Group 2 Elements - Inorganic Chemistry - Edexcel PDF

4. Inorganic Chemistry and the Periodic Table 4A: Group 2 Melting points Down the group the melting points decrease. The Atomic radius...

4. Inorganic Chemistry and the Periodic Table 4A: Group 2 Melting points Down the group the melting points decrease. The Atomic radius metallic bonding weakens as the atomic size Atomic radius increases down the Group. increases. The distance between the positive ions and As one goes down the group, the atoms have more delocalized electrons increases. Therefore the shells of electrons making the atom bigger. electrostatic attractive forces between the positive ions and the delocalized electrons weaken. 1st ionisation energy The outermost electrons are held more weakly because they are successively further from the nucleus in additional shells. In addition, the outer shell electrons become more shielded from the attraction of the nucleus by the repulsive force of inner shell electrons Group 2 reactions Reactivity of group 2 metals increases down the group The reactivity increases down the group as the atomic radii increase there is more shielding. The nuclear attraction decreases and it is easier to remove (outer) electrons and so cations form more easily Reactions with oxygen. Mg will also react slowly with oxygen without a flame. The group 2 metals will burn in oxygen. Mg ribbon will often have a thin layer of magnesium oxide on it Mg burns with a bright white flame. formed by reaction with oxygen. 2Mg + O2  2MgO 2Mg + O2  2MgO This needs to be cleaned off by emery paper before doing MgO is a white solid with a high melting reactions with Mg ribbon. point due to its ionic bonding. If testing for reaction rates with Mg and acid, an un-cleaned Mg ribbon would give a false result because both the Mg and MgO would react but at different rates. Mg + 2HCl  MgCl2 + H2 MgO + 2HCl  MgCl2 + H2O Reactions with chlorine The group 2 metals will react with chlorine Mg + Cl2  MgCl2 Reactions with water. Magnesium reacts in steam to produce Mg will also react with warm water, giving a different magnesium oxide and hydrogen. The Mg magnesium hydroxide product. would burn with a bright white flame. Mg + 2 H2O  Mg(OH)2 + H2 Mg (s) + H2O (g)  MgO (s) + H2 (g) This is a much slower reaction than the reaction with steam and there is no flame. The hydroxides produced make the water alkaline The other group 2 metals will react with cold water with increasing vigour down the group to form hydroxides. One would observe: Ca + 2 H2O (l) Ca(OH)2 (aq) + H2 (g) fizzing, (more vigorous down group) Sr + 2 H2O (l) Sr(OH)2 (aq) + H2 (g) the metal dissolving, (faster down group) Ba + 2 H2O (l) Ba(OH)2 (aq) + H2 (g) the solution heating up (more down group) and with calcium a white precipitate appearing (less precipitate forms down group) N Goalby chemrevise.org 1 Reactions of the Oxides of Group 2 elements with water Group 2 ionic oxides react with water to form hydroxides The ionic oxides are basic as the oxide ions accept protons to become hydroxide ions in this reaction (acting as a bronsted lowry base) MgO (s) + H2O (l)  Mg(OH)2 (s) pH 9 Mg(OH)2 is only slightly soluble in water so fewer free OH- ions are produced and so lower pH CaO (s) + H2O (l)  Ca(OH)2 (aq) pH 12 Reactions of the Oxides of Group 2 elements with Acids MgO (s) + 2 HCl (aq)  MgCl2 (aq) + H2O (l) SrO (s) + 2 HCl (aq)  SrCl2 (aq) + H2O (l) CaO (s) + H2SO4 (aq)  CaSO4 (aq) + H2O (l) Reactions of the hydroxides of Group 2 elements with Acids 2HNO3 (aq) + Mg(OH)2 (aq)  Mg(NO3)2 (aq) + 2H2O (l) 2HCl (aq) + Mg(OH)2 (aq)  MgCl2 (aq) + 2H2O (l) Solubility of hydroxides Group II hydroxides become more soluble down the group. All Group II hydroxides when not soluble appear as white precipitates. Magnesium hydroxide is classed as insoluble in water. Calcium hydroxide is reasonably soluble in water. It is used in agriculture to Simplest Ionic Equation for formation of Mg(OH)2 (s) neutralise acidic soils. Mg2+ (aq) + 2OH-(aq)  Mg(OH)2 (s). An aqueous solution of calcium hydroxide A suspension of magnesium hydroxide in water will appear is called lime water and can be used a test slightly alkaline (pH 9) so some hydroxide ions must therefore for carbon dioxide. The limewater turns have been produced by a very slight dissolving. cloudy as white calcium carbonate is produced. Magnesium hydroxide is used in medicine (in suspension as milk of magnesia) to neutralise excess acid in the stomach and Ca(OH)2 (aq) + CO2 (g)  CaCO3 (s) + H2O(l) to treat constipation. Mg(OH)2 + 2HCl  MgCl2 + 2H2O Barium hydroxide would easily dissolve in water. The hydroxide ions present It is safe to use because it so weakly alkaline. It is preferable to would make the solution strongly using calcium carbonate as it will not produce carbon dioxide alkaline. gas. Ba(OH)2 (S) + aq  Ba2+ (aq) + 2OH- (aq) Solubility of Sulfates Group II sulfates become less soluble down the group. BaSO4 is the least soluble. If Barium metal is reacted with sufuric acid it will only react slowly as the insoluble barium sulfate produced will cover the surface of the metal and act as a barrier to further attack. Ba + H2SO4  BaSO4 + H2 The same effect will happen to a lesser extent with metals going up the group as the solubility increases. The same effect does not happen with other acids like hydrochloric or nitric as they form soluble group 2 salts. N Goalby chemrevise.org 2 Thermal decomposition of group 2 carbonates Thermal decomposition is defined as the use Group 2 carbonates decompose on heating to produce of heat to break down a reactant into more than group 2 oxides and carbon dioxide gas. one product MgCO3(s)  MgO(s) + CO2(g) CaCO3(s)  CaO(s) + CO2(g) The ease of thermal decomposition decreases down the group Group 2 carbonates become more thermally stable going down the group. As the cations get bigger they have less of a polarising effect and distort the carbonate ion less. The C-O bond is weakened less so it less easily breaks down Group 1 carbonates do not decompose with the exception of lithium. As they only have +1 charges they don’t have a big enough charge density to polarise the carbonate ion. Lithium is the exception because its ion is small enough to have a polarising effect Li2CO3(s)  Li2O(s) + CO2(g) There are a number of experiments that can be done to investigate the ease of decomposition. One is to heat a known mass of carbonate in a side arm boiling tube lime water and pass the gas produced through lime water. Time for the first permanent cloudiness to appear in the limewater. Repeat for different carbonates using the same moles of carbonate/same volume of limewater/same Bunsen flame and height of tube above flame. Heat Thermal decomposition of group 2 nitrates (V) Group 2 nitrates decompose on heating to produce group 2 oxides, oxygen and nitrogen dioxide gas. 2Mg(NO3)2 → 2MgO + 4NO2 + O2 You would observe brown gas evolving (NO2) and the The ease of thermal decomposition White nitrate solid is seen to melt to a colourless solution decreases down the group and then re-solidify. The explanation for change in thermal stability is the same as for carbonates Magnesium nitrate decomposes the easiest because the Mg2+ ion is smallest and has the greater charge density. It causes more polarisation of the nitrate anion and weakens the N―O bond Group 1 nitrates, with the exception of lithium nitrate, do not decompose in the same way as group 2 nitrates. They decompose to give a nitrate (III) salt and oxygen. 2NaNO3 → 2NaNO2 + O2 Lithium nitrate decomposes in the Sodium Sodium same way as group 2 nitrates nitrate(V) nitrate(III) 4 LiNO3 → 2Li2O + 4NO2 + O2 N Goalby chemrevise.org 3 Flame tests Method Lithium : Scarlet red Use a nichrome wire ( nichrome is an unreactive metal Sodium : Yellow and will not give out any flame colour) Potassium : lilac Clean the wire by dipping in concentrated hydrochloric Rubidium : red Caesium: blue acid and then heating in Bunsen flame Magnesium: no flame colour (energy emitted If the sample is not powdered then grind it up. of a wavelength outside visible spectrum) Dip wire in solid and put in Bunsen flame and observe Calcium: brick red flame Strontium: red Barium: apple green Explanation for occurrence of flame In a flame test the heat causes the electron to move to a higher energy level. The electron is unstable at the higher energy level and so drops back down. As it drops back down from the higher to a lower energy level, energy is emitted in the form of visible light energy with the wavelength of the observed light N Goalby chemrevise.org 4  4B Halogens Fluorine (F2): very pale yellow gas. It is highly reactive Chlorine : (Cl2) greenish, reactive gas, poisonous in high concentrations Bromine (Br2) : red liquid, that gives off dense brown/orange poisonous fumes Iodine (I2) : shiny grey solid sublimes to purple gas. Trend in melting point and boiling point Trend in electronegativity Increase down the group Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself. As the molecules become larger they have more electrons and so have larger London As one goes down the group the electronegativity of the forces between the molecules. As the elements decreases. intermolecular forces get larger more energy As one goes down the group the atomic radii increases due has to be put into break the forces. This to the increasing number of shells. The nucleus is therefore increases the melting and boiling points less able to attract the bonding pair of electrons The reactivity of the halogens decreases down the group as the atoms get bigger with more shielding so they less easily attract and accept electrons. They therefore form -1 ions less easily down the group 1. The oxidation reactions of halide ions by halogens. A halogen that is a strong oxidising agent will The oxidising strength decreases down the group. displace a halogen that has a lower oxidising Oxidising agents are electron acceptors. power from one of its compounds know these Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions observations ! Chlorine (aq) Bromine (aq) Iodine (aq) The colour of the solution in the test tube shows which free potassium Very pale green Yellow solution, no Brown solution, halogen is present in solution. chloride (aq) solution, no reaction no reaction Chlorine =very pale green reaction solution (often colourless), potassium Yellow solution, Cl Yellow solution, no Brown solution, Bromine = yellow solution bromide (aq) has displaced Br reaction no reaction Iodine = brown solution (sometimes black solid potassium Brown solution, Cl Brown Solution, Br Brown Solution, present) iodide (aq) has displaced I has displaced I no reaction Observations if an organic solvent is added Chlorine (aq) Bromine (aq) Iodine (aq) The colour of the organic potassium colourless, no yellow, no purple, no solvent layer in the test tube chloride (aq) reaction reaction reaction shows which free halogen is present in solution. potassium yellow, Cl has yellow, no purple, no Chlorine = colourless bromide (aq) displaced Br reaction reaction Bromine = yellow Iodine = purple potassium purple, Cl has purple, Br has purple, no iodide (aq) displaced I displaced I reaction Cl2(aq) + 2Br – (aq)  2Cl – (aq) + Br2(aq) Cl2(aq) + 2I – (aq)  2Cl – (aq) + I2(aq) Br2(aq) + 2I – (aq)  2Br – (aq) + I2(aq) N Goalby chemrevise.org 5  The oxidation reactions of metals and metal ion by halogens. In all reactions where halogens 2Na  2Na+ + 2e- are reacting with metals, the Br2(l) + 2Na (s)  2NaBr (s) Br2 + 2e-  2Br - metals are being oxidised 3Cl2(g) + 2 Fe (s)  2 FeCl3 (s) Br2(l) + Mg (s)  MgBr2 (s) Cl2(g) + 2Fe2+ (aq)  2 Cl- (aq) + 2Fe3+ (aq) Chlorine and Bromine can oxidise Fe2+ to Fe3+. Iodine is not strong enough oxidising agent to do this reaction. 2I- (aq) + 2Fe3+ (aq)  I2 (aq) + 2Fe2+ (aq) The reaction is reversed for Iodine The disproportionation reactions of chlorine. Disproportionation is the name for a reaction where Chlorine is both simultaneously reducing and an element simultaneously oxidises and reduces. oxidising changing its oxidation number from 0 to -1 and 0 to +1 Chlorine with water: Cl2(g) + H2O(l)  HClO(aq) + HCl (aq) If some universal indicator is added to the solution it will The pale greenish colour of these solutions first turn red due to the acidity of both reaction products. It is due to the Cl2 will then turn colourless as the HClO bleaches the colour. Chlorine is used in water treatment to kill bacteria. It has been used to treat drinking water and the water in swimming pools. The benefits to health of water treatment by chlorine outweigh its toxic effects. Reaction of halogens with cold dilute NaOH solution: Cl2, Br2, and I2 in aqueous solutions will react with cold sodium hydroxide. The colour of the halogen solution will fade to colourless Cl2(aq) + 2NaOH(aq)  NaCl (aq) + NaClO (aq) + H2O(l) The mixture of NaCl and NaClO is used as Bleach and to disinfect/ kill bacteria Reaction of halogens with hot dilute NaOH solution: With hot alkali disproportionation also occurs but the halogen that is oxidised goes to a higher oxidation state. 3Cl2 (aq) + 6 NaOH(aq)  5 NaCl (aq) + NaClO3 (aq) + 3H2O (l) 3I2 (aq) + 6NaOH (aq)  5 NaI (aq) + NaIO3 (aq) + 3H2O (l) 3I2 (aq) + 6OH- (aq)  5 I- (aq) + IO3- (aq) + 3H2O (l) In IUPAC convention the various forms of sulfur and chlorine compounds where oxygen is combined are all called sulfates and chlorates with relevant oxidation number given in roman numerals. If asked to name these compounds remember to add the oxidation number. NaClO: sodium chlorate(I) NaClO3: sodium chlorate(V) K2SO4 potassium sulfate(VI) K2SO3 potassium sulfate(IV) N Goalby chemrevise.org 6 The reaction of halide salts with concentrated sulfuric acid. The halides show increasing power as Explanation of differing reducing power of halides reducing agents as one goes down the A reducing agent donates electrons. group. This can be clearly demonstrated in The reducing power of the halides increases down group 7 the various reactions of the solid halides with They have a greater tendency to donate electrons. concentrated sulfuric acid. This is because as the ions get bigger it is easier for the Know the equations and observations of outer electrons to be given away as the pull from the nucleus these reactions very well. on them becomes smaller. Fluoride and Chloride The H2SO4 is not strong enough an oxidising reagent to oxidise the chloride and fluoride ions. No redox reactions occur. Only acid-base reactions occur. NaF(s) + H2SO4(l) NaHSO4(s) + HF(g) These are acid –base reactions and Observations: White steamy fumes of HF are evolved. not redox reactions. H2SO4 plays the NaCl(s) + H2SO4(l)  NaHSO4(s) + HCl(g) role of an acid (proton donor). Observations: White steamy fumes of HCl are evolved. Bromide Br- ions are stronger reducing agents than Cl- and F- and after the initial acid- base reaction reduce the sulfur in H2SO4 from +6 to + 4 in SO2 Observations: White steamy Acid- base step: NaBr(s) + H SO (l)  NaHSO (s) + HBr(g) fumes of HBr are evolved. 2 4 4 Redox step: 2HBr + H2SO4  Br2(g) + SO2(g) + 2H2O(l) Red fumes of Bromine are also evolved and a colourless, acidic gas SO2 Ox ½ equation 2Br -  Br2 + 2e- Re ½ equation H2SO4 + 2 H+ + 2 e- SO2 + 2 H2O Reduction product = sulfur dioxide Note the H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step. Iodide I- ions are the strongest halide reducing agents. They can reduce the sulfur from +6 in H2SO4 to + 4 in SO2, to 0 in S and -2 in H2S. NaI(s) + H2SO4(l)  NaHSO4(s) + HI(g) Observations: 2HI + H2SO4  I2(s) + SO2(g) + 2H2O(l) White steamy fumes of HI are evolved. 6HI + H2SO4 3 I2 + S (s) + 4 H2O (l) Black solid and purple fumes of Iodine are 8HI + H2SO4 4I2(s) + H2S(g) + 4H2O(l) also evolved A colourless, acidic gas SO2 A yellow solid of Sulphur Ox ½ equation 2I -  I2 + 2e- H2S (Hydrogen Sulphide), a gas with a bad egg Re ½ equation H2SO4 + 2 H+ + 2 e- SO2 + 2 H2O smell, Re ½ equation H2SO4 + 6 H+ + 6 e- S + 4 H2O Reduction products = sulfur dioxide, sulfur Re ½ equation H2SO4 + 8 H+ + 8 e- H2S + 4 H2O and hydrogen sulfide Note the H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the three redox steps Often in exam questions these redox reactions are worked out after first making the half-equations N Goalby chemrevise.org 7 The reactions of halide ions with silver nitrate. The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate This reaction is used as a test to identify which halide ion Ag2CO3. This would mask the desired observations is present. The test solution is made acidic with nitric 2 HNO3 + Na2CO3  2 NaNO3 + H2O + CO2 acid, and then silver nitrate solution is added dropwise. Fluorides produce no precipitate Chlorides produce a white precipitate Effect Of Light on Silver Halides Ag+(aq) + Cl- (aq)  AgCl(s) The precipitates ( except AgI) darken in sunlight forming Bromides produce a cream precipitate silver. This reaction is used in photography to form the Ag+(aq) + Br- (aq)  AgBr(s) dark bits on photographic film Iodides produce a pale yellow precipitate Ag+(aq) + I- (aq)  AgI(s) Silver chloride dissolves in dilute ammonia to form a Effect of ammonia on silver halides complex ion AgCl(s) + 2NH3(aq) [Ag(NH3)2]+ (aq) + Cl- (aq) Colourless solution The silver halide precipitates can be Silver bromide dissolves in concentrated ammonia to form a treated with ammonia solution to help complex ion differentiate between them if the AgBr(s) + 2NH3(aq) [Ag(NH3)2]+ (aq) + Br - (aq) colours look similar: Colourless solution Silver iodide does not react with ammonia – it is too insoluble. Producing hydrogen halides Hydrogen halides are made by the reaction of solid sodium halide salts with phosphoric acid This is the apparatus used to make the NaCl(s) + H3PO4(l)  NaH2PO4(s) + HCl(g) hydrogen halide using phosphoric acid. Observations: White steamy fumes of the hydrogen Notice the downward halides are evolved. delivery which is used The Steamy fumes of HCl are produced when the HCl because the hydrogen meets the air because it dissolves in the moisture in the halides are more dense air than air Phosphoric acid is not an oxidising agent and so does not oxidise HBr and HI. Phosphoric acid is more suitable for producing hydrogen halides than using concentrated sulfuric acid to make HCl, HBr, and HI. This is because there are no extra redox reactions taking place and no other products formed. Solubility in water : Hydrogen Halide All the hydrogen halides react readily with The hydrogen halides ammonia to give the white smoke of the are all soluble in ammonium halide The water quickly water. They dissolve rises up the tube to form acidic HCl(g) + NH3 (g)  NH4Cl (s) solutions. HBr(g) + NH3 (g)  NH4Br (s) HCl (g)+ H2O(l)  H3O+(aq)+ Cl-(aq) HI(g) + NH3 (g)  NH4I (s) This can be used as a test for the presence of hydrogen halides N Goalby chemrevise.org 8 4C Analysis of Inorganic Compounds Testing for Negative ions (anions) Testing for presence of a carbonate CO32- and hydrogencarbonates HCO3- Add any dilute acid and observe effervescence. Bubble gas through limewater to test for CO2 – will turn limewater cloudy Fizzing due to CO2 would be observed if a carbonate or a 2HCl + Na2CO3  2NaCl + H2O + CO2 2H+ + CO32-  H2O + CO2 hydrogencarbonate was present HCl + NaHCO3  NaCl + H2O + CO2 H+ + HCO3-  H2O + CO2 Testing for presence of a sulfate Ba2+ (aq) + SO42-(aq)  BaSO4 (s). Acidified BaCl2 solution is used as a reagent to test for sulfate ions If Barium Chloride is added to a solution that contains sulfate ions a Other anions should give a negative white precipitate forms result which is no precipitate forming The acid is needed to react with carbonate impurities that Sulfuric acid cannot be used to acidify the are often found in salts which would form a white Barium mixture because it contains sulfate ions carbonate precipitate and so give a false result which would form a precipitate Testing for halide ions with silver nitrate. The role of nitric acid is to react with any This reaction is used as a test to identify which halide ion carbonates present to prevent formation of the is present. The test solution is made acidic with nitric precipitate Ag2CO3. This would mask the acid, and then silver nitrate solution is added dropwise. desired observations 2 HNO3 + Na2CO3  2 NaNO3 + H2O + CO2 Fluorides produce no precipitate Chlorides produce a white precipitate Ag+(aq) + Cl- (aq)  AgCl(s) Hydrochloric acid cannot be used to Bromides produce a cream precipitate acidify the mixture because it contains Ag+(aq) + Br- (aq)  AgBr(s) chloride ions which would form a Iodides produce a pale yellow precipitate precipitate Ag+(aq) + I- (aq)  AgI(s) The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar: Silver chloride dissolves in dilute ammonia to form a complex ion AgCl(s) + 2NH3(aq)  [Ag(NH3)2]+ (aq) + Cl- (aq) Colourless solution Silver bromide dissolves in concentrated ammonia to form a complex ion AgBr(s) + 2NH3(aq)  [Ag(NH3)2]+ (aq) + Br - (aq) Colourless solution Silver iodide does not react with ammonia – it is too insoluble. Testing for positive ions (cations) Test for ammonium ion NH4+, by reaction with warm Ammonia gas can be identified by its pungent smell or by turning damp red NaOH(aq) forming NH3 litmus paper blue NH4+ +OH-  NH3 + H2O See flame tests on page 4 for more cation tests N Goalby chemrevise.org 9