Inorganic Chemistry

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Oxidation and Reduction

If electrons are transferred, it's a redox reaction. A loss of electrons is called oxidation, and a gain of electrons is called reduction. Reduction and oxidation happen simultaneously - hence the term "redox" reaction. An oxidising agent accepts electrons and gets reduced. A reducing agent donates electrons and gets oxidised. Sometimes it's easier to talk about oxidation numbers. There are lots of rules. All atoms are treated as ions, even if covalently bonded. Uncombined elements and elements bonded to identical atoms have an oxidation number of 0. The oxidation number of a simple monatomic ion is the same as its charge. In compouns or compound ions, the overall oxidation number is just the ion charge. The sum of oxidation numbers for a neutral compound is 0. Combined oxygen is nearly always -2, except in peroxides, where it's -1. (And in the flourides OF2, where it's +2, and O2F2, where it's +1.) Combined hydrogen is +1, except in metal hydrides where it is -1. In HF, it is +1, and in NaH, it is -1. Roman numerals give oxidation numbers. If you see Roman numerals in a chemical name, it's an oxidation number. Oxidation numbers go up or down as electrons are lost or gained. The oxidation for an atom will increase by 1 for each electron lost, and decrease by 1 for each electron gained. Elements can also be oxidised and reduced at the same time, known as disproportionation. You can seperate redox reactions into half-equations. Ionic half-equations show oxidation or reduction. You can combine half-equations for different oxidising or reducing agents together to make full equations for reactions.

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Group 2

Ionisation energy decreases down the group. Each element down Group 2 has an extra electron shell compared to the one above. The extra inner shells shield the outer electrons from the attraction of the nucleus. Also, the extra shell means the outer electrons are further away from the nucleus, which reduces the nucleus's attraction. These factors result in a lower ionisation energy. Group 2 elements react with water, oxygen and chlorine. When Group 2 elements react, they're oxidised from a state of 0 to +2, forming M 2+ ions. This is because Group 2 atoms have two outer shell electrons. They react with water to produce hydroxides. The Group 2 metals react with water to give a metal hyrdoxide and hydrogen. They get increasingly reactive down the group the ionisation energies decrease. They burn inm oxygen with characteristic flame colours to form solid white oxides. Magnesium burns with a white flame, and the others burn with their characteristic flame colours (see Group 1 and 2 Compounds). They react with chlorine forming white solid chlorides. The oxides and hydroxides are bases. The oxides of the Group 2 metals react readily with water to form metal hydroxides, which dissolve. The hydroxide, OH-, make these solutions alkaline. Magnesium oxide is an exception - it reacts slowly and the hydroxide isn't very soluble. The oxides form more strongly alkaline solutions as you go down the group, because the hydroxides get more soluble. Because they're bases, both the oxides and hydroxides will neutralise dilute acids, forming solutions of the corresponding salts. The oxides and hydroxides form alkaline solutions and neutralise solutions. A metal oxide reacting with water makes the metal hydroxide. They also react with acids to form the metal chloride and water. The metal hydroxides react with water to form the metal ion and two hydroxide ions. They also react with acids to form the metal chloride and 2 moles of water. Solubility trends depend on the compound anion. Generally, compounds of Group 2 elements that contain singly charged negative ions increase in solubility down the group, whereas compounds that contain doubly charged negative ions decrease in solubility down the group. The hydroxides get more soluble down the group, but for the sulfates it increases up the group. Compounds with very low solubilities are said to be sparingly soluble.

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Group 1 and 2 Compounds

Thermal stability of carbonates and nitrates changes down the group. Thermal decomposition is when a substance breaks down when heated. The more thermally stable a substance is, the more heat it needs to decompose it. Thermal stability increases down a group. The carbonate and nitrate ions are large and can be made unstable by the presence of a positively charged ion (a cation). The cation polarises the anion, distorting it. The greater the distortion, the less stable the anion. Larger cations cause less distortion than smaller cations. So the further down the group, the larger the cations, the less distortion caused and the more stable the carbonate/nitrate anion. Group 2 compounds are less thermally stable than Group 1 compounds. The greater the charge on the cation, the greater the distortion and the less stable the carbonate/nitrate ion becomes. Group 2 ions have a 2+ charge compared to a 1+ charge for Group 1 cations. So Group 2 carbonates and nitrates are less stable than those of Group 1. Group 1 carbonates are thermally stable - they wont decompose with a bunsen heating it, but they do decompose at higher temperatures (except lithium carbonate, which decomposes to lithium oxide and carbon dioxide). Group 2 carbonates decompose to form the oxide and carbon dioxide. Group 1 nitrates decompose to form the nitrite and oxygen, except lithium nitrate, which decomposes to form lithium oxide, nitrogen dioxide and oxygen. Group 2 nitrates decompose to form the oxide, nitrogen dioxide and oxygen. You test the thermal stability of nitrates either by measuring how long it takes until oxygen is produced, how long it takes for nitrogen dioxide to form (this needs to be done in a fume cupboard, because nitrogen dioxide is toxic). For carbonates, you measure how long it takes for carbon dioxide to be produced. The classic test for this is lime water (saturated calcium hydroxide solution), which turns cloudy with carbon dioxide. Group 1 and 2 compounds burn with distinctive flame colours, but not all of them. For compounds containing the ions below, flame tests can help identify them. Li - red, Na -  orange/yellow, K - lilac, Rb - red, Cs - blue, Ca - brick-red, Sr - crimson, and Ba - green. You do a flame test by mixing the compound with a few drops of hydrochloric acid, heating a piece of platinum/nichrome wire, and dipping the wire into the coumpound/acid mixture and holding it in a bunsen flame. This is the explanation for flame colours: The energy absorbed from the flame causes electrons to move to higher energy levels. The colours are seen as electrons fall back down to lower energy levels, releasing energy in the form of light. The difference in energy between the higher and lower levels determines the wavelength of light released - which determines the colour of the light. (This will also come up in Quantum Phenomena in AS Physics). 

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The Halogens

Halogens are the highly reactive non-metals of Group 7. Fluorine - pale yellow gas, chlorine - green gas, bromine - red-brown liquid, iodine - grey solid. Electronegativity increases up the group. Halogens in their natural state exist as covalent diatomic molecules. Because they're covalent, they have low solubility in water. They do dissolve easily in organic compiunds like hexane though. Chlorine - colourless in water and hexane, bromine - yellow/orange in water, orange/red in hexane, iodine - brown in water, pink/violet in hexane. Halogens get less reactive down the group. Halogen atoms react by gaining an electron in their outer p sub-shell. This means they're reduced. As they're reduced, they oxidise another substance (redox reaction) - so they're oxidising agents. As you go down the group, the atoms become larger so the outer electrons are further from the nucleus. The outer electrons are also shielded more from the attraction of the positive nucleus, because there are more inner electrons. This makes it harder for larger atoms to attract the electron needed to form an ion, so larger atoms are less reactive. Another way of saying that the halogens get less reactive down the group is to say that they become less oxidising. You can use patterns to predict properties. The smalles halogen, flourine, is the most reactive non-metal element. You can predict it's properties by looking at those of the other halogens. The melting points and boiling points increase down the group, so you can predict that flourine is a gas like chlorine. It should also be coloured like the others are (it's pale yellow). Astatine is a solid. You'd expect it to be the least reactive halogen, but it's properties are hard to study since astatine is highly radioactive and decays quickly. Halogens undergo disproportionation with alkalis. The halogens react with hot and cold alkali solutions. In these reactions, the halogen is simultaneously oxidised and reduced (called disproportionation). If the alkali is cold, the halogen reacts with it to make the metal halogen -ate ion, the metal salt and water. If the alkali is hot, the halogen reacts with it to form the metal halogen -ate ion, the metal salt, and water, like if it is cold, but more moles of products form because theere are more moles of reactants. All the halogen -ate ions have a single halogen atom and a charge of -1. Halogens (except flourine) can exist in a wide range of oxidation states. Chloride - -1, chlorine - 0, chlorate(I)/bromate(I) - +1, bromate(III) - +3, iodate(V) - +5, and iodate(VII) is +7. Halogens oxidise metals. For example, flourine and chlorine react with hot iron to form iron(III) halides. Iron is taken to its highest oxidation state, +3, because these halogens are very strong oxidising agents. For bromine, iron(II) and iron(III) bromide is produced. For iodine it is only iron(II) iodide - no Fe 3+ ions form. They oxidise non-metals. E.g chlorine reacts with sulfur to form sulfur(I) chloride. Sulfur is oxidised to +1 and chlorine is reduced to -1. They oxidise some ions aswell. For example, all the halogens except iodine will oxidise iron(II) ions to iron(III) ions in solution. The solution will change colour from green to orange. Iodine wont because it is not a strong enough oxidising agent.

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Reactions of the Halides

The reducing power of halides increases down the group. A halide ion can act as a reducing agent by losing an electron from its outer shell. How easy this is depends on the attraction between the halide's nucleus and the outer electrons. As you go down the group the attraction gets weaker because: the ions get bigger, so the electrons are further away from the +ve nucleus, and there are extra inner electron shells, so there's a greater shielding effect. This explains their reactions with sulfuric acid. All the halides react with concentrated sulfuric acid to give a hydrogen halide as a product to start with. But what happens next depends on which halide you've got. Reaction of KF/KCl with sulfuric acid forms hydrogen flouride/chloride gas. You'll see misty fumes as the gas comes into contact with moisture in the air. But HF and HCl aren't strong enough reducing agents to reduce the sulfuric acid. So the reaction ends there. It's not a redox reaction - the oxidation states of the halide and sulfur remain the same. Reaction of KBr with sulfuric acid forms hydrogen bromide gas. But HBr is a stronger reducing agent than HCl and reacts with the sulfuric acid in a redox reaction. Sulfur dioxide and bromine gas is produced. Reaction of KI with sulfuric acid forms the gas, then the HI reduces the sulfuric acid as above, but HI keeps going and reduces the sulfur dioxide to hydrogen sulfide. Hydrogen halides are acidic gases. They're very soluble, dissolving in water to make strong acids. Hydrogen chloride forms hydrochloric acid, hydrogen bromide produces hydrobromic acid, and hydrogen iodide produces hydroiodic acid. They react with ammonia gas to give white fumes. Hydrogen chloride will give ammonia chloride. Halide ions are displaced from solution by more reactive halogens. The halogens' relative oxidising strengths can be seen in their displacement reactions with halide ions, where the halogen displaces the other halogen. Chlorine water oxidises KBr and KI to make an orange and brown solution respectively. Bromine makes a brown solution with KI, and iodine doesn't react at all. Using an organic solvent, the halogen that's present will dissolve readily in the organic solvent, which settles out as a distinct layer above the aqueous solution. These displacement reactions can be used to help identify which halogen/halide is present in a solution. Chlorine displaces bromine and iodine, bromine displaces iodine, and iodine displaces none of them. You can also say a halogen will oxidise another one if it is below it in the periodic table. Halides give coloured precipitates with silver nitrate solution. You add dilute nitric acid to remove ions that may interfere with the test. Then you add silver nitrate solution, and a silver halide forms. The colour of this precipitate determines the halide. You can be extra sure by adding ammonia solution. Flouride - no precipitate, chloride - white precipitate, dissolves in dilute ammonia solution, bromine - cream precipitate, dissolves in concentrated ammonia solution, iodide - yellow precipitate, insoluble in ammonia solution. Silver halides react with sunlight. Silver halides decompose when light shines on them, producing silver and the halogen. You might have to make predictions. Halide ions are reducing agents, and get more reducing down the group. Fe 3+ ions don't form with iodide ions because they reduce Fe 3+ ions to Fe 2+ ions. Iron(III) chloride is stable though because the Cl- ions are not powerful enough reducing agents. You could predict, because iodide ions will reduce Fe 3+, then astatide ions would do the same, so iron(III) astatide would be unstable.

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Acid-Base Titrations and Uncertainty

Titrations need to be done accurately. Titrations allow you to find out exactly how much acid is needed to neutralise a quantity of alkali. You measure out some alkali using a pipette and put it in a flask, along with some indicator like phenolphthalein. Do a rough titration to get an idea of where the end point is (the point where the alkali is neutralised and the indicator changes colour). Add the acid to the alkali using a burette - giving the flask a regular swirl. Now do an accurate titration. RUn the acid in to within 2 cm^3 of the end point, then add the acid dropwise. If you don't notice exactly when the solution changed colour you've overshot and the result wont be accurate. Record the amount of acid needed to neutralise the alkali. It's best to repeat this process a few times to make sure you get the same answer each time. Indicators show you when the reaction has just finished. In titrations, indicators that change colour quickly over a very small pH range are used so you know exactly when the reaction has ended. The main two for acid/alkali reactions are: methyl orange - turns yellow to red when adding acid to alkali, and phenolphthalein - turns red to colourless when adding acid to alkali. You can calculate conecntrations from titrations. First write a balanced equation and decide what you know and what you need to know. Then use no. of moles = concentration x volume (cm^3) divided by 1000. Then use the value you've calculated to work out the concentration of the alkali, using conc. of alkali = moles of alkali x 1000 divided by the volume (cm^3). Uncertainty is the amount of error your measurements might have. The results from a titration won't be exactly right. The maximum possible error is a useful measure of uncertainty (in your measurements). The uncertainty in your measurements varies on different equipment. The maximum uncertainty in measuring on scales or burettes is half the accuracy of the measuring device. Uncertainty values in things like fixed-volume pipettes and volumetric flasks are provided by the manafacturers. You can minimise some uncertainties. One obvious way to reduce errors in your measurements is to buy the most precise equipment available. There are other ways aswell (since your stuck with whatever your schoo, or college has got). Using a burette, working out a titre could have 0.1 cm^3 uncertainty because each measurement could be up to 0.05 cm^3 out, and a titre is the second reading minus the first reading. The percentage uncertainty can be calulated by: precentage uncertainty = uncertainty/reading x 100. If you use a burette to measure 10 cm^3 of liquid the percentage is (0.1/10) x 100 = 1%, but using 20 cm^3 results in (0.1/20) x 100 = 0.5%. The percentage uncertainty can be reduced by planning a titration so that a larger volume will be measured by the burette. The same principle can applied to other measurements like weighing solids. Errors can be systematic or random. Systematic errors are the same each time the experiment is repeated. E.g a pipette that reads 0.05cm^3 too low you measurement will be that much lower every time you repeat the experiment. Random errors vary - they're what make the results different each time you repeat an experiment. The errors when you make a reading from a burette are random. You have to estimate the value, so it may be above or below the true value. Repeating an experiment and finding the mean can help eliminate random errors. The results that are a little too high are cancelld out by those that are a little too low. Systematic errors will always alter your experiment though, repeating makes no difference. The total uncertainty in a result should be calculated. Knowing the uncertainty in a result can be really important. If a driver's blodd has just enough alcohol to be above the limit, then the uncertainty could result it to be below the limit if it was large enough. In titrations, to find out the total uncertainty, you find the percentage uncertainty for each bit of equipment, then add the individual precentage uncertainties together. This gives the percentage uncertainty in the final result. Use this to work out the actual total uncertainty in the final result. Once you have this, you multiply the final result by the uncertainty to give the number uncertainty.

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Iodine-Sodium Thiosulfate Titration

 Iodine-sodium thiosulfate titrations are dead handy. They are a way of finding the concentration of an oxidising agent. The more concentrated an oxidising agent is, the more ions will be oxidised by a certain volume of it. So here's how you can find out the concentration of a solution of the oxidising agent potassium iodate(V): Stage 1: Use a sample of oxidising agent to oxidise as much iodide as possible. Measure out a certain volume of potassium iodate(V), then add this to an excess of acidic potassium iodide solution. The iodate(V) ions in the potassium iodate(V) solution oxidise some of the iodide ions to iodine. Stage 2: Find out how many moles of iodine have been produced. You do this by titrating the resulting solution with sodium thiosulfate. (You need to know the concentration of sodium thiosulfate solution.) The iodine in the solution reacts with the thiosulfate ions to form 2I- and S4O6 2-. Put all the solution produced in stage 1 in a flask. From the burette, add sodium thiosulfate solution to the solution in the flask. When the iodine colour fades to pale yellow, add 2 cm^3 of starch solution. The solution in the conical flask will go dark blue, showing there's still some iodine. Add sodium thiosulfate one drop at a time until the blue colour disappears. When this happens, it means all the iodine has just been reacted. Here's the calculation to find the number of moles of iodine produced in stage 1. E.g The iodine in the solution produced in stage 1 reacted fully with 11.1 cm^3 of 0.12 mol dm^-3 thiosulfate solution. Number of moles of thiosulfate = concentration x volume (cm^3) divided by 1000, = 0.12x11.1/1000 = 1.332x10^-3 moles. 1 mole of iodine reacts with 2 moles of thiosulfate, so number of moles of iodine in the solution = 1.332x10^-3 divided by 2 gives 6.66 x 10^-4 moles. Stage 3: Calculate the concentration of the oxidising agent. Now look at your original equation, which shows that one mole of iodate(V) ions produces three moles of iodine. 25 cm^3 of potassium iodate(V) solution produced 6.66x10^-4 moles of iodine. So there must have been 6.66x10^-4 divided by 3 = 2.22x10^-4 moles of iodate(V) ions. There would be the same number of moles of potassium iodate(V) in the solution. So now to find the concentration of the potassium iodate(V) solution: Number of moles = concentration x volume (cm^3) divided by 1000 ---> 2.22x10^-4 = concentration x 25 divided by 1000. Concentration of potassium iodate(V) solution = 0.00888 mol dm^-3. You have to be able to evaulate the titration procedure. Titrations like the one on the previous page can give very accurate results, but there are a few ways things could go wrong. Using contaminated apparatus could make your results inaccurate - so make sure the burette is clean, and rinse it out with sodium thiosulfate before you start. (Water dilutes the solution). Read the burette correctly (from the bottom of the meniscus, with your eyes level with the liquid). To reduce random errors, repeat and take an average. But remember to wash the flask between experiments or use a new, clean one. This particular experiment has a few problems: The solutions react with the air, so they need to be made up as freshly as possible. If you add the starch too soon, the iodine will stick to the starch and won't react as expected with the thiosulfate, making the result unreliable. Only add the starch when the solution is pale yellow.

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