# OCR Advanced GCE in Chemistry/Printable version

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/OCR_Advanced_GCE_in_Chemistry

# Arenes

Benzene is the most common Arene. It has a hexagonal shape. The electrons in a benzene molecule are in 'pi' bonds. Each of the carbon atoms contribute one electron into a 'pi' bond. The electrons are 'delocalised' as they are spread out over all 6 'pi' bonds within the benzene molecule.
To break the 'pi' bonds in benzene compounds requires a lot of energy and therefore they do not always react easily. The majority of their reactions are substitution reactions and there are 4 groups known as the 4 ations that are important to this module.
The first of the 4 ations are nitroarenes. The nitrogroups are substituted for hydrogen atoms on the benzene ring. They are used for drugs and dyes and also explosives such as TNT.

# Lattice enthalpy

Lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent.

## Lattice enthalpy

Lattice enthalpy is simply the change in Enthalpy associated with the formation of one mole of an ionic compound from its gaseous ions under standard conditions.
e.g. Mg$^{2+}$(g) + 2Cl$^-$(g) ---> MgCl$_2$(s)ΔH$_{latt}$ = -2526 kj mol$^{-1}$
Therefore:
* The enthalpy change will always be exothermic (negative)
* A more negative value shows greater electrostatic attraction and therefore a stronger bond in the solid
Definition: The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions
Lattice enthalpy cannot be measured because gaseous ions do not combine directly to form a compound.

## Born-Haber cycles

As lattice enthalpy cannot be measured, another way of working out lattice enthalpy values is needed. To do this we must show the complete pathway for the reaction, including lattice enthalpy (Route 2 below) and compare this to the enthalpy change of formation (Route 1). We can do this because Hess's law states:
enthalpy change of formation = atomisation enthalpies + electron affinity + lattice enthalpy
The lattice enthalpy is step 5 in route 2 of the example. First, the elements must be atomised and ionised in steps 1-3:
Step 1 - Atomisation of solid Lithium to the gaseous state
Definition: ΔH$_{at}$The standard enthalpy change of atomisation of an element is the enthalpy change when one mole of gaseous atoms are formed from the element in its standard state.
Step 2 - Ionisation of gaseous Lithium atoms to 1+ ions
Definition: ΔH$_{i.e.}$The standard enthalpy change accompanying the removal of one electron from an atom in the gas phase.
Step 3 - Atomisation of fluorine gas (Hint: fluorine still needs to be atomised here even though it is gaseous because it is an Fl$_2$ molecule)
Step 4 - Electron affinity of fluorine. This is basically making the gaseous fluorine atoms into ions by adding an electron.
Definition: ΔH$_{ea1}$The first electron affinity is the enthalpy change when one electron is added to each gaseous atom in one mole, to form one mole of 1- ions.
Step 5 - Lattice enthalpy! can now be worked out. The easiest way to understand how the cycle works is to remember that route 1 and route 2 are equal. So using Hess's law (as above) lattice enthalpy can be calculated when the other values are known.
Things to watch out for:
* Ions requiring two additions or subtractions of electrons e.g. Mg$^{2+}$
For these use the values for each, DO NOT simply double the first value.
* Ionic solids that have more than one of a single ion e.g. MgCl$_2$
Also try and learn the definitions because these are easy marks.

## Trends in Lattice enthalpy

There are two factors that affect the lattice enthalpy of an ionic compound:
* The size of the ions
* The charge of the ions
This is because these are also the factors that affect charge density. Charge density of an ion will determine its "attracting power" and therefore the greater the charge density of the ions, the greater the electrostatic forces between them. So a larger ion will decrease the charge density and a more highly charged ion will increase the charge density.

## Thermal decomposition of group 2 carbonates

The Group 2 carbonates MgCO$_3$, CaCO$_3$, SrCO$_3$ and BaCO$_3$ all react in the same way when heated:
MgCO$_3$(s) ---> MgO(s) + CO$_2$(g)
CaCO$_3$(s) ---> CaO(s) + CO$_2$(g)
SrCO$_3$(s) ---> SrO(s) + CO$_2$(g)
BaCO$_3$(s) ---> BaO(s) + CO$_2$(g)
The decomposition temperatures for these reactions are as follows:
MgCO$_3$ = 350 °C
CaCO$_3$ = 832 °C
SrCO$_3$ = 1340 °C
BaCO$_3$ = 1450 °C
A trend of increasing temperature down the group is evident. This may seem confusing because the magnesium cation is the smallest and they all have the same charge. Magnesium therefore has the greatest lattice enthalpy and the greatest attraction to the carbonate ion. However, this does not result in a higher decomposition temperature - the stronger pull of a smaller cation combined with the diffuse electron cloud of a carbonate ion causes much more polarisation of the carbonate ion to occur.
The change of shape is where the electrons are pulled onto one of the Oxygen atoms causing distortion in the carbonate ion. This change in shape reduces strength and causes the carbonate to break up when heated.
Magnesium Oxide is a particularly favourable product due to its high lattice enthalpy, this contributes to the ease of decomposition of magnesium carbonate and is why magnesium oxide is used as a refractory lining.

## Questions

1. Explain why the lattice enthalpy of NaBr is much less exothermic than MgCl$_2$
2. How is Hess's law used to work out lattice enthalpy from born-haber cycle?
3. What are the factors affecting lattice enthalpy?
4. What is the definition of 1st electron affinity?
5. What is the trend in decomposition temperatures of group two carbonates? why is this?
Answers

# Periodic Table: Period 3

In this section there is a large amount of information and unfortunately most of it just has to be learnt. Read the suggestions at the end of each sub-section for best ways of learning.

## Period 3 Oxides

You are required to know about the formation and reactions of MgO, Al$_2$O$_3$ and SO$_2$

### Magnesium oxide

MgO is formed simply by heating in oxygen:
Mg(s) + 1/2 O$_2$(g) ---> MgO(s)
Observations: Burns with bright white flame to form white solid. (used in fireworks for this reason)
Basic oxide: Magnesium oxide is a basic oxide, meaning that when it dissolves in water an alkaline solution will be made:
MgO(s) + H$_2$O (l) ---> Mg(OH)$_2$ (aq)
Mg(OH)$_2$ is an weak alkali due to partial dissociation of ions.
Structure: Magnesium oxide forms a giant lattice structure with ionic bonding

### Aluminium Oxide

Aluminium reacts very easily in oxygen, so much so that at room temperature it forms a layer around aluminium. Once this layer has formed, it protects the aluminium and prevents it from readily reacting. However, if the aluminium is in powdered form the aluminium burns. The formation reaction is as follows:
4Al(s) + 3O$_2$(g)---> 2Al$_2$O$_3$(s)
Observations: Burns when powdered - white solid product
Amphoteric oxide: This means that it can react as either an acid or a base (see below). However, aluminium oxide will not react with water (insoluble).
Structure: Aluminium Oxide has 'intermediate bonding'- This basically means that it has a Giant lattice structure but also has some covalent bonding too. The intermediate bonding in aluminium oxide explains why it is insoluble.

### Sulphur dioxide

Sulphur burns in oxygen to produce sulphur dioxide:
S(s) + O$_2$(g) ---> SO$_2$(g)
Observations: Burns with blue flame to give gas
When heated with a platinum catalyst SO$_3$(l) is formed
Acidic oxide: Sulphur dioxide reacts with water leaving an acidic solution:
SO$_2$(l) + H$_2$O(l) ---> H$_2$SO$_3$(aq)
and similarly:
SO$_3$(l) + H$_2$O(l) ---> H$_2$SO$_4$(aq)
Structure: Simple molecular

### Acid-base reactions

Some questions ask you to give examples of "acid-base reactions" they may or may not say oxides specifically. Basically an acid-base reaction is one where an acid and a base react with each other. We have already discussed the nature of these oxides (e.g. acidic oxide - sulphur) and this refers to the role that it plays in the reaction:
MgO(s) + 2HCl(aq) ---> MgCl$_2$(aq) + H$_2$O(l)
Here, magnesium oxide acts as a base in a reaction with hydrochloric acid.
Aluminium oxide is amphoteric so acts as a base or an acid. Its reaction as a base is similar to above but its reaction as an acid occurs when it is hydrated:
Al(OH)$_3$(s) + OH$^-$(aq) ---> [Al(OH)$_4$]$^-$(aq)
It is also worth knowing the acid-base reaction for sulphur because it is a little different to what may be expected. We already know that sulphur is an acidic oxide and that it is able to form SO$_3$, and now we can see this in action:
SO$_3$(g) + NaOH(aq) ---> Na$_2$SO$_4$(aq) + H$_2$O(l)
The OH group is removed in the reaction as SO$_3$ is acting as an acid.
Remember:
* Basic oxides (left of periodic table) have giant lattice structures and form alkaline solutions in water
* Amphoteric oxides (aluminium oxide - metalloid) have giant lattice structure with some covalent bonding so don't dissolve
* Acidic oxides (right of periodic table) are simple molecules and react with water to produce acid

## Period 3 Chlorides

The chlorides that you need to know about are NaCl, MgCl$_2$, AlCl$_3$, SiCl$_4$ and PCl$_5$

### Sodium and Magnesium Chlorides

2Na(s) + Cl$_2$(g) ---> 2NaCl(s)
Mg(g) + Cl$_2$(s) ---> MgCl$_2$(s)
There is nothing fancy about these reactions or products - you simply heat the metals and lower into chlorine gas to form white solids.
Structure and bonding - Both of these products have giant ionic lattice structures with ionic bonding
Reactions with water (products) - Sodium chloride dissolves in water, leaving a neutral solution. Magnesium chloride also dissolves but leaves a slightly acidic solution.

### Aluminium Chloride

Some text books miss out Aluminium chloride but the specification includes it with the other chlorides so it is important. Aluminium Chloride can be prepared under anhydrous conditions by heating aluminium in dry chlorine gas:
2Al(s) + 3Cl$_2$(g) ---> 2AlCl$_3$(s)
When AlCl$_3$ is heated above its melting point of 178 °C (451 K) it forms a dimer with formula Al$_2$Cl$_6$ where the chlorine atoms donate electron pairs forming dative covalent bonds. At higher temperatures the Al$_2$Cl$_6$ dimer dissociates into trigonal planar AlCl$_3$
Structure - Aluminium trichloride has simple molecular structure with covalent bonding
Reaction with water - Hydrolysis reaction evolving hydrogen chloride gas, some dissolves to leave acidic solution:
Al$_2$Cl$_6$(s) + 6H$_2$O(l) ---> 2Al(OH)$_3$(s) + 6HCl(g)

### Silicon tetrachloride

Silicon tetrachloride is prepared in a similar way to aluminium chloride except a side arm is needed because silicon tetrachloride is a liquid at room temperature.
Si(s) + 2Cl$_2$(g) ---> SiCl$_4$(l)
Structure - Molecular
Reaction with water - Hydrolysis:
SiCl$_4$(l) + 2H$_2$O(l) ---> SiO$_2$(s) + 4HCl(g)
Observations - white fumes of hydrogen chloride gas

### Phosphorus pentachloride

Phosphorus trichloride combines directly with chlorine without heating:
P$_4$(s) + 6Cl$_2$(g) ---> 4PCl$_3$(l)
However, we are not really interested in the trichloride but the solid pentachloride that is formed in excess chlorine:
PCl$_3$(l) + Cl$_2$(g) ---> PCl$_5$(s)
Stucture - Covalent molecular in gaseous state
Reaction with water - Hydrolysis, leaving acidic solution:
PCl$_5$(s) + H$_2$O(l) ---> 5HCl(g) + H$_3$PO$_4$(aq)
Remember:
* Neutral chlorides (left of periodic table) have giant lattice structures and dissolve to give neutral solutions (or very slightly acidic)
* Acidic chlorides (right of periodic table) have simple molecular structure and undergo hydrolysis reactions to leave acidic solutions

## Reactions of period 3 elements with water
### Sodium

Sodium reacts vigorously with water, fizzing around the surface and sometimes catching fire:
2Na(s) + 2H$_2$O(l) ---> 2NaOH(aq) + H$_2$(g)
The product sodium hydroxide is a strong alkali with a high PH (>12) sodium is a very reactive element in the periodic table

### Magnesium

The reaction of magnesium with water is much slower than sodium, hydrogen bubbles form very slowly:
Mg(s) + 2H$_2$O(l) ---> Mg(OH)$_2$ + H$_2$(g)
This alkali is much weaker than sodium hydroxide (PH 9-11) This is because sodium hydroxide dissolves more easily - more ions in the solution.
However, if magnesium is reacted with steam the reaction is much faster in forming the oxide:
Mg(s) + H$_2$O(g) ---> MgO(s) + H$_2$(g)

## Questions

(These questions are easy if you look at the page so do it without to check you know everything)
1. Give an example of hydrolysis and acid-base reactions. Show equations and state observations
2. State the type of bonding in Aluminium oxide, does it dissolve?
3. Which reacts most readily with water - sodium or magnesium?
4. How can magnesium be made to react with water?
5. How does one make PCl$_5$?
6. Write out all 21 equations shown on this page...(no seriously)
Answers

# Periodic Table: Transition elements
## Electronic structures

A transitional element is defined as a d-block element forming one or more stable ions with incompletely filled d-orbitals.
If we look at some examples of the transitional elements electronic configurations we can see how this works:
Scandium (Sc) - [Ar] 3d$^1$ 4s$^2$
titanium (Ti) - [Ar] 3d$^2$ 4s$^2$
I don't get this?
Remembering that electrons are first removed from the 4s sub-shell:
titanium can form 2+ 3+ (and 4+) both of which result in ions with incomplete d-orbitals
However, scandium only forms 3+ ions and they have an empty d sub-shell. Therefore scandium is not classified as a transitional element. Zinc is the same because it only form 2+ ions, loosing two electrons from the 4s sub-shell and maintaining a full d sub-shell.
So now we have a nice pattern of increasing electron number in the 3d sub-shell in the group 4 transition elements. However as always there are exceptions to this rule:
One would expect chromium to have an electronic structure as follows:
[Ar] 3d$^4$ 4s$^2$
but actually it has the structure:
[Ar] 3d$^5$ 4s$^1$
This is because half filled sub-shells and completely filled sub-shells are the most stable so they are favoured. This is the same in copper:
[Ar] 3d$^{10}$ 4s$^1$
Where having a complete d sub-shell is preferred to having 9 electrons and 2 in the 4s sub-shell

## Oxidation states, coloured ions and catalytic behaviour

The transitional elements can form compounds in multiple oxidation sates. Most elements only form compounds in one oxidation state so transitional elements are rare. The reason that this happens is to do with the energy levels between the 4s sub-shell and within the d sub-shell; they are all very similar so movement of multiple electrons doesn't require too much energy.
The main oxidation states of iron and copper compounds (the ones that you must know) are as follows:
Fe +2 +3 +4 +5 +6
Cu +1 +2 +3
With the most common oxidation states in bold
Transitional metal ions are coloured in aqueous solutions because they have partly filled d orbitals (this is discovered more in the transitional elements exam). This therefore suggests that compounds of zinc and scandium will not have colour - this is true.
The most important coloured ions to learn are:
Fe$^{2+}$ green
Fe$^{3+}$ yellow
Cu$^{2+}$ blue
By having partly fill d orbitals transitional metals can absorb light in our visible spectrum. This means that the light that is reflected (the light we see) will have the absorbed wavelength of light missing. Therefore the wavelength and colour of absorbed light is different to the wavelength and colour of the light we see.
This colour wheel demonstrates which colour a compound will appear if it only has one absorption in the visible spectrum. For example, if the compound absorbs red light, it will appear green.
λ absorbed versus colour observed 400nm Violet absorbed, Green-yellow observed (λ 560nm) 450nm Blue absorbed, Yellow observed (λ 600nm) 490nm Blue-green absorbed, Red observed (λ 620nm) 570nm Yellow-green absorbed, Violet observed (λ 410nm) 580nm Yellow absorbed, Dark blue observed (λ 430nm) 600nm Orange absorbed, Blue observed (λ 450nm) 650nm Red absorbed, Green observed (λ 520nm)

## Precipitation reactions

There are 3 precipitation reactions in this section which must be learnt with appropriate observations remembered. They are all very similar with a solution of transition metal ions reacting with aqueous sodium hydroxide:
* Cu$^{2+}$(aq) + 2OH$^-$(aq) ---> Cu(OH)$_2$(s)
(blue) --------> (pale blue)
* Fe$^{2+}$(aq) + 2OH$^-$(aq) ---> Fe(OH)$_2$(s)
(green) --------> (green)
* Fe$^{3+}$(aq) + 3OH$^-$(aq) ---> Fe(OH)$_3$(s)
(yellow) --------> (rust)
These reactions produce solid 'gelatinous' precipitates (called this because they are jelly-like)

## Complexes
### What is a complex ion?

Transition metals have a tendency to form complexes (coordination compounds) this is due to their partially filled d sub-shell accepting donated electron pairs from other ions or molecules.
A complex ion is an ion containing a central atom or ion to which other atoms, ions or molecules are bonded. The atoms, ions or molecules are bonded to it with dative covalent bonds and are called ligands
On the right is an example of the complex ion Hexaaquavanadium(III). The central vanadium 3+ metal ion has formed dative bonds from the lone pairs of electrons provided by the water molecule ligands. As water is neutral the overall charge will stay as 3+ but with other ligands that are charged such as Cl$^-$ and CN$^-$ the overall charge will be affected.
The number of dative covalent (coordinate) bonds that are formed between the central metal and ligands is called the coordination number.

### Shapes and Names

Naming complexes can seem tricky at first but once you get used to the systematic way of naming the compounds it is easy to stick to the rules:
1. Identify the number of ligands and use the appropriate prefix - 1 mono, 2 di, 3 tri, 4 tetra, 5 penta, 6 hexa
2. Identify the name of the ligands:
* If the ligand is an anion the name will end with an 'o' e.g. cloro (Cl$^-$), cyano (CN$^-$), hydroxo (OH$^-$)
* Most neutral ligands use their usual names but water is aqua, ammonia is ammine and carbon monoxide is carbonyl.
3. Identify the metal in the centre and consider the overall charge of the complex:
* If neutral or positive use the normal name of the metal e.g. copper
* If negative the name will end in 'ate' e.g. ferrate (iron), cuprate (copper), plumbate (lead)
4. Identify the oxidation state of the central metal ion and show in roman numerals in brackets at the end.
On the left we have an example of a complex with an overall negative charge (2-) so cuprate is used instead of copper.
On the right there is an example of a complex with two different types of ligands, this does not affect the naming procedure other than having to do steps 1 and 2 twice.
Here are some more examples of complexes:
[Co(H$_2$O)$_6$]$^{2+}$ hexaaquacobalt(ll) ion
[CuCl$_4$]$^{2-}$ tetrachlorocuprate(ll) ion
[FeCN$_6$]$^{4-}$ hexacyanoferrate(ll) ion
[CrCl$_2$(NH$_3$)$_4$]$^+$ dichlorotetraamminechromium(lll) ion
Shapes - The shape of Hexaaquavanadium(III) as seen above is octahedral and the shape of Tetrachlorocuprate(II) is tetrahedral

### Ligand substitution reactions

When a more stable complex is possible, ligands can be substituted by other ligands. For the exam 3 ligand substitutions must be known:
[Cu(H$_2$O)$_6$]$^{2+}$(aq) + 4Cl$^-$(aq) ---> [CuCl$_4$]$^{2-}$(aq) + 6H$_2$O(l)
In this reaction blue hexaaquacopper(ll) undergoes ligand substitution when concentrated hydrochloric acid is added drop by drop and yellow tetrachlorocuprate(ll) is formed.
[CuCl$_4$]$^{2-}$(aq) + 4NH$_3$(aq) + 2H$_2$O(l) ---> [Cu(NH$_3$)$_4$(H$_2$O)$_2$]$^{2+}$(aq) + 4Cl$^-$(aq)
The yellow tetrachlorocuprate(ll) will then react with conc ammonia and water to form deep-blue tetraamminediaquacopper(ll)
Therefore out of the three complex ions that we have mentioned tetraamminediaquacopper(ll) is the most stable and hexaaquacopper(ll) is the least stable
[Fe(H$_2$O)$_6$]$^{3+}$(aq) + SCN$^-$(aq) ---> [Fe(H$_2$O)$_5$SCN]$^{2+}$(aq) + H$_2$O(l)
Here yellow hexaaquairon(lll) becomes blood-red pentaaquathiocyanoiron(lll)

### Colorimetry

We can use the colour changes that we have just learned about in the ligand substitution reactions to work out the ratio of ligand to metal in a complex. This can be done by adding different volumes of the new ligand to the original complex. A narrow beam of light can then be shone through a filter which allows only one wavelength of light to be absorbed by the solution (otherwise white light would be absorbed by all colours and the colour change would make no difference). The remaining light is then detected by a photocell the other side of the solution and a meter records the results.
The colour of the filter should be that of the colour that the new complex will absorb (opposite to the colour reflected). This means that as the reaction proceeds, more and more light will be absorbed.
Providing enough samples are used, a pattern should appear where there is a limit to the absorbance after a certain volume is reached. This is because all of the ligands have been substituted. Using n=cv the number of moles of both the new ligand used and the original complex for the first of the maximum absorbance values can be calculated. The ratio between these two values is the ration of ligand to metal in a complex.

## Redox behaviour
### Redox reactions

A redox reaction is one where reduction and oxidation both occur simultaneously. Transition metals have a variety of different oxidation states so can be reduced or oxidised easily. Constructing redox equations isn't very difficult once you have two half-equations:
e.g. Redox reaction between iron(ll) ions and manganate(Vll) ions in acidified aqueous solution
The half-equation for the oxidation of Fe(ll)
Fe$^{2+}$(aq) ---> Fe$^{3+}$(aq) + e$^-$
The half-equation for the reduction of potassium manganate(Vll) in acidified solution
MnO$_4$$^-$(aq) + 8H$^+$(aq) + 5e$^-$ ---> Mn$^{2+}$(aq) + 4H$_2$O(l)
To construct the full equation the number of electrons must be equal so the Fe equation must be multiplied by 5:
5Fe$^{2+}$(aq) ---> 5Fe$^{3+}$(aq) + 5e$^-$
The full equation can now be constructed by adding both sides from the two equations:
5Fe$^{2+}$(aq) + MnO$_4$$^-$(aq) + 8H$^+$(aq) + 5e$^-$ ---> 5Fe$^{3+}$(aq) + 5e$^-$ + Mn$^{2+}$(aq) + 4H$_2$O(l)
And without the electrons:
5Fe$^{2+}$(aq) + MnO$_4$$^-$(aq) + 8H$^+$(aq) ---> 5Fe$^{3+}$(aq) + Mn$^{2+}$(aq) + 4H$_2$O(l)
Observations: Colour change from purple to very pale pink (almost colourless) and the Fe$^{3+}$ ions cannot be seen as they are too faint.

### Redox titrations

Using the equation learnt in the previous section we can work out (for example) the percentage by mass of iron in iron tablets bought from a chemist:
In a titration reaction 2.13g of iron tablets were dissolved in sulphuric acid, filtered and washed with water into a standard flask making 250cm$^3$. Then 25cm$^3$ was titrated with 0.01M potassium manganate(Vll). The average titre value after 3 consistent titrations was 12.0cm$^3$.
Remembering that the equation below will confirm the reactions completion at the first permanent pink tinge.
5Fe$^{2+}$(aq) + MnO$_4$$^-$(aq) + 8H$^+$(aq) ---> 5Fe$^{3+}$(aq) + Mn$^{2+}$(aq) + 4H$_2$O(l)
First the number of moles of MnO$_4$$^-$ must be calculated using n=cv:
0.01 x 12 = 1.2x10$^{-4}$ moles
Then the number of Fe$^{2+}$ can be calculated by looking at the molar ratios in the equation:
1.2x10$^{-4}$ x 5 = 6 x 10$^{-4}$
This is the number of moles for the 25cm$^3$ so to find the total for all the tablets (in the 250cm$^3$) the value must be multiplied by 10 to give 6 x 10$^{-3}$ moles
Finally the number of moles in the solution is multiplied by the mass of Iron 56 and divided by 5 to give the value of one tablet only:
6 x 10$^{-3}$ x 56 x 1/5 = 0.00672
Therefore: 0.0672/0.425 x 100 = 15.77%
Unfortunately this value is not the same as on the box and this is due to inaccuracies in the execution of the procedure

## Questions

1. Why do the transition metals Cu and Cr not follow the usual electronic structure pattern?
2. What is a transitional metal? which period 4 d-block elements aren't classified as transitional metals? why?
3. Why are transitional metal ions coloured?
4. What colour are aqueous Fe$^{2+}$ ions? Which colour and wavelength do they therefore absorb most of?
5. Fe$^{3+}$(aq) + 3OH$^-$(aq) ---> Fe(OH)$_3$(s) what do you see?
6. name this : [CrCl$_2$(NH$_3$)$_4$]$^+$
7. Write down two ligand substitution reactions involving Cu and describe how colorimetry could be used to determine the ligand:metal ratio
8. 1.7g of steel was dissolved in dilute sulphuric acid and the solution was made up to 250cm$^3$ in a standard flask with distilled water. 25cm$^3$ samples were titrated with 0.02M potassium manganate(Vll) solution. 26.05cm$^3$ of potassium manganate(Vll) solution was required for complete reaction. Calculate the percentage of iron in the steel sample. (A$_r$: Fe = 56)
Answers

# Electrode potentials
## Electrode potentials

Redox reactions such as this:
Zn(s) + Cu$^{2+}$(aq) ---> Cu(s) + Zn$^{2+}$(aq)
Can be split into two half equations:
Cu$^{2+}$(aq) + 2e$^-$ <---> Cu(s)
Zn$^{2+}$(aq) + 2e$^-$ <---> Zn(s)
But as you may notice the second reaction must occur in reverse for the redox reaction to happen and for zinc to be oxidised (loss of electrons). If we set up this reaction as an electrochemical cell we can measure the voltage of the reaction in the cell.
An electrochemical cell is an exothermic reaction set up as two half-cells in two separate containers so that the energy released can produce current between them
A half-cell is half of an electrochemical cell. One half-cell supplies electrons, the other half-cell receives electrons
The more 'easily' the reduction reactions happen (as above in the half-equations) the higher the electrode potential value in volts. In a cell, the higher electrode potential half-cell will undergo the forward reaction and the lower the reverse. The voltage for the reduction of copper 2+ ions is +0.34V and for the reduction of Zinc 2+ ions is -0.76V. Therefore, the copper half-cell is higher and the zinc reaction is in reverse.
The voltage of the cell can be calculated from the difference in the electrode potentials e.g. 0.34V - (-0.76V) = 1.1V With this reaction always work out the difference and don't worry about minus signs.

## Measuring Standard electrode potentials E$^O$

Electrode potential - The voltage measured for a half-cell. Another half-cell is essential for this measurement to be made.
The standard electrode potential of a half cell is measured by making a cell with a 'standard hydrogen electrode' This is used as a reference electrode so has an electrode potential of 0V
Standard hydrogen electrode (the reference electrode) - A half-cell in which hydrogen gas at a pressure of one atmosphere bubbles into a solution of 1M H+ ions. Electrical contact is made with a platinum wire. Each half-cell is given a standard electrode potential of 0V; all other standard electrode potentials are measured relative to it
(Looking at the diagram of the standard Zn2+/Zn half-cell diagram)
* The salt bridge allows movement of ions between the two half-cells completing the circuit. This can simply be filter paper soaked in potassium nitrate solution
* With zinc the negative sign is to the left of the voltage because the standard electrode potential is a negative value
* If there are two ions in the solution without a rod then a platinum electrode is used
* To measure the standard electrode potential of non-metals a platinum electrode must also be used (as they are unlikely to conduct electricity)
* If gases are in the reaction the apparatus is the same as for hydrogen
Standard electrode potential - The voltage measured under standard conditions when the half-cell is incorporated into an electrochemical cell with the other half-cell being a standard hydrogen electrode

## What standard electrode potentials mean

Now that we have looked at measuring electrode potentials we can now use them to look at what they can actually be used for. If we had two half-cells:
Ag$^+$ + e$^-$ <---> Ag
E$^O$ = +0.80V
Zn$^{2+}$ + 2e$^-$ <---> Zn
E$^O$ = -0.76V
We can see that the silver reaction has a higher electrode potential value than the zinc equation. This means that the following can be suggested:
* The silver reaction will have a higher tendency to proceed in the forward direction
* Zn$^{2+}$/Zn has a greater tendency to proceed in the backward direction
* Ag$^+$ is a stronger oxidising agent than Zn$^{2+}$ because it has a higher tendency to proceed in the forward direction (reduction reaction - causing oxidation to other reactants)

## Using standard electrode potentials

We have looked at how standard electrode potentials can be compared to give the voltage of a half-cell. This is useful because we can predict whether or not an oxidising agent is powerful enough to oxidise another substance.
Remember for these to work standard conditions are needed:
* temp of 298K
* pressure of one atmosphere
* 1.00 mol dm$^{−3}$
When the conditions are changed in any way Le Chatelier's Principle can be used to predict whether the E$^O$ value will increase or decrease.
e.g. With the following reaction:
MnO4-(aq) + 8H+(aq) + 5e- ---> Mn2+(aq) + 4H2O(l)
An increase in H$^+$ will shift the equilibrium to the right and therefore increase the E$^O$ value
If the difference between the two half cells is less than 0.30V (including negative numbers) What? then the reaction may have an unexpected outcome. If the difference between the half-cells is greater than 0.30V the reaction predicted by the E$^O$ value is nearly always the one that occurs.
Remember E$^O$ values only say whether or not a reaction is likely to happen and NOT how fast the rate of the reaction will be

## Questions

1. Define an electrochemical cell and a half-cell
2. What is the voltage of this cell?:
2H$^+$ + 2e$^-$ <---> H$_2$
E$^O$ = 0.00V
Zn + 2H$^+$ + 2e$^-$ <---> H$_2$ + Zn$^{2+}$ + 2e$^-$
E$^O$ = -0.76V
3. Why have a salt brige in an electrochemical cell?
4. Draw out a diagram for measuring the electrode potential of a Zn$^{2+}$/Zn half-cell
5. How can the likelihood of a reaction happening in a certain way be changed?
6. What do E$^O$ values not tell us?
Answers

# Ligands and complexes

If you are not sure what a complex is or would like to recap, look here.
Using the concepts learnt in the trends and patterns module we can now develop those and look further into ligands and complexes

## Complexes and shape

For the trends and patterns module we learnt about complexes consisting of ligands donating a lone pair of electrons to a central metal ion. This is called a monodentate ligand, meaning that each of the ligands donate one lone pair of electrons. The total number of coordinate bonds made by the donating of electron pairs is called the coordination number and the overall charge is a combination of the ligand charges and the metal ion charge. In this module it is required to know more about the different types of ligands which make up complexes:
* Monodentate ligands - Donate one pair of electrons to the central metal ion e.g. H$_2$O, NH$_3$, CO, Cl$^-$
* Bi dentate ligands - Donate two pairs of electrons to the central metal ion e.g. NH$_2$CH$_2$CH$_2$NH$_2$ - ethane-1,2-diamine (en)
* If a ligand is donating more than one pair of electrons it will usually be known as polydentate e.g. en, EDTA
ethane-1,2-diamine is usually shortened to en when it is a ligand (this is acceptable in exams). (Looking right) There are two ammine groups (blue) that each have a lone pair of electrons. They both donate electrons in complexes so one molecule of en will form two coordinate bonds, hence bidentate. en can be put in a complex as any other ligand:
e.g. [Ni(en)$_3$]$^{2+}$
When the coordination number is four there are two different shapes:
* Square planar
* Tetrahedral
When the coordination number is six the complex is octahedral

## Isomerism

As with the organic chemistry modules, in complexes there is stereoisomers and optical isomers:

### Stereoisomerism

In octahedral complexes (with 4 of one ligand and 2 of another) and square planar complexes (with 2 of one ligand and two of another) there are two different arrangements of the same atoms with the same bonds. These different arrangements are called cis and trans:
* In cis molecules the 2 ligands are on the same side of the complex. This means that in both octahedral and square planar complexes the bond angle between the two like ligands will be 90 degrees.
* In trans molecules the similar ligands are on the opposite sides of the molecules so the bond angle is 180 degrees.
Tetrahedral molecules do not show stereoisomerism
Cis-platin - This chemical is used as an anti-cancer drug because it can bind to the DNA of fast growing cells and prevent correct replication. The trans form of this chemical is not as effective at binding to the DNA so it is very important that the cis form is used.

### Optical Isomerism

You should be familiar with the idea of optical isomerism from AS organic chemistry. When two molecules have the same atoms and bonds but are a mirror image of each other, they are not super imposable and are said to be optical isomers.
If you shine light through pure solutions of each optical isomer they would rotate the plane of polarised light in opposite directions. When they are together this effect is cancelled out.

## Questions

1. What is a monodentate ligand?
2. How many coordinate bonds does an en ligand form in a complex?
3. What is cis-platin used for? why not trans?
4. Draw a the shape of a tetrahedral complex
5. What is the bond angle between two of the same ligand if there are 4 different ligands in a trans octahedral complex?
6. If i have two samples of a complex how can i test to find if they are the same optical isomers?
Answers

# Colour

First review the section on colour in the trends and patterns module:

## 3d Subshell

All five orbitals in the 3d subshell have two electrons each and are all at the same energy level (degenerate). However as we are about to find out this is only when the atom or ion involved is isolated (not bonded with anything else). Electrons will fill any of these orbitals in transitional atoms or ions and will fill each one before pairing.
d-orbital splitting: This is where the d-orbitals are in an (octahedral) transition metal complex so are no longer isolated. Six coordinate bonds are formed along the same axis as the 3d$_{z2}$ and 3d$_{x2y2}$ orbitals. Consequently, these orbitals are of higher energy than the other orbitals. This creates a higher tier energy level orbitals with an energy gap between them and the lower energy orbitals and is labelled ΔE.

## Electron promotion

If a transition element complex has at least one electron in a lower energy orbital and at least one space in a higher energy orbital the electron can jump to the higher energy orbital. This requires energy though, that's why light is needed. The energy gap ΔE previously described is the amount of energy from absorbed light required to allow a promotion. According to quantum theory and the equation e=hf, the energy of light is proportional to its wavelength. Therefore the amount of energy for a particular promotion will correspond to a certain wavelength (and therefore colour) of light. Transitional metals absorb wavelengths of light which are in humans visible range so we see colour.
If a particular wavelength of light is absorbed by the complex for electron promotion then the light reflected will be without that wavelength so will appear a different colour to white (all wavelengths).
Using the colour wheel from the trends and patterns module we can work out the colour of a complex solution that requires 4*10$^{-19}$J for a promotion:
green light has energy 4*10$^{-19}$J so green is absorbed and reflected light appears violet.
Compounds that do not fit the criteria explained above will be white e.g. Cu+ compounds are colourless because the electron configuration is [Ar]3d$^{10}$4s$^0$. As the 3d subshell is full there cannot be electron promotion.

## Changes in colour from ligand exchange

If ligands are substituted the E gap may increase or decrease, this will change the frequency of light absorbed from white light:
* If the energy gap becomes smaller the energy, and therefore frequency, of light absorbed will go down. This light will have a greater wavelength e.g. blue-->green light absorbed so yellow--->red viewed.
* If the energy gap increases the frequency of light needed increases. The wavelength of absorbed light decreases e.g. orange ---> blue absorbed, blue ---> yellow viewed
Remember: With all waves Velocity = Frequency X wavelength
Also, when a reactant is added to a complex a reduction or oxidation reaction could also occur with the transitional metal. This could also change the colour of the solution.

## Metal complex uses and spectra

Transitional elements are used for their colours in pigments for their brilliance in colour and for pure white:
* TiO$_2$ is used in white paint as it is excellent at hiding other colours beneath
* Phthalocyanine Blue BN is used for brilliant blue in paint and dyes - it is very stable
Visible spectrometry can be used to predict the colour of a transition metal complex. A graph can be drawn using a visible spectrometer for a complex and the relative absorbance of each wavelength can be identified.

## Questions

1. What can we call the orbitals of an isolated transition metal atom/ion?
2. Why does d-orbital splitting happen?
3. Which d-orbitals are the higher energy ones? draw them
4. What does the energy gap correspond to?
5. How can i change the energy gap and what will it do?
Answers

# Chemistry of Transition metals
## Vanadium

Vanadium is a silver-grey metal which is very resistant to corrosion.
It can form compounds in the +5,+4,+3 and +2 oxidation states

## Cobalt

Cobalt (Atomic no 27 relative mass no 59) is an unreactive white metal with a slight blue appearance and forms compounds with +2 and +3 oxidation states. The +2 state is usually the most stable.

### Co$^{2+}$

Co$^{2+}$ forms the complex molecule [Co(H$_2$O)$_6$]$^{2+}$ which is a pink octahedral complex ion.
Co$^{2+}$ also forms the tetrahedral complex ion [CoCl$_4$]$^{2-}$ which is blue.
Cobalt chloride paper can therefore be used to test for water because the complex [Co(H$_2$O)$_6$]$^{2+}$ is formed. This causes the blue cobalt chloride paper to turn pink.
This reaction is an equilibrium reaction:
[CoCl$_4$]$^{2-}$(aq) + 6H$_2$O(l) <---> [Co(H$_2$O)$_6$]$^{2+}$(aq) + 4Cl$^-$(aq)
The reaction is exothermic in the forward reaction. By adding either heat or a high concentration of Cl$^-$ the reverse ligand substitution reaction will occur.

### Co$^{3+}$

Co$^{3+}$ forms the aqueous complex [Co(H$_2$O)$_6$]$^{3+}$ which is blue but is so easily reduced to [Co(H$_2$O)$_6$]$^{2+}$ that we do not really look at it. Other simple compound of Co$^{3+}$ become hydrated and are reduced in the same way.
Co$^{3+}$ does however make other complex ions with different ligands and they prove to be very stable indeed:
e.g.
[Co(H$_2$O)$_6$]$^{3+}$ + e$^-$ <---> [Co(H$_2$O)$_6$]$^{2+}$; E= +1.81V
[Co(NH$_3$)$_6$]$^{3+}$ + e$^-$ <---> [Co(NH$_3$)$_6$]$^{2+}$; E= +0.11V
Here the Electrode potential values show that the Co$^{3+}$ complex without H$_2$O ligands (Hexaamminecobalt ll) is the more stable. This is because ammonia has only one lone pair of electrons resulting in stronger dative bonds.