The problems are on the group 2 and halogens worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Group 2 metals are reactive metals that lose two outer electrons to form 2+ ions, and their reactivity rises down the group. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on group 2 and halogens, from group 2 basics to explain the halogen trends: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the formula, the substitution and the result with its unit. Check your method first, then your number, and make sure every equation balances for atoms and charge.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.2.2-3.2.3
Problem 1 Group 2 basics
1. Two outer electrons (ns²).2. Sr²⁺.
3. Barium is more reactive: its larger atom loses electrons more easily.
Problem 2 Halogen states
1. Chlorine is a pale green gas; iodine is a grey-black solid.2. I₂ has many more electrons than Cl₂, so its London forces between molecules are stronger and more energy is needed to separate them.
Problem 3 Balance the equations
1. Sr + 2H₂O → Sr(OH)₂ + H₂2. Mg + 2HCl → MgCl₂ + H₂
3. Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
Problem 4 Identify the halide
1. A: chloride (white), B: iodide (yellow), C: bromide (cream).2. Dilute ammonia does not dissolve AgI or AgBr; concentrated ammonia dissolves AgBr (C) but not AgI (B).
3. It removes carbonate and hydroxide ions that would give a false precipitate.
Problem 5 Hydrogen from magnesium
1. Mg + 2HCl → MgCl₂ + H₂.2. n = 0.486 ÷ 24.3 = 0.0200 mol.
3. n(H₂) = 0.0200 mol, V = 0.0200 × 24.0 = 0.480 dm³ = 480 cm³.
Problem 6 Oxidised or reduced
1. Chlorine goes from 0 to −1.2. Bromide ions (−1 → 0) are oxidised; chlorine is the oxidising agent.
3. The colourless solution turns orange as bromine forms.
Problem 7 Alkalinity of limewater
1. Each Ca(OH)₂ gives 2 OH⁻, so [OH⁻] = 2 × 0.020 = 0.040 mol dm⁻³.2. pOH = −log 0.040 = 1.40, pH = 14.00 − 1.40 = 12.6.
3. Higher: barium hydroxide is more soluble, so [OH⁻] is larger.
Problem 8 Liming acid soil
1. Ca(OH)₂ + 2H⁺ → Ca²⁺ + 2H₂O.2. n(Ca(OH)₂) = 2.40 ÷ 2 = 1.20 mol; mass = 1.20 × 74.1 = 88.9 g.
Problem 9 Barium hydroxide titration
1. Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O.2. n(HCl) = 0.02140 × 0.100 = 2.140 × 10⁻³ mol; n(Ba(OH)₂) = 1.070 × 10⁻³ mol; c = 1.070 × 10⁻³ ÷ 0.0250 = 0.0428 mol dm⁻³.
Problem 10 Which halogens react?
1. No reaction (bromine is less reactive than chlorine).2. Reacts: Cl₂ + 2I⁻ → 2Cl⁻ + I₂ (brown).
3. No reaction.
4. Reacts: Br₂ + 2I⁻ → 2Br⁻ + I₂ (brown).
Problem 11 Cold alkali and chlorine
1. Cl₂ + 2NaOH → NaCl + NaClO + H₂O.2. 0, −1 and +1.
3. The same element, chlorine, is reduced (0 → −1) and oxidised (0 → +1) in one reaction.
Problem 12 Chloride in a water sample
1. n = 0.287 ÷ 143.4 = 2.00 × 10⁻³ mol.2. n(Cl⁻) = 2.00 × 10⁻³ mol; c = 2.00 × 10⁻³ ÷ 0.0200 = 0.100 mol dm⁻³.
Problem 13 Boiling points of halogens
1. 184 − (−34) = 218 °C.2. Molecules get larger with more electrons, so London forces between them strengthen and more energy is needed to separate them.
3. Astatine has even more electrons than iodine, so its London forces are stronger and it will be a solid at room temperature.
Problem 14 Hydrogen halides and sulfuric acid
1. 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O.2. Sulfur goes from +6 to −2, gaining 8 electrons.
3. Iodide ions are larger and give up electrons more easily, so they are stronger reducing agents. Chloride only produces HCl and sulfur does not change oxidation state.
Problem 15 Spot the errors
1. Magnesium reacts very slowly with cold water; it reacts quickly with steam to give MgO and H₂.2. Barium’s first ionisation energy is lower, which makes it more reactive.
3. AgI is insoluble even in concentrated ammonia; only AgCl and AgBr dissolve.
Problem 16 Barium meal
1. BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O; soluble BaCl₂ releases toxic Ba²⁺.2. BaSO₄ is insoluble and does not react with acid, so almost no Ba²⁺ is released.
3. n(BaCl₂) = 0.0250 × 0.200 = 5.00 × 10⁻³ mol, so n(BaSO₄) = 5.00 × 10⁻³ mol; mass = 5.00 × 10⁻³ × 233.4 = 1.17 g.
Problem 17 Identify the chloride
1. Y is barium chloride.2. Ba²⁺ + SO₄²⁻ → BaSO₄(s); Ag⁺ + Cl⁻ → AgCl(s).
3. Calcium sulfate is slightly soluble, so only a faint, slow precipitate would appear; magnesium sulfate is soluble.
Problem 18 Which group 2 metal
1. n(H₂) = 0.274 ÷ 24.0 = 0.01142 mol.2. X + 2H₂O → X(OH)₂ + H₂, so n(X) = 0.01142 mol; Ar = 1.00 ÷ 0.01142 = 87.6, so X is strontium.
3. n(OH⁻) = 2 × 0.01142 = 0.02283 mol = n(H⁺); V = 0.02283 ÷ 0.100 = 0.2283 dm³ = 228 cm³.
Problem 19 Bleach analysis
1. n = 0.02450 × 0.0500 = 1.225 × 10⁻³ mol.2. n(I₂) = 6.125 × 10⁻⁴ mol and n(ClO⁻) = 6.125 × 10⁻⁴ mol.
3. Diluted conc = 6.125 × 10⁻⁴ ÷ 0.0100 = 0.06125 mol dm⁻³; original = 10 × this = 0.613 mol dm⁻³; mass conc = 0.6125 × 74.5 = 45.6 g dm⁻³.
Problem 20 Explain the halogen trends
1. Chlorine atoms are smaller with less shielding, so the nucleus attracts an incoming electron more strongly, making Cl₂ more strongly oxidising.2. Iodide has its outer electron further from the nucleus and more shielded, so it is lost more easily: I⁻ is a stronger reducing agent.
3. Bond length increases and bond enthalpy falls from H–Cl to H–I, so stability is HCl > HBr > HI.
