May/June 2025 Paper 43 Worked Answers (IGCSE Chemistry 0620 Extended)

49 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksCompound with a giant covalent structure
   Step 1: A giant covalent structure has many atoms joined by a network of covalent bonds, and a compound is made of two or more different elements. Step 2: Silicon(IV) oxide is a network of silicon and oxygen atoms joined by covalent bonds, so it is a giant covalent compound. Step 3: Graphite is giant covalent but an element, calcium oxide is ionic and propene is a simple molecule, so the answer is silicon(IV) oxide.
   Method:

   Pick the substance that is both a giant covalent network and a compound.
   Examiner tips

   • Giant covalent + more than one element = giant covalent compound (e.g. silicon(IV) oxide).

2. Question 1(b)

   1 marksIdentifying an unsaturated hydrocarbon
   Step 1: A hydrocarbon contains only carbon and hydrogen, and unsaturated means it contains a carbon-carbon double bond. Step 2: Propene is an alkene with a C=C double bond, so it is an unsaturated hydrocarbon. Step 3: Propane is saturated, ethanol contains oxygen (not a hydrocarbon) and graphite is an element, so the answer is propene.
   Method:

   Find the substance with only C and H that contains a C=C double bond.
   Examiner tips

   • Hydrocarbon = C and H only; unsaturated = has a C=C double bond.

3. Question 1(c)

   1 marksIdentifying an amphoteric oxide
   Step 1: An amphoteric oxide reacts with both acids and alkalis to form a salt and water. Step 2: Aluminium oxide reacts with both acids and bases, so it is amphoteric. Step 3: Calcium oxide is basic, silicon(IV) oxide is acidic and oxygen is an element, so the amphoteric oxide is aluminium oxide.
   Method:

   Recall which oxide reacts with both acids and bases.
   Examiner tips

   • Aluminium oxide and zinc oxide are the common amphoteric oxides.

4. Question 1(d)

   1 marksConductor of electricity when solid
   Step 1: Conducting electricity when solid needs charged particles that are free to move, such as delocalised electrons. Step 2: Graphite has one delocalised electron per carbon atom that can move between its layers, so solid graphite conducts electricity. Step 3: Silicon(IV) oxide has no free electrons, and ionic calcium oxide only conducts when molten or dissolved, so the answer is graphite.
   Method:

   Find the solid with free-moving charged particles (delocalised electrons).
   Examiner tips

   • Graphite is the non-metal that conducts because of delocalised electrons.

5. Question 1(e)

   1 marksSimple molecule containing 9 atoms
   Step 1: Count the atoms in each molecule from its formula. Step 2: Propene is $\text{C}_3\text{H}_6$, which has 3 carbon atoms plus 6 hydrogen atoms = 9 atoms (ethanol, $\text{C}_2\text{H}_6\text{O}$, also has 9 atoms). Step 3: Chlorine, nitrogen and oxygen are diatomic (2 atoms each), so the molecule with 9 atoms is propene.
   Method:

   Work out the number of atoms in each molecular formula and pick the one with 9.
   Examiner tips

   • Count every atom in the molecular formula: $\text{C}_3\text{H}_6$ = 9 atoms.

6. Question 1(f)

   2 marksSubstances that react to form slag
   Step 1: In the blast furnace, limestone decomposes to calcium oxide, a basic oxide. Step 2: Calcium oxide reacts with the acidic sandy impurity silicon(IV) oxide to form molten slag (calcium silicate), which is run off. Step 3: So the two substances that react to form slag are silicon(IV) oxide and calcium oxide.
   Method:

   Match the basic oxide with the acidic sandy impurity that react to form slag.
   Examiner tips

   • Slag = calcium oxide (basic) + silicon(IV) oxide (acidic) → calcium silicate.

7. Question 1(g)

   1 marksHomologous series with general formula CnH2n+2
   Step 1: The general formula $\text{C}_n\text{H}_{2n+2}$ is the formula of the alkanes. Step 2: Propane, $\text{C}_3\text{H}_8$, is an alkane, so it fits $\text{C}_n\text{H}_{2n+2}$. Step 3: Propene is an alkene ($\text{C}_n\text{H}_{2n}$), ethanol is an alcohol and graphite is an element, so the answer is propane.
   Method:

   Match the general formula to the correct homologous series and pick its member.
   Examiner tips

   • Alkanes: $\text{C}_n\text{H}_{2n+2}$; alkenes: $\text{C}_n\text{H}_{2n}$.

8. Question 1(h)

   1 marksManufacture by catalytic addition of steam to ethene
   Step 1: Ethene reacts with steam in the presence of a catalyst (an addition reaction) to form ethanol. Step 2: This catalytic hydration of ethene is one of the two industrial methods of making ethanol. Step 3: Propane and propene come from cracking and calcium oxide from heating limestone, so the answer is ethanol.
   Method:

   Recall the industrial reaction of ethene with steam and name the product.
   Examiner tips

   • Catalytic addition of steam to ethene is the industrial route to ethanol.

9. Question 1(i)

   1 marksMain gas in clean dry air
   Step 1: Clean dry air is about 78% nitrogen and 21% oxygen, with small amounts of other gases. Step 2: The gas at roughly 78% is therefore nitrogen. Step 3: Oxygen is about 21%, and chlorine and propane are not normal components of clean dry air, so the answer is nitrogen.
   Method:

   Recall the composition of clean dry air and pick the gas at about 78%.
   Examiner tips

   • Remember the 78% nitrogen / 21% oxygen split of clean dry air.

10. Question 2(a)

    1 marksParticles in the nucleus of an atom
    Step 1: The nucleus is the dense central part of the atom. Step 2: Protons and neutrons are found together in the nucleus, while electrons orbit it in shells. Step 3: So the particles in the nucleus are protons and neutrons.
    Method:

    Recall which sub-atomic particles sit in the nucleus.
    Examiner tips

    • Nucleus = protons + neutrons; electrons are in the surrounding shells.

11. Question 2(b)(i)

    3 marksSub-atomic particles in isotopes of sulfur
    Step 1: Protons = proton number = 16 for both species; neutrons = mass number − protons. Step 2: For $^{34}\text{S}$: neutrons = 34 − 16 = 18, and as a neutral atom electrons = protons = 16. Step 3: For $^{32}\text{S}^{2-}$: neutrons = 32 − 16 = 16, and the 2- charge means 2 extra electrons, so electrons = 16 + 2 = 18.
    Method:

    Use the proton number and mass number, then adjust the electrons for the ionic charge.
    Examiner tips

    • Protons = proton number; neutrons = mass − protons; add electrons for a negative ion.

12. Question 2(b)(ii)

    2 marksRelative atomic mass from isotopic abundances
    Step 1: Relative atomic mass = [(mass × % abundance) summed] ÷ 100. Step 2: $(32 \times 95) + (34 \times 5) = 3040 + 170 = 3210$. Step 3: $3210 \div 100 = 32.1$, so the relative atomic mass is 32.1.
    Method:

    Take the abundance-weighted average of the isotope masses.
    Examiner tips

    • Weighted average: $\frac{(32 \times 95)+(34 \times 5)}{100}$.

13. Question 2(b)(iii)

    1 marksStandard isotope for relative atomic mass
    Step 1: Relative atomic mass is defined on a scale where one atom of a chosen isotope is the standard. Step 2: The agreed standard is the carbon-12 isotope, $^{12}\text{C}$. Step 3: Hydrogen-1 and oxygen-16 were older standards, and carbon-14 is a different isotope, so the standard is carbon-12.
    Method:

    Recall the agreed reference isotope for the relative atomic mass scale.
    Examiner tips

    • The modern relative-mass scale is based on $^{12}\text{C}$.

14. Question 2(c)

    2 marksSpecies with the same electronic configuration as an oxide ion
    Step 1: An oxygen atom is 2,6; gaining 2 electrons to form $\text{O}^{2-}$ gives 2,8 (10 electrons). Step 2: The atom with 10 electrons (2,8) is neon. A positive ion with 2,8 is $\text{Na}^+$ (sodium loses one electron), or $\text{Mg}^{2+}$ or $\text{Al}^{3+}$. Step 3: So the atom is neon and a suitable positive ion is $\text{Na}^+$.
    Method:

    Count the electrons in $\text{O}^{2-}$, then find an atom and a positive ion with the same number.
    Examiner tips

    • $\text{O}^{2-}$ is 2,8 (10 electrons), the same as Ne, $\text{Na}^+$, $\text{Mg}^{2+}$ and $\text{Al}^{3+}$.

15. Question 3(a)(i)

    1 marksType of bonding in zinc
    Step 1: Zinc is a metal, and metals are held together by metallic bonding. Step 2: Metallic bonding is the electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons. Step 3: So the bonding in zinc is metallic bonding.
    Method:

    Identify the bonding type found in a pure metal.
    Examiner tips

    • Pure metals are held together by metallic bonding.

16. Question 3(a)(ii)

    2 marksParticles held together in metallic bonding
    Step 1: In a metal, the atoms lose their outer electrons, which become delocalised. Step 2: This leaves a lattice of positive metal ions surrounded by a sea of delocalised electrons. Step 3: Metallic bonding is the attraction between these positive ions and delocalised electrons, so the two particles are positive ions and (delocalised) electrons.
    Method:

    Recall the two particles described by the metallic-bonding model.
    Examiner tips

    • Picture positive ions fixed in place with delocalised electrons moving between them.

17. Question 3(a)(iii)

    1 marksParticle responsible for electrical conduction in a metal
    Step 1: Electrical conduction needs charged particles that are free to move. Step 2: In a metal the delocalised electrons can move through the lattice when a voltage is applied. Step 3: The positive ions stay fixed in the lattice, so it is the movement of delocalised electrons that allows zinc to conduct electricity.
    Method:

    Identify the mobile charged particle in a metal.
    Examiner tips

    • Free-moving delocalised electrons carry the current in a metal.

18. Question 3(b)(i)

    1 marksMeaning of the term alloy
    Step 1: An alloy is made by mixing a metal with other elements (usually other metals). Step 2: The atoms are mixed but not chemically bonded into a compound, so an alloy is a mixture. Step 3: So an alloy is a mixture of a metal with one or more other elements.
    Method:

    State the definition of an alloy as a mixture.
    Examiner tips

    • Alloy = a metal mixed with one or more other elements.

19. Question 3(b)(ii)

    1 marksOther metal present in brass
    Step 1: Brass is a common alloy of two metals. Step 2: Brass is made from copper and zinc. Step 3: Since zinc is already given, the other metal in brass is copper.
    Method:

    Recall the two metals that make up brass.
    Examiner tips

    • Brass is copper and zinc; bronze is copper and tin.

20. Question 3(c)(i)

    2 marksObservations showing a solid is in excess
    Step 1: Zinc carbonate reacts with the acid, fizzing as carbon dioxide is released, until the acid is used up. Step 2: When the acid has all reacted, any further zinc carbonate cannot react, so solid remains undissolved and the fizzing stops. Step 3: So the two observations are that solid stops dissolving (remains) and no more bubbles are produced.
    Method:

    Describe what is seen once the acid is fully used up.
    Examiner tips

    • Acid used up → solid stops dissolving and fizzing stops.

21. Question 3(c)(ii)

    1 marksNaming the filtrate after removing excess solid
    Step 1: The reaction makes zinc sulfate, which is soluble and stays in solution. Step 2: Filtration removes the unreacted solid zinc carbonate, leaving the dissolved product to pass through as the filtrate. Step 3: So the filtrate is aqueous zinc sulfate.
    Method:

    Identify the dissolved product that passes through the filter paper.
    Examiner tips

    • Filtrate = the liquid that passes through (the soluble salt solution).

22. Question 3(c)(iii)

    1 marksOther compound that reacts with sulfuric acid to give zinc sulfate
    Step 1: Zinc sulfate is the salt of zinc with sulfuric acid, so the added compound must be a base or carbonate of zinc. Step 2: Zinc oxide (or zinc hydroxide) is a base that reacts with dilute sulfuric acid to form zinc sulfate and water. Step 3: So a suitable compound, other than zinc carbonate, is zinc oxide.
    Method:

    Choose a zinc base/carbonate that gives zinc sulfate with sulfuric acid.
    Examiner tips

    • To make zinc sulfate, react sulfuric acid with a zinc base: zinc oxide or zinc hydroxide.

23. Question 3(c)(iv)

    2 marksMeaning of a saturated solution
    Step 1: A saturated solution has dissolved as much solute as it can at a given temperature. Step 2: Any extra solute added will not dissolve and stays as solid. Step 3: So a saturated solution contains the maximum concentration of dissolved solute at that specified temperature.
    Method:

    Define a saturated solution including the temperature condition.
    Examiner tips

    • Mention both "maximum dissolved" and "at a specified temperature".

24. Question 3(c)(v)

    2 marksEffect of particle size on rate
    Step 1: Reaction rate depends on how often reacting particles collide. Step 2: Large pieces have a smaller surface area than the same mass of powder, so fewer acid particles can collide with the solid at any moment. Step 3: The frequency of collisions decreases, so the rate of reaction decreases.
    Method:

    Connect surface area to collision frequency to explain the slower rate.
    Examiner tips

    • Link smaller surface area → lower collision frequency → slower rate.

25. Question 3(c)(vi)

    1 marksMeaning of the term hydrated
    Step 1: Hydrated refers to a substance that contains water of crystallisation. Step 2: This water is chemically combined within the crystal structure in a fixed proportion. Step 3: So hydrated means the crystals contain water that is chemically combined in their structure.
    Method:

    Define hydrated in terms of chemically combined water of crystallisation.
    Examiner tips

    • Hydrated = water of crystallisation chemically combined in the solid.

26. Question 4(a)(i)

    1 marksBalancing a symbol equation
    Step 1: Balance iron: 4 Fe on the left needs $c = 2$ ($2\text{Fe}_2\text{O}_3$ gives 4 Fe). Step 2: Balance sulfur: $4\text{FeS}_2$ has 8 S, so $b = 8$ ($8\text{SO}_2$). Step 3: Balance oxygen: products have $(8 \times 2) + (2 \times 3) = 22$ O atoms, so $a = 22 \div 2 = 11$. So a = 11, b = 8, c = 2.
    Method:

    Balance Fe and S, count the oxygen in the products, then find the $\text{O}_2$ coefficient.
    Examiner tips

    • Balance metals and non-metals first, leaving oxygen until last.

27. Question 4(a)(ii)

    2 marksDot-and-cross diagram of an oxygen molecule
    Step 1: Each oxygen atom has 6 outer electrons and needs 2 more to fill its shell. Step 2: The two atoms share two pairs of electrons, forming an O=O double bond (two dots and two crosses between them). Step 3: This leaves each oxygen with two non-bonding (lone) pairs, so the diagram shows a double bond plus two lone pairs on each oxygen.
    Method:

    Work out how many electron pairs must be shared so both oxygen atoms have full shells.
    Examiner tips

    • Two oxygen atoms each need 2 electrons → share 2 pairs → O=O double bond.

28. Question 4(b)(i)

    3 marksConditions for the Contact process (stage 2)
    Step 1: The Contact process needs a temperature high enough for a good rate but not so high that the yield falls; about 450 °C is used. Step 2: A pressure of about 2 atm is enough, and a vanadium(V) oxide catalyst speeds up the reaction. Step 3: So the three conditions are about 450 °C, about 2 atm, and a vanadium(V) oxide catalyst.
    Method:

    Recall the standard Contact-process conditions for the $\text{SO}_2$ + $\text{O}_2$ reaction.
    Examiner tips

    • Memorise the Contact-process conditions: ~450 °C, ~2 atm, $\text{V}_2\text{O}_5$ catalyst.

29. Question 4(b)(ii)

    1 marksEquation for the formation of sulfur trioxide
    Step 1: Sulfur dioxide ($\text{SO}_2$) reacts with oxygen ($\text{O}_2$) to form sulfur trioxide ($\text{SO}_3$). Step 2: Balancing gives 2 $\text{SO}_2$ + 1 $\text{O}_2$ → 2 $\text{SO}_3$ (4 S? no, 2 S each side; O: left 4+2=6, right 6). Step 3: So the balanced equation is $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$.
    Method:

    Write the reactants and product, then balance the equation.
    Examiner tips

    • Sulfur dioxide + oxygen → sulfur trioxide; balance to $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$.

30. Question 4(b)(iii)

    1 marksEquation for the formation of oleum
    Step 1: Oleum is formed when sulfur trioxide is absorbed into concentrated sulfuric acid. Step 2: One $\text{SO}_3$ combines with one $\text{H}_2\text{SO}_4$ to give $\text{H}_2\text{S}_2\text{O}_7$. Step 3: Checking atoms confirms $\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7$.
    Method:

    Combine sulfur trioxide with sulfuric acid to give the oleum formula.
    Examiner tips

    • $\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7$ (oleum).

31. Question 4(c)

    2 marksOxidation number of carbon
    Step 1: An element on its own always has an oxidation number of 0, so carbon as C is 0. Step 2: In $\text{CO}_2$ each oxygen is −2, giving −4 in total, so carbon must be +4 to make the molecule neutral. Step 3: So carbon is 0 in C and +4 in $\text{CO}_2$.
    Method:

    Assign 0 to the element and use the oxygen values to find carbon in $\text{CO}_2$.
    Examiner tips

    • Uncombined element = 0; balance the oxidation numbers to zero for a neutral molecule.

32. Question 4(d)

    3 marksVolume of gas from a mole calculation
    Step 1: Moles of $\text{NaHCO}_3$ = mass ÷ $M_r$ = 4.20 ÷ 84 = 0.0500 mol. Step 2: The equation shows 2 $\text{NaHCO}_3$ give 2 $\text{CO}_2$ (a 1:1 ratio), so moles of $\text{CO}_2$ = 0.0500 mol. Step 3: Volume = moles × 24 000 = 0.0500 × 24 000 = 1200 cm³.
    Method:

    Convert mass to moles, use the equation ratio, then convert moles of gas to volume.
    Examiner tips

    • Always: mass→moles, apply the mole ratio, then moles→volume (×24 000 cm³).

33. Question 5(a)

    2 marksCharacteristics of equilibrium
    Step 1: At equilibrium the forward and reverse reactions are still happening, but at the same rate. Step 2: Because the rates are equal, the concentrations of reactants and products no longer change. Step 3: So two characteristics are: rate of forward = rate of reverse, and concentrations stay constant.
    Method:

    State the two defining features of a dynamic equilibrium.
    Examiner tips

    • Dynamic equilibrium: equal rates + constant (not equal) concentrations.

34. Question 5(b)(i)

    2 marksEffect of temperature on rate and yield
    Step 1: A lower temperature slows particles down, so below 300 °C the rate of reaction is too low. Step 2: The forward reaction is exothermic, so a higher temperature shifts the equilibrium backwards, lowering the yield. Step 3: So below 300 °C the rate is too slow, and above 300 °C the yield of ethanoic acid decreases.
    Method:

    Link low temperature to rate and high temperature to the equilibrium yield.
    Examiner tips

    • Exothermic forward reaction: heat reduces yield; cold reduces rate.

35. Question 5(b)(ii)

    3 marksEffect of catalyst and pressure on equilibrium and rate
    Step 1: A catalyst speeds up both directions equally, so it does not change the position of equilibrium: no change to the concentration of $\text{CH}_3\text{COOH}$. Step 2: The forward reaction goes from 2 moles of gas to 1 mole of gas, so increasing the pressure shifts the equilibrium towards the product, increasing the concentration of $\text{CH}_3\text{COOH}$. Step 3: Higher pressure also pushes particles closer together, so the rate of the forward reaction increases.
    Method:

    Apply catalyst and pressure rules to both the equilibrium position and the rate.
    Examiner tips

    • Count gas moles each side: more pressure favours the fewer-moles side; catalysts never shift equilibrium.

36. Question 5(b)(iii)

    2 marksChoosing a transition element as a catalyst
    Step 1: Transition elements and their compounds are widely used as catalysts. Step 2: From the list, cobalt is the transition element, so it is the suitable catalyst. Step 3: Sodium and magnesium are reactive main-group metals and carbon is a non-metal, so the best answer is cobalt because it is a transition element.
    Method:

    Identify the transition element in the list and link it to catalytic behaviour.
    Examiner tips

    • Pick the transition element from the list; they are the usual catalysts.

37. Question 5(c)(i)

    1 marksCarboxylic acid with one carbon atom
    Step 1: Carboxylic acids are named by the number of carbon atoms: meth- (1), eth- (2), prop- (3), but- (4). Step 2: The acid with one carbon atom uses the prefix meth-, giving methanoic acid (HCOOH). Step 3: So the carboxylic acid with one carbon atom is methanoic acid.
    Method:

    Use the carbon-count naming prefixes to find the one-carbon acid.
    Examiner tips

    • Carbon-count prefixes: meth-(1), eth-(2), prop-(3), but-(4).

38. Question 5(c)(ii)

    1 marksMolecular formula of a four-carbon carboxylic acid
    Step 1: Butanoic acid is $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$, which has 4 carbon atoms. Step 2: Counting atoms gives 4 C, 8 H and 2 O. Step 3: So the molecular formula is $\text{C}_4\text{H}_8\text{O}_2$.
    Method:

    Build the four-carbon carboxylic acid and count its atoms.
    Examiner tips

    • A carboxylic acid contains the -COOH group, so it has 2 oxygen atoms.

39. Question 5(d)(i)

    2 marksDisplayed formula of an ester with two carbon atoms
    Step 1: An ester has the linkage -COO- joining an acid part to an alcohol part. Step 2: The smallest ester with two carbon atoms is methyl methanoate, from methanoic acid (HCOOH, 1 C) and methanol ($\text{CH}_3\text{OH}$, 1 C). Step 3: Its displayed formula shows H-C with a C=O double bond and a single bond to O, which joins to a $\text{CH}_3$ group (H-CO-O-$\text{CH}_3$).
    Method:

    Identify the smallest two-carbon ester and show its ester linkage.
    Examiner tips

    • Esters contain the -CO-O- linkage; count carbons to get methyl methanoate.

40. Question 5(d)(ii)

    2 marksAcid and alcohol that form a named ester
    Step 1: An ester is named as [alcohol part][acid part]: the first word comes from the alcohol and the -oate from the acid. Step 2: "Propyl" comes from propanol (propan-1-ol), and "butanoate" comes from butanoic acid. Step 3: So propyl butanoate is made from butanoic acid and propan-1-ol.
    Method:

    Split the ester name into the alcohol part and the acid part.
    Examiner tips

    • Ester name = alcohol part + acid-oate; split the name to find the reactants.

41. Question 5(e)

    3 marksEmpirical formula from percentage composition
    Step 1: Divide each percentage by its $A_r$: C 58.82 ÷ 12 = 4.90; H 9.80 ÷ 1 = 9.80; O 31.38 ÷ 16 = 1.96. Step 2: Divide by the smallest (1.96): C 4.90 ÷ 1.96 = 2.5; H 9.80 ÷ 1.96 = 5; O 1.96 ÷ 1.96 = 1, giving the ratio 2.5 : 5 : 1. Step 3: Multiply by 2 to get whole numbers: 5 : 10 : 2, so the empirical formula is $\text{C}_5\text{H}_{10}\text{O}_2$.
    Method:

    Convert masses to moles, simplify the ratio, and scale to whole numbers.
    Examiner tips

    • Empirical formula = simplest whole-number ratio; scale 2.5 : 5 : 1 by 2.

42. Question 6(a)

    1 marksName given to Group I elements
    Step 1: Group I elements (lithium, sodium, potassium and so on) react with water to form alkaline solutions. Step 2: Because of this they are called the alkali metals. Step 3: The halogens are Group VII, the noble gases are Group VIII/0 and the transition elements are the central block, so Group I are the alkali metals.
    Method:

    Recall the family name of the Group I elements.
    Examiner tips

    • Group I are the alkali metals (they form alkalis with water).

43. Question 6(b)

    1 marksLeast reactive Group I element
    Step 1: In Group I, reactivity increases going down the group. Step 2: Lithium is at the top of Group I, so it is the least reactive. Step 3: Potassium (lower down) is more reactive than sodium, which is more reactive than lithium, so lithium is the least reactive.
    Method:

    Use the Group I reactivity trend to find the least reactive element.
    Examiner tips

    • Down Group I: reactivity increases, so lithium (top) is least reactive.

44. Question 6(c)(i)

    2 marksObservations when lithium reacts with water
    Step 1: Lithium is less dense than water, so it floats and moves around on the surface. Step 2: The reaction produces hydrogen gas, so bubbles (fizzing) are seen and the lithium gradually gets smaller as it dissolves/disappears. Step 3: So two valid observations are that the lithium floats and moves on the surface and that bubbles (fizzing) are produced.
    Method:

    Describe what is seen as a Group I metal reacts with water.
    Examiner tips

    • Look for: floats, moves, fizzes/bubbles, gradually disappears.

45. Question 6(c)(ii)

    2 marksEquation for lithium reacting with water
    Step 1: Lithium reacts with water to give lithium hydroxide (LiOH) and hydrogen ($\text{H}_2$). Step 2: Balancing: 2 Li and 2 $\text{H}_2\text{O}$ give 2 LiOH and 1 $\text{H}_2$. Step 3: Checking atoms confirms $2\text{Li} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH} + \text{H}_2$.
    Method:

    Write the correct products, then balance lithium, hydrogen and oxygen.
    Examiner tips

    • Group I + water → metal hydroxide + hydrogen; balance the metal and the water.

46. Question 6(d)

    2 marksPhysical properties of Group I vs transition elements
    Step 1: Group I metals are soft enough to cut with a knife, unlike the hard transition metals. Step 2: Group I metals also have low densities (some float on water), much lower than the dense transition metals. Step 3: So two differences are that Group I metals are softer and have lower densities than transition elements.
    Method:

    Contrast the physical properties of Group I metals with transition metals.
    Examiner tips

    • Group I: soft, low density, low melting points; transition metals: the opposite.

47. Question 6(e)(i)

    2 marksState and colour of chlorine
    Step 1: Down Group VII the state changes: chlorine gas, bromine liquid, iodine solid. Step 2: Chlorine at room temperature is a gas, and its colour is pale yellow-green. Step 3: So chlorine is a pale yellow-green gas.
    Method:

    Recall the state and colour of chlorine among the halogens.
    Examiner tips

    • Chlorine: pale yellow-green gas; bromine: red-brown liquid; iodine: grey-black solid.

48. Question 6(e)(ii)

    2 marksIonic equation for a halogen displacement
    Step 1: The potassium ions ($\text{K}^+$) are spectator ions, so they are left out of the ionic equation. Step 2: Chlorine gains electrons and bromide ions lose electrons: $\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$. Step 3: This is balanced for atoms and charge, so it is the correct ionic equation.
    Method:

    Cancel the spectator ions and balance the remaining ionic equation.
    Examiner tips

    • Remove spectator ions ($\text{K}^+$); keep the species that actually change.

49. Question 6(e)(iii)

    3 marksEnthalpy change from bond energies
    Step 1: Energy to break bonds = I-I + Cl-Cl = 150 + 242 = 392 kJ. Step 2: Energy released forming bonds = 2 × I-Cl = 2 × 218 = 436 kJ. Step 3: $\Delta H$ = bonds broken − bonds formed = 392 − 436 = −44 kJ/mol (the negative sign shows it is exothermic).
    Method:

    Add the reactant bond energies, add the product bond energies, then subtract.
    Examiner tips

    • Break = reactant bonds (in); form = product bonds (out); $\Delta H$ = in − out.