ICSE Class 10 Chemistry (Science Paper – 2) Solved Paper 2025

(a) Sulphur dioxide

Explanation:

Sulphur dioxide gas turns acidified potassium permanganate from pink to clear colourless.

2KMnO₄ + 2H₂O + 5SO₄ ⟶ K₂SO₄ + MnSO₄ + H₂SO₄

(b) C₅H₁₂

Explanation:

C₅H₁₂ is an alkane that represents a saturated hydrocarbon. It is a simplest open chain hydrocarbon represented by the formula C_nH_2n+2.

(a) Copper

Explanation:

Copper lies below hydrogen in the reactivity (activity) series, so its oxide is not very stable. Therefore, common chemical reducing agents such as carbon, carbon monoxide, or hydrogen can readily reduce copper(II) oxide to metallic copper, e.g.

CuO + H₂ → Cu + H₂O

In contrast, the oxides of sodium, aluminium and potassium are extremely stable because those metals are highly reactive and have a very strong affinity for oxygen. Their oxides can be reduced only by electrolysis and not by ordinary chemical reducing agents.

(d) C₃H₈

Explanation:

Given, Vapour density of organic compound = 22

Molecular weight = 2 × V.D. = 2 × 22 = 44

Hence, C₃H₈ is a organic compound that has the molecular mass of 44.

(c) Oxidising agent

Explanation:

Oxidising agents are substances which add oxygen atoms to others compounds or remove hydrogen atoms from others compounds. Sulphuric acid accepts electrons and donates oxygen atoms to sulphur, leading to the formation of SO₂ and water (H₂O). Thus, sulphuric acid acts as an oxidising agent.

(c) (A) is true but (R) is false.

Explanation:

As we move down any group in the periodic table, atomic size increases and the outermost electron is held less tightly by the nucleus. Therefore the loss of this electron becomes easier, so the tendency to lose electrons increases; Assertion (A) is true. In group 1 the reactivity of metals increases down the group for the same reason. Consequently, the most reactive metal is found at the bottom, not at the top; Reason (R) is false.

(b) Haematite

Explanation:

Magnetic separation works when either the ore or the gangue shows magnetic behaviour. Haematite, an iron ore, is weakly magnetic, whereas its common impurity (silica) is non-magnetic. When the crushed mixture is passed over a high-intensity magnetic separator, the haematite particles are attracted and collected separately from the non-magnetic gangue, concentrating the ore.

(d) A: 2, 6; B: 2, 8, 7

Explanation:

Atom A has six valence electrons (configuration 2, 6) and therefore needs two more electrons to complete its octet.

Each atom B has seven valence electrons (configuration 2, 8, 7) and needs one additional electron to attain a stable, noble-gas configuration.

In the AB₂ molecule, atom A forms a single covalent bond with each of two B atoms. Through this sharing, atom A gains two electrons (one from each B), while each B gains one electron from A, so all three atoms achieve an octet.

(a) CH₄, C₂H₆, C₃H₈

Explanation:

These three molecules are consecutive members of the alkane series, which has the general formula C_nH_2n+2. Each successive member differs from the previous one by a CH₂ group, satisfying the defining characteristic of a homologous series.

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

Explanation:

When sulphur trioxide is brought into direct contact with water,

SO₃(g) + H₂O(l) ⟶ H₂SO₄(l)

the reaction is highly exothermic. The heat released vaporises some of the acid and produces a dense fog of minute H₂SO₄ droplets, which are difficult to condense and collect. Therefore, in the Contact Process the SO₃ is first absorbed in concentrated sulphuric acid to form oleum, which is later diluted with water to give the required concentration of H₂SO₄.

(c) Cl⁻ & K⁺

Explanation:

Chlorine has atomic number 17 and its electronic configuration is 2, 8, 7. A chloride ion, Cl⁻, has gained one electron and therefore has 18 electrons with the configuration 2, 8, 8.

Potassium has atomic number 19 and its electronic configuration is 2, 8, 8, 1. A potassium ion, K⁺, has lost one electron and therefore has 18 electrons with the configuration 2, 8, 8.

Because both Cl⁻ and K⁺ have 18 electrons, they have the same electronic configuration.

(c) Sodium sulphate + Lead nitrate

Explanation:

Both sodium sulphate (Na₂SO₄) and lead(II) nitrate [Pb(NO₃)₂] are readily soluble in water. When their aqueous solutions are mixed, a double-decomposition reaction occurs, forming insoluble lead(II) sulphate (PbSO₄) as a white precipitate:

Pb(NO₃)₂(aq) + Na₂SO₄(aq) ⟶ PbSO₄(s) ↓ + 2NaNO₃(aq)

Because the desired product is obtained as an insoluble precipitate, this pair of reactants is best used to produce lead(II) sulphate.

(d) The number of copper (II) ions in the electrolyte decreases.

Explanation:

Cu²⁺ ions deposited at the cathode are replaced by Cu²⁺ ions produced at the anode, so the overall Cu²⁺ concentration stays constant.

(c) Z

Explanation:

The greater the tendency to lose electrons, the greater the metallic character and the stronger the reducing power of a metal. As we move down a group, atomic size increases and ionisation energy decreases, so the tendency to lose electrons and thus the reducing power also increases. Therefore, of the three atoms, Z (the lowest member of the group) is the best reducing agent.

(b) 40 cm³

Explanation:

From the equation, 1 volume of CH₄ reacts with 2 volumes of O₂.

For 40 cm³ CH₄, required O₂ = 40 × 2 = 80 cm³.

Since only 60 cm³ O₂ is available, oxygen is the limiting reactant.

Amount of CH₄ that reacts = 60\over2 = 30 cm³.

Excess CH₄ = 40 − 30 = 10 cm³.

2 volumes of O₂ produce 1 volume of CO₂.

CO₂ formed from 60 cm³ O₂ = 60/2 = 30 cm³.

The remaining gases at room temperature are 10 cm³ CH₄ and 30 cm³ CO₂.

Total volume = 10 + 30 = 40 cm³.

(a) X — Water

(b) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃↑

(c) Y — Quicklime (CaO)

(d) Substance Y — Quicklime (CaO) is used as a drying agent

(e) Ammonia gas is highly soluble in water so it cannot be collected over water hence it is collected by the downward displacement of air.

(a) Spurious alcohol

(b) Basic salt

(c) Isomers

(d) Electronegativity

(e) Polar covalent bond.

(a) Copper nitrate solution forms a coloured precipitate with ammonium hydroxide which is soluble in excess of ammonium hydroxide.

(b) Zinc blende is converted to zinc oxide by roasting.

(c) Molten sodium chloride conducts electricity by the movement of ions.

(d) The reaction that takes place at the anode during the electrolysis of aqueous Sodium argentocyanide with silver electrodes is Ag ⟶ Ag⁺ + e^–

(e) The salt formed when ZnO reacts with hot concentrated NaOH is sodium zincate.

(a)

1. 2-methyl propene

2. butanal

(b) IUPAC names

1. 1,1,2,2-Tetrachloroethane

2. Butanoic acid

3. Pent-2-ene

(a) The atomic number of X is 14

Electronic configuration ⟶ 2, 8, 4

Number of electron shells = 3, hence X belongs to period 3.

(b) The atomic number of Y is 8

Electronic configuration ⟶ 2, 6

X has a valency of 4 and Y has valency of 2.

∴ Simplified formula of the compound is XY₂.

(a) Anode is known as the oxidizing electrode because oxidation takes place at the anode. The anions lose electrons at the anode.

(b) Graphite electrodes are preferred in the electrolysis of molten lead bromide because graphite is a good conductor of electricity and is not affected by the bromine vapours liberated at the anode.

Mg + 2H₂SO₄ ⟶ MgSO₄ + 2H₂O + SO₂

(a) 24 g of Mg reacts with 196 g of H₂SO₄.

60 g of Mg reacts with = 196\over24 × 60 = 490 g of H₂SO₄.

Hence, mass of acid required = 490 g.

(b) 24 g of Mg gives 22.4 L of SO₂ at STP.

60 g of Mg gives = 22.4\over24 = 56 L of SO₂ at STP.

Hence, volume of SO₂ at STP = 56 L.

(a) When barium chloride solution is added to zinc sulphate solution, a white precipitate of barium sulphate is formed which is insoluble in dilute hydrochloric acid or nitric acid.

ZnSO₄ + BaCl₂ ⟶ BaSO₄↓ [white ppt.] + ZnCl₂

(b) When lead nitrate is heated strongly, brown fumes of nitrogen dioxide are evolved and a yellow residue of lead oxide is left behind.

2Pb(NO₃)₂ ⟶ 2PbO + 4NO₂ + O₂

(c) Moist starch iodide paper turns blue-black.

Cl₂ + 2KI ⟶ 2KCl + I₂

(a) V.D. = mass of certain volume of gas under similar conditions / mass of equal volume of H₂ under similar conditions

= 450\over150 = 3

(b) Molecular weight = 2 × V.D.

= 2 × 3 = 6 a.m.u.

(a) \mathrm{CH_3COONa + NaOH} \xrightarrow{\mathrm{CaO,\ heat}} \mathrm{CH_4 + Na_2CO_3}

(b) \mathrm{2CH_3COOH + Mg} \rightarrow \mathrm{(CH_3COO)_2Mg + H_2} \uparrow

(a) Nitrogen dioxide (NO₂)

(b) Nitrogen (N₂)

(c) Hydrogen sulphide (H₂S)

(a) Element G

Explanation – Element G belongs to group VA (group 15). Group 15 elements have 5 valence electrons in their outermost shell they can gain 3 electrons to complete the octet hence, the valency will be 3.

(b) Element E

Explanation – Catenation (ability to form long chains of its own atoms) is strongest in the group IVA (14) element shown, i.e., E.

(c) Element M

Explanation –  M is in group IIA they have 2 valence electrons and forms M²⁺ and P is in group VIIA have 7 valence electrons and forms P⁻, so one molecule of M combines with two molecules of P. Hence, the neutral formula is MP₂.

(a) Cl

(b) Ne

Explanation:

(a) Cl — Among the three elements, Cl is the rightmost in period 3. As ionisation energy increases when moving across a period from left to right hence ionisation energy is highest for Cl.

(b) Ne — It is the rightmost element in period 2 and an inert gas. Hence it has the highest ionisation potential.

(a) Nitrogen (N₂)

(b) Reddish-brown

Explanation

(a) When ammonia gas is passed over heated copper (II) oxide in a combustion tube nitrogen gas is evolved.

\mathrm{2NH_3 + 3CuO} \rightarrow \mathrm{3Cu + 3H_2O + N_2} \uparrow

(b) The residue is reddish-brown copper metal.

(a) \mathrm{(NH_4)_2CO_3 + 2HCl} \rightarrow \mathrm{2NH_4Cl + H_2O + CO_2}

(b) \mathrm{S + 6HNO_3} \xrightarrow{\mathrm{conc.,\ hot}} \mathrm{H_2SO_4 + 6NO_2 + 2H_2O}

(c) \mathrm{C + 2H_2SO_4} \xrightarrow{\mathrm{conc.}} \mathrm{CO_2 + 2H_2O + 2SO_2}

(a) In each test tube a white precipitate is formed.

(b) 1. In test tube C, the white precipitate of silver chloride dissolves in excess ammonium hydroxide, giving a clear solution.

2. In test tube D, the white precipitate of lead(II) chloride does not dissolve in excess ammonium hydroxide and remains unchanged.

Explanation :

(a) When dilute hydrochloric acid is added to silver nitrate (AgNO₃) solution in test tube C, silver chloride (AgCl) is formed, which is a white precipitate. Similarly, when dilute HCl is added to lead(II) nitrate [Pb(NO₃)₂] solution in test tube D, lead(II) chloride (PbCl₂) is formed, which is also a white precipitate. Therefore, a common observation is the formation of a white precipitate in both test tubes.

(b) In test tube C, the white precipitate of silver chloride dissolves in excess ammonium hydroxide whereas, in test tube D, the white precipitate of lead chloride does not dissolve in excess ammonium hydroxide.

(a) Amphoteric oxides

(b) Copper(II) oxide is placed incorrectly. It is a basic oxide, but it is shown under acidic categories.

(c) Sodium oxide (Na₂O) is the oxide that will form an alkali when dissolved in water:

\mathrm{Na_2O + H_2O \rightarrow 2NaOH}

(a) C

(b) D and F

(c) A and B

Explanation:

(a) C reacts with bromine to give a single product 1,2-dibromoethane.

\mathrm{C_2H_4 + Br_2 \xrightarrow{CCl_4\, inert\, solvent} C_2H_4Br_2}

(b) D is straight-chain butane and F is branched isobutane or 2-methylpropane. They have the same molecular formula C₄H₁₀ and are structural isomers.

(c) A, CH₃COOH, and B, CH₃CH₂OH, undergo esterification in the presence of concentrated H₂SO₄ to give ethyl ethanoate, which has a fruity smell, and water.

Simplest ratio C : H : O = \frac{3}{2}:4:1=3:8:2

Empirical formula = C₃H₈O₂

Empirical formula weight = (3 × 12) + (8 × 1) + (2 × 16) = 76

Relative molecular mass = 76

Molecular formula = C₃H₈O₂

(a) The acid P is concentrated nitric acid (HNO₃).

(b) \mathrm{Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2H_2O + 2NO_2}

(a) Bayer’s process is used to concentrate bauxite ore to alumina.

(b) \mathrm{2Al(OH)_3 \xrightarrow{1000^\circ C} Al_2O_3 + 3H_2O}

(a) When Mg reacts with dilute HCl, MgCl₂ is formed.

(b) The covalent compound formed when H₂ reacts with Cl₂ is HCl.

(c) The positive ion produced when ammonia gas is dissolved in water is the ammonium ion.

(a) Water in pure state consists almost entirely of molecules. It is a polar covalent compound and can form ions when traces of dilute sulphuric acid is added.

(b) \mathrm{4H^+ + 4e^- \rightarrow 4H}

\mathrm{2H + 2H \rightarrow 2H_2}

(c) At the anode hydroxide ions are preferentially discharged, producing oxygen gas.

\mathrm{4OH^- – 4e^- \rightarrow 4OH};

\mathrm{4OH \rightarrow 2H_2O + O_2}

(a) Avogadro’s Law — Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(b) Co-ordinate bond — The bond formed between two atoms by sharing a pair of electrons, provided entirely by one of the combining atoms but shared by both is called a coordinate bond.

(a) When ammonium chloride is reacted with an alkali like NaOH, the colourless gas ammonia having a sharp pungent characteristic smell is evolved. Whereas, sodium chloride (NaCl) does not show any reaction with alkali.

\mathrm{NH_4Cl + NaOH \rightarrow NH_3 \uparrow + NaCl + H_2O}

\mathrm{NaCl + NaOH \rightarrow No\ reaction}

(b) Add sodium hydroxide (NaOH) solution first drop by drop and then in excess.

Zinc Nitrate solution: A white gelatinous precipitate of zinc hydroxide forms, which dissolves in excess NaOH to give a clear solution.

Calcium Nitrate solution: A white precipitate of calcium hydroxide forms and remains insoluble even in excess NaOH.

(a) SO₂

Explanation – Sulphur dioxide is a colourless gas. It is a reducing agent and turns acidified potassium dichromate solution from orange to green.

(b) NCl₃

Explanation – Nitrogen trichloride is a yellow explosive oily liquid formed when excess chlorine reacts with ammonia.

\mathrm{NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl}

(c) PbO

Explanation – Lead(II) oxide is yellow and is produced when lead carbonate (PbCO₃) undergoes thermal decomposition.

\mathrm{PbCO_3 \xrightarrow{\Delta} PbO + CO_2}

(a) PbCl₂ — R: Precipitation

(b) FeCl₃ — S: Direct combination

(c) Na₂SO₄ — Q: Neutralisation by titration