iGCSE Chemistry 0620 Extended - 8.3 Group VII properties- Exam Style Questions Paper 4- New Syllabus
Question
This question is about the Periodic Table.
(a) State the name given to Group VII elements.
(b) State which Group VII element is most reactive.
(c) Give the physical state and colour of iodine at room temperature and pressure.
(d) When bromine is added to aqueous potassium iodide a displacement reaction occurs.
The equation for the reaction is shown.
$\mathrm{Br_2 + 2KI \rightarrow 2KBr + I_2}$
(i) Write an ionic equation for the reaction.
(ii) Iodine and bromine react at high temperatures to form iodine monobromide, IBr.
The equation is shown.
$\mathrm{I_2(g) + Br_2(g) \rightarrow 2IBr(g)}$
The structures of the molecules involved in the reaction are I–I, Br–Br and I–Br.
Calculate the enthalpy change, $\Delta H$, for the reaction using the bond energies in Table 6.1.
Use the following steps.
• Calculate the total amount of energy required to break the bonds in 1 mol of $\mathrm{I_2(g)}$ and 1 mol of $\mathrm{Br_2(g)}$.
• Calculate the total amount of energy released when the bonds in 2 mol of $\mathrm{IBr(g)}$ are formed.
• Calculate the enthalpy change, $\Delta H$, for the reaction. Your answer should include a sign.
(e) Sodium is in Group I of the Periodic Table. When sodium is added to water a chemical reaction occurs.
(i) Give two observations when sodium is added to water.
(ii) Thymolphthalein is added to the solution when the reaction has finished. State the final colour of the thymolphthalein in the solution.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 8.3 — Group VII properties (Parts (a), (b), (c), (d)(i))
• Topic 6.4 — Redox (Part (d)(i) — oxidation and reduction in displacement)
• Topic 5.1 — Exothermic and endothermic reactions / bond energy calculations (Part (d)(ii))
• Topic 8.2 — Group I properties (Parts (e)(i), (e)(ii))
▶️ Answer/Explanation
(a) The name given to Group VII elements is Halogens.
The term halogen means “salt-former” — these elements readily form salts when they react with metals.
(b) The most reactive Group VII element is fluorine ($\mathrm{F_2}$).
Reactivity in Group VII decreases down the group. Fluorine, at the top of the group, has the highest electronegativity and smallest atomic radius, making it the most powerful oxidising agent among the halogens.
(c) Iodine at room temperature and pressure:
State: solid
Colour: grey-black
Iodine is a diatomic non-metal ($\mathrm{I_2}$) that exists as a shiny grey-black solid at r.t.p. When heated, it sublimes to produce a purple vapour.
(d)(i) The ionic equation for the displacement reaction is:
$\mathrm{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$
Potassium ions ($\mathrm{K^+}$) are spectator ions — they appear unchanged on both sides of the equation and are therefore omitted from the ionic equation.
(d)(ii) Enthalpy change calculation using bond energies:
Step 1: Energy required to break bonds (endothermic, positive):
Breaking 1 mol of I–I bonds: $1 \times 150 = 150$ kJ
Breaking 1 mol of Br–Br bonds: $1 \times 193 = 193$ kJ
Total energy required = $150 + 193 = \mathbf{343}$ kJ
Step 2: Energy released when bonds are formed (exothermic, negative):
Forming 2 mol of I–Br bonds: $2 \times 175 = \mathbf{350}$ kJ
Step 3: Enthalpy change of the reaction:
$\Delta H = \text{Energy required to break bonds} – \text{Energy released when bonds are formed}$
$\Delta H = 343 – 350 = \mathbf{-7\ \mathrm{kJ/mol}}$
The negative sign indicates that the reaction is exothermic — more energy is released in forming the new I–Br bonds than is required to break the original I–I and Br–Br bonds.
(e)(i) Two observations when sodium is added to water (any two):
- The solid dissolves / disappears (sodium reacts and is consumed).
- Bubbling / effervescence / fizzing (hydrogen gas is produced).
- The sodium melts / forms a silvery ball (the reaction is exothermic and sodium has a low melting point).
- The sodium floats on the water surface (sodium is less dense than water).
- The sodium moves around on the water surface (propelled by the escaping hydrogen gas).
The reaction equation is: $2\mathrm{Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)}$.
(e)(ii) The final colour of thymolphthalein in the solution is blue.
Sodium reacts with water to form sodium hydroxide ($\mathrm{NaOH}$), a strong alkali. Thymolphthalein is an indicator that is colourless in acidic and neutral solutions but turns blue in alkaline solutions (pH range approximately 9.3–10.5).