Edexcel International A Level Chemistry: Inorganic chemistry of Group 7 (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers Topic 8C, Inorganic chemistry of Group 7, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 8.24 to 8.28, taught at IAS level within Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Study the content first in the Group 7 study guide and the Group 7 revision notes, and record progress on the printable checklist. Other topics are on the course hub, and a free 10-minute diagnostic shows your gaps.

Use these relative atomic masses: Ag 107.9, Ca 40.1, Cl 35.5, H 1.0, I 126.9, K 39.1, Mg 24.3, Na 23.0, O 16.0. Take the gas molar volume at room temperature and pressure as 24.0 dm³ mol⁻¹.

Questions

1. Answer these about how the halogens look.

(a) State the colour and physical state of iodine at room temperature. [1] (b) State the colour of iodine dissolved in hexane. [1] (c) State the colour of a dilute aqueous solution of bromine. [1]

2. Use these values for the halogens.

	Cl₂	Br₂	I₂
Melting temperature / °C	−101	−7	114
Boiling temperature / °C	−34	59	184

(a) Use the data to deduce the physical state of bromine at 25 °C. Give a reason. [1] (b) Explain the trend in boiling temperature from chlorine to iodine. [3]

3. A few drops of bromine water are added to two tubes: one holds sodium chloride solution, the other sodium iodide solution. Both tubes are then shaken with hexane.

(a) State the colour of the hexane layer in each tube. [2] (b) Write the ionic equation for the reaction that takes place. [1] (c) Identify the oxidising agent in this reaction and explain your answer in terms of electrons. [2]

4. Magnesium ribbon burns in iodine vapour to form magnesium iodide.

(a) Write the equation and give the oxidation numbers of magnesium and iodine before and after. [2] (b) Calculate the mass of iodine that reacts with 0.729 g of magnesium, and the mass of magnesium iodide formed. [2]

5. Bleach is made by bubbling chlorine into 120 cm³ of cold, dilute 0.850 mol dm⁻³ sodium hydroxide until all the alkali has reacted.

(a) Give the equation for this reaction. [1] (b) Use oxidation numbers to explain why this is a disproportionation reaction. [2] (c) Calculate the mass of chlorine that reacts. [2] (d) The solution is made up to 150 cm³ with water. Calculate the concentration of sodium chlorate(I). [1]

6. A colourless solution contains calcium ions and one type of halide ion. Adding dilute nitric acid, then silver nitrate, produces a white precipitate that dissolves in dilute aqueous ammonia.

(a) Identify the halide ion. [1] (b) Explain why dilute nitric acid is added before the silver nitrate. [1] (c) Write the ionic equation, with state symbols, for the precipitation. [1] (d) Exactly 18.0 cm³ of 0.100 mol dm⁻³ silver nitrate reacts with all the halide ions in 15.0 cm³ of the solution. Calculate the concentration of the calcium halide. [2]

7. Concentrated sulfuric acid is added drop by drop to solid potassium iodide in a fume cupboard.

(a) Write the equation for the first step, which forms hydrogen iodide. [1] (b) Describe three observations. [3] (c) One reaction forms hydrogen sulfide. Write its equation and state the change in oxidation number of sulfur. [2] (d) Explain why solid potassium chloride gives no redox reaction with concentrated sulfuric acid. [2] (e) Assume all the iodide in 3.32 g of potassium iodide ends up as iodine by the hydrogen sulfide reaction. Calculate the mass of iodine formed. [2]

8. Chlorine is passed into hot potassium hydroxide solution.

(a) Write a balanced equation for this reaction. [1] (b) Calculate the mass of potassium chlorate(V) formed from 4.26 g of chlorine. [3] (c) Calculate the percentage of the chlorine atoms that end up in potassium chlorate(V). [1] (d) Calculate the minimum volume of 2.00 mol dm⁻³ potassium hydroxide needed. [1]

9. 2.34 g of sodium chloride is warmed with excess concentrated sulfuric acid.

(a) Give a balanced equation, with the sodium-containing product. [1] (b) Calculate the volume of hydrogen chloride formed, in cm³, at room temperature and pressure. [2] (c) All the gas is absorbed in water, then made up to 250 cm³ of solution. Calculate the concentration of the acid. [1] (d) State what you would see if the gas met ammonia gas, and name the product. [1]

10. This question asks you to make predictions about fluorine and astatine.

(a) Predict the physical state of astatine at room temperature. [1] (b) Predict whether chlorine water reacts with sodium astatide solution. Write an ionic equation if it does. [2] (c) Predict the appearance of silver astatide and its behaviour with concentrated aqueous ammonia. [2] (d) Explain why fluorine is expected to be the strongest oxidising agent in the group. [2] (e) Suggest why fluorine does not form a chlorate(V)-type ion when it reacts with hot alkali. [1]

11. Chlorine is added to the water supply of a town.

(a) Write the equation for chlorine reacting with water. [1] (b) Give the oxidation number of chlorine in each chlorine-containing product. [1] (c) Evaluate the use of chlorine in water treatment. [4]

Answers

1. (a) Grey-black (shiny) solid [1] (b) Violet (purple) [1] (c) Yellow (to orange) [1] [3] Examiner insight: Colour answers must match the conditions asked for; “brown” for iodine in hexane describes the aqueous colour, so it is unlikely to gain credit.

2. (a) Liquid: 25 °C is above its melting temperature (−7 °C) but below its boiling temperature (59 °C) [1] (b) All are simple molecular with London forces between molecules [1]. Molecules have more electrons down the group, so London forces are stronger [1]. Separating the molecules therefore takes more energy, and the boiling temperature rises [1] [4] Examiner insight: Mentioning broken Cl–Cl or I–I bonds during boiling contradicts the answer and usually costs credit.

3. (a) Sodium chloride tube: orange (no reaction) [1]; sodium iodide tube: violet [1] (b) Br₂ + 2I⁻ → 2Br⁻ + I₂ [1] (c) The oxidising agent is bromine [1]; it accepts electrons from iodide ions, becoming Br⁻ [1] [5] Examiner insight: For the chloride tube, say the orange colour is the bromine itself; “no change” alone is too vague.

4. (a) Mg + I₂ → MgI₂ [1]; Mg 0 → +2 and I 0 → −1 [1] (b) n(Mg) = 0.729 ÷ 24.3 = 0.0300 mol; mass of I₂ = 0.0300 × 253.8 = 7.61 g [1]; mass of MgI₂ = 0.0300 × 278.1 = 8.34 g [1] [4] Examiner insight: Use Mr(I₂) = 253.8, not Ar(I); the factor-of-two slip is the usual source of a wrong mass here.

5. (a) Cl₂ + 2NaOH → NaCl + NaClO + H₂O [1] (b) In NaCl chlorine is −1, reduced from 0; in NaClO it is +1, oxidised from 0 [1]; the same element in one species is oxidised and reduced at once [1] (c) n(NaOH) = 0.120 × 0.850 = 0.102 mol, so n(Cl₂) = 0.0510 mol [1]; mass = 0.0510 × 71.0 = 3.62 g [1] (d) n(NaClO) = 0.0510 mol; concentration = 0.0510 ÷ 0.150 = 0.340 mol dm⁻³ [1] [6] Examiner insight: Both oxidation number changes must be stated for disproportionation; giving only the reduction to −1 does not show the definition.

6. (a) Chloride, Cl⁻ [1] (b) To remove carbonate (or other) ions that would also form a precipitate with silver ions [1] (c) Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1] (d) n(Cl⁻) = n(Ag⁺) = 0.0180 × 0.100 = 1.80 × 10⁻³ mol, so n(CaCl₂) = 9.00 × 10⁻⁴ mol [1]; concentration = 9.00 × 10⁻⁴ ÷ 0.0150 = 0.0600 mol dm⁻³ [1] [5] Examiner insight: Calcium chloride provides two chloride ions per formula unit; missing the factor of 2 doubles the answer.

7. (a) KI + H₂SO₄ → KHSO₄ + HI [1] (b) Any three from: steamy fumes [1]; purple vapour or black solid (iodine) [1]; yellow solid (sulfur) or smell of bad eggs (hydrogen sulfide) [1] (c) 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O [1]; sulfur goes from +6 to −2, a change of 8 [1] (d) HCl is the weakest reducing agent of the hydrogen halides [1]; the small chloride ion holds its outer electrons tightly and does not lose them to sulfuric acid [1] (e) n(KI) = 3.32 ÷ 166.0 = 0.0200 mol, so n(I₂) = 0.0100 mol [1]; mass = 0.0100 × 253.8 = 2.54 g [1] [10] Examiner insight: Observations must say what is seen or smelled; “hydrogen sulfide” without “bad-egg smell” may not count.

8. (a) 3Cl₂ + 6KOH → 5KCl + KClO₃ + 3H₂O [1] (b) n(Cl₂) = 4.26 ÷ 71.0 = 0.0600 mol [1]; n(KClO₃) = 0.0600 ÷ 3 = 0.0200 mol [1]; mass = 0.0200 × 122.6 = 2.45 g [1] (c) One of the six chlorine atoms: 16.7% [1] (d) n(KOH) = 2 × 0.0600 = 0.120 mol; volume = 0.120 ÷ 2.00 = 0.0600 dm³ = 60.0 cm³ [1] [6] Examiner insight: Using the cold-alkali equation spoils every later step; check the 3 : 1 ratio of Cl₂ to KClO₃ first.

9. (a) NaCl + H₂SO₄ → NaHSO₄ + HCl [1] (b) n(NaCl) = 2.34 ÷ 58.5 = 0.0400 mol = n(HCl) [1]; volume = 0.0400 × 24 000 = 960 cm³ [1] (c) 0.0400 ÷ 0.250 = 0.160 mol dm⁻³ [1] (d) White smoke of ammonium chloride, NH₄Cl [1] [5] Examiner insight: The question asks for cm³, so multiply by 24 000; an answer of 0.960 with no unit does not answer the question as set.

10. (a) Solid [1] (b) Yes, chlorine is the stronger oxidising agent [1]; Cl₂ + 2At⁻ → 2Cl⁻ + At₂ [1] (c) Dark (darker than yellow) precipitate [1]; insoluble in concentrated ammonia [1] (d) Fluorine atoms are the smallest, with the least shielding [1]; so an incoming electron is attracted most strongly and fluorine gains electrons most readily [1] (e) Fluorine is the most electronegative element, so it cannot take a positive oxidation number such as +5 [1] [8] Examiner insight: Tie each prediction to a trend; a reason that contradicts the prediction, such as weaker London forces, can lose the mark.

11. (a) Cl₂ + H₂O ⇌ HCl + HClO [1] (b) −1 in HCl and +1 in HClO [1] (c) Chloric(I) acid kills bacteria [1]; this cuts waterborne disease such as cholera [1]; chlorine is toxic, so the dose must be controlled [1]; it can form possibly harmful chlorinated organic compounds with organic matter, a risk to weigh against the benefit [1] [6] Examiner insight: A judgement backed by points for and against is expected; a list of benefits alone limits the credit available.

Where marks are usually lost

• Writing NaClO₃ for cold alkali.
• Stating that bromine displaces chloride ions.
• Calling iodine in hexane brown.
• Forgetting the acid–base first step with concentrated sulfuric acid.
• Using Ar(I) where Mr(I₂) is needed.
• Missing the factor of 2 for halide ions in calcium or magnesium halides.
• Explaining boiling temperatures with covalent bond strength.
• Giving only the reduction when explaining disproportionation.

Next steps

• Revise with the Group 7 revision notes.
• Re-read the Group 7 study guide.
• Browse the Edexcel IAL Chemistry course hub.
• Book a free trial class.
• Record progress on the printable checklist.
• Take a free 10-minute diagnostic.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Scope: Topic 8C, outcomes 8.24 to 8.28.