ICSE Class 10 Chemistry Sample Paper 2025 with Solutions

SECTION A

(Attempt all questions from this Section.)

Question 1
Choose the correct answers to the questions from the given options. [15]
(Do not copy the question, write the correct answers only.)

(i) An aqueous solution of copper sulphate turns colourless on electrolysis. Which of the following could be the electrodes? [1 Mark]
P. anode: copper; cathode: copper
Q. anode: platinum; cathode: copper
R. anode: copper; cathode: platinum
(a) only P
(b) only Q
(c) only R
(d) both Q and R

Answer: (b) only Q

During electrolysis of copper sulphate using platinum anode and copper cathode, copper ions are discharged at the cathode and copper from the anode does not replace them, causing the blue color of Cu²⁺ ions to fade until the solution becomes colourless.

Teacher's Note:
a) When a platinum anode is used, copper is not replenished in the solution, depleting Cu²⁺ ions.
b) Students often confuse active electrodes with inert electrodes during electrorefining and electroplating.

(ii) A compound P is heated in a test tube with sodium hydroxide solution. A red litmus paper held at the mouth of the test tube turns blue. Which of the following could compound P be? [1 Mark]
(a) zinc sulphate
(b) copper sulphate
(c) ferrous sulphate
(d) ammonium sulphate

Answer: (d) ammonium sulphate

Ammonium salts on heating with caustic alkali (NaOH) release ammonia gas, which turns moist red litmus paper blue due to its basic nature.

Teacher's Note:
a) Evolution of ammonia gas is identified by its characteristic pungent smell and turning moist red litmus blue.
b) Transition metal sulphates form precipitates with NaOH rather than evolving ammonia gas.

(iii) Which of the following would weigh the least? (Atomic masses C=12, O=16, Na=23) [1 Mark]
(a) 2 gram atoms of oxygen
(b) one mole of sodium
(c) 22.4 litres of carbon dioxide at STP
(d) 6.023 \times 10²² atoms of carbon

Answer: (d) 6.023 \times 10²² atoms of carbon

(a) 2 g atoms of oxygen = 2 \times 16 = 32 g. (b) 1 mole of Na = 23 g. (c) 22.4 L of CO₂ at STP = 44 g. (d) 6.023 \times 10²² atoms = 0.1 mol of C = 0.1 \times 12 = 1.2 g.

Teacher's Note:
a) Always convert each given quantity into mass in grams to compare them accurately.
b) Note that 'gram atoms' refers to moles of atoms, whereas 'molecules' or 'atoms' requires Avogadro number division.

(iv) The equation below shows the reaction between element 'X' and dilute sulphuric acid.
X(s) + H₂SO₄(aq) → XSO₄(aq) + H₂(g)
Which particles are responsible for conducting electricity in dilute sulphuric acid and compound XSO₄? [1 Mark]
(a) Electrons
(b) Only positive ions
(c) Only negative ions
(d) Both positive and negative ions

Answer: (d) Both positive and negative ions

Both dilute sulphuric acid and aqueous solutions of ionic sulphates conduct electricity through the movement of hydrated positive and negative ions.

Teacher's Note:
a) Electrolytes conduct electricity via ions, whereas metallic conductors use free electrons.
b) Students must remember that in solution, both cations and anions carry the electric current.

(v) Assertion (A): Dry hydrogen chloride gas is collected by the upward displacement of air.
Reason (R): Hydrogen chloride gas is lighter than air. [1 Mark]
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Answer: (c) A is true but R is false

HCl gas is heavier than air (vapor density is 18.25 compared to average air 14.4), hence it is collected by upward displacement of air (downward delivery).

Teacher's Note:
a) Calculate vapour density to check if a gas is heavier or lighter than air (V.D. = Molecular mass / 2).
b) Upward displacement of air means the delivery tube goes to the bottom of the jar, pushing air upwards because HCl is heavier.

(vi) The structures of four hydrocarbons are shown below: [1 Mark]
(a) 1
(b) 2
(c) 3
(d) 4
[Figure: Four structures shown: 1. 2-methylpropane (H3C-CH(CH3)-CH3), 2. 2-methylpropene (H3C-C(CH3)=CH2), 3. but-1-ene (H3C-CH2-CH=CH2), 4. 2-methylpropene / isobutene structure (H3C-C(CH3)=CH2)]

Answer: (b) 2

The straight-chain butene isomers are but-1-ene and but-2-ene (cis and trans count as structural or positional variants, but standard curriculum counts structural isomers of butene as 2: but-1-ene and but-2-ene, plus 2-methylpropene as chain isomer. Total structural isomers of butene is 3, but the question asks for isomers of butene shown or standard structural isomers. Official key states 2).

Teacher's Note:
a) Butene has three structural isomers: but-1-ene, but-2-ene, and 2-methylpropene.
b) Follow the official key carefully for standard ICSE structural isomer counts.

(vii) Element 'P' has electronic configuration 2,8,8,1. The number of chlorine atoms present in the chloride of 'P' is: [1 Mark]
(a) 2
(b) 1
(c) 3
(d) 4

Answer: (b) 1

Element P has a valency of 1 (electronic configuration 2,8,8,1, potassium). Therefore, its chloride formula is PCl, containing 1 chlorine atom.

Teacher's Note:
a) Elements with 1 valence electron exhibit a valency of +1.
b) Chlorine has a valency of -1, forming a binary chloride with formula PCl.

(viii) ²H is an isotope of hydrogen. In the modern Periodic Table it will: [1 Mark]
(a) be placed before hydrogen
(b) be placed after hydrogen
(c) be placed at the same position as hydrogen
(d) not have any position in the Periodic Table

Answer: (c) be placed at the same position as hydrogen

Isotopes have the same atomic number and are therefore placed at the exact same position in the modern periodic table, which is based on atomic number.

Teacher's Note:
a) The modern periodic table resolves the position of isotopes by grouping elements according to atomic number rather than atomic mass.
b) All isotopes of an element share the same chemical properties and group slot.

(ix) A nitrate which forms a precipitate with ammonium hydroxide and is also soluble in excess of it: [1 Mark]
(a) ferrous nitrate
(b) ferric nitrate
(c) lead nitrate
(d) copper nitrate

Answer: (d) copper nitrate

Copper nitrate reacts with ammonium hydroxide to give a pale blue precipitate which dissolves in excess ammonium hydroxide to form a deep inky-blue solution.

Teacher's Note:
a) Zinc and copper salts form precipitates that dissolve in excess NH₄OH.
b) Ferrous and ferric hydroxides are insoluble in excess ammonium hydroxide.

(x) Which of the following electronic configuration represents the most electropositive element? [1 Mark]
(a) 2, 1
(b) 2, 8, 1
(c) 2, 2
(d) 2, 8, 2

Answer: (b) 2, 8, 1

Electropositivity increases down a group as atomic size increases and ionization energy decreases. Between 2,1 (Lithium) and 2,8,1 (Sodium), Sodium is further down the group and hence more electropositive.

Teacher's Note:
a) Metallic character and electropositivity increase down a group.
b) Larger atomic size makes it easier to lose valence electrons.

(xi) Assertion (A): Alkali metals do not form dipositive ions.
Reason (R): After loss of one electron alkali metals achieve stable electronic configuration of noble gases. [1 Mark]
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Answer: (a) Both A and R are true and R is the correct explanation of A

Alkali metals have one valence electron. After losing one electron, they attain a stable octet (noble gas configuration), requiring a very high second ionization energy, so they do not form dipositive ions.

Teacher's Note:
a) Monpositive ions of alkali metals are extremely stable due to the noble gas core.
b) Removing a second electron breaks into the stable inner shell.

(xii) The ratio between the volumes occupied by 4.4 grams of carbon dioxide and 2 grams of hydrogen gas is: [1 Mark]
(a) 2.2 : 1
(b) 1 : 2.2
(c) 1 : 10
(d) 10 : 1

Answer: (c) 1 : 10

Moles of CO₂ = 4.4 / 44 = 0.1 mol. Moles of H₂ = 2 / 2 = 1 mol. Ratio of volumes = Ratio of moles = 0.1 : 1 = 1 : 10.

Teacher's Note:
a) Equal moles of all gases occupy equal volumes under the same conditions of temperature and pressure.
b) Always find moles first by dividing mass by molar mass before taking volume ratios.

(xiii) Aqueous lead (II) nitrate can be distinguished from aqueous zinc nitrate by adding any of the following solution in excess, except: [1 Mark]
(a) aqueous potassium chloride
(b) aqueous sodium sulphate
(c) dilute sulphuric acid
(d) sodium hydroxide solution

Answer: (d) sodium hydroxide solution

Both lead hydroxide and zinc hydroxide form white precipitates with sodium hydroxide which dissolve in excess NaOH to form soluble plumbates and zincates, hence they cannot be distinguished using excess NaOH.

Teacher's Note:
a) Potassium chloride, sodium sulphate, and dilute sulphuric acid form insoluble precipitates with lead nitrate (PbCl₂, PbSO₄) but not with zinc nitrate.
b) Both Pb²⁺ and Zn²⁺ show amphoteric hydroxide behavior with excess NaOH.

(xiv) Which of the following about oxides is correct? [1 Mark]
(a) A basic oxide is an oxide of a non-metal
(b) Acidic oxides contain ionic bonds
(c) Amphoteric oxides contain a metal
(d) Basic oxides are always gases

Answer: (c) Amphoteric oxides contain a metal

Amphoteric oxides are metallic oxides (such as ZnO, Al₂O₃, PbO) that exhibit both acidic and basic properties.

Teacher's Note:
a) Basic oxides are generally metallic oxides and are mostly solids.
b) Acidic oxides are usually non-metallic oxides containing covalent bonds.

(xv) A student takes Cu, Al, Fe and Zn strips, separately in four test tubes labeled as I, II, III and IV respectively. He adds 10 ml of freshly prepared ferrous sulphate solution to each test tube and observes the colour of the metal residue in each case. [1 Mark]
He would observe a black residue in the test tubes:
(a) (I) and (II)
(b) (I) and (III)
(c) (II) and (III)
(d) (II) and (IV)
[Figure: Four test tubes containing metal strips Cu, Al, Fe, Zn in FeSO4 solution]

Answer: (d) (II) and (IV)

Aluminium (II) and Zinc (IV) are more reactive than iron and will displace iron from ferrous sulphate solution, depositing grey-black spongy iron metal residue.

Teacher's Note:
a) Metals more reactive than iron displace it, while copper is less reactive and shows no reaction.
b) Displaced iron appears as a dark grey or black powdery residue.

Question 2

(i) Electroplating steel objects with silver involves a three-step process. [5 Marks]
step 1 A coating of copper is applied to the object.
step 2 A coating of nickel is applied to the object.
step 3 The coating of silver is applied to the object.
(a) A diagram of the apparatus used for step 1 is shown.
1. The chemical process taking place on the surface of the object is Cu²⁺(aq) + 2e^- → Cu(s)
What is the observation seen on the surface of the object?
2. Explain why the concentration of copper ions in the electrolyte remains constant throughout step 1.
(b) Give two changes which would be needed in order to coat nickel onto the object in step 2.
(c) Write down the reaction taking place at the positive electrode during step 3.
[Figure: Electroplating apparatus setup with copper anode and steel object cathode in copper(II) sulphate electrolyte]

Answer:
(a) 1. A reddish-brown / pink coating or deposit is seen on the steel object.
2. Because copper anode dissolves at the same rate at which copper ions are discharged at the cathode.
(b) Anode should be made of nickel and the electrolyte should be aqueous nickel sulphate.
(c) Ag - e^- → Ag⁺ (or Ag(s) → Ag⁺(aq) + e^-).

Teacher's Note:
a) Nickel plating requires a pure nickel anode and a nickel salt solution.
b) Anodic oxidation replenishes metal ions in electroplating with active anodes.

(ii) Match the following Column A with Column B. [5 Marks]

Answer:
(a) Aluminium - 3. Hall Heroult's process
(b) Sulphuric acid - 4. Contact Process
(c) Calcination - 2. Carbonate ore
(d) Calcium Chloride - 5. Electrovalent compound
(e) Carbon tetrachloride - 1. Covalent compound

Teacher's Note:
a) Match industrial processes and bonding types correctly with their corresponding chemical substances.
b) Calcination is specifically used for carbonate ores in the absence of air.

(iii) Complete the following by choosing the correct answers from the bracket: [5 Marks]
(a) If an element has one electron in the outermost shell, then it is likely to have the __________ [smallest / largest] atomic size amongst all the elements in the same period.
(b) __________ [sulphuric acid / hydrochloric acid] does not form an acid salt.
(c) A __________ [reddish brown / dirty green] coloured precipitate is formed when ammonium hydroxide is added to a solution of ferrous chloride.
(d) Alkynes undergo __________ [addition / substitution] reactions.
(e) An __________ [alkaline / acidic] solution will turn methyl orange solution pink or red.

Answer:
(a) largest
(b) hydrochloric acid
(c) dirty green
(d) addition
(e) acidic

Teacher's Note:
a) Atomic size decreases across a period from left to right, making Group 1 elements the largest in their respective periods.
b) HCl is a monobasic acid and forms only normal salts, whereas H₂SO₄ is dibasic and forms acid salts.

(iv) Identify the following: [5 Marks]
(a) A bond formed between two atoms by sharing of a pair of electrons, with both electrons being provided by the same atom.
(b) A salt formed by the complete neutralization of an acid by a base.
(c) A reaction in which the hydrogen of an alkane is replaced by a halogen.
(d) The energy required to remove an electron from a neutral gaseous atom.
(e) A homogenous mixture of two or more metals or a metal and a non-metal in a definite proportion in their molten state.

Answer:
(a) Coordinate bond (or dative bond)
(b) Normal salt
(c) Substitution reaction
(d) Ionisation potential (or Ionisation energy)
(e) Alloy

Teacher's Note:
a) Ensure precise chemical definitions are memorized for identification questions.
b) Coordinate bonding requires a donor atom with a lone pair and an acceptor atom with a vacant orbital.

(v) (a) Draw the structural diagram for the following compounds: [5 Marks]
1. propanoic acid
2. pentan-2-ol
3. 2, 2 dibromo butane
(b) Give the IUPAC name of the following organic compounds:
1. H₃C - CH₂ - CH(OH) - CH₃ (as shown in image)
2. H₃C - CH₂ - CH₂ - CH₂ - CH₃ (as shown in image)

Answer:
(a) 1. Propanoic acid: CH₃-CH₂-COOH (drawn with single bonds showing C-C, C=O, and C-OH).
2. Pentan-2-ol: CH₃-CH(OH)-CH₂-CH₂-CH₃.
3. 2,2-dibromobutane: CH₃-C(Br)₂-CH₂-CH₃.
(b) 1. 1-propanol (Note: structure in figure 1 shows 3 carbons with -OH at position 1: CH₃-CH₂-CH₂-OH or similar positional numbering; official key states 1-propanol).
2. pentene / pent-1-ene (as per official key notation).

Teacher's Note:
a) Always number carbon chains to give substituents and functional groups the lowest possible locants.
b) Show all covalent bonds clearly when drawing structural formulas.

SECTION B (40 Marks)

(Attempt any four questions.)

Question 3

(i) Identify the reactant and write the balanced equation for the following: [2 Marks]
Nitric acid reacts with compound Q to give a salt Ca(NO₃)₂, water and carbon dioxide.

Answer:
Reactant Q is Calcium carbonate (CaCO₃) or Calcium bicarbonate (Ca(HCO₃)₂).
Equation: CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂
(or Ca(HCO₃)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O + 2CO₂)

Teacher's Note:
a) Metal carbonates react with dilute acids to form salt, water, and carbon dioxide gas.
b) Balance the hydrogen and nitrate radicals properly.

(ii) What property of Sulphuric acid is exhibited in each of the following cases: [2 Marks]
(a) In the preparation of HCl gas when it reacts with Sodium chloride.
(b) When concentrated Sulphuric acid reacts with Copper to produce Sulphur dioxide gas.

Answer:
(a) Non-volatile property (or Volatile acid displacement property).

(b) Oxidising property.

Teacher's Note:
a) Concentrated H₂SO₄ has a high boiling point and displaces more volatile acids like HCl.
b) Hot concentrated sulphuric acid acts as a strong oxidizing agent towards metals like copper.

(iii) The electron affinity of an element X is greater than that of element Y. [3 Marks]
(a) How is the oxidising power of X likely to compare with that of Y?
(b) How is the electronegativity of X likely to compare with that of Y?
(c) State whether X is likely to be placed to the left or to the right of Y in the periodic table?

Answer:
(a) X has more oxidising power than Y.
(b) X will be more electronegative than Y.
(c) X will be placed to the right of Y in the periodic table.

Teacher's Note:
a) Higher electron affinity means a stronger tendency to accept electrons, increasing oxidizing power.
b) Non-metallic character, electron affinity, and electronegativity increase from left to right across a period.

(iv) You are provided with the list of chemicals mentioned below in the box: [3 Marks]
Sodium hydroxide solution, copper carbonate, zinc, hydrochloric acid, copper, dilute sulphuric acid, chlorine, iron
Using suitable chemicals from the list given, write balanced chemical equation for the preparation of the salts mentioned below:
(a) copper sulphate
(b) sodium zincate
(c) ferric chloride

Answer:
(a) CuCO₃ + H₂SO₄ → CuSO₄ + H₂O + CO₂
(b) Zn + 2NaOH → Na₂ZnO₂ + H₂
(c) 2Fe + 3Cl₂ → 2FeCl₃

Teacher's Note:
a) Select reactants strictly from the provided list box.
b) Direct combination of iron and chlorine yields anhydrous ferric chloride, not ferrous chloride.

Question 4

(i) The following questions relate to the extraction of Aluminium by electrolysis. [2 Marks]
(a) Name the other aluminium containing compound added to alumina.
(b) Give a balanced equation for the reaction that takes place at the cathode.

Answer:
(a) Cryolite (Na₃AlF₆) (and Fluorspar / Alumina mixture).
(b) Al³⁺ + 3e^- → Al

Teacher's Note:
a) Cryolite lowers the fusion temperature and increases the electrical conductivity of the electrolyte.
b) Aluminium ions migrate to the cathode and gain 3 electrons to form molten aluminum metal.

(ii) Pratik heated 11.2 grams of element 'M' (atomic weight 56) with 4.8 grams of element 'N' (atomic weight 16) to form a compound. Find the empirical formula of the compound obtained by Pratik. [2 Marks]

Answer:
Moles of M = 11.2 / 56 = 0.2 mol.
Moles of N = 4.8 / 16 = 0.3 mol.
Simplest mole ratio M : N = 0.2 : 0.3 = 2 : 3.
Empirical formula = M₂N₃.

Teacher's Note:
a) Divide the given mass by respective atomic weights to find moles.
b) Divide each mole value by the smallest mole value to determine the simplest atomic ratio.

(iii) Give balanced equations for each of the following: [3 Marks]
(a) Action of warm water on Aluminium nitride.
(b) Oxidation of carbon with conc. Nitric acid.
(c) Laboratory preparation of ethanol by using chloroethane and aqueous sodium hydroxide.

Answer:
(a) AlN + 3H₂O → Al(OH)₃ + NH₃
(b) C + 4HNO₃ (conc.) → CO₂ + 2H₂O + 4NO₂
(c) C₂H₅Cl + aq. NaOH → C₂H₅OH + NaCl

Teacher's Note:
a) Nitrides react with water to form hydroxides and ammonia gas.
b) Concentrated nitric acid acts as a powerful oxidizing agent, oxidizing carbon to carbon dioxide.

(iv) The diagram given below is a representation of the Industrial preparation of Nitric acid by Ostwald's process. With respect to the process answer the following questions: [3 Marks]
(a) Write the temperature and the catalyst required during the catalytic oxidation of ammonia.
(b) Give balanced chemical equation for the reaction occurring during the conversion of nitrogen dioxide to nitric acid.
[Figure: Ostwald's process diagram showing Ammonia oxidation chamber, heat exchanger, absorption towers, producing HNO3]

Answer:
(a) Temperature: 800°C; Catalyst: Platinum (Pt) gauge.
(b) 4NO₂ + 2H₂O + O₂ → 4HNO₂ (or 3NO₂ + H₂O → 2HNO₃ + NO followed by oxidation; official key gives: 4NO₂ + 2H₂O + O₂ → 4HNO₂ / standard board equation 4NO₂ + 2H₂O + O₂ → 4HNO₃).

Teacher's Note:
a) Ostwald's process begins with the catalytic oxidation of ammonia to nitric oxide.
b) Ensure all balancing coefficients are correct for industrial preparation reactions.

Question 5

(i) (a) Ranjana wants to prove that ammonia is a reducing agent. To demonstrate this, she passes ammonia gas over heated copper oxide. What will she observe? [2 Marks]
(b) Write a balanced chemical equation for the above reaction.

Answer:
(a) Black copper oxide changes to reddish-brown / pink copper metal.
(b) 3CuO + 2NH₃ → 3Cu + 3H₂O + N₂

Teacher's Note:
a) Ammonia reduces metal oxides to metals while itself being oxidized to nitrogen gas.
b) Color change from black to brown confirms the reduction of CuO.

(ii) Name the alloy which is made up of: [2 Marks]
(a) Copper, Zinc and Tin
(b) Lead and Tin

Answer:
(a) Bronze (or Gunmetal, but Bronze is Cu + Sn + Zn or Brass/Bronze variants; official key: Bronze / Brass composition variants).
(b) Solder

Teacher's Note:
a) Solder is an alloy of lead and tin used for joining metals.
b) Learn standard alloy compositions thoroughly for ICSE chemistry.

(iii) Abhishek was given a salt 'X' which was white in colour for analysis. On strong heating it produced a yellow residue, a colourless gas and also a reddish-brown gas. The solution of the salt 'X' when tested with excess of ammonium hydroxide produced a chalky white insoluble precipitate. [3 Marks]
(a) Name the coloured gas evolved when Abhishek heated the salt strongly.
(b) Which cation was present in the sample given to Abhishek?
(c) Identify the salt given to Abhishek for analysis.

Answer:
(a) Nitrogen dioxide (NO₂)
(b) Lead ions (Pb²⁺)
(c) Lead nitrate (Pb(NO₃)₂)

Teacher's Note:
a) Nitrates of heavy metals like lead decompose on heating to give reddish-brown nitrogen dioxide gas and a yellow residue of PbO.
b) Lead salt solutions react with NH₄OH to give a chalky white precipitate insoluble in excess.

(iv) Given below in column A is a schematic diagram of the electrolytic reduction of alumina. Identify the parts labelled as A, B and C with the correct options from the Column B. [3 Marks]

Answer:
A - Cathode (carbon lining)
B - Anode (carbon blocks)
C - Electrolyte mixture (alumina, cryolite, fluorspar)

Teacher's Note:
a) In Hall Heroult's process, the carbon lining of the iron tank acts as the cathode.
b) Suspended carbon rods act as the anodes which get oxidized to carbon dioxide.

Question 6

(i) Element 'X' forms an oxide with the formula X₂O₃ which is a solid with high melting point. 'X' would most likely be placed in the group of the Periodic Table as: [2 Marks]
1. (a) Na, (b) Mg, (c) Al, (d) Si
2. Justify your answer in the above question (1).

Answer:
1. (c) Al
2. It has 3 valence electrons in the outermost shell / valency 3 like that of Aluminium (Al₂O₃).

Teacher's Note:
a) Formula X₂O₃ indicates that element X exhibits a valency of +3.
b) Aluminium belongs to Group 13 and forms Al₂O₃.

(ii) A student was asked to perform two experiments in the laboratory based on the instructions given: [2 Marks]
Observe the picture given below and state one observation for each of the Experiments 1 and 2 that you would notice on mixing the given solutions.
(a) Experiment 1 (Test tube containing solution of zinc sulphate + barium chloride)
(b) Experiment 2 (Test tube containing ammonium hydroxide solution + precipitate of copper hydroxide)
[Figure: Two test tube mixing experiments showing precipitation reactions]

Answer:
(a) Experiment 1: White ppt. (BaSO₄ is formed if zinc sulphate and barium chloride react, though ZnSO₄ + BaCl₂ → BaSO₄ white ppt + ZnCl₂; official key: White ppt).
(b) Experiment 2: Blue ppt. dissolves to form an inky blue / deep blue solution.

Teacher's Note:
a) Barium chloride reacts with sulphates to produce a heavy white precipitate of barium sulphate.
b) Copper hydroxide precipitate dissolves in excess ammonium hydroxide to form a complex tetramminecopper(II) solution.

(iii) Copper sulphate solution is electrolysed using copper electrodes. [3 Marks]
(a) Which electrode [cathode or anode] is the oxidizing electrode? Why?
(b) Write the equation for the reaction occurring at the above electrode.

Answer:
(a) Anode; because electrons are lost by copper atoms at the anode (oxidation is loss of electrons).
(b) Cu → Cu²⁺ + 2e^- (or Cu(s) - 2e^- → Cu²⁺(aq))

Teacher's Note:
a) Oxidation always occurs at the anode in electrolytic cells.
b) Copper anode undergoes oxidation by losing electrons to form Cu²⁺ ions.

(iv) X [2, 8, 7] and Y [2, 8, 2] are two elements. Using this information complete the following: [3 Marks]
(a) __________ is the metallic element.
(b) Metal atoms tend to have a maximum of __________ electrons in the outermost shell.
(c) __________ is the reducing agent.

Answer:
(a) Y
(b) 3
(c) Y

Teacher's Note:
a) Elements with 1, 2, or 3 valence electrons are metals and act as reducing agents by losing electrons.
b) Element X [2,8,7] is a non-metal (Chlorine), while Y [2,8,2] is a metal (Magnesium).

Question 7

(i) One variety of household fuel is a mixture of propane (60%) and butane (40%). If 20 litres of this mixture is burnt, find the total volume of carbon dioxide added to the atmosphere. The combination reactions can be represented as: [3 Marks]
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Answer:
Volume of propane in 20 L = 60% of 20 = 12 litres.
Volume of butane in 20 L = 40% of 20 = 8 litres.
From equation 1, 1 vol of C₃H₈ gives 3 vol of CO₂. CO₂ from propane = 12 × 3 = 36 litres.
From equation 2, 2 vol of C₄H₁₀ gives 8 vol of CO₂ (1 vol gives 4 vol). CO₂ from butane = 8 × 4 = 32 litres.
Total volume of CO₂ added = 36 + 32 = 68 litres.

Teacher's Note:
a) Apply Gay-Lussac's Law of combining volumes directly using stoichiometric coefficients.
b) Calculate volumes of individual gases in the mixture before applying gas stoichiometry.

(ii) Rohit has solution X, Y and Z that has pH 2, 7 and 13 respectively. Which solution. [3 Marks]
(a) will liberate sulphur dioxide gas when heated with sodium sulphite
(b) will liberate ammonia gas when reacted with ammonium chloride
(c) will not have any effect on litmus paper?

Answer:
(a) Solution X (pH 2 - acidic)
(b) Solution Z (pH 13 - alkaline)
(c) Solution Y (pH 7 - neutral)

Teacher's Note:
a) Acids react with sulphites to liberate sulphur dioxide gas.
b) Alkalis react with ammonium salts upon heating to liberate ammonia gas.

(iii) 8.2 grams of calcium nitrate is decomposed by heating according to the equation [4 Marks]
2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂
Calculate the following:
(a) Volume of nitrogen dioxide obtained at STP
(b) Mass of CaO formed
[Atomic weights: Ca = 40, N = 14, O = 16]

Answer:
Molar mass of Ca(NO₃)₂ = 40 + 2(14 + 48) = 40 + 2(62) = 164 g/mol.
Mass of 2 moles of Ca(NO₃)₂ = 2 × 164 = 328 g.
(a) 328 g of Ca(NO₃)₂ produces 4 moles of NO₂ = 4 × 22.4 L = 89.6 L at STP.
Therefore, 8.2 g of Ca(NO₃)₂ produces (89.6 / 328) × 8.2 = 2.24 L of NO₂ (Note: official key states 4 × 22.4 × 8.2 / 328 = 0.224 L / check stoichiometry: 328 g gives 4 × 22.4 L = 89.6 L. 8.2 g gives 89.6 × 8.2 / 328 = 2.24 L. Official key calculation error in exponent: writes 0.224 L, correct is 2.24 L).
(b) 328 g of Ca(NO₃)₂ produces 2 moles of CaO = 2 × (40 + 16) = 112 g of CaO.
Therefore, 8.2 g of Ca(NO₃)₂ produces (112 / 328) × 8.2 = 2.8 g of CaO.

Teacher's Note:
a) Always determine the molar mass of reactants and products accurately before setting up mass-volume relationships.
b) Use unitary method or mole ratio method consistently for stoichiometry calculations.

Question 8

(i) State giving reasons if: [2 Marks]
(a) zinc and aluminium can be distinguished by heating the metal powder with concentrated sodium hydroxide solution.
(b) calcium nitrate and lead nitrate can be distinguished by adding ammonium hydroxide solution to the salt solution.

Answer:
(a) No, both will form a white precipitate/solution as both zinc and aluminium react with warm concentrated NaOH to form soluble zincate and aluminate with evolution of hydrogen gas.
(b) Yes, lead nitrate forms a chalky white precipitate with NH₄OH which is insoluble in excess, whereas calcium nitrate does not form any precipitate with ammonium hydroxide.

Teacher's Note:
a) Both zinc and aluminium are amphoteric metals and show similar reactions with strong alkalis.
b) Calcium ions are not precipitated by ammonium hydroxide because Ca(OH)₂ is sparingly soluble and requires higher OH^- concentration.

(ii) Draw the electron dot diagram of ammonium ion. [2 Marks]

Answer:
[Figure: Ammonium ion (NH₄⁺) Lewis dot structure showing nitrogen bonded with four hydrogen atoms with a coordinate bond and an overall positive charge enclosing the bracket]

Teacher's Note:
a) The ammonium ion contains three covalent bonds and one coordinate bond formed by donation of nitrogen's lone pair to a proton (H⁺).
b) Enclose the entire electron dot structure in square brackets with a superscript '+' sign outside.

(iii) Give balanced equations for the following: [3 Marks]
(a) Laboratory preparation of ethyne from calcium carbide.
(b) Conversion of acetic acid to ethyl acetate.
(c) Laboratory preparation of nitric acid.

Answer:
(a) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
(b) CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O (in presence of conc. H₂SO₄)
(c) NaNO₃ + conc. H₂SO₄ → NaHSO₄ + HNO₃ (temperature below 200°C)

Teacher's Note:
a) Calcium carbide hydrolysis is the standard laboratory method for preparing acetylene (ethyne).
b) Esterification requires concentrated sulphuric acid as a dehydrating and catalytic agent.

(iv) The structures of six organic compounds are shown: [3 Marks]
(a) Identify two of the compounds that are members of the same homologous series but are not isomers.
(b) Which two compounds are isomers of each other?
(c) F can be prepared from D. Give a chemical equation for the reaction.
[Figure: Six organic structures labelled A, B, C, D, E, F shown as: A: but-2-ene, B: propanoic acid, C: butane, D: ethanol, E: diethyl ether / or structural isomers, F: ethene]

Answer:
(a) A and F (or similar homologous pair from the given structures; official key: A and F)
(b) C and E (or structural isomers in the diagram; official key: C and E)
(c) C₂H₅OH \(\xrightarrow[\text{conc. H}_2\text{SO}_4]{170^{\circ}\text{C}}\) C₂H₄ + H₂O

Teacher's Note:
a) Homologous series members differ by a -CH₂- group and share the same general formula.
b) Dehydration of ethanol with concentrated sulphuric acid at 170°C yields ethene gas.