Questions on Group 7: trends down the group and displacement reactions

Question 1

Explain how the reactions of chlorine water, bromine water and iodine solution with solutions of potassium chloride, bromide and iodide show the order of oxidising power of the halogens. Include observations and equations.

Accepted Answer

Indicative content

chlorine water turns potassium bromide orange and potassium iodide brown, so chlorine oxidises both bromide and iodide ions
bromine water turns potassium iodide brown, so bromine oxidises iodide, but it causes no colour change with potassium chloride
iodine solution reacts with neither chloride nor bromide, and the mixtures stay the colour of the iodine solution added
a halogen displaces only the halogens below it, so oxidising power decreases in the order Cl₂ > Br₂ > I₂
a correct ionic equation with state symbols, such as Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq), with no potassium ions shown
the trend arises because the electron gained enters a shell further from the nucleus and more shielded down the group, so it is less strongly attracted

Levels

Level 3 (5–6 marks) all the observations are given for the three halogens with the three halide solutions, including where nothing happens, the order of oxidising power is deduced from them, a correct ionic equation is written, and the trend is explained through the distance and shielding of the electron gained
Level 2 (3–4 marks) the chlorine and bromine results are given and the order deduced with an equation; the iodine results, the state symbols, or the reason the oxidising power falls down the group is absent
Level 1 (1–2 marks) chlorine is said to displace bromide or iodide with a colour change; the trend is usually explained by the halogen losing electrons more easily down the group

Question 2

State the physical state and colour at room temperature of chlorine, bromine and iodine, and the colour of iodine vapour.

Accepted Answer

B1 chlorine is a pale green gas; allow yellow-green B1 bromine is a red-brown (dark orange-brown) liquid B1 iodine is a grey-black solid; allow shiny black B1 iodine vapour is purple; allow violet

Question 3

Explain why the boiling points of the halogens increase from fluorine to iodine, and why the Br–Br bond is not broken when bromine boils.

Accepted Answer

B1 the molecules get larger, with more electrons, down the group B1 so the van der Waals forces between molecules get stronger; accept induced dipole–dipole, London or dispersion forces B1 more energy is needed to overcome these forces between molecules B1 boiling separates whole Br₂ molecules from one another, and the covalent bond within each molecule is far stronger than the forces between them

Question 4

Explain the trends in atomic radius and in electronegativity from fluorine to iodine.

Accepted Answer

B1 atomic radius increases down the group because each element has one more occupied shell than the one above B1 so the bonding pair of electrons is further from the nucleus B1 there are more inner shells, so the bonding pair is more shielded from the nucleus B1 the attraction between the nucleus and the bonding pair is weaker, so electronegativity decreases; reject an answer about attracting a single electron to form an ion

Question 5

Give the ionic equation, with state symbols, for the reaction of chlorine water with potassium bromide solution, the two half-equations, and the species that is oxidised.

Accepted Answer

B1 Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq); reject equations that show potassium ions B1 reduction: Cl₂(aq) + 2e⁻ → 2Cl⁻(aq) B1 oxidation: 2Br⁻(aq) → Br₂(aq) + 2e⁻; allow electrons subtracted on the left B1 bromide ions are oxidised, from −1 to 0, and chlorine is the oxidising agent

Question 6

Suggest the physical state and colour of astatine at room temperature, and how its electronegativity compares with iodine's. Explain each suggestion.

Accepted Answer

B1 astatine is a solid, because At₂ molecules have more electrons than I₂, so the van der Waals forces are stronger still B1 it is black or very dark, continuing the darkening of colour down the group B1 its electronegativity is lower than iodine's B1 the bonding pair is further from the nucleus and more shielded in the larger astatine atom, so it is attracted less strongly

Question 7

Explain why chlorine is a stronger oxidising agent than iodine.

Accepted Answer

B1 an oxidising agent gains electrons, and the electron gained by a halogen enters its outer shell B1 in chlorine that shell is closer to the nucleus and less shielded than in iodine B1 so an incoming electron is more strongly attracted, and chlorine gains electrons more readily

Question 8

Describe what is seen when bromine water is added to potassium iodide solution and the mixture is then shaken with cyclohexane, and explain why the organic layer is coloured.

Accepted Answer

B1 the solution turns brown as iodine forms B1 after shaking with cyclohexane the upper organic layer is purple B1 iodine is non-polar and dissolves far better in non-polar cyclohexane than in water, so it moves into the organic layer

Question 9

Suggest why fluorine is never used in school displacement tests, and what happens when fluorine is bubbled into water. Give an equation for the reaction with water.

Accepted Answer

B1 fluorine is the strongest oxidising agent of the halogens, too reactive and toxic to store or handle in a school B1 it oxidises water itself, releasing oxygen, so a solution of fluorine in water cannot be made B1 2F₂(g) + 2H₂O(l) → 4HF(aq) + O₂(g)

Question 10

Compare the colours of bromine and iodine in water and in cyclohexane, and use the comparison to explain why cyclohexane is added in displacement tests.

Accepted Answer

B1 in water bromine is orange (yellow when dilute) and iodine is brown (yellow when dilute), so they look similar B1 in cyclohexane bromine is orange and iodine is purple, which are clearly different B1 so shaking with cyclohexane shows which halogen has been formed when the colour in water is ambiguous

Question 11

Calculate the mass of iodine formed when excess chlorine water is added to 25.0 cm³ of 0.100 mol dm⁻³ potassium iodide solution. Ar of I = 126.9.

Accepted Answer

M1 amount of I⁻ = 25.0 ÷ 1000 × 0.100 = 2.50 × 10⁻³ mol M1 Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq), so amount of I₂ = 2.50 × 10⁻³ ÷ 2 = 1.25 × 10⁻³ mol A1 mass = 1.25 × 10⁻³ × 253.8 = 0.317 g, to three significant figures; error carried forward from M2

Question 12

Calculate the volume of chlorine gas, in cm³ at room temperature and pressure, that reacts with 50.0 cm³ of 0.200 mol dm⁻³ sodium bromide solution. Molar gas volume = 24.0 dm³ mol⁻¹.

Accepted Answer

M1 amount of Br⁻ = 50.0 ÷ 1000 × 0.200 = 1.00 × 10⁻² mol M1 Cl₂ reacts with Br⁻ in a 1 : 2 ratio, so amount of Cl₂ = 5.00 × 10⁻³ mol A1 volume = 5.00 × 10⁻³ × 24.0 = 0.120 dm³ = 120 cm³, to three significant figures

Question 13

Evaluate a student's conclusion that bromine water reacted with potassium chloride solution, based on the mixture being orange.

Accepted Answer

B1 the orange colour is that of the bromine water added, diluted, not evidence of a product B1 bromine is a weaker oxidising agent than chlorine, so it cannot oxidise chloride ions and no reaction is expected B1 the conclusion is not valid; a control of bromine water with pure water would show the same colour

Question 14

Explain why the F–F bond (158 kJ mol⁻¹) is weaker than the Cl–Cl bond (243 kJ mol⁻¹), although the Cl–Cl, Br–Br and I–I bonds weaken down the group.

Accepted Answer

B1 from chlorine to iodine the atoms get larger, so the bonding pair is further from the nuclei and the bond weakens B1 fluorine atoms are very small, so the F–F bond is very short and the lone pairs on the two atoms are close together B1 the lone pairs repel each other strongly, which offsets much of the attraction of the bonding pair, so the bond is weak

Question 15

Calculate the enthalpy change for H₂(g) + F₂(g) → 2HF(g) from the bond enthalpies H–H 436, F–F 158 and H–F 562 kJ mol⁻¹.

Accepted Answer

M1 bonds broken: H–H + F–F = 436 + 158 = 594 kJ mol⁻¹ M1 bonds made: 2 × H–F = 2 × 562 = 1124 kJ mol⁻¹ A1 ΔH = 594 − 1124 = −530 kJ mol⁻¹, to three significant figures; the minus sign is needed for the mark

Question 16

Suggest why a red-hot wire placed in hydrogen iodide produces purple fumes, while one placed in hydrogen chloride produces no change. Give the equation for the change that is seen.

Accepted Answer

B1 the H–I bond (298 kJ mol⁻¹) is much weaker than the H–Cl bond (432 kJ mol⁻¹), because the larger iodine atom holds the shared pair further from its nucleus B1 the hot wire supplies enough energy to break H–I bonds, so hydrogen iodide decomposes to hydrogen and purple iodine vapour, but not enough to break H–Cl bonds B1 2HI(g) ⇌ H₂(g) + I₂(g); allow a single arrow

Question 17

Justify the conclusion that a colourless solution contains iodide ions and not bromide ions, from the observation that it turns brown when chlorine water is added and that the colour moves into a purple layer when cyclohexane is added and the tube is shaken.

Accepted Answer

B1 chlorine is a stronger oxidising agent than either bromine or iodine, so it displaces either halide; the colour of the halogen formed decides which halide was present B1 iodine is brown in water and purple in cyclohexane, whereas bromine would be orange in water and orange-red in cyclohexane, so the two colours identify iodine B1 Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq), the iodide ions being oxidised; the cyclohexane layer is the deciding observation because dilute iodine and bromine in water can look alike

Question 18

Suggest which atom carries the partial negative charge in the polar molecule iodine monochloride, ICl, and explain why.

Accepted Answer

B1 the chlorine atom carries the partial negative charge, δ− B1 chlorine is more electronegative than iodine, being a smaller atom, so it attracts the bonding pair more strongly