 Trends and Patterns

Note: From 2008, most of the stuff here will be found in Unit 5: Equilibria, Energetics and Elements.

Trends and Patterns, as the title goes, is all about the patterns of chemical reactions. Bringing back quite a bit of How Far, How Fast?, it might be good revision to have a quick look through that topic before moving to this.

Lattice Enthalpy

We've met some enthalpy changes, formation and combustion, before. Lattice enthalpy is another type, defined as the enthalpy change when one mole of ionic compound is formed from its gaseous ions under standard conditions. In other words, the energy released when gaseous ions form a giant ionic lattice.

For example: Mg2+(g) + O2-(g) -----> MgO(s). Note that both components are ions - no O2 molecules!

Four things you need to know:

1. Lattice enthalpies are always exothermic.
2. They are represented by ΔHlatt.
3. They measure the ionic lattice. A large ΔHlatt means strong electrostatic forces between ions in the compound.
4. ΔHlatt cannot be measured experimentally because gaseous ions don't combine directly to form solids.

Born Haber Cycle

Because ΔHlatt can't be measured experimentally, we need to calculate it by using reactions that we can measure. These form a born haber cycle, which looks like this: It might look complicated to first look at, but it is quite easy to use once you get used to it. All you need is to fill in the boxes with the correct label and then fill in the numbers in the correct places. Add all these numbers together will result in the box that is missing, whether it be lattice enthalpy or something else.

For example: Use the following data to calculate a value for the lattice enthalpy of strontium chloride.

 Atomisation enthalpy of strontium +164.4 kJ mol-1 Atomisation enthalpy of chlorine +121kJ mol-1 First ionisation energy of strontium +550.0 kJ mol-1 Second ionisation energy of strontium +1064.0 kJ mol-1 Electron affinity of chlorine -348.8 kJ mol-1 Enthalpy of formation for strontium chloride -828.9 kJ mol-1

First of all, construct the diagram and fill in the labels: Notice that the atmoisation energy of clorine and the ionisation of clorine are doubled because we are making two moles of it in the reaction.

Also notice that both 1st ionisation and 2nd ionisation energies are included for Sr. This is because of the two stages of ionisation:

Sr -----> Sr+ + e-

Sr+ ------> Sr2+ + e-

Each has a different energy as showed in the table. It's much harder to remove or add an electron to a charged particle rather than a neutral one and this makes the second ionisation or second electron affinity much more endothermic. For ionisation, this is because the remaining electrons are held more strongly by the positive nucleus; for affinity, it is caused by additional repulsion of the negative ion.

Once all the numbers are in place, we know that formation enery = all the other ethalpies added together.

So, we do the maths:

-828.9 = +164.4 + 242 + 550.0 + 1064 + -697.6 + ΔHlatt

-828.9 = + 1322.8 + ΔHlatt

ΔHlatt = - 2151.7

It's as easy as that! Make sure you're careful with your signs and you multipy if ever you see one mole go to more than one mole, but other than that, it's straightforward.

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Group II

Or specifically, we're looking at the thermal decompositions of carbonates from group two. MgCO3, CaCO3 and BaCO3 all decompose when heated, e.g.:

MgCO3 ------> MgO + CO2
CaCO3 ------> CaO + CO2

However, the temperatures at which these thermal decompositions take place are not equal: The trend shows that temperature rises as we move down the group. Why? It's nothing to do with MgCO3 having the largest lattice enthalpy. There are two reasons for the increase.

Firstly, as you move down the group the cation gets larger. Mg2+ is the smallest and has the highest charge density. This causes distortion on the CO32- ion, because the anion is so large and easily polarised, and all the electrons move to one oxygen. The lattice is weakened, causing MgO and CO2 to form.

The second reason is due to the products formed. MgO has a huge exothermic lattice enthalpy, because of the high charge density of the small Mg2+ ion, much larger than the other group two oxides. This increases its stability and makes its formation more favourable.

That's all there is to this little bit! Make sure you learn it!

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Period Three You might recall that we looked at Period Three back in AS Foundation. It's back again, but this time with a lot more detail. It's essential that you learn everything here.

Reactions with water

 Na Mg With water, Na fizzes and produces hydrogen gas. It may burn with an orange flame. Equation: 2Na(s) + 2H2O -----> 2NaOH(aq) + H2(g) Because of the sodium hydroxide, the product is alkali: pH = 12-14. With water, the reaction is very slow. To make it faster H2O gas is used. Equations: Mg(s) + 2H2O(l) -----> Mg(OH)2(aq) + H2 Mg(s) + H2O(g) -----> MgO(s) + H2 The ph = 9-11.

The sodium reacts much more vigorously than magnesium, because Group I elements are more reactive than Group II elements. Sodium also forms a stronger alkali solution because NaOH is more soluable than Mg(OH)2 and therefore more aqueous OH- ions are present.

Reactions with oxygen

 Mg Al S The white crystalline solid burns to produce a brilliant white flame. Equation: Mg(s) + ½O2(g) -----> MgO(s) MgO can also react with water, though it is a slow reaction: MgO(s) + H2O(l) -----> Mg(OH)2(aq) The resulting pH is 9-11. Al only burns as a powder (without oxide layer). Forms a white powder. Equation: 2Al(s) + 1½O2(g) -----> Al2O3(s) Al2O3 does not react with water, so the pH is still 7. Sulphur burns readily with a blue flame to produce the irritant gas SO2. Equation: S(g) + O2(g) -----> SO2(g) SO2 reacts with water to form sulphurous acid: SO2(g) + H2O(l) -----> H2SO3(aq) SO3 sometimes forms instead and this reacts to form sulphuric acid: SO3(g) + H2O(l) -----> H2SO4 (aq) The resulting pH is a mix 1-3.

Metal oxides on the left of the period form alkali solutions with water (excluding Al2O3, which doesn't react with water!) and non-metals form acid solutions on the right.

Reactions with chlorine

 Na Mg Sodium burns with chlorine to produce a white solid - common salt. Equation: Na(s) + ½Cl2(g) -----> NaCl(s) NaCl dissolves in water. Equation: NaCl -----> Na+(aq) + Cl-(aq) The ph = 7. Mg burns to produce a white solid. Equation: Mg(s) + Cl2 -----> MgCl2 This also dissolves in water. Equation: MgCl -----> Mg+(aq) + 2Cl-(aq) This is slightly acidic. Si P When heated, it forms a liquid. Equation: Si + 2Cl2 -----> SiCl4(l) Adding water, a solid precipitate forms in a vigorous hydrolysis. White fumes of HCl gas are formed. Equation: SiCl4(l) + 2H2O -----> Si(OH)4 + HCl The pH=2. No heat is required for the reaction, only excess chlorine. Equations: P4(s) + 6Cl2(g) ------> 4PCl3(l) PCl3(l) + Cl2(s) ------> PCl5(s) Hydrolysis occurs if water is added and white fumes of HCl gas are formed. Equation: PCl3 + 3H2O -----> H3PO3 + 3HCl The pH=2.

The metal chlorides on the left give neutral or slightly acidic solutions when water is added, while the non-metal chlorides on the right hydrolysise and form acidic solutions.

Covalent or ionic?

 Formula NaCl MgCl2 AlCl3 SiCl4 PCl5 Bonding ionic ionic covalent covalent covalent Structure giant lattice giant lattice simple simple simple Electrical conductor? (free ions in solution) (free ions in solution)   They follow the general rule of ionic lattices for the metals on the left and simple covalent molecules for the non-metals on the right.

Be careful with Al. As you can see in the table above, the Al3+ cation disorts the large Cl- anion that it becomes more like a simple covalent molecule than a giant ionic lattice. However, in its oxide state below, it acts as a giant ionic lattice.

 Formula Na2O MgO Al2O3 SiO2 P4O10 SO2 Bonding ionic ionic ionic covalent covalent covalent Structure giant lattice giant lattice giant lattice giant lattice simple simple

Be careful of SiO2 too. Although it is still covalently bonded, the covalent bonds form a giant lattice.

That's it for this section. Deep breath and get learning it...ALL.

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Oxidation Numbers

Remember oxidation numbers from AS Foundation? You need to be able to calculate the oxidation numbers in this module, too, so to quickly remind you of the rules:

1. The ON of a free element is always zero. e.g. the ON of Mg = 0.
2. The combined ON of atoms in a compound is also 0. e.g. NaCl = 0.
3. In compounds, ON in Group I = +1, Group II = +2,etc..
4. In compounds, H = +1 (except in metal hydrates, such as NaH where it would be -1).
5. In compounds, O = -2 (except in peroxides).
6. In compounds, F/Cl/Br/I = -1 (except when combined with O or F).
7. The ON of a singular ion is just its charge. e.g. Na+= +1 and Cl- = -1.
8. The sum of all atoms in a polyatomic ion (ion made of more than one atom) is equal to the charge on the ion. e.g. CO3-. O has an ON of -2 so 3x-2=-6. To leave a -1 ion, we need C to be +5.

For example, work out the ON of N in N2O4.

We know that O has ON of -2, and there are 4 of them. (-2 x 4) = -8.
In order to get an overall state of 0, N2 must be +8.
To get N, simply divide by two, to get +4.

Work out the ON of Cr in CrO4-2.

It's important to notice that the overall charge is -2.
Again, we have 4 O's at -2, so (-2 x 4) = -8.
In order to leave a -2 charge, the Cr must have ON of +6.

Not too hard, huh?

Don't forget the redox rules either:

• Oxidation occurs when the Oxidation Number increases.
• Reduction occurs when the Oxidation Number decreases.

Work out what has been oxidised and reduced in the following equation:
2Mn2+ + 5BiO3- + 14H+ ------> 2MnO4- + 5Bi3+ + 7H2O.

 In Mn2+, Mn has ON of 2+. In 2MnO4-, we have 4 O's at -2 (-2 x 4) = -8. The overall charge is -1. Therefore, Mn's ON must be +7. The ON has increased, so Mn has been oxidised. In 5BiO3-, there are 3 O's at -2, so (-2 x 3) = -6. The overall charge is -1. Therefore, Bi must have ON of +5. In 5Bi3+, Bi has ON of 3+. The ON has decreased, so Bi has been reduced.

Easy!

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Transition Elements

A great thing about doing the option module Transition Metals is that most of it overlaps with this modules, meaning there will be a lot less to revise in the coming weeks. Something to smile about. Just what is a transition element? Broadly speaking, it is a d-block elements in the periodic table: However, zinc and scandium, though they have some similar properties, are not usually classed as transitition elements. A better definition of a transition element is an element with one or more ions that have partially filled d oritals.

Only one ion is formed from zinc: Zn2+. If we consider the ion's electronic configuration, it has: 1s22s22p63s23p63d104s2. The d orbital is full and therefore, it isn't a transition element.

Scandium only forms one ion: Sc3+. Looking at its electronic configuration, we see 1s22s22p63s23p63d04s2. The d orbital is empty and so it isn't a transition element.

 Symbol Electron configuration Scandium Sc 1s22s22p63s23p63d14s2 Titanium Ti 1s22s22p63s23p63d24s2 Vanadium V 1s22s22p63s23p63d34s2 Chromium Cr 1s22s22p63s23p63d54s1 Manganese Mn 1s22s22p63s23p63d54s2 Iron Fe 1s22s22p63s23p63d64s2 Cobolt Co 1s22s22p63s23p63d74s2 Nickel Ni 1s22s22p63s23p63d84s2 Copper Cu 1s22s22p63s23p63d104s1 Zinc Zn 1s22s22p63s23p63d104s2

Chromium and Copper are unusual because their 4s sub-shell only contains one electron. This is because it's more stable to have either a full d-orbital (d10 for copper), or a half filled d-orbital (d5 for chromium).

Remember that the 4s electrons are removed from the elements before the 3d electrons, when they form positive ions in ionic bonding.

Another characteristic of tranistion metals is that they form a variety of oxidation states. This is due to the d orbital. It's as easy to remove several electrons from one orbital as it is to remove one, and because there are upto 10 electrons in the d orbital, this means there's a potential for oxidation states of +1 all the way up to +9. In elements that aren't transition, there's a large energy gap between the s and p orbitals and so only one ion usually forms.

• Iron can have oxidation states of: +2, +3, +4, +5 and +6. +2 and +3 are the most common (Fe2+ and Fe3+).
• Copper can have oxidation states of: +1, +2 and +3. +1 and +2 are the most common (Cu+ and Cu2+).

Catalysts

Because of their different oxidation states, transition elements can lose and gain electrons easily, allowing them to transfer electrons to speed a reaction. In other words, they offer sites and alternative paths for reactions to take place, and so are great catalysts.

A few examples you should know:

• Fe in the Haber Process (production of ammonia).
Equation: N2 + 3H2 -----> 2NH3.
• V2O5 in the contact process (production of sulphuric acid).
Equation: SO3 + H2O -----> H2SO4.
• Ni in the halogenation of alkenes.
Equation: CH2=CH2 + H2 -----> CH3CH3.

Differentiating

How can we determine a transition element when it's in a solution? If we add hydroxide ions, we result in a transition hydroxide solid. Each one is a different colour, and by remembering the colours produced, it forms a great identification test.

For example:

Cu2+(aq) + 2OH-(aq) -----> Cu(OH)2(s)
Solid Cu(OH)2(s) is blue.

Fe2+(aq) + 2OH-(aq) -----> Fe(OH)2(s)
Solid Fe(OH)2(s) is green.

Fe3+(aq) + 3OH-(aq) -----> Fe(OH)3(s)
Solid Fe(OH)3(s) is rust.

An example of a full equation is CuSO4(aq) + 2NaOH(aq) -----> Cu(OH)2(s) + Na2SO4(aq)

These are the only ones you need to know for the exam, but make sure you can predict the reaction between any transition element and OH-.

Coloured ions

Another cool property of transition elements is that they're often coloured as ions in solutions. This is because of the electrons in the d orbital. All we need to know here is that colour is formed due to the partially filled 3d orbitals... (see Transition Metals for more details).

The ions absorb certain wavelengths of light, and only transmit (let through) or reflect certain colours. White light contains a number of colours (the spectrum): red, orange, yellow, blue, green and violet. When the transmitted or reflected colours reach the eyes, they mix, and give transition elements their colours.

In solution:

• Fe2+ is green [Ar] = 3p63d6
• Fe3+ is yellow [Ar] = 3p63d5
• Cu2+ is blue [Ar] = 3p63d9

Don't get confused with Fe3+ in solution (yellow) and Fe3+ in hydroxide (rust)!

There are a few exceptions...

• Cu+ has electronic configuration [Ar] = 3p63d10. The 3d subshell is full and so it is a colourless solution, unlike other transition elements.
• Zn2+ also has electronic configuration of [Ar] = 3p63d10. Like Cu+, it isn't coloured.
• Sn3+ has electronic configuration of [Ar] = 3s2p6. Its 3d subshell is empty and so it is also colourless in solution.
• As solids, Cu+, Zn2+ and Sn3+ are white.

If a solution is coloured, we can use a technique called spectroscopy to measure the light absorbed, which often also incorporates uv-light as well as visible light. A spectrometer is a machine that measures the light absorbed, and it produces a graph reading called a spectrum.

For example, a reading might look like this: The wavelength of light is shown along the x-axis, measured in nanometers, (nm). The colour spectrometer shows how the nms correspond to the colours. In the graph, we see large absorption around the blue, yellow and green colours, meaning the colour will be a purple (as it transmits violet and red).

For the exam, you need to know three colours:

• 450 nm wavelengths are blue.
• 580 nm wavelengths are yellow.
• 650 nm wavelengths are red.

Complex ions

A complex ion is a transition metal ion surrounded by ligands, and a ligand is a molecule or negative ion that donates at least a pair of electrons to the central metal ion, to form a coordinating covalent bond.

For example, [Cu(H2O)6]2+, which has a positive Cu2+ ion and six H2Os as ligands. The charge goes on the outside of the square brackets.

If a negative ion acts as a ligand, the overall charge of the complex ion is the change of the metal ion + the charge of the negative ion. For example, if 4Cl- are bonded to Cu2+, the charge is: (+2 + -4) = -2. [CuCl4]2-.

Complex ions have two different shapes: tetrahedral and octahedral.

Tetrahedral complex ions as the 'tetra' indicates, have four ligands, and octahedral complex ions have six ligands.

 Though it has only six ligands, if you build up an octahedral complex ion, you'll see it's diamond shaped, with eight faces, thus 'octa'. [Cu(H2O)]2+ [CuCl4]2- Octahedral Bond angles of 90o and 180o. Tetrahedral Bond angles of 109.5o.

If a ligand donates one pair of electrons, it is called monodentate. If it donates two pairs of electrons, it is called bidentate. For more than two pairs of electrons, it is called polydentate.

Some common ligands are:

 Ligand Formula Type of ligand Water H2O monodentate Ammonia NH3 monodentate Chloride ion Cl- monodentate Cyanide ion CN- monodentate Hydroxide ion OH- monodentate Thiocyanate ion SCN- monodentate Ethane-1,2-diamine (en) CH2(NH2)CH2(NH2) bidentate Edta (ethylenediaminetetraacentic acid) polydentate

Some ligands bind more strongly with transition ions than others. A ligand that binds more strongly can displace a weaker ligand, something called ligand substitution. Ligand substitution can be seen in experiments by colour changes.

Ligand strength follows this trend:

 edta > NH3 > Cl- > H2O strongest weakest

There are a few important ligand substitutions that you need to learn:

If excess aqueous ammonia is added dropwise to a solution of copper sulphate, the solution turns from blue to deep blue. This is because the following ligand substitution occurs:

 [Cu(H2O)6]2+ + 4NH3 -----> [Cu(H2O)2(NH3)4]2+ + 4H2O Blue Deep blue

If excess hydrocholoric acid is added dropwise to copper sulphate, the solution turns a yellow colour. This is because of the following ligand substitution:

 [Cu(H2O)6]2+ + 4HCl -----> [CuCl4]2- + 6H2O Blue Yellow

Notice that the charge of the complex ion changes from 2+ to 2-, due to the minus charge of the Cl- ions.

If ammonia and water is added to the copper complex, it will turn deep blue. A blue precipitate may form that dissolves (Cu(OH)2(s)).

 [CuCl4]2- + 2H2O + 4NH3 -----> [Cu(H2O)2(NH3)4]2+ + 4Cl- Yellow Deep blue

If potassium thiosulphate is added to a solution of Fe3+, the solution turns from yellow to blood red in the following substitution:

 [Fe(H2O)6]3+ + SCN- ------> Fe(H2O)5SCN]2+ + H2O Yellow Blood red

The charge changes here again, as the negative ion is added.

Colorimetry

In order to calculate the formula of a complex ion, scientists use a technique called colorimetry. It uses a colorimeter.

In a colorimeter, a beam of light passes through the solution and hits a photocell to record the transmitted light and work out the colour. You should know the method:

1. Around ten test tubes are used, and the first of these is filled with only water - the zero test tube.
2. Next a solution of metal ion and a solution of ligand are added to each of the remaining test tubes, each solution having a different volume. The overall volume of each test tube must be equal, and water can be added if necessary. Usually, the number of moles of metal ion remain consistant, while the ligand's moles increase.
3. The solution with the deepest colour is placed inthe colorimeter and the needle is adjusted so that it is at the maximum end. This will increase accuracy of other readings.
4. Plot a graph from the results. Absorbance is on the vertical axis and test tube is of horizontal axis.
5. The test tube with the greatest absorbance will be the one from which the complex can be calculated. Work out the number of moles in the metal ion and the ligand, and the ratio of the moles forms the complex ratio.

A reading may look like this:

 Test tube 1 2 3 4 5 6 7 8 9 10 11 cm3 of Ni2+(aq) (0.01 mol dm-3) 0 0.5 0.1 1.5 2 2.5 3 3.5 4 4.5 5 cm3 of edta(aq) (0.01 mol dm-3) 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 Absorbance 0.2 0.4 0.61 0.79 0.91 1.01 0.89 0.81 0.6 0.41 0.19

We can plot on a graph, and get two lines of best fit. Where they intercept, we have the greatest absorbance: Therefore, the maximum absorbance occurs on test tube 6. This has 2.5 cm3 of 0.1 mol dm-3 Ni2+ and 2.5 cm3 of 0.1 mol dm-3 edta. So, there are 0.00025 mol of Ni2+ and 0.00025 mol of edta. They have a ratio of 1:1, so the formula would be [Ni(edta)]2+.

Redox reactions

You should be familiar with redox reactions from the AS year. Redox reactions are where one reactant is oxidised while another is reduced. Redox is important for transition metals because those can often have multiple oxidation states, due to their partially filled d orbitals.

Iron's half equation looks like this:

 Fe2+(aq)-----> Fe3+(aq) + e- green yellow

Iron can change from ON of 2+ to 3+ if an oxidising agent is added. The reverse can happen from ON 3+ to 2+ if a reducing agent is added.

For example, if I- is added, the iron will be reduced from a yellow solution to green. However, the colour of I2 is brownish, and so will mask the green. The colour will change from yellow to brown.

The question might ask you to construct the overall equation from the half equations. We have iron's above, and iodine's looks like this:

 2I-(aq) -----> I2(aq) + 2e-

The half equation is always written in the reduction form, but for Fe3+, oxidation is occuring, so we need to flip the equation around.

 Fe3+(aq) + e- -----> Fe2+(aq)

If we try to put the two half equations together, there's a problem: the electrons don't cancel out. Therefore, we need to multiply Fe's equation by two (or half iodine's).

 2Fe3+(aq) + 2e- -----> 2Fe2+(aq)

Next, we put the two sides of the equation together and cancel the electrons:

2I-(aq) + 2Fe3+(aq) + 2e- -----> I2(aq) + 2e- + 2Fe2+(aq)

2I-(aq) + 2Fe3+(aq) -----> I2(aq) + 2Fe2+(aq)

Magnanese also has multiple oxidation states. Its half equation is:

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

The reduction can only occur in acidic conditions.

In MnO4-, MN has oxidation number of +7, while in Mn2+, it has ON of 2+.

If we add iron solution, the manganese solution will act as an oxidising agent:

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

The colour change here is from purple to yellow, because the Fe3+ masks the pale pink of the manganese.

If we add C2O42-, the manganese will act as an oxidising agent:

MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) -----> Mn2+(aq) + 8H2O(l) + 10CO2(g)

This goes from purple to colourless as the very pale pink of Mn2+ doesn't show.

As long as you can construct half equations and work with ON numbers from the AS section, you'll have this section cracked!

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That's everything for Trends and Patterns! Let me know if I've missed anything.