ICSE Class 10 Chemistry — 2025 Paper

Sulphur dioxide

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

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

C₅H₁₂

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

Copper

Reason — 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.

C₃H₈

Reason

Given,

Vapour density of organic compound = 22

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

Compound	Molecular mass
1. CH4	16
2. C2H4	28
3. C2H6	30
4. C3H8	44

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

Oxidising agent

Reason — 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.

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

Reason — 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.

Haematite

Reason — 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.

4

2, 6

2, 8, 7

Reason

• 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.

CH₄, C₂H₆, C₃H₈

Reason — 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.

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

Reason — 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.

Cl^- & K⁺

Reason

• Chlorine has atomic number 17. Its neutral atom contains 17 electrons with the configuration 2, 8, 7. When it gains one electron to form Cl⁻, it has 18 electrons, giving the configuration 2, 8, 8 (the same as argon).
• Potassium has atomic number 19. A neutral K atom has the configuration 2, 8, 8, 1. When it loses one electron to form K⁺, it is left with 18 electrons, also giving the configuration 2, 8, 8.

Because both Cl⁻ and K⁺ possess 18 electrons and share the electron configuration 2, 8, 8, they form the required pair with identical electronic configurations.

Sodium sulphate + Lead nitrate

Reason — 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 precipitates out directly and the by-product (sodium nitrate) remains in solution, this pair of reactants is the most convenient for preparing pure lead(II) sulphate.

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

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

Z

Reason — 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.

40 cm³

Reason — From the equation :

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O

1 volume of CH₄ reacts with 2 volumes of O₂

For 40 cm³ of CH₄ the required O₂ is 40 cm³ x 2 = 80 cm³

Since only 60 cm³ of O₂ is available.

The amount of CH₄ that reacts with 60 cm³ of O₂ is = $\frac{60}{2}$ = 30 cm³

∴ The excess CH₄ remaining is 40 cm³ - 30 cm³ = 10 cm³

From the balanced equation, 2 volumes of O₂ produce 1 volume of CO₂.

∴ The volume of CO₂ produced from 60 cm of O₂ is $\frac{60}{2}$ = 30 cm³

The remaining gases are excess CH₄ and produced CO₂.

∴ Total volume = 10 cm³ (excess CH₄) + 30 cm³ (CO₂) = 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

$$\begin{aligned}\overset{\phantom{4}{4}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{Y}} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{4}{4}}{\text{Y}} \\[1em] \text{X}_2\text{Y}_4\end{aligned}$$

∴ Simplified formula of the compound is XY₂

(a) Oxidation is the loss of electrons. In an electrolytic cell the anode is the positive electrode that attracts anions. These anions give up electrons to the anode, so oxidation takes place there. Because the anode accepts electrons and thereby causes the ions to be oxidised, it is called the oxidising electrode.

(b) As graphite is unaffected by the reactive bromine vapours released at the anode hence, a graphite anode is preferred to other inert electrodes like platinum during the electrolysis of molten lead bromide.

$$\begin{matrix} \text{Mg} & + & 2\text{H}_2 \text{SO}_4 & \longrightarrow & \text{MgSO}_4 & + & 2\text{H}_2\text{O} & + & \text{SO}_2 \\ 1 \text{ mole} && 2 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 24 && 2[2(1) + 32 + 4(16)] & & 24 + 32 + 4(16) & & 2[2(1) + 16] && 32 + 2(16) \\ = 24 \text{ g} & & = 196 \text{ g} & & = 120 \text{ g} & & 36 \text{ g} & & = 64\text{ g} \\ \end{matrix}$$

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

∴ 60 g of Mg reacts with $\frac{196}{24}$ x 60 = 490 g of H₂SO₄

Hence, 490 g of H₂SO₄ is required.

(b) 24 g of Mg gives 22.4 L volume of SO₂

∴ 60 g of Mg gives $\frac{22.4}{24}$ x 60 = 56 L

Hence, the volume of sulphur dioxide gas liberated at S.T.P. is 56 L.

Thus, 490 g of sulphuric acid are needed and 56 L of SO₂ will be liberated at S.T.P.

(a) When barium chloride solution is added to a solution of Zinc sulphate, white ppt. of barium sulphate is obtained which is insoluble in dil. HCl or nitric acid.

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

(b) When lead nitrate is heated in a test tube, brown fumes of nitrogen dioxide (NO₂) are evolved and a yellow residue of lead(II) oxide remains.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

(c) The moist starch-iodide paper turns blue-black because chlorine liberates iodine from iodide.

Cl₂ + 2KI ⟶ 2KCl + I₂

(a ) V.D. = $\frac{\text{mass of certain vol. of gas under similar conditions}}{\text{mass of equal vol. of H}_2\text{ under similar conditions}}$ = $\frac{450}{150}$ = 3

(b) Molecular weight = 2 x V.D. = 2 x 3 = 6 a.m.u.

(a) CH₃COONa + NaOH $\xrightarrow{\text{CaO, heat}}$ CH₄ + Na₂CO₃

(b) 2CH₃COOH + Mg ⟶ (CH₃COO)₂Mg + H₂ ↑

(a) Nitrogen dioxide (NO₂)

(b) Nitrogen (N₂)

(c) Hydrogen sulphide (H₂S)

(a) Element G

Reason — 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

Reason — 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

Reason — 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.

$$\begin{aligned}\text{M}^{2+} \space \text{P}^{1-} \\[1em] \overset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{P}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{P}} \\[1em] \text{M}\text{P}_2\end{aligned}$$

(a) Cl

(b) Ne

Reason

(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

Reason

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

2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

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

​(a) (NH₄)₂CO₃ + 2HCl ⟶ 2NH₄Cl + H₂O + CO₂

(b) S + 6HNO₃(conc, hot) ⟶ H₂SO₄ + 6NO₂ + 2H₂O

(c) C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

(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.

Reason

(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:
Na₂O + H₂O ⟶ 2NaOH

(a) C

(b) D and F

(c) A and B

Reason

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

$\underset{\text{ ethene}}{\text{C}_2\text{H}_4} + \text{Br}_2 \xrightarrow[\text{[inert solvent]}]{\text{CCl}_4} \underset{\text{1,2, dibromooethane [ethylene bromide]}}{\text{C}_2\text{H}_4\text{Br}_2}$

(b) D is a straight-chain butane and F is a branched isobutane or 2-methylpropane, they have the same molecular formula C₄H₁₀ and are structural isomers of each other.

(c) A is a carboxylic acid (CH₃COOH) and B is an alcohol (CH₃CH₂OH). In presence of concentrated H₂SO₄ they undergo esterification to give ethyl ethanoate (ethyl acetate), which has a fruity smell.

$\underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \text{CH}_3\text{COOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{{\text{CH}_3\text{COOC}_2\text{H}_5}} + \text{H}_2\text{O}$

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	47.4	12	$\frac{47.4 }{12}$ = 3.95	$\frac{3.95}{2.63 }$ = $\frac{3 }{2}$
Hydrogen	10.5	1	$\frac{10.5}{1}$ = 10.5	$\frac{10.5}{2.63 }$ = 4
Oxygen	42.1	16	$\frac{42.1}{16}$ = 2.63	$\frac{2.63}{2.63 }$ = 1

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

Hence, empirical formula is C₃H₈O₂

Empirical formula weight = (3 x 12) + (8 x 1) + (2 x 16)= 36 + 8 + 32 = 76

Empirical formula weight = Relative molecular mass = 76

So, molecular formula = C₃H₈O₂

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

(b) Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

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

(b)

$2\text{Al(OH)}_3 \xrightarrow{1000 ^{\circ}\text{C}} \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O}$

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

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

$$\begin{aligned}\overset{\phantom{2}{2}}{\text{H}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{Cl}} \Rightarrow \underset{\phantom{2}{2}}{\text{H}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{Cl}} \\[1em] \text{H}_2\text{Cl}_2 \Rightarrow \text{HCl}\end{aligned}$$

(c) The positive ion produced when ammonia gas is dissolved in water is 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) 4H⁺ + 4e^- ⟶ 4H
2H + 2H ⟶ 2H₂

(c) At the anode (positive electrode) hydroxide ions are preferentially discharged, producing oxygen gas:

4OH^- - 4e^- ⟶ 4OH
4OH ⟶ 2H₂O + O₂

(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.

NH₄Cl + NaOH ⟶ NH₃ ↑ + NaCl + H₂O

NaCl + NaOH ⟶ 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

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

(b) NCl₃

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

NH₃ + 3Cl₂ ​⟶ NCl₃ + 3HCl

(c) PbO

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

PbCO₃ ⟶ PbO + CO₂

(a) PbCl₂ — R: Precipitation

(b) FeCl₃ — S: Direct combination

(c) Na₂SO₄ — Q: Neutralisation by titration