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3.2.3 GROUP 7(17), THE HALOGENS

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Q: Some scientists thought that the waste water from a waste disposal factory contained two sodium halides. They tested a sample of the waste water. They added three reagents, one after the other, to the same test tube containing the waste water. The table below shows their results. Identify the yellow precipitate that did not dissolve in concentrated ammonia solution. Write the simplest ionic equation for the formation of this precipitate from silver ions and the correct halide ion. Identify the other sodium halide that must be present in this mixture of two sodium halides. (3 MARKS)

- (yellow precipitate is) silver iodide OR AgI (which may be awarded from the equation) - Ag+ + I-→ AgI - sodium chloride OR NaCl

Q: Write an equation for the redox reaction of sodium bromide with concentrated sulphuric acid. (2 MARKS)

- 2 H2SO4 + 2 Br- → SO2 + Br2 + 2 H2O + SO4 2- - Allow 2 H2SO4 + 2 NaBr → SO2 + Br2 + 2 H2O + Na2SO4 - H2SO4 + 2 HBr → 2 H2O + Br2 + SO2 etc

EQ: This question is about the chemical properties of chlorine, sodium chloride and sodium bromide. Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride. Write an equation for this reaction of sodium bromide and explain why bromide ions react differently from chloride ions. (3 MARKS)

- 2NaBr + 2H2SO4 → Na2SO4 + Br2 + SO2 + 2H2O - Br- ions are bigger than Cl - ions - Therefore Br- ions more easily oxidised / lose an electron more easily (than Cl- ons)

State the half equations of the reaction of sodium bromide with conc. sulfuric acid.

- 2X-(g) → X2 (s) + 2e- - H2SO4 + 2H+ + 2e- → SO2 + 2H2O X represents any halogen, here you can replace it with bromine.

Explain the differing reducing powers of halides.

- A reducing agent donates electrons. - The reducing power of the halides increases down group 7 - They have a greater tendency to donate electrons. - This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller.

How does group 1 react with inorganic ions?

- ALL SOLUBLE! (no precipitate) - Every single one of them; OH-, SO4 2-, CO3 2-, NO3-, Cl-, Br-, I-

Q: Strontium chloride is used in toothpaste for sensitive teeth. Both strontium carbonate and strontium sulfate are white solids that are insoluble in water. Strontium carbonate reacts with nitric acid to produce a solution of strontium nitrate. Strontium sulfate does not react with nitric acid. Describe briefly how you could obtain strontium sulfate from a mixture of strontium carbonate and strontium sulfate. You are not required to describe the purification of the strontium sulfate. (2 MARKS)

- Add nitric acid to the mixture (until in excess) - Filter (to isolate strontium sulfate)

Q: Medicines for the treatment of nervous disorders often contain calcium bromide. Silver nitrate, acidified with dilute nitric acid, can be used together with another reagent to test for the presence of bromide ions in a solution of a medicine. Describe briefly how you would carry out this test and state what you would observe. (3 MARKS)

- Add silver nitrate, then dilute ammonia (solution) - Cream precipitate - No visible change or precipitate dissolves slightly in dilute ammonia

EQ: The table below shows observations of changes from some test-tube reactions of aqueous solutions of compounds Q, R and S with five different aqueous reagents. The initial colours of the solutions are not given. Write ionic equations for each of the positive observations with S. (4 MARKS)

- Ba2+ + SO42- → BaSO4 - [Fe(H2O)6]3+ + 3OH → Fe(H2O)3(OH)3 + 3H2O - 2[Fe(H2O)6]3+ + 3CO32 → 2Fe(H2O)3(OH)3 + 3H2O + 3CO2 - [Fe(H2O)6]3+ + 4Cl- → [FeCl4]- + 6H2O

EQ: This question is about ion testing. Describe how a student could distinguish between aqueous solutions of potassium nitrate, KNO3, and potassium sulfate, K2SO4, using one simple test-tube reaction. (3 MARKS)

- BaCl2 / Ba(OH)2 / Ba(NO3)2 / BaX2 or names - colourless solution / no (visible) change (nvc) / no ppt / no (visible) reaction - white precipitate / white solid

Q: Copper(II) sulfate solution, together with copper(II) carbonate (CuCO3) powder, can be used to determine the identity of three solutions A, B and C. The three solutions are known to be hydrochloric acid, barium chloride, and sodium chloride. In Experiment 1 a small amount of copper(II) carbonate powder was added to each of the three solutions. In Experiment 2 a dropping pipette was used to add 2 cm3 of copper(II) sulfate solution to each of the three solutions. The results of these experiments are shown in the table below. Explain why a precipitate was formed when copper(II) sulfate solution was added to solution A. Write an equation for the reaction that occurred. (2 MARKS)

- Barium sulfate is insoluble - CuSO4 + BaCl2 → BaSO4 + CuCl2

Q: Explain why chloride ions are weaker reducing agents than bromide ions. (2 MARKS)

- Chloride ions are smaller than bromide ions OR chloride ion electron(s) are closer to the nucleus OR chloride ion has fewer (electron) shells/levels OR chloride ion has less shielding (or converse for bromide ion) - Outer shell/level electron(s) OR electron(s) lost from a chloride ion is more strongly held by the nucleus compared with that lost from a bromide ion (or converse for bromide ion)

EQ: Explain why chlorine has a lower melting point than bromine. (2 MARKS)

- Chlorine is smaller than bromine - OR has fewer electrons/electron shells - OR It is smaller / It has a smaller atomic radius / it is a smaller molecule / atom (or converse) - The forces between chlorine / Cl2 molecules are weaker (than the forces between bromine / Br2 molecules) - (or converse for bromine) - OR chlorine / Cl, has weaker / fewer/ less (VdW) intermolecular forces / forces between molecules

Q: When chlorine gas dissolves in cold water, a pale green solution is formed. In this solution, the following equilibrium is established. Cl2(g) + H2O(l) ⇌ H+ (aq) + Cl- (aq) + HClO(aq) Give the formula of the species responsible for the pale green colour in the solution of chlorine in water. Use Le Chatelier's principle to explain why the green colour disappears when sodium hydroxide solution is added to this solution. (3 MARKS)

- Cl2 (provides the pale green colour) - NaOH reacts with the acid(s)/the HCl/the HClO/H+ - Equilibrium shifts (from left ) to right

Q: Chlorine is a powerful oxidising agent. Write the simplest ionic equation for the reaction between chlorine and aqueous potassium bromide. State what is observed when this reaction occurs. (2 MARKS)

- Cl2 + 2Br- --> 2Cl- + Br2 - Solution goes orange / yellow ( from colourless)

EQ: This question is about the chemical properties of chlorine, sodium chloride and sodium bromide. Write an ionic equation for the reaction between chlorine and cold dilute sodium hydroxide solution. Give the oxidation state of chlorine in each of the chlorine-containing ions formed. (2 MARKS)

- Cl2 + 2HO- → OCl- + Cl- + H2O - OCl- is +1 - Cl- is -1

Q: Write an equation for the reaction between chlorine and cold, dilute aqueous sodium hydroxide. Give two uses of the resulting solution. (3 MARKS)

- Cl2 + 2NaOH → NaCl + NaOCl +H2O - bleach - disinfectant /steriliser/kills bacteria

Q: How can the addition of an aqueous solution of chlorine be used to distinguish between aqueous solutions of sodium bromide and sodium iodide? State any observations you would make and write equations for the reactions occurring. (4 MARKS)

- Cl2(aq) to Br- (aq); yellow-orange or yellow-red - 2Br- + Cl2 → 2Cl- + Br2 - Cl2(aq) to I- (aq); brown/black solution formed - 2I- + Cl2 → 2Cl- + I2

EQ: Solution Y contains two different negative ions. To a sample of solution Y in a test tube a student adds: • silver nitrate solution • then an excess of dilute nitric acid • finally an excess of concentrated ammonia solution. The observations after each addition are recorded in the table. Give the formulas of D, E and F. Give an ionic equation to show the formation of E. Give an equation to show the conversion of D into G. (6 MARKS)

- D: AgBr - E: Ag2CO3 - F: CO2 - 2Ag+ + CO32- → Ag2CO3 - AgBr + 2 NH3 → Ag(NH3)2+ + Br -

Q: Chlorine is used in water treatment. When chlorine is added to cold water it reacts to form the acids HCl and HClO. The following equilibrium is established. Cl2(aq) + H2O(I) ⇌ H+ (aq) + Cl- (aq) + HClO(aq) Deduce what happens to this equilibrium as the HClO reacts with bacteria in the water supply. Explain your answer. (2 MARKS)

- Equilibrium will shift/move to the right - To oppose the loss of HClO

Q: Some scientists thought that the waste water from a waste disposal factory contained two sodium halides. They tested a sample of the waste water. They added three reagents, one after the other, to the same test tube containing the waste water. The table below shows their results. The method that the scientists used could not detect one type of halide ion. Identify this halide ion. Give one reason for your answer. (2 MARKS)

- F- OR Flouride (ion) - Silver fluoride / AgF is soluble / dissolves (in water) - No precipitate would form / no visible /observable change

EQ: Silver cyanide can be precipitated from sodium cyanide solution by adding an excess of silver nitrate solution. Describe how you would obtain a pure dry sample of silver cyanide from this mixture. (3 MARKS)

- Filter - Wash (the residue) with water - Dry by pressing between filter paper or in air

Q: When vanadium reacts with chlorine at 400°C, a brown compound is obtained. When an aqueous solution containing 0.193 g of this compound was treated with aqueous silver nitrate all the chlorine in the compound was precipitated as silver chloride. The mass of silver chloride (AgCl) produced was 0.574 g. Which one of the following could be the formula of the brown compound? (1 MARK) A: VCl B: VCl2 C: VCl3 D: VCl4

- Find number of moles of AgCl - Find number of moles of chlorine - Find mass of chlorine - Find mass of vanadium - Find the formula. (VCl4) It's a moles question! Calculate the mass of chlorine in the AgCl = 0.574 x 35.5/143.5 = 0.142g This came from 0.193g of the vanadium chloride therefore 0.193 - 0.142g is vanadium = 0.051g - Vanadium (RAM=51) - Chlorine (RAM= 35.5) ratio of moles = 0.051/51 to 0.142/35.5 = 0.001 : 0.004 Therefore the formula is VCl4

EQ: In terms of electrons, state the meaning of the term oxidising agent. (1 MARK)

- Gains / takes / accepts electrons - Or causes another species to lose electrons

EQ: A solution of sodium chlorate(l) was added to a colourless solution of potassium iodide. Suggest what is observed. Explain the reaction that leads to this observation. (3 MARKS)

- Goes brown (or shades of brown) - Due to iodine or I3− - Because I− oxidised

Q: In two further tests, the student made the following observations. Write the simplest ionic equation for the reaction that has taken place in Test 3. Identify the species responsible for the misty white fumes produced in Test 3. (2 MARKS)

- H2SO4 + 2CL- --> 2HCL + SO4 2- - OR H2SO4 + Cl- --> HCl + H2SO4 - - OR H+ + Cl- ---> HCl - Hydrogen chloride OR HCl

Why is silver nitrate acidified before reacting it with halides?

- HNO3 (aq) added to test solution first to remove (neutralise) any CO3 2- or OH- impurities present - CO3 2- / OH- interfere with test because: - AgCO3 = white precipitate - AgOH = brown precipitate Tip: You can't use hydrochloric acid instead of nitric acid because the silver nitrate would just react with the chloride ions from the HCI - and that would mess up your results completely.

Q: Copper(II) sulfate solution, together with copper(II) carbonate (CuCO3) powder, can be used to determine the identity of three solutions A, B and C. The three solutions are known to be hydrochloric acid, barium chloride, and sodium chloride. In Experiment 1 a small amount of copper(II) carbonate powder was added to each of the three solutions. In Experiment 2 a dropping pipette was used to add 2 cm3 of copper(II) sulfate solution to each of the three solutions. The results of these experiments are shown in the table below. Use the observations in the table to deduce which of the solutions, A, B or C is hydrochloric acid and barium chloride. (2 MARKS)

- Hydrochloric acid = C - Barium chloride = A

SME: The boiling points of the hydrogen halides are represented in the graph in Figure 1. Explain why hydrogen fluoride has the highest boiling point out of the hydrogen halides. (3 MARKS)

- Hydrogen fluoride / HF / exhibits hydrogen bonding; - The other hydrogen halides only contain permanent dipole dipole forces; - Hydrogen bonding is stronger than permanent dipole dipole forces;

SME: Explain why the reaction between solid potassium iodide and concentrated sulfuric acid produces hydrogen sulfide whereas the reaction between solid potassium bromide, and concentrated sulfuric acid does not. (4 MARKS)

- Iodide ion / I stronger reducing agent than bromide ion / Br; - Able to reduce S / sulfur in H2SO4 from +6 to -2; - Iodide ion / I much bigger than bromide ion / Br; - Readily donates an electron (compared to Br) (as further away from the nucleus / more shielding);

Q: Explain why iodide ions are stronger reducing agents than chloride ions. (2 MARKS)

- Iodide ions are larger/have more (electron) shells/levels than chloride ions (or converse for chloride ion) OR electron(s) to be lost/outer shell/level is further from the nucleus (or converse for chloride ion) OR greater/more shielding - Electron(s) lost from an iodide ion is less strongly held by the nucleus compared with that lost from a chloride ion

Q: Explain why iodine has a higher melting point than fluorine. (2 MARKS)

- Iodine has more electrons / iodine is bigger (atom or molecule) / iodine has bigger Mr / bigger surface area - Stronger more van der Waals forces / vdw / London / temporarily induced dipole / dispersion forces between molecules

EQ: State a reason why chlorine is added to drinking water, and suggest a disadvantage of treating water in this way. (2 MARKS)

- Kills bacteria - Wasteful as most potable water not used for drinking - used in washing clothes etc - OR Some people suffer eye irritation / Some people find the taste unpleasant - OR can react with organic compounds to produce harmful substances

Q: In Peru, chlorine was removed from the water supply due to concerns about it reacting with organic chemicals in the water to produce toxic substances. This resulted in the death of ten thousand people due to cholera. The cholera epidemic ceased when chlorination of the water supply was restarted. State why chlorine is added to the water supply and give a reason why the amount of chlorine must be carefully monitored. Write an equation for the reaction of chlorine with water. (3 MARKS)

- Kills bacteria / prevents bacterial diseases - Chlorine is a toxic substance - Cl2 + H2O → HCl + HClO

Q: The student had read in a textbook that the equation for one of the reactions in Test 4 is as follows. 8H+ + 8I- + H2SO4 → 4I2 + H2S + 4H2O Write the two half-equations for this reaction. State the role of the sulfuric acid and identify the yellow solid that is also observed in Test 4. (4 MARKS)

- M1 2I- → I2 +2e- - OR 8I- → 4I2 + 8e- - M2 H2SO4 + 8H+ + 8e- → H2S + 4H2O - OR SO4 2- + 10H+ + 8e- → H2S + 4H2O - M3 oxidising agent / oxidises the iodide (ions) - OR electron acceptor - M4 sulfur OR S OR S, OR S, OR sulphur

Q: The following pairs of compounds can be distinguished by simple test-tube reactions. For each pair, give a suitable reagent that could be added separately to each compound to distinguish between them. Describe what you would observe in each case. HCl(aq) and HNO3(aq). (3 MARKS)

- M1 AgNO3 OR silver nitrate OR any soluble silver salt - M2 white precipitate or white solid / white suspension - M3 remains colourless OR no reaction OR no (observed) change OR no precipitate

Q: (iii) The student knew that bromine can be used for killing microorganisms in swimming pool water. The following equilibrium is established when bromine is added to cold water. BR2(l) + H2O(l) ⇌ HBrO (aq) + H+ (aq) + Br- (aq) Use Le Chatelier's principle to explain why this equilibrium moves to the right when sodium hydroxide solution is added to a solution containing dissolved bromine. (2 MARKS)

- M1 The NaOH / OH- / (sodium) hydroxide reacts with / neutralises the H+ / acid / HBr (lowering its concentration) - OR a correct neutralisation equation for H+ or HBr with NaOH or with hydroxide ion - M2 The (position of) equilibrium moves / shifts(from L to R) - to replace the H+ / acid / HBr that has been removed / lost - OR to increase the H+ / acid / HBr concentration - OR to make more H+ / acid / HBr / product(s) - OR to oppose the loss of H+ / loss of product(s) - OR to oppose the decrease in concentration of product(s)

What type of agents are halogens? Explain why.

- Oxidising agents - Have 7 electrons so need to gain 1 - Therefore it gains / accepts electrons and is reduced itself - That electron must have been taken from something else - Thereby it is oxidising something else and being reduced

Q: Samples of solid sodium fluoride, sodium chloride, sodium bromide and sodium iodide are each warmed separately with concentrated sulphuric acid. All four compounds react with concentrated sulphuric acid but only two can reduce it. In addition to sulphur dioxide, two further reduction products are formed when one of these two halides reacts with concentrated sulphuric acid. Identify the two reduction products and write a half-equation to show the formation of one of them from concentrated sulphuric acid. (3 MARKS)

- Products Sulphur (or S8 not S4), Hydrogen sulphide - Equation:- H2SO4 (or 2H+ + SO4 2-) +6H+ + 6e- → S + 4H2O - OR H2SO4 (or 2H+ + SO4 2-) + 8H+ + 8e- → H2S + 4H2O

EQ: The table below shows observations of changes from some test-tube reactions of aqueous solutions of compounds Q, R and S with five different aqueous reagents. The initial colours of the solutions are not given. Identify each of compounds Q, R and S. You are not required to explain your answers. (6 MARKS)

- Q is calcium or magnesium - bromide - R is aluminium - chloride - S is iron(III) - sulfate

EQ: The following pairs of compounds, each in aqueous solution, can be distinguished by simple test-tube reactions. Give a reagent, or combination of reagents, that can be added to the solutions in each pair to distinguish between them in a single reaction. State what is observed in each case. NaCl(aq) and Na2CO3(aq). (3 MARKS)

- Reagent: H2SO4 / HCl / HNO3 - Observation with NaCl: no (visible) change - Observation with Na2CO3: effervescence/bubbles/fizzing OR - Reagent: acidified AgNO3 - Observation with NaCl: white ppt / white solid formed - Observation with Na2CO3: effervescence/bubbles/fizzing

- Reagent: H2SO4 / Na2SO4 / any soluble sulfate - Observation with NaCl: no (visible) change - Observation with BaCl2: white ppt / white solid formed

What type of agents are halides? Explain why.

- Reducing agents - As the lose electrons more easily going down the group

Q: Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid. Give the formula for each of the following in this reaction. - Formula of the solid reduction product - Formula of the oxidation product (2 MARKS)

- S OR S8 OR S2 - I2

Q: Identify two sulphur-containing reduction products formed when concentrated sulphuric acid oxidises iodide ions. For each reduction product, write a half-equation to illustrate its formation from sulphuric acid. (4 MARKS)

- SO2; - SO4 2- + 4H+ + 2e- → SO2 + 2H2O - S (also H2S); - SO4 2- + 8H+ + 6e- → S + 4H2O - OR SO4 2- + 10H+ + 6e- → H2S + 4H2O

EQ: This question is about elements in Group 7 of the Periodic Table and their compounds. Bromine (Br2), strontium chloride (SrCl2) and iodine monochloride (ICl) all have similar Mr values. Suggest, with reasons, the order of melting points for these three substances. (6 MARKS)

- SrCl2> ICl > Br> - SrCl2 strong ionic bonds / (strong electrostatic attraction between opposite ions) - Lattice so many strong bonds to overcome - ICl has dipole-dipole between molecules - weaker than ionic bonds - Br2 has van der Waals forces between molecules - much weaker

Q: Chlorine is toxic to humans. This toxicity does not prevent the large-scale use of chlorine in water treatment. Explain why the toxicity of chlorine does not prevent this use. (1 MARK)

- The (health) benefit outweighs the risk - OR a clear statement that once it has done its job, little of it remains - OR used in (very) dilute concentrations / small amounts / low doses

SME: A compound that exhibits a high thermal stability can withstand high temperatures before breaking down. Hydrogen iodide, HI, has a lower thermal stability than hydrogen chloride, HCI. In terms of bond length, explain why hydrogen iodide has a lower thermal stability than hydrogen chloride. (3 MARKS)

- The atomic radius of the halogens increase as you go down the group; - The hydrogen halide bond / H-X bond is (therefore) longer; - Therefore the bonding pair of electrons are further from the nucleus and is easier to break;

Describe the test for ammonium (NH4+)

- Warm with NaOH - NH3 (g) released. - Use damp, red litmus to test - Red → blue (positive test)

Q: When chlorine reacts with water in the absence of sunlight the chlorine is both oxidised and reduced and equilibrium is reached. The pool manager maintains the water at a pH slightly greater than 7.0 Explain how this affects the equilibrium established when chlorine is added to water

- When Cl2 is added to water, an equilibrium is established - If MORE CI2 is added, that means the equilibrium will shift to the right (le chatelier's) - As the equilibrium shifts to the right, more HCl is produced. HCl is acidic which DECREASES the pH.

EQ: The halogens are the elements in Group 7. The electronegativities of the halogens are shown in Table 4. Explain the trend in electronegativities shown by the halogens. (2 MARKS)

- increasing atomic radius / shielding / number of shells / size (down group) or reverse argument - decreasing attraction of nucleus/protons for shared (electron) pair / bond electrons

SME: Place the following halogens in order of increasing bond strength. Chlorine, Cl2, Bromine, Br2 and iodine, I2,

- lodine / I2< Bromine / Br2< Chlorine / Cl2; - As you go down group 7 the bond strength decreases because: o The atomic radius increases o Therefore the bond length becomes longer The exception is fluorine, F2, which actually has a weaker bond than chlorine, Cl2, due to the repulsion between lone pairs in the molecule

Q: State why the silver nitrate solution is acidified when testing for iodide ions. (1 MARK)

- react with / remove (an)ions that would interfere with the test - prevent the formation of other silver precipitates / insoluble silver compounds that would interfere with the test - remove (other) ions that react with the silver nitrate - react with / remove carbonate / hydroxide / sulfite (ions)

Q: Chlorine is toxic to humans. This toxicity does not prevent the large-scale use of chlorine in water treatment. Give one reason why water is treated with chlorine. (1 MARK)

- to sterilise / disinfect water - to destroy / kill microorganisms / bacteria / microbes / pathogens

What are the general steps for testing for halide ions?

1 - Acidified silver nitrate (H+/ AgNO3 (aq)) 2 - Aqueous ammonia (NH3 (aq))

Describe the first step of testing halide ions.

1 - add AgNO3 (aq) dropwise: - Ag+(aq) + F-(aq) → AgF(aq) = no precipitate - Ag+(aq) + Cl-(aq) → AgCl(s) = white precipitate - Ag+(aq) + Br-(aq) → AgBr(s) = cream precipitate - Ag+(aq) + I-(aq) → AgI(s) = yellow precipitate

EQ: The following tests were carried out to identify an unknown green salt Y. An aqueous solution of Y gave a cream precipitate of compound A when reacted with silver nitrate solution. Compound A gave a colourless solution when reacted with concentrated ammonia solution. Another aqueous solution of Y gave a green precipitate B when reacted with sodium carbonate solution. The green precipitate B was filtered and dried and then reacted with sulfuric acid to give a pale green solution containing compound C and a colourless gas D. 1. Identify by name or formula the compounds A, B, C, D and Y. (5 MARKS) 2. Write the simplest ionic equation for the reaction that occurs between the green precipitate B and sulfuric acid. (1 MARK)

1. - A Silver bromide / AgBr - B Iron(II) carbonate / FeCO3 - C Iron(II) sulphate / FeSO4 - D Carbon dioxide / CO2 - Y Iron(II) bromide / FeBr2 2. 2H+ + CO32− ⟶ H2O + CO2

1. -Reagent: H2SO4 / Na2SO4 / any soluble sulfate - Observation with NaCl: no (visible) change - Observation with BaCl2: white ppt / white solid formed 2. - Reagent: H2SO4/ HCl / HNO3 - Observation with NaCl: no (visible) change - Observation with Na2CO3: effervescence/bubbles/fizzing OR - Reagent: acidified AgNO3 - Observation with NaCl: white ppt / white solid formed - Observation with Na2CO3: effervescence/bubbles/fizzing

Q: Iodine reacts with concentrated nitric acid to produce nitrogen dioxide (NO2). 1. Complete the balancing of the following equation. (1 MARK) I2 + 10HNO3 → __HIO3 + __NO2 + __H2O 2. In industry, iodine is produced from the NaIO3 that remains after sodium nitrate has been crystallised from the mineral Chile saltpetre. The final stage involves the reaction between NaIO3 and NaI in acidic solution. Half-equations for the redox processes are given below. IO3- + 5e- + 6H+ → 3H2O + ½ I2 I- → ½ I2 + e- Use these half-equations to deduce an overall ionic equation for the production of iodine by this process. Identify the oxidising agent. (2 MARKS)

1. 2, 10, 4 2. - IO3- + 6H+ + 5I- → 3I2 + 3H2O - NalO3 OR IO3- OR iodate ions OR iodate(V) ions etc. (Accept "the iodine in iodate ions" but NOT "iodine" alone)

Q: Solubility data for barium hydroxide and calcium hydroxide are given in the table below. 1. Use the data given in the table to calculate the concentration, in mol dm-3, of a saturated solution of calcium hydroxide (Mr = 74.1) at 20°C. (1 MARK) 2. Suggest one reason why calcium hydroxide solution is not used in the titration of a 0.200 mol dm-3 solution of an acid. (1 MARK)

1. 2.3(3) × 10-2 2. Dilution of acid needed / may react with carbon dioxide in air

EQ: Chlorine is an important industrial chemical. Chlorine is formed when KMNO4 reacts with hydrochloric acid. The ionic equation for this redox reaction is: 16H+ + 2MnO4- + 10CI- → 2MN 2+ + 8H2O + 5Cl2 1. Deduce the half-equation for the oxidation of chloride ions to chlorine. (1 MARK) 2. Deduce the half-equation for the reduction of the MnO4- ions in acidified solution to manganese(II) ions and water. (1 MARK)

1. 2Cl- → Cl2 + 2e- 2. MnO4- + 8H+ + 5e- → Mn 2+ + 4H2O

EQ: A student oxidised a solution of hydrochloric acid with a few drops of sodium chlorate(l) solution. The reaction mixture effervesced and turned pale green. The gas formed bleached universal indicator paper. 1. Write a half-equation for the oxidation of chloride ions. (1 MARK) 2. Write a half-equation for the reduction of chlorate(I) ions to chlorine in acidic conditions. (1 MARK) 3. Write an overall equation for the redox reaction of chlorate(I) ions with hydrochloric acid. (1 MARK)

1. 2Cl− → Cl2 + 2e(−) 2. 2ClO− + 4H+ + 2e− → Cl2 + 2H2O 3. ClO− + Cl− + 2H+ → Cl2 + H2O

Q: Chlorine behaves as an oxidising agent in the extraction of bromine from seawater. In this process, chlorine gas is bubbled through a solution containing bromide ions. 1. Write the simplest ionic equation for the reaction of chlorine with bromide ions. (1 MARK) 2. Give one observation that would be made during this reaction. (1 MARK)

1. Cl2 + 2Br- → 2Cl- + Br2 2. (Turns to) yellow / orange / brown (solution)

Q: Chlorine reacts with water to form an equilibrium mixture containing hydrochloric acid and chloric(I) acid. 1. Write an equation for the formation of this equilibrium mixture. (1 MARK) 2. Household bleach contains sodium chlorate(1) and sodium chloride. State and explain, with reference to your equation in part (1), why it is dangerous to acidify an aqueous mixture of sodium chlorate(1) and sodium chloride. (2 MARKS)

1. Cl2 + H2O ⇌ HCl + HClO 2. - equilibrium shifts/moves left - (producing) chlorine (which) is toxic/poisonous

Q: For many years, swimming pool water has been treated with chlorine gas. The chlorine is added to kill any harmful bacteria unintentionally introduced by swimmers. Pool managers are required to check that the chlorine concentration is high enough to kill the bacteria without being a health hazard to the swimmers. When chlorine reacts with water in the absence of sunlight, the chlorine is both oxidised and reduced and an equilibrium is established. 1. Write an equation for this equilibrium. (1 MARK) 2. The pool manager maintains the water at a pH slightly greater than 7.0 Explain how this affects the equilibrium established when chlorine is added to water. (2 MARKS)

1. Cl2 + H2O ⇌ HOCl + HCl 2. - Hydroxide / alkali ions react with the acids - Equilibrium moves to the right

Q: Solid sodium halides react with concentrated sulfuric acid. A sample of solid sodium iodide is reacted with concentrated sulfuric acid. A black solid forms and hydrogen sulfide gas is produced. 1. Write a half-equation for the reaction of sulfuric acid to form hydrogen sulfide. (1 MARK) 2. Write a half-equation for the formation of the black solid. (1 MARK) 3. Use your answers to parts (1) and (2) to write an overall equation for the reaction of sodium iodide with concentrated sulfuric acid. (1 MARK)

1. H2SO4 + 8H+ + 8e- → H2S + 4H2O 2. 2I- → I2 + 2e- 3. H2SO4 + 8H+ + 8I- → H2S + 4H2O + 4I2

EQ: Chlorine is used to decrease the numbers of microorganisms in water. When chlorine is added to water, there is a redox reaction, as shown by the equation Cl2 + H2O ⇌ HClO + HCl 1. Deduce the oxidation state of chlorine in HClO and the oxidation state of chlorine in HCl. (1 MARK) 2. Give two half-equations to show the oxidation and reduction processes that occur in this redox reaction. (2 MARKS)

1. HClO = +1 HCl = -1 2. - Oxidation: Cl2 + 2H2O → 2HClO + 2H+ +2e- - Reduction: Cl2 + 2H+ + 2e- → 2HCl

Q: Magnesium hydroxide and magnesium carbonate are used to reduce acidity in the stomach. Magnesium hydroxide can be prepared by the reaction of solutions of magnesium chloride and sodium hydroxide. 1. Write the simplest ionic equation for the reaction that occurs between magnesium chloride and sodium hydroxide. Include state symbols in your equation. (1 MARK) 2. Other than cost, explain one advantage of using magnesium hydroxide rather than magnesium carbonate to reduce acidity in the stomach. (1 MARK)

1. Mg2+ (aq) + 2OH- (aq) --> Mg(OH)2 (s) 2. Does not produce CO2 / gas which distends stomach / does not produce wind / does not increase pressure in stomach

Describe the reaction of sodium bromide and conc. sulfuric acid.

1. NaBr + H2SO4 → HBr + NaHSO4 - Acid base reaction - HBr is steamy fumes - Br- ions much stronger reducing agent so secondary redox reaction occurs 2. 2HBr + H2SO4 → Br2 + SO2 + 2H2O - Redox reaction - Br2 observed as brown fumes - SO2 is colourless gas - Sulfur has been reduced = +6 → +4 - Bromine has been oxidised = -1 → 0 - As bromine is strong enough reducing agent (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.)

Describe the reaction of sodium iodide and conc. sulfuric acid.

1. NaI + H2SO4 → HI + NaHSO4 - Acid base reaction - HI is steamy fumes - I- ions strongest reducing agent so secondary redox reaction occur 2. 2HI + H2SO4 → I2 + SO2 + 2H2O - Redox reaction - I2 observed as purple fumes (or maybe black solid) - SO2 is colourless gas - Sulfur has been reduced = +6 → +4 - Iodine has been oxidised = -1 → 0 - I- such a strong reducing agent that two further reactions happen. 3. 6HI + H2SO4 → 3I2 + S + 4H2O - Redox reaction - I2 observed as purple fumes (or maybe black solid) - S shown as yellow solid - Sulfur has been reduced = +6 → 0 4. 8HI + H2SO4 → 4I2 + H2S + 4H2O - I2 observed as purple fumes (or maybe black solid) - H2S is pungent gas (smells like rotten eggs) - Sulfur has been reduced = +6 → -2 (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.) Dont think of these reactions as sequential, think of them all happening simultaneously.

Q: Anhydrous strontium chloride is not used in toothpaste because it absorbs water from the atmosphere. The hexahydrate, SrCl2.6H2O, is preferred. A chemist was asked to determine the purity of a sample of strontium chloride hexahydrate. The chemist weighed out 2.25 g of the sample and added it to 100 cm³ of water. The mixture was warmed and stirred for several minutes to dissolve all of the strontium chloride in the sample. The mixture was then filtered into a conical flask. An excess of silver nitrate solution was added to the flask and the contents swirled for 1 minute to make sure that the precipitation was complete. The silver chloride precipitate was separated from the mixture by filtration. The precipitate was washed several times with deionised water and dried carefully. The chemist weighed the dry precipitate and recorded a mass of 1.55 g. The equation for the reaction between strontium chloride and silver nitrate is SrCl2 + 2AgNO3 → 2AgCl + Sr(NO3)2 Several steps in the practical procedure were designed to ensure an accurate value for the percentage by mass of strontium chloride hexahydrate in the sample. 1. Explain why the solution of strontium chloride was filtered to remove insoluble impurities before the addition of silver nitrate. (1 MARK) 2. Explain why the precipitate of silver chloride was washed several times with deionised water. (1 MARK)

1. Would give an incorrect / too large mass (of silver chloride) 2. To remove soluble impurities / excess silver nitrate (solution) / strontium nitrate (solution)

Describe the second additional step of testing halide ions.

2 - followed by NH3 (aq): precipitates formed may not be clear to distinguish so they can be tested further - AgCl(s) dissolves in DILUTE NH3 (aq) - Forms complex ion; colourless solution - AgCl(S) + 2NH3 (aq) → [Ag(NH3)2]+(aq) + Cl-(aq) - AgBr(s) dissolved in CONC NH3 (aq) - Forms complex ion; colourless solution - AgBr(S) + 2NH3 (aq) → [Ag(NH3)2]+(aq) + Br-(aq) - AgI(s) will not dissolve → too insoluble (so doesnt react)

Describe and state the reaction of chlorine with water in the presence of U.V. (usually in the form of sunlight).

2Cl2 + 2H2O ⇌ 4HCl + O2 - So outdoor pools need chlorinating more often due to presence of U.V. - An alternative to direct chlorination is to add solid sodium (or calcium) chlorate(1). This dissolve in water to form chloric(1) acid, HCLO, in a reversible reaction NaClO(s) + H2O ⇌ Na+(aq) + OH-(aq) + HClO(aq) In an alkaline solution, this equilibrium moves to the left, to prevent this swimming pools need to be kept slightly acidic.

SME: The reaction between chlorine and an alkali is an example of a disproportionation reaction. Write an equation, including state symbols, for the reaction between dilute sodium hydroxide, NaOH (aq) and chlorine, Cl2 (g), at 70°C. (3 MARKS)

3CI2 (g) + 6NaOH (aq) → 5NaCI (aq) + NaCIO3 (aq) + 3H2O (l)

Q: When concentrated sulfuric acid is added to potassium iodide, iodine is formed in the following redox equations. ......KI + ......H2SO4 → ......KHSO4 + ......I2 + S + ......H2O 8KI + 9H2SO4 → 8KHSO4 + 4I2 + H2S + 4H2O Balance the equation for the reaction that forms sulfur. (1 MARK)

6, 7, 6, 3, 4

EQ: Potassium chlorate(VII), KClO4, is used in fireworks. When potassium chlorate(VII) decomposes, it produces potassium chloride and oxygen. Give an equation for the decomposition of potassium chlorate(VII). Use the data in Table 3 to calculate the enthalpy change for this reaction. (2 MARKS)

= KClO4 → KCl + 2O2 ∆H = - 436 - - 434 = - 2 kJ mol-1

EQ: Which statement is not correct about the trends in properties of the hydrogen halides from HCl to HI ? (1 MARK) A: The boiling points decrease. B: The bond dissociation energy of H−X decreases. C: The polarity of the H−X bond decreases. D: They are more easily oxidised in aqueous solutions.

Q: Which one of the following statements is true? A: A blue solution containing the ion [CoCl4]2- turns pink when added to an excess of water. B: A purple solution is formed when chlorine is bubbled into aqueous sodium bromide. C: A yellow precipitate is formed when aqueous silver nitrate is added to aqueous sodium chloride. D: A green solution containing the ion [CuCl4]2- turns blue when added to an excess of concentrated hydrochloric acid. (1 MARK)

How does group 2 react with CO3 2-?

ALL insoluble: all white precipitates

How does group 2 react with Br-?

ALL soluble (no precipitate)

How does group 2 react with Cl-?

ALL soluble (no precipitate)

How does group 2 react with I-?

ALL soluble (no precipitate)

How does group 2 react with NO3-?

ALL soluble (no precipitate)

How does group 2 react with halides?

ALL soluble (no precipitate)

Q: A yellow precipitate is formed when silver nitrate solution, acidified with dilute nitric acid, is added to an aqueous solution containing iodide ions. Write the simplest ionic equation for the formation of the yellow precipitate. (1 MARK)

Ag+ + I- → AgI

Dilute ammonia solution CANNOT be used to distinguish between...

AgBr(s) and AgI(s)

Q: Explain why it is not possible to distinguish between separate solutions of sodium nitrate and sodium fluoride by the addition of silver nitrate solution. (1 MARK)

AgF is soluble

How does NH4 + react with all ions? (OH-, SO4 2-, CO3 2-, NO3-, Cl-, Br-, I)

All soluble

Q: Give one large-scale application of the use of chlorine in water. (1 MARK)

Any one from • in swimming pools • in drinking water • to sterilise/disinfect/sanitise water • in water treatment

EQ: What is the most suitable reagent for detecting the presence of carbonate ions in the presence of an excess of sulfate ions? A: dilute NaOH(aq) B: dilute H2SO4(aq) C: BaCl2(aq) D: NaCl(aq) (1 MARK)

Q: Which one of the following statements is correct? (1 MARK) A: The first ionisation energies of the elements in Period 3 show a general decrease from sodium to chlorine. B: The electronegativities of Group 2 elements decrease from magnesium to barium. C: The strength of the intermolecular forces increases from hydrogen fluoride to hydrogen chloride. D: The ability of a halide ion to act as a reducing agent decreases from fluoride to iodide.

Q: Give the formula of the substance responsible for the orange colour when chlorine gas is bubbled through an aqueous solution of sodium bromide. (1 MARK)

Br2

What is the state of bromine in room temperature? describe what it looks like.

Br2 (l) - Red / orange liquid - Gives off dense brown / orange poisonous fumes

Q: Explain why bromine does not react with aqueous chloride ions. (1 MARK)

Br2 is a weaker oxidising agent than Cl2

Q: How can reactions with concentrated sulphuric acid be used to distinguish between solid samples of sodium bromide and sodium iodide? State the observations you would make and give all the oxidation and reduction products formed in both reactions. Using half-equations, construct an overall equation for one of these redox reactions. (11 MARKS)

Bromide: - Brown/orange fumes - Bromine produced - Sulphur dioxide produced Iodide: - Purple fumes or black/brown/grey solid QWC or smell of bad eggs - Iodine produced - SO2, S, H2S produced (one mark each) Half-equations: 2Br- → Br2 + 2e-1 OR 2I- →I2 + 2e- 1 - H2SO4 + 2e- + 2H+ → SO2 + 4H2O - OR H2SO4 + 6e- + 6H+ → S + 4H2O - OR H2SO4 + 8e- + 8H+ → H2S + 4H2O - Overall equation Any correct equation based on half-equations QWC

How does Ag+ (AgNO3 (aq)) react with hydroxides (OH-)?

Brown precipitate: AgOH(s)

Q: Which is the correct equation for the reaction of hot, dilute aqueous sodium hydroxide, NaOH, and chlorine, Cl2? A: Cl2 + 2NaOH → NaCl + NaClO + H2O B: ½ Cl2 + 2OH- → CIO- + H2O + e- C: ½ Cl2 + 6OH- → CIO3- + 3H2O + e- D: 3Cl2 + 6NaOH → 3NaCl + 3NaCIO3 + H2O

What is the state of chlorine in room temperature? describe what it looks like.

Cl2 (g) - Greenish reactive gas - Poisonous in high concentrations

Describe and state the reaction of chlorine with cold water (not in presence of U.V. light).

Cl2 + H2O ⇌ HCl + HClO - Disproportionation: where an element is simultaneously oxidised and reduced - Oxygen is more electronegative than chlorine so oxidation state of Cl in HClO is +1 and not -1 - HClO is known as chloric (1) acid - HClO is an antibacterial (oxidising agent), hence the addition of chlorine to swimming pools and in water treatment for mains water supply - Poses some risk as Cl is toxic, but benefits of clean, treated water outweigh risks - Also used in very dilute concentrations

EQ: Which one of the following reactions does not involve donation of an electron pair? A H+ + CH3NH2 → CH3NH B AlCl3 + Cl− → AlCl C CH3Cl + CN− → CH3CN + Cl− D 1/2 Cl2 + I− → Cl− + 1/2 I2

Q: An aqueous solution of a white solid gives a yellow precipitate with aqueous silver nitrate. The formula of the white solid could be: A: AgBr B: AgI C: NaBr D: NaI (1 MARK)

Q: Which of these substances reacts most rapidly to produce a silver halide precipitate with acidified silver nitrate? (1 MARK) A: CH3Br B: CH3CI C: CH3F D: CH3I

Q: Which one of the following is the electron arrangement of the strongest reducing agent? (1 MARK) A: 1s2 2s2 2p5 B: 1s2 2s2 2p6 3s2 C: 1s2 2s2 2p6 3s2 3p5 D: 1s2 2s2 2p6 3s2 3p6 4s2

Q: Which one of the following can act as an oxidising agent but not as a reducing agent? A: CH3CHO В: Fe2+ C: I- D: MnO4-

D - Maganese's oxidation is the highest it can be here (+7) - So it cant be oxidised any further - So cant act as a reducing agent - However it can only be reduced back down to +5, +2 - In this its acting like a oxidising agent

Describe the trend in oxidising power down group 7 halogens.

Decreases down the group: - F2 is the strongest oxidising agent because its most easily reduced. This is because it attracts an electron really strongly. - High effective nuclear charge (low shielding) - Small atomic radius - I2 is the weakest oxidising agent (most difficult to reduce): - Lowest effective nuclear charge due to high shielding - Large atomic radius

Describe the trend in reactivity going down group 7.

Decreases: - Need to gain an electron - Atomic radius increases - So harder to attract electron due to increased shielding - As weaker electrostatic attraction between nucleus and outermost shell electrons

Describe the trend in electronegativity going down group 7.

Descreases: - F most electronegative - Has the greatest effective charge on the nucleus due to low shielding - Increases its ability to attract a bonding pair of electrons - Down the group atomic radius decreases as shielding increases - Nucleus less able to attract bonding pair of electrons - Less attraction for bond (pair electrons)

What is the state of fluorine in room temperature? describe what it looks like.

F2 (g) - Pale yellow gas - Highly reactive

Q: Suggest why iodide ions are stronger reducing agents than chloride ions. (2 MARKS)

For M1 and M2, iodide ions are stronger reducing agents than chloride ions, because M1 Relative size of ions: - Iodide ions / they are larger /have more electron levels(shells) (than chloride ions) / larger atomic / ionic radius - OR electron to be lost/outer shell/level (of the iodide ion) is further the nucleus - OR iodide ion(s) / they have greater / more shielding - OR converse for chloride ion M2 Strength of attraction for electron(s): - The electron(s) lost /outer shell/level electron from (an) iodide ion(s) less strongly held by the nucleus compared with that lost from a chloride ion - OR converse for a chloride ion

Q: Copper(II) sulfate is toxic. Suggest one safety precaution you would take to minimise this hazard when wiping up a spillage of copper(II) sulfate solution. (1 MARK)

Gloves / wash hands after use

Q: A different solid sodium halide reacts with concentrated sulphuric acid without reduction forming a halogen-containing product X. Write an equation for the reaction with concentrated sulphuric acid in which X is formed. (1 MARK)

H+ + F- → HF

Q: Give an example of a reagent which could be used to show that the reducing ability of bromide ions is different from that of chloride ions. (2 MARKS)

H2SO4

Q: When concentrated sulfuric acid is added to potassium iodide, solid sulfur and a black solid are formed. Deduce the half-equation for the formation of sulfur from concentrated sulfuric acid. (1 MARK)

H2SO4 + 6H+ + 6e- → S + 4H2O OR SO42- + 8H+ + 6e- → S + 4H2O

Q: When concentrated sulfuric acid is added to potassium iodide, iodine is formed in the following redox equation. 8KI + 9H2SO4 → 8KHSO4 + 4I2 + H2S + 4H2O Deduce the half-equation for the formation of hydrogen sulfide from concentrated sulfuric acid. (1 MARK)

H2SO4 + 8H+ + 8e- → H2S + 4H2O OR SO4 2- + 10H+ + 8e- → H2S + 4H2O

Q: Explain why dilute hydrochloric acid is not used to acidify the silver nitrate solution in this test for iodide ions. (1 MARK)

HCl would form a (white) precipitate / (white) solid (with silver nitrate and this would interfere with the test)

Q: Samples of solid sodium fluoride, sodium chloride, sodium bromide and sodium iodide are each warmed separately with concentrated sulphuric acid. All four compounds react with concentrated sulphuric acid but only two can reduce it. Identify the two halides which reduce concentrated sulphuric acid to sulphur dioxide. Using half-equations for the oxidation and reduction processes, deduce an overall equation for the formation of sulphur dioxide when concentrated sulphuric acid reacts with one of these halides. (3 MARKS)

Halides:- Bromide and iodide (1) Equation:- H2SO4 (or 2H+ + SO4 2-) + 2H+ +2e- → SO2 + 2H2O - 2Br- → Br2 + 2e- - H2SO4 + 2H+ + 2Br - (or 2HBr) → Br2 + SO2 + 2H2O (1)

Q: Samples of solid sodium fluoride, sodium chloride, sodium bromide and sodium iodide are each warmed separately with concentrated sulphuric acid. All four compounds react with concentrated sulphuric acid but only two can reduce it. Identify the two halides which do not reduce concentrated sulphuric acid. Write an equation for the reaction which does occur with one of these two halides. (3 MARKS)

Halides:- Fluoride, Chloride Equation:- H+ + F- → HF (or molecular / for a correct halide)

What is the state of iodine in room temperature? describe what it looks like.

I2 (s) - Shiny grey solid - (aq) and liquid = brown - Gas = purple

Why wont I2 and F- react?

I2 is the weakest oxidising agent (most difficult to reduce) and F- is the weakest reducing agent (most difficult to oxidise)

How do you name chlorates / sulfates?

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)

Describe the trend in melting and boiling points going down group 7.

Increases: - Halogens exist as non-polar diatomic molecules (simple covalent molecules) - Give rise to weak I.D / VDW's forces exist between the molecules (IMFs) - These are broken when melted or boiled - I.D forces become stronger down the group as Mr increases - This is because more e's are involved - So more energy required to overcome the VDW's forces

Q: A solution of magnesium sulfate is sometimes given as first aid to someone who has swallowed barium chloride. Explain why drinking magnesium sulfate solution is effective in the treatment of barium poisoning. (1 MARK)

Insoluble barium sulfate is formed

How does group 2 react with OH-?

Insoluble → soluble (solubility increases down the group) Eg, Be(OH)2 (s) (TIP: hydroxide= higher)

How does concentrated NH3 (aq) react with I-?

Insoluble; NVR (no visible reaction)

How does dilute NH3 (aq) react with Br-?

Insoluble; NVR (no visible reaction)

How does dilute NH3 (aq) react with I-?

Insoluble; NVR (no visible reaction)

SME: Explain why I, cannot convert Fe2+ to Fe3+ (1 MARK)

Iodine / I2 is not a strong enough oxidising agent;

Q: The following pairs of compounds can be distinguished by simple test-tube reactions. For each pair of compounds, give a reagent (or combination of reagents) that, when added separately to each compound, could be used to distinguish between them. State what is observed in each case. HCI(aq) and HNO3 (aq). (3 MARKS)

M1 AGNO, OR silver nitrate OR any soluble silver salt M2 white precipitate or white solid / white suspension M3 remains colourless OR no reaction OR no (observed) change OR no precipitate

M1 Any soluble chloride including hydrochloric acid (ignore concentration) M2 white precipitate or white solid / white suspension M3 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear OR as an alternative M1 Any soluble iodide including HI M2 yellow precipitate or yellow solid / yellow suspension M3 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear OR as an alternative M1 Any soluble bromide including HBr M2 cream precipitate or cream solid / cream suspension M3 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear OR as an alternative M1 NaOH or KOH or any soluble carbonate M2 brown precipitate or brown solid / brown suspension with NaOH / KOH (white precipitate / solid / suspension with carbonate) M3 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear

EQ: The following pairs of compounds can be distinguished by simple test-tube reactions. For each pair of compounds, give a reagent (or combination of reagents) that, when added separately to each compound, could be used to distinguish between them. State what is observed in each case. Aqueous magnesium chloride and aqueous barium chloride. (3 MARKS)

M1 Any soluble sulfate including (dilute or aqueous) sulfuric acid M2 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear M3 white precipitate or white solid / white suspension OR M1 NaOH or KOH M2 white precipitate or white solid / white suspension M3 remains colourless or no reaction or no (observed) change or no precipitate or clear solution or it remains clear

EQ: The concentration of ClO- ions in bleach solution can be found by reaction with iodide ions. The overall equation for this reaction is shown. ClO- + 2I- + 2H+ → I2 + Cl - + H2O A sample of bleach solution was found to contain ClO- ions with a concentration of 0.0109 mol dm-3 Potassium iodide is added to a 20.0 cm3 portion of this bleach solution. Calculate the mass, in mg, of potassium iodide needed to react with all of the ClO- ions in the sample of bleach. Give your answer to the appropriate number of significant figures. Give one observation during this reaction. (4 MARKS)

Mol ClO - = conc x vol = 0.0109 x 0.02 = 0.000218 mol Mol KI = 0.000218 x 2 = 0.000436 Mass KI = Mr x mol = 166 x 0.000436 = 0.072376 g 72.4 (mg) - black solid/ppt appears/forms (in a colourless solution) or (colourless solution) turns brown (solution)

SME: When aqueous chlorine is added to cold, aqueous sodium hydroxide to produce a key ingredient to produce bleach. Explain why it is important not to add any household cleaning agents that contain hydrochloric acid to a solution of bleach. (1 MARK)

NaCIO reacts with HCl to produce (toxic) chlorine gas;

Describe the reaction of sodium chloride and conc. sulfuric acid.

NaCl + H2SO4 → HCl + NaHSO4 - Acid base reaction (not redox) - HCl is steamy fumes - Acid is H2SO4 as donates H+ ions to Cl- ions - Base is Cl- ion as it is accepting the hydrogen ion - NO REDOX REACTION: Cl- too weak reducing agents to reduce S - No change in oxidation states

Describe the reaction of sodium fluoride and conc. sulfuric acid.

NaF + H2SO4 → HF + NaHSO4 - Acid / base reaction - HF is steamy fumes - Acid is H2SO4 as donates H+ ions to F- ions - Base is F- ion as it is accepting the hydrogen ion - NO REDOX REACTION: F- too weak reducing agents to reduce S - No change in oxidation states

State the overall equation of the reaction of sodium iodide with conc. sulfuric acid.

NaI + H2SO4 ---> Na2SO4 + I2 + SO2 + H2O + S + H2S (This isnt balanced yet, have a go at balancing it yourself!)

Describe and state the reaction of chlorine with NaOH. State the conditions neededd.

NaOH: cold, aqeous / dilute Cl2 + 2NaOH → NaCl + NaClO + H2O - Disproportionation: chlorine is both oxidised and reduced - NaClO (sodium chlorate (I)) is an active ingredient in bleach and kills bacteria / microorganisms

How does Ag+ (AgNO3 (aq)) react with Br-?

OFFICIAL TEST FOR X-(aq): - Cream precipitate - AgBr(s)

How does Ag+ (AgNO3 (aq)) react with Cl-?

OFFICIAL TEST FOR X-(aq): - White precipitate - AgCl(s)

How does Ag+ (AgNO3 (aq)) react with I-?

OFFICIAL TEST FOR X-(aq): - Yellow precipitate - AgI(s)

State the overall equation of the reaction of sodium bromide with conc. sulfuric acid.

Overall equation: combining the two equations: 2NaBr + 3H2SO4 → 2NaHSO4 + SO2 + Br2 + 2H2O OR 2NaBr + 2H2SO4 → Na2SO4 + SO2 + Br2 + 2H2O

What is a base?

Proton ACCEPTOR

What is an acid?

Proton DONOR

Why is silver nitrate solution used to identify halide ions?

Reacts to form different colored precipitates depending on the ion present

How does dilute NH3 (aq) react with Cl-?

Redissolve

How does concentrated NH3 (aq) react with Br-?

Redissolves

How does concentrated NH3 (aq) react with Cl-?

Redissolves

Describe displacement reactions in group 7.

Redox reactions between halogen (aq) and halide (aq): - E.g: Cl2 (aq) + KBr (aq) → 2KCl(aq) + Br2 (aq) - Chlorine is a better oxidising agent than bromine - Bromine is a better reducing agent than chlorine - Ionic equation: Cl2 (aq) + 2Br -(aq) → 2Cl-(aq) + Br2 (aq)

Describe the trend in reducing power down group 7 halides.

Reducing power increases down the group: - Iodide is the strongest reducing agent (most easily oxidised) - Lowest effective nuclear charge (high shielding) - So electron gained is very easily lost - Fluoride is the weakest reducing agent (most difficult to oxidised) - High effective nuclear charge - Low shielding

How does group 2 react with SO4 2-?

Soluble → insoluble (solubility decreases down the group) (TIP: sulfate = smaller)

Q: Strontium chloride is used in toothpaste for sensitive teeth. Both strontium carbonate and strontium sulfate are white solids that are insoluble in water. Write an equation for the reaction between strontium chloride solution and sodium sulfate solution. Include state symbols in your equation. (1 MARK)

SrCl2 (aq) + Na2SO4 (aq) --> SrSO4 (s) + 2NaCl (aq)

EQ: Separate unlabelled solid samples of three anhydrous sodium compounds are provided for a student to identify. These compounds are known to be sodium carbonate, sodium fluoride and sodium chloride but it is not known which sample is which. Outline a logical sequence of test-tube reactions that the student could carry out to identify each of these compounds. Include the observations the student would expect to make. Give equations, including state symbols, for any reactions that would take place. (6 MARKS)

Stage 1 Suggested tests: 1a Add named acid to all 3 (e.g. HCl) 1b Add water / make into a solution 1c Add AgNO3 (Ignore addition of NH3 / Ignore additional test for CO2 produced) Stage 2 Expected observations - conclusions 2a Na2CO3 will fizz with acid 2b NaCl gives white ppt with AgNO3 2c NaF shows no (visible) change / no ppt (Additional incorrect observations loses point) Stage 3 Equations - state symbols must match method 3a Na2CO3 + 2HNO3 → 2NaNO3 + CO2 + H2O ... or ionic 3b AgNO3 + NaCl → AgCl + NaNO3 ... or ionic 3c correct state symbols

EQ: This question is about the chemical properties of chlorine, sodium chloride and sodium bromide. A colourless solution contains a mixture of sodium chloride and sodium bromide. Using aqueous silver nitrate and any other reagents of your choice, develop a procedure to prepare a pure sample of silver bromide from this mixture. Explain each step in the procedure and illustrate your explanations with equations, where appropriate. (6 MARKS)

Stage 1: formation of precipitates - Add silver nitrate - To form precipitates of AgCl and AgBr - AgNO3 + NaCI → AgCl + NaNO3 - AgNO3 + NaBr → AgBr + NANO3 Stage 2: selective dissolving of AgCl - Add excess of dilute ammonia to the mixture of precipitates - The silver chloride precipitate dissolves - AgCl + 2NH3 → Ag(NH3)2^+ + Cl- Stage 3: separation and purification of AgBr - Filter off the remaining silver bromide precipitate - Wash to remove soluble compounds - Dry to remove water

How do you test for hydroxides?

Test for OH- ions: - pH indicator: blue (as alkaline) - E.g. Red litmus paper turns blue if present

SME: The uric acid in urine can react with one of the products in the reaction in part (a) producing a toxic compound known as cyanogen chloride. Explain why this should not be a concern for the swimming pool users. (1 MARK)

The compound is not produced in enough quantities;

TIP!

The production of NaHSO4 OR Na2SO4 is acceptable when it comes to MS's. I see it all the time. In the old spec, they used to prefer Na2SO4. Now, for your spec it's NaHSO4. However, that being said you can use either.

Q: State why the silver nitrate is acidified when testing for iodide ions. (1 MARK)

The silver nitrate is acidified to: - react with / remove (an)ions that would interfere with the test - prevent the formation of other silver precipitates / insoluble silver compounds that would interfere with the test - remove (other) ions that react with the silver nitrate - react with / remove carbonate / hydroxide / sulfite (ions)

Q: The presence of halide ions in solution can be detected by adding silver nitrate solution and dilute nitric acid. State the purpose of the nitric acid in this test. (1 MARK)

To ensure that other (an)ions do not interfere

What can displacement reactions of group 7 be used for?

To help identify which halogen (or halide) is present in a solution. E.g., look at the image attached. You NEED to learn these displacement reaction observations as they are often asked.

Why is fluorine not used in the school laboratory?

Too dangerous

How does Ag+ (AgNO3 (aq)) react with carbonates (CO3 2-)?

White precipitate: Ag2CO3 (s)

How does Ag+ (AgNO3 (aq)) react with sulfuric acid (SO4 2-)?

White precipitate: Ag2SO4(s)

List displacement reactions of group 7 alongside their observations.

You don't need to know about fluorine here (because it will oxidise the water instead of forming a solution).

EQ: The halogens can all behave as oxidising agents in reactions. Explain, in terms of electron transfer, the meaning of the term oxidising agent. (1 MARK)

electron acceptor / species that accepts electrons / species that gains electrons

3.2.3 GROUP 7(17), THE HALOGENS

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