is important to understand that the electrons will only absorb the frequency they need to move from the ground state to the excited state, no MORE, no LESS. Hence the thing that defines just how far the distance is between the ground state and the excited state is the complex ion itself. This difference between the ground state and excited state is known as the d-d transition and changes as charge, oxidation state and ligand change, therefore different complexes have different colours.By finding out the frequency of light that gave maximum absorption of energy (we focus on maximum absorption so as to reduce percentage error); we can determine the concentration of a transition metal complex.Hence as you can see, the amount of maximum absorbance at a certain frequency tells you the concentration of complex. Hence, the higher the maximum absorbance is, the higher the concentration of the transition metal it is. This is similar to how a spectrophotometer works. It measures light transmission (opposite of absorbance). The higher the transmission is, the lower the concentration. However, all OTHER products must be not be coloured.

But it is important to remember that ligands are only one of many ways for a chemical to gain colour. But it is always caused by electrons absorbing a frequency of light and moving to a higher energy level, sometimes, such as in Fluorescence, electrons absorb ultra-violet light frequency and yet these electrons give out the energy they have gained in the form VISIBLE LIGHT. How is this possible? This is because after gaining energy from the frequency of Ultra-Violet light, the electrons get excited and move to a higher energy level, however, they then de-excite and move back down to a slightly lower energy level, which is in the visible light range.

A change in any of the following can cause a complex ion to change colour:a) Ligand (always remember ligands intensify colour, which helps a spectrophotometer a lot)b) Coordination Numberc) Oxidation State

Overall, the meaning of the colour of transition-metal complexes is that electrons get excited in the d-shell by ABSORBING ENERGY that is in the VISIBLE LIGHT RANGE.

Lewis Base and Bronstead Lowry BaseThere are 6 metals whose aqua complexes you must remember and the reaction of these aqua complexes with three different reagents. These 6 metals are:Cu2+, Co2+, Fe2+, Fe3+, Cr3+, Al3+Always remember these General Rules: Lewis Base = Electron Pair Donor = Ligand SubstitutionBronstead Lowry Base = Proton (usually H+) Acceptor = Acidity Reaction = Hydrated HydroxideIf it has a [ ] around the complex, it is a solution, if it does not have a [ ], then it is a precipitate. Any mention of the word AQUEOUS (in water), should automatically tell you that the aqua complex for that metal is present as metals form complex ions (make dative covalent bonds) with water as soon as they fall into water. Even if it says aqueous Cobalt Sulphate, it will still be [Cu(H2O)]2+. Do NOT write CoSO4.In these reactions, when we are making a hydrated hydroxide, we must ensure that the hydrated hydroxide has charge 0. Hence when a 3+ such as Fe3+ ion reacts with a 2- Bronstead Lowry Base such as CO32-, we must ensure that 2Fe3+ reacts with 3CO32-. This will make a 0 charge hydrated hydroxide. Think about it, a Bronstead Lowry Base accepts protons, hence 3 x (-2) = -6, therefore CO32- has 6 extra electrons (-), however 2 x (+3) = +6, therefore Fe3+ has 6 extra protons (+), hence the CO32- accepts the 6 protons, making the overall charge zero and creating a hydrating hydroxide.Polarisation: It is the distortion of positive charge by a negative charge. Hence a small negative ion is more polarising than a larger negative ion. It is the size to charge ratio, i.e. something that is very small but has a high charge is very polarising (has a very high charge density). This strengthens lattices but in transition metals it weakens the complex. This means that the compound becomes weaker. It is through polarisation that we can explain that M3+ ions are more acidic than M2+ ions. HOW?Well, if we added Fe3+ ions and Fe2+ ions to water, we would get the complexes [Fe(H2O)6]3+ and [Fe(H2O)6]2+. However, when added to water the Fe3+ distorts the H2O bounds surrounding it as does Fe2+. This is because the Fe3+ and Fe2+ are small ions with a charge greater than +1. In the Fe2+, the ligands around it (H2O) are being squeezed a little around it, hence M2+ ions such as Fe2+ are a little polarising and when added to water the distorted bond between O-H breaks and a H+ is released, this H+ does not sit around, it attaches itself via a co ordinate bond to the H2O of water and forms H3O+. Hence the H3O+ is the actual way of writing H+. Hence H+ is an acid and therefore M2+ ions when dissolved in water are slightly acidic. (pH = 3.5-4.5)

In the Fe3+, the ligands around it (H2O) are being squeezed A LOT around it, hence M3+ ions such as Fe3+ are a VERY polarising and when added to water the EXTREMELY DISTORTED bond between O-H breaks and a H+ is released, this H+ does not sit around, it attaches itself via a co ordinate bond to the H2O of water and forms H3O+. Hence the H3O+ is the actual way of writing H+. Hence H+ is an acid and therefore M3+ ions when dissolved in water are VERY acidic due to high polarisation producing higher concentration of H3O+. (pH = 3.5-4.5)NOTE: Fe3+ and Fe2+ were just examples. All M2+ and M3+ will follow the same procedure, for example Cr3+/ [Cr(H2O)6]3+ is also highly acidic for the above reasons.So overall why are M3+ ions more acidic than M2+ ions?M3+ ions are smaller and are therefore more polarising and therefore more O-H bonds are broken, making more H+.But always remember, in lattices, a high charge density strengthen the lattice, such as Sodium Chloride (Na+) being weaker than Aluminium Chloride (Al3+).

As a general rule: remember H+ is acid, hence if a substance is added to water and it gives you H+ as a product, then it is an acid.remember OH- is alkali, hence if a substance is added to water and it gives you OH- as a product, then it is an alkali.Always remember, complexes without charge such as hydrated hydroxides are insoluble in water.

The idea is that add OH- (alkali) to a metal aqua complex such as adding 2OH- [Fe(H2O)]2+ and you will get a hydrated hydroxide as the 2OH- will take one H+ each resulting in: Fe(H2O)4(OH)2, now if we add 2H3O+ (acid) to this Fe(H2O)4(OH)2 it will go back to [Fe(H2O)]2+ as the 2H3O+ will give back a H+ each to Fe(H2O)4(OH)2. Hence hydrated hydroxides are bases. However Al and Cr are amphoteric, this means that their hydrated hydroxide reacts with acid (as all others do) and bases too. If the hydrated hydroxide Al(H2O)3(OH)3 is reacted with a base again, we would get [Al(OH)4]- (in exam they may present this as [Al(OH)6]3- and [Cr(OH)6]3-.

Our first reagent: NaOH (Sodium Hydroxide); so we add our metal ion into water and this forms the aqua complex shown in the first box below, then we add NaOH into the water and this reaction takes place: NaOH + H2O Na+ + OH- , hence it is fully ionised and we get a strong concentration (amount) of OH- ions and hence these ions form the following complexes with the Aqua Complexes.ComplexpHFew Drops of Sodium Hydroxide (OH-)Excess Sodium Hydroxide (OH-)

[Cu(H2O)6]2+Blue Octahedral6.5Cu(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Pale Blue precipitate (Octahedral)[Cu(H2O)6]2+ + 2OH- Cu(H2O)4(OH)2 + 2H2ONO FURTHER CHANGE SAME PRODUCTS

[Co(H2O)6]2+Pink Octahedral6.5Co(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Pink/Blue precipitate (Octahedral)[Co(H2O)6]2+ + 2OH- Co(H2O)4(OH)2 + 2H2ONO FURTHER CHANGE SAME PRODUCTS

[Fe(H2O)6]2+Green Octahedral6.5Fe(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Green precipitate (Octahedral)[Fe(H2O)6]2+ + 2OH- Fe(H2O)4(OH)2 + 2H2ONO FURTHER CHANGE SAME PRODUCTSBut when left standing, it is oxidised (loses electrons) by air to Fe3+, and therefore turns dark brown.

[Fe(H2O)6]3+Orange/VioletOctahedral3.5 to 4.5Fe(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Orange/Brown precipitate (Octahedral)[Fe(H2O)6]3+ + 3OH- Fe(H2O)3(OH)3 + 3H2ONO FURTHER CHANGE SAME PRODUCTS

[Cr(H2O)6]3+Green (Ruby)Octahedral3.5 to 4.5Cr(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Green/Grey precipitate (Octahedral)[Cr(H2O)6]3+ + 3OH- Cr(H2O)3(OH)3 + 3H2O[Cr(OH)4]- Acidity Reaction (Lewis Base)Green Solution (Tetrahedral because OH- is a big ligand)Cr(H2O)3(OH)3 + OH- [Cr(OH)4]- + 3H2O

[Al(H2O)6]3+ColourlessOctahedral3.5 to 4.5Al(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)White precipitate (Octahedral)[Al(H2O)6]3+ + 3OH- Al(H2O)3(OH)3 + 3H2O[Al(OH)4]- Acidity Reaction (Lewis Base)Colourless Solution (Tetrahedral because OH- is a big ligand)Al(H2O)3(OH)3 + OH- [Al(OH)4]- + 3H2O

N.B. Every time a precipitate becomes a solution, we say that it is because it dissolved in that reagent, for example Al(H2O)3(OH)3 becomes [Al(OH)4]- because it DISSOLVED in excess Sodium Hydroxide

Our second reagent: NH3 (Ammonia); so we add our metal ion into water and this forms the aqua complex shown in the first box below, then we add NH3 into the water and this reaction takes place: NH3 + H2O NH4+ + OH-. Hence this tells us that as OH- is produced, the same reactions will occur when a few drops are added to the solution of water containing metal aqua complexes, so again with the OH- acting as a Bronstead Lowry Base (H+ acceptor) but well also get NH4+ as each NH3 molecule accepts a proton (H+) and becomes NH4+. Except, the difference, is that when excess NH3 is added, the NH3 begins to react with the complexes acting as a Lewis Base and causes Ligand Substitution. Also notice that Aluminium now has no difference when excess reagent is added, this is because NH3 produces a low concentration of OH- ions in comparison to NaOH, and the evidence for this is that this reaction has a , whereas NaOH has . (Any reaction with goes to completion which is an advantage; any reaction with does not go to completion which is not advantageous.ComplexpHFew Drops of Ammonia (NH3)Excess Ammonia (NH3)

[Cu(H2O)6]2+Blue Octahedral6.5Cu(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Pale Blue precipitate (Octahedral) [Cu(H2O)6]2+ + 2NH3 Cu(H2O)4(OH)2 + 2NH4+[Cu(H2O)2(NH3)4]2+ Ligand Substitution (Lewis Base) as you can see, not complete ligand substitution this ones a bit weird. Deep Blue Solution (Octahedral)Cu(H2O)4(OH)2 + 4NH3 [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-

[Co(H2O)6]2+Pink Octahedral6.5Co(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Pink/Blue precipitate (Octahedral)[Co(H2O)6]2+ + 2NH3 Co(H2O)4(OH)2 + 2NH4+[Co(NH3)6]2+ Ligand Substitution (Lewis Base)Straw/Yellow solution (Octahedral)Co(H2O)4(OH)2 + 6NH3 [Co(NH3)6]2+ + 4H2O + 2OH-

[Fe(H2O)6]2+Green Octahedral6.5Fe(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Green precipitate (Octahedral)[Fe(H2O)6]2+ + 2NH3 Fe(H2O)4(OH)2 + 2NH4+NO FURTHER CHANGE SAME PRODUCTS

[Fe(H2O)6]3+Orange/VioletOctahedral3.5 to 4.5Fe(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Orange/Brown precipitate (Octahedral)[Fe(H2O)6]3+ + 3NH3 Fe(H2O)3(OH)3 + 3NH4+NO FURTHER CHANGE SAME PRODUCTS

[Cr(H2O)6]3+Green (Ruby)Octahedral3.5 to 4.5Cr(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Green/Grey precipitate (Octahedral)[Cr(H2O)6]3+ + 3NH3 Cr(H2O)3(OH)3 + 3NH4+[Cr(NH3)6]3+ Ligand Substitution (Lewis Base)Purple SLOWLY (Octahedral)Cr(H2O)3(OH)3 + 6NH3 [Cr(NH3)6]3+ + 3H2O + 3OH-

[Al(H2O)6]3+ColourlessOctahedral3.5 to 4.5Al(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)White precipitate (Octahedral)[Al(H2O)6]3+ + 3NH3 Al(H2O)3(OH)3 + 3NH4+NO FURTHER CHANGE SAME PRODUCTS

Once the [Co(NH3)6]2+ is formed, when left on standing it gets oxidised (loses electron) and turns to [Co(NH3)6]3+ which is brown. The Oxidising agent (gains electron) is obviously O2 (oxygen).This is how it happens:[Co(NH3)6]2+ [Co(NH3)6]3+ + e- (x4)4e- + O2 + 4H+ 2H2O (x1)

OVERALL EQUATION: 4[Co(NH3)6]2+ + O2 + 4H+ [Co(NH3)6]3+ + 2H2O

However this reaction took place in alkaline solution, (can you not see when we put NH3 in water you get OH- and OH- is alkali), therefore we should have no H+, but instead OH-. So we do this be putting the same amount of OH- as there are H+ on the opposite side of the H+Hence: 4[Co(NH3)6]2+ + O2 [Co(NH3)6]3+ + 2H2O + 4OH-Then balance by adding H2O molecules wherever necessary; 4[Co(NH3)6]2+ + O2 + 4H2O [Co(NH3)6]3+ + 2H2O + 4OH-Then by cancelling the H2O molecules in the product and H2O molecules in reactant we get: 4[Co(NH3)6]2+ + O2 + 2H2O [Co(NH3)6]3+ + 4OH-

Also remember that this [Co(NH3)6]3+ is a very powerful oxidising agent and can take electrons very easily, for example it will easily turn 2I- to I2 and become Co2+ again. Ironically by adding H+ to [Co(NH3)6]3+ which will turn it into [Co(H2O)6]3+, and as mentioned this is a powerful oxidising agent and will turn back to [Co(H2O)]2+.

Our third reagent: Na2CO3; so we add our metal ion into water and this forms the aqua complex shown in the first box below, then we add Na2CO3 into the water and this reaction takes place: CO32- + H2O HCO3- + OH- (dont think where did the Na2 go? remember its an ionic compound, add it to water and it fully ionises). Hence this tells us that as OH- is produced, the same reactions will occur when a few drops are added to the solution of water containing metal aqua complexes, so again with the OH- acting as a Bronstead Lowry Base (H+ acceptor) but well also get HCO3- as each CO32- molecule accepts a proton (H+) and becomes HCO3-.However, a more major product is made in these reactions that involve a 2+ metal ion, a metal carbonate. This is because the 2+ metal ions react with the CO32- to make MCO3. This does happen with 3+ metal ions, this is because in acidic conditions, CO32- becomes CO2, and as we earlier mentioned 3+ metal ions are HIGHLY ACIDIC due to being very polarising.

ComplexpHFew Drops of Sodium CarbonateExcess Sodium Carbonate

[Cu(H2O)6]2+Blue Octahedral6.5Cu(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Pale green precipitate, this is the CuCO3[Cu(H2O)6]2+ + 2CO32- Cu(H2O)4(OH)2 + 2HCO3-Cu2+ + CO32- CuCO3 = MAJOR PRODUCTNO FURTHER CHANGE SAME PRODUCTS

[Co(H2O)6]2+Pink Octahedral6.5Co(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Purple precipitate, this is the CoCO3[Co(H2O)6]2+ + 2CO32- Co(H2O)4(OH)2 + 2HCO3-Co2+ + CO32- CoCO3 = MAJOR PRODUCTNO FURTHER CHANGE SAME PRODUCTS

[Fe(H2O)6]2+Green Octahedral6.5Fe(H2O)4(OH)2 Acidity Reaction (Bronstead Lowry Base)Green precipitate, this is the FeCO3[Fe(H2O)6]2+ + 2CO32- Fe(H2O)4(OH)2 + 2HCO3-Fe2+ + CO32- FeCO3 = MAJOR PRODUCTNO FURTHER CHANGE SAME PRODUCTS

[Fe(H2O)6]3+Orange/VioletOctahedral3.5 to 4.5Fe(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Orange/Brown precipitate and fizzing (CO2)2[Fe(H2O)6]3+ + 3CO32- 2Fe(H2O)3(OH)3 + 3CO2 +3H2ONO FURTHER CHANGE SAME PRODUCTS

[Cr(H2O)6]3+Green (Ruby)Octahedral3.5 to 4.5Cr(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)Green/Grey precipitate and fizzing (CO2)2[Cr(H2O)6]3+ + 3CO32- 2Cr(H2O)3(OH)3 + 3CO2 + 3H2ONO FURTHER CHANGE SAME PRODUCTSBut if you added some NaOH, it can become

[Al(H2O)6]3+ColourlessOctahedral3.5 to 4.5Al(H2O)3(OH)3 Acidity Reaction (Bronstead Lowry Base)White precipitate and fizzing (CO2)2[Al(H2O)6]3+ + 3CO32- 2Al(H2O)3(OH)3 + 3CO2 + 3H2ONO FURTHER CHANGE SAME PRODUCTS

N.B. Always remember that in the examination they may give you a different reagent but tell you that it reacts just like one of the above reagents, hence you will use the same method as you would for the above reagent, i.e. same type of products and observations, etc.

We can use these laws and apply it to metal extraction of Cu from Cu2+ using scrap Fe. Cu2+(aq) + Fe Cu + Fe2+(aq), this is better than extraction of Cu from CuO using CO (Carbon Monoxide. WHY? There is a low energy requirement, CO2 harmful gas not produced and scrap iron is cheap.

N.B. In all these reactions we have used transition metal aqua complexes as [M(H2O)6]2+/3+, whereas really it should really be [M(OH2)6]2+/3+ because remember, it is the Oxygen that forms a coordinate bond to the metal, not the Hydrogen.

There is one more reagent: HCl, but we only need to know it for two transition metals: Co2+, Cu2+[Cu(H2O)6]2+ + 4Cl- [CuCl4]2- + 6H2O[Co(H2O)6]2+ + 4Cl- [CoCl4]2- + 6H2OCatalysts

Transition metals can also act as catalysts; this is due to the fact that they can change their oxidation states. This is different from other blocks such as the S block that have only one oxidation state.They act as catalyst intermediates. They speed up the reaction but have no effect on the position of the equilibrium, this is because forward and backward rate changes by the same amount, hence concentration of reactants and products remain constant and therefore yield does not change. These catalysts are made more efficient by increasing their surface area; therefore they are placed over a coated mesh or made into a powder or placed on a ceramic support, doing this also reduces the cost of the catalyst. ONLY transition metals can be catalysts.

A + catalyst Y + reactive intermediate Reactive intermediate + B catalyst + ZA + B Y + Z

Homogenous Catalyst is when the catalyst is in the same phase (solid, liquid, gas) as the reactants. They form an intermediate.

EXAMPLE 1: Catalyst: Fe3+(aq)I-(aq) + S2O82-(aq) has very high activation energy; this is because they are both negatively charged, therefore repulsion, but when they do react, the products are I2 and 2SO42- . Hence we use Fe3+ as an intermediate and this lowers the activation energy1) 2Fe3+ + 2I- 2Fe2+ + I22) 2Fe2+ + S2O82- 2Fe3+ + 2SO42-

2I- + S2O82- I2 + 2SO42-P.S. We can also use Fe2+ as the catalyst and react with S2O82- first, hence equation 1 and 2 can occur in any order. The fact that equation 1 and 2 can occur in any order is evidence that Fe2+ and Fe3+ are equally effective as the catalyst here.

EXAMPLE 2: Catalyst: Mn2+MnO4-(aq) + C2O42- has very high activation energy; this is because they are both negatively charged, therefore repulsion, but when they do react, the products are CO2 and Mn2+. Hence we use a Mn2+(aq) catalyst as then activation energy would be low as a positively charged ion would be reacting with a negatively charged ion. But Mn2+(aq) is also a products of the reaction, therefore autocatalysis takes place where the product of the reaction was the catalyst for the reaction.

HENCE THE REACTION THAT TAKES PLACE IS:1) 8H+ + 5e- + MnO4- Mn2+ + 4H2O (x2)2) C2O42- 2CO2 + 2e- (x5)

16H+ + 2MnO4- + 5C2O42- 10CO2 + 2Mn2+ + 8H2O

But the real reaction that has happened with the catalyst is:1) 4Mn2+ + 8H+ + MnO4- 5Mn3+ + 4H2O2) 2Mn3+ + C2O42- 2CO2 + 2Mn2+

Hence MnO4- is purple coloured, so we can track its progress in becoming Mn2+ (colourless) via a spectrophotometer, this is because all other substances in the above reactions are colourless. Hence our spectrometer would turn from purple (MnO4-) to colourless (Mn2+). Also remember that a spectrophotometer is the best method to measure the concentration of coloured ions as it gives you a rapid determination of concentration and does not interfere with the reaction.

This is a very important graph that appears frequently in examinations.Therefore as you can see, the reaction is slow at first, but then speeds up as more of Mn2+ is made.

EXAMPLE 3: Catalyst Co2+SO32- is turned into SO42- in the process of flue-gas desulphurisation. The oxidising agent (takes electrons) is Oxygen.

HENCE THE REACTION THAT TAKES PLACE:1) SO32-

Heterogeneous Catalyst is when the catalyst is in a different phase (solid, liquid, gas) to the reactants. Their catalytic action occurs on the surface. They lower the activation energy by forming bonds between the molecule and the surface whilst breaking them, hence they make bonds whilst breaking them, balancing out the enthalpies. The strength of adsorption helps to determine the activity of the catalyst. If adsorption (bringing reactants together at the surface and orientating them) is too weak, the reactants are not brought together at the surface (e.g. silver).If adsorption is too strong, products cannot move around the surface and block the active sites, also they cannot be desorbed from the surface (e.g. tungsten).

EXAMPLE 1: Haber Process; Catalyst: Fe (Iron)N2(g) + 3H2(g) 2NH3 (g)H = - S = -

EXAMPLE 2: The Contact Process ; Catalyst: V2O5 (gains electrons/reduced as V has oxidation state +5)1) V2O5 + SO2 V2O4 + SO32) V2O4 + 1/2O2 V2O5H = - S = -SO2 + 1/2O2 SO3The Sulphur Trioxide is converted to Sulphuric Acid

EXAMPLE 3: Making Methanol from CO and H2; Catalyst: Cr2O3CO + H2 CH3OHH = - S = -

Sometimes, we use a weak catalyst (such as Silver which is a weak catalyst as reactants are adsorbed very weakly) even though we know that a more powerful catalyst could be used, this is because the reaction is very exothermic and using a better catalyst may cause an explosion.

On the surface of the catalyst, there are active sites and particles migrate to and from the active site. If there is some sort of impurity inside the catalyst, POISON could attach to the active sites irreversibly. To ensure this does not happen, the catalyst must be purified properly. The poison will reduce the effectiveness of the catalyst as it attaches to the surface. However Catalyst Poisoning does happen, such as lead in petrol will poison the Catalytic Converter. This is extremely dangerous as if the Catalytic Converter is not working, then more harmful gases would be produced by the car.

Just to recap Unit 1, the reaction in the Catalytic Converter is: 2NO + 2CO N2 + 2CO2The reducing agent in this reaction is CO (Carbon Monoxide). The catalyst is Platinum which is coated over a mesh.

Usually all sulphur compounds are excellent poisons, sulphur actually poisons the iron catalyst in the Haber Process, this is why sulphur compounds must be removed from oil/gas before catalytic reactions.

Always remember, adsorption is fuelled by enthalpy, but desorption is fuelled by entropy.

Reactions Involving Changing Valency

The most important reactions you must remember showing the changing of valency for transition metals involve the reduction (gain electrons) of the transition metals: Chromium and Vanadium.

The reducing agents (lose electrons) are: Zinc (Zn) and Hydrochloric Acid (HCl)/Sulphuric Acid (H2SO4)

For Chromium reduction, it goes like this: Cr6+ (Cr2O72-/CrO42-) = orange Cr3+ ([Cr(H2O)6]3+) = green Cr2+ = blueThe first reaction:Cr2O72- + 6e- +14H+ 2Cr3+ + 7H2OThe second reaction:Cr3+ + e- Cr2+The Zinc (Zn) and Hydrochloric Acid (HCl) oxidation reaction will be:Zn Zn2+ + 2e-

For Vanadium reduction, it goes like this:V5+ (VO2+) = yellow V4+ (VO2+) = blue V3+ = green V2+ = violetThe first reaction:VO2+ + 2H+ + e- VO2+ + H2OThe second reaction:VO2+ + e- + 2H+ V3+ + H2OThe third reaction:V3+ V2+ + e-The Zinc (Zn) and Hydrochloric Acid (HCl) reaction oxidation will be:Zn Zn2+ + 2e-

Be wary though, because at times they can ask you to go from V5+ straight to V2+ or from Cr6+ straight to Cr2+.

Now, in the above, the species which had metals in them were all gaining electrons (e.g. Cr3+ to Cr2+). However, there is one special case that we are required to learn that involves oxidation (lose electrons) of the transition metals Cr3+ and Co2+ to Cr6+ and Co3+ respectively.

The oxidising agents (gain electrons) will be H2O2 and NaOH

For Chromium oxidation it goes like this: 4H2O + Cr3+ CrO42- + 3e- +8H+The H2O2 and NaOH will be:H2O2 + 2e- + 2H+ 2H2O

But remember, that once again, this reaction will be in alkaline conditions, therefore, in the overall reaction, you must replace the H+ with OH-.

Also remember the following equilibrium:2CrO42- + 2H+ Cr2O72- + H2OyelloworangeHence in alkaline conditions, you will get CrO42-, and hence you get a yellow colour, but when in acid conditions, you will get Cr2O72-, and hence you get an orange colour.

For Cobalt oxidation it goes like this:Co2+ Co3+ +e-The H2O2 and NaOH will be:H2O2 + 2e- + 2H+ 2H2O

TitrationsThe main titrations the examination will ask are (i.e. the questions regarding moles the examination will ask):1) Fe2+ (after acidification) with Cr2O72- (oxidation number of Cr is +6)2) Fe2+ (after acidification) with MnO4- (oxidation number of Mn is +7)

Above, we have mentioned that Fe2+ is acidified first, but with MnO4-, the acidification should not be done with Hydrochloric Acid, this is because MnO4- is a powerful oxidising agent and therefore would oxidise the 2Cl- to Cl2(aq). The acidification should also not be done with Nitric Acid as NO3- can oxidise Fe2+ to Fe3+.Always remember in both of these, you should write the half-equations first, this will gain you a mark.The fact that Fe2+ is acidifies tells us that this reaction takes place in acidic conditions, therefore H+ in equations.

1) Half-equations of the first reaction:Fe2+ Fe3+ + e- (x6)Cr2O72- + 6e- +14H+ 2Cr3+ + 7H2O (x1)

Therefore, overall reaction: 6Fe2+ + Cr2O72- +14H+ 2Cr3+ + 7H2O + 6Fe3+

2) Half-equations of the second reaction:Fe2+ Fe3+ + e- (x5)MnO4- + 5e- + 8H+ Mn2+ + 4H2O (x1)

Therefore, overall equation reaction: 5Fe2+ + MnO4- + 8H+ Mn2+ + 4H2O Fe3+

But beware as they may at times ask the titrations of Cr2O72-, MnO4- with H2O2 instead of Fe2+. You should then use the same procedure of using half-equations, but this time involving H2O2.Now as we know H2O2 can be oxidised (gives electrons) to form Oxygen, and H2O2 can also be reduced (takes electrons) to form Hydrogen. Here we will have to use the oxidation version as reduction is already happening to the Cr2O72-, MnO4- (theyre both gaining electrons). Hence, our product will be Oxygen.

The half-equation for H2O2 is: H2O2 O2 + 2e- + 2H+

Therefore from above we know the half equations for both Cr2O72-, MnO4-. Hence we balance electrons in the same way that we did for Fe2+.

Remember the following for titrations:SOLIDSOLUTIONGAS

Moles=Mass/MrMoles=(Concentration x volume)/1000Moles=Volume/24000

Moles= Particles/6.02 1023

Atom Economy: Mr of wanted product/Mr of total productIf atom economy is very low, the company tries to sell the unwanted product to gain financially. Percentage Yield: (Mass of actual yield/Mass of theoretical yield) 100Theoretical yield done like this: Mass/Mr=Mass/MrIdeal Gas Equation: pV=nRTP must be in Pa, not KPaV must be in m3, (if in cm3 multiple by 10-6 or divide by 1000000) (if in dm3 multiply by 10-3 or divide by 1000)N will be in MolR will be 8.31T will always be in K (Kelvin), if not, add 273 to the oC.If ever asked for the concentration in g dm-3, work out the concentration in mol dm-3 like normal and then multiply by the Mr.

A FEW TIPS:

Remember HCl(g) is corrosive Remember reactions that are highly exothermic are vigorous and therefore could be dangerous Hydrogen gas is collected with a gas syringe Remember AgCl is a white precipitate, AgBr is a cream precipitate and AgI is a yellow precipitate. In every J = m x 4.18 x T equation, remember that when an enthalpy is given it is given per mol, hence when you attempt to get your J value, work out the moles of the reactant, multiply it by the enthalpy of the reactant (as enthalpy is KJmol-1), then multiply by 1000 (to get it into Joules (J)). In the same way remember when it is mentioned that 3.49kJ of heat energy to convert to 1.53g of liquid water into steam, hence work out enthalpy. Now, first work out the moles of water, then remember that enthalpy is kJ per mole (kJ moles), therefore now you will do 3.49 moles of liquid water. Remember simple reactions such as Fe + 2HCl FeCl2 + H2 FeC2O4 Fe3+ + 2CO2 + 3e- can be separated into two equations: Fe2+ Fe3+ + e-C2O42- 2CO2 + 2e- Remember, oxidation of alcohols is alcohol aldehyde carboxylic acid Also aldehydes have a low boiling point and can be distilled readily In metal extraction, you can only separate it a substance via electrolysis if it is able to conduct electricity, i.e. if the substance is a molecular covalent structure, you cannot use electrolysis. Finally, read up on all AS notes, especially metal extraction (unit 2), alcohols (unit 2) and bonding (unit 1)