ICSE Class 10 Chemistry (Science Paper – 2) Solved Paper 2026

(b) Carbon

Explanation –

Carbon (a non-metal) reacts with hot concentrated sulphuric acid to form two gases — carbon dioxide (CO₂) and sulphur dioxide (SO₂). Both these gases turn lime water milky.

\mathrm{C} + 2\mathrm{H_2SO_4 [conc.]} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O} + 2\mathrm{SO_2}

(c) R

Explanation –

An ionic bond is formed between a metal (low ionisation potential) and a non-metal (high electron affinity). Group 2 elements are metals which lose 2 electrons, while Group 17 elements are non-metals which gain 1 electron. Hence, the elements of Group 2 and Group 17 form ionic bond (e.g., MgCl₂, CaCl₂).

Pair P consists of two metals (no bond formation), Q consists of two non-metals (covalent bond), and S consists of a non-metal and a noble gas (no bond formed).

(b) 2

Explanation –

An element with electronic configuration 2, 8, 2 is magnesium (Mg) with valency 2. Its hydroxide is Mg(OH)₂, which contains 2 hydroxyl (OH⁻) ions per molecule.

\mathrm{Mg(OH)_2} \rightarrow \mathrm{Mg^{2+}} + 2\mathrm{OH^-}

(a) Copper metal is deposited at the negative electrode.

Explanation –

During electrolysis of copper(II) sulphate solution using copper electrodes:

At the cathode (negative electrode): \mathrm{Cu^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Cu}, so copper is deposited here and the cathode increases in mass.

At the anode (positive electrode): \mathrm{Cu} – 2\mathrm{e^-} \rightarrow \mathrm{Cu^{2+}}, so copper of the anode dissolves and the anode decreases in mass.

No oxygen gas is produced because the copper is preferentially oxidised over the OH⁻ ions.

(c) C₂H₆ + Cl₂ ⟶ C₂H₅Cl + HCl

Explanation –

In the presence of ultraviolet (UV) light, ethane (a saturated hydrocarbon) undergoes a substitution reaction with chlorine. One hydrogen atom of ethane is replaced by a chlorine atom forming chloroethane (C₂H₅Cl) and hydrogen chloride (HCl) is released.

\mathrm{C_2H_6 + Cl_2} \xrightarrow{\mathrm{UV\ light}} \mathrm{C_2H_5Cl + HCl}

(b) Y only

Explanation –

The reaction shown is cracking (thermal decomposition) of butane (X = C₄H₁₀), a saturated hydrocarbon, into propene (Y = C₃H₆) and methane (Z = CH₄).

X (butane, C₄H₁₀) — saturated (only single bonds).

Y (propene, C₃H₆) — unsaturated (contains one C=C double bond).

Z (methane, CH₄) — saturated (only single bonds).

Hence, only Y is unsaturated.

(d) Both 2 and 4

Explanation –

The reactivity of these metals with dilute HCl depends on their position in the activity series:

Copper lies below hydrogen in the activity series, so it does not react with dilute HCl. Hence, no gas is given off — observation (1) is incorrect.

Iron is above hydrogen in the activity series, hence it reacts with dilute HCl liberating hydrogen gas — observation (2) is correct.

\mathrm{Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow}

Magnesium is highly reactive and reacts vigorously with dilute HCl liberating hydrogen gas — observation (3) is incorrect.

\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2 \uparrow}

Zinc is above hydrogen in the activity series and reacts with dilute HCl liberating hydrogen gas — observation (4) is correct.

\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2 \uparrow}

Hence, observations 2 and 4 are correct.

(c) 12

Explanation –

An element with atomic number 12 has the electronic configuration 2, 8, 2. It loses its 2 valence electrons easily to attain a stable octet configuration (2, 8) and forms a divalent positive ion X²⁺. This element is magnesium (Mg).

\mathrm{Mg \rightarrow Mg^{2+} + 2e^-}

Whereas, atomic number 16 (Sulphur) gains 2 electrons to form S²⁻, atomic number 10 (Neon) is a noble gas which doesn’t form ions, and atomic number 14 (Silicon) shares electrons.

(c) 3

Explanation –

Copper lies below hydrogen in the activity series, so it does not displace hydrogen from dilute acids. Hence, copper salts cannot be prepared by the action of dilute acids on the metal copper.

However, copper salts can be prepared from copper bases like Cu(OH)₂, copper carbonate (CuCO₃), and copper sulphite by the action of dilute acid.

(b) Sodium chloride

Explanation: 

Sodium chloride (NaCl) is an ionic compound. Ionic compounds have very strong electrostatic forces of attraction between oppositely charged ions, requiring a large amount of energy to break, so they have very high melting points.

Whereas, methane, ammonia and ethanol are covalent compounds with weak intermolecular forces, so they have low melting points.

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

Explanation –

Dilute sulphuric acid is a stronger electrolyte because the presence of water causes greater ionisation, producing more mobile ions. Concentrated sulphuric acid contains comparatively fewer free ions, hence Assertion (A) is true.

The strength of an electrolyte depends on the concentration of mobile ions. In dilute sulphuric acid greater ionisation and ion mobility make it a better conductor of electricity, hence Reason (R) is true and is the correct explanation of (A).

\mathrm{H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}}

(b) 15 litres

Explanation –

According to Gay-Lussac’s law of gaseous volumes, when gases react they do so in volumes which bear a simple whole number ratio to one another, given temperature and pressure remain constant.

From the equation: 1 volume of C₃H₈ produces 3 volumes of CO₂.

∴ 5 litres of C₃H₈ will produce 5 \times 3 = 15\text{ litres of }\mathrm{CO_2}.

(c) propene

Explanation –

Propene is an alkene with molecular formula C₃H₆, which contains 3 carbon atoms and 6 hydrogen atoms with a C=C double bond, making it unsaturated. It follows the general formula of alkenes (C_nH_2n).

\mathrm{CH_2{=}CH{-}CH_3}\;\text{(propene)}

Whereas, propyne (C₃H₄) has 4 hydrogens, propane (C₃H₈) is saturated, and propanol (C₃H₈O) is an alcohol containing oxygen.

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

Explanation –

During the electrolysis of acidified water:

At cathode: 4\mathrm{H^+} + 4\mathrm{e^-} \rightarrow 2\mathrm{H_2}

At anode: 4\mathrm{OH^-} – 4\mathrm{e^-} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}

The ratio of H₂ to O₂ liberated is 2:1, so the volume of hydrogen liberated is twice the volume of oxygen formed. Hence, assertion (A) is true.

However, on electrolysis water gives hydrogen and oxygen in the ratio of 2:1 by volume, and not 1:2 as stated. Hence, reason (R) is false.

(a) 1

Explanation –

The negative electrode is the cathode where reduction takes place. At the cathode, the positively charged Al³⁺ ions get attracted, gain 3 electrons, and get deposited as aluminium metal.

\mathrm{Al^{3+} + 3e^- \rightarrow Al}

Reaction (2) takes place at the anode (oxidation), reaction (3) is the reaction of carbon anode with oxygen, and reaction (4) is the dissociation of alumina.

(a) Haber’s process.

(b) Finely divided iron.

(c) The mixture of ammonia and unreacted gases (N₂ and H₂) coming out of the catalytic chamber is passed through the cooling pipes (condenser). Ammonia liquefies first, whereas nitrogen and hydrogen do not liquefy easily. The ammonia produced is stored as liquid under pressure.

(d) The two properties of ammonia gas demonstrated by the Fountain Experiment are:

Ammonia gas is highly soluble in water.

Ammonia is a basic gas (its aqueous solution turns red litmus blue / phenolphthalein turns pink).

(a) Calcium carbide (CaC₂)

(b) Ethane (C₂H₆)

(c) Ferric hydroxide [Fe(OH)₃]

(d) Lone pair of electrons

(e) Gram molecular mass

(a) The oxide that dissolves in caustic potash (KOH) is ZnO.

Explanation : ZnO is amphoteric in nature, so it dissolves in both acids and alkalis like KOH.

ZnO + 2KOH ⟶ K₂ZnO₂ + H₂O

(b) The reaction that takes place at the anode during the electrolysis of molten lead bromide is 2Br⁻ − 2e⁻ ⟶ Br₂.

(c) The volume occupied by 8 grams of oxygen gas at STP is 5.6 litres.

Explanation :

Molar mass of O₂ = 32 g

32 g of O₂ at STP occupies 22.4 L

∴ 8 g of O₂ occupies {22.4\over32} \times 8 = 5.6 L

(d) Cu does not give hydrogen gas when it reacts with cold and very dilute nitric acid.

Explanation : Copper lies below hydrogen in the activity series, so it does not liberate hydrogen with nitric acid.

(e) HCl is a polar covalent compound.

Explanation : HCl is formed by sharing of electrons between H and Cl, but due to large electronegativity difference, the bonding electrons are unequally shared, giving the molecule a polar character. CCl₄ is a non-polar covalent compound due to symmetric structure.

(a) IUPAC names of the given compounds:

1. Butan-2-ol

2. 3-methylpentane

(b) Structural diagrams:

1. butanal

2. Pent-2-yne (CH₃—C≡C—CH₂—CH₃)

3. The isomer of n-butane is 2-methylpropane (iso-butane).

(a) Anion — Chloride (Cl⁻)

(b) Cation — Copper(II) (Cu²⁺)

Explanation –

(a) The white precipitate formed with silver nitrate solution which is soluble in ammonium hydroxide is silver chloride (AgCl). This indicates the presence of chloride ion (Cl⁻) as the anion.

AgNO₃ + Cl⁻ ⟶ AgCl ↓ [white ppt.] + NO₃⁻

AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂]Cl + 2H₂O

(b) The pale blue precipitate formed with sodium hydroxide solution is copper(II) hydroxide [Cu(OH)₂]. This indicates the presence of copper(II) ion (Cu²⁺) as the cation.

Cu²⁺ + 2NaOH ⟶ Cu(OH)₂ ↓ [pale blue ppt.] + 2Na⁺

Hence, the salt Y is copper(II) chloride (CuCl₂).

(a) The temperature is maintained below 200°C because nitric acid decomposes at higher temperature to give nitrogen dioxide, water and oxygen.

4HNO₃ ⟶ 4NO₂ + 2H₂O + O₂

(b) Concentrated sulphuric acid is used because it is a non-volatile acid with a high boiling point. It displaces the more volatile nitric acid from sodium nitrate and does not distil over with nitric acid vapours.

(a) Element A is Nitrogen (N) with electronic configuration 2, 5.

(b) Nitrogen has a valency of 3 and combines with hydrogen (valency 1) to form ammonia (NH₃).

(c) Dot (•) and cross (×) structure of NH₃:

(a) C₂H₄Br₂ + 2KOH (alcoholic) ⟶ C₂H₂ + 2KBr + 2H₂O

(b) 8NH₃ (excess) + 3Cl₂ ⟶ N₂ + 6NH₄Cl

(c) \mathrm{C_12H_22O_11} \xrightarrow{\mathrm{conc. H_2SO_4}} \mathrm{12C + 11H_2O}

(a)

(b)

According to Avogadro’s Law, equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

Hence, the number of molecules of chlorine gas = number of molecules of ammonia gas.

Molar mass of NH₃ = 14 + 3(1) = 17 g

Number of moles of NH₃ = \frac{\mathrm{Mass}}{\mathrm{Molar mass}}

= \frac{34}{17}=2

Number of molecules in 1 mole = 6.022 × 10²³

∴ Number of molecules in 2 moles of NH₃ = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules

Hence, the number of molecules in chlorine gas = 1.2044 × 10²⁴ molecules.

P loses one valence electron, so P is a Group 1 metal (most electropositive). R has high electronegativity, so R is a non-metal on the right side of the periodic table. Q shares electrons (covalent), so Q is in Group 14 (between P and R).

(a) P would be most difficult to reduce.

Explanation : Reduction means gain of electrons. Element P readily loses its valence electron and is the most electropositive (strongest reducing agent itself). Hence, its tendency to gain electrons (be reduced) is the lowest, so it is most difficult to reduce.

(b) R has the smallest atomic radius.

Explanation : Across a period from left to right, the atomic radius decreases due to increasing nuclear charge. R is on the rightmost side, so it has the smallest atomic radius.

(c) Decreasing ionisation potential: R > Q > P

Explanation : Across a period, ionisation potential increases from left to right because atomic size decreases and the effective nuclear charge increases.

(a) Pb(NO₃)₂ + Na₂CO₃ ⟶ PbCO₃ ↓ + 2NaNO₃

(b) CuCO₃ + 2HCl ⟶ CuCl₂ + H₂O + CO₂ ↑

(c) Fe + 2HCl ⟶ FeCl₂ + H₂ ↑

(a) The type of reaction that occurred is addition.

(b) Nickel (Ni)

(c) \mathrm{C_2H_2 + 2H_2} \xrightarrow{\mathrm{Ni}} \mathrm{C_2H_6}

(a) Sulphur trioxide

(b) Sulphuric acid (H₂SO₄)

(c) Sulphur trioxide (SO₃) does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid which are difficult to condense and escape into the atmosphere. Hence, it is not directly dissolved in water.

(a) Ag⁺ + e⁻ ⟶ Ag

(b) Silver nitrate solution is not preferred because it ionises very readily releasing a high concentration of Ag⁺ ions. This causes silver to be deposited too rapidly on the spoon, producing a non-uniform, loose, and rough coating with poor adhesion. Sodium argentocyanide, on the other hand, releases Ag⁺ ions slowly, producing a smooth, uniform, and shiny coating.

(c) Alternating current (AC) is not used because in AC, the polarity of electrodes keeps changing. As a result, silver deposited on the spoon at one moment would dissolve back into the electrolyte at the next moment. Hence, direct current (DC) is used to ensure uni-directional flow of ions and continuous deposition of silver on the spoon.

(d) The silver anode gradually dissolves into the electrolyte as silver ions:

Ag − e⁻ ⟶ Ag⁺

The mass of the anode decreases during the process.

(a) Hydrochloric acid (HCl) is a monobasic acid, i.e., it has only one replaceable hydrogen ion per molecule. Acid salts are formed by the partial replacement of replaceable hydrogen ions of an acid by a metal, which requires the acid to be at least dibasic (with two or more replaceable H⁺). Since HCl has only one replaceable hydrogen, it can only form normal salts and not acid salts.

(b) On moving across a period from left to right, the nuclear charge increases while the number of shells (energy levels) remains the same. As a result, the atomic radius decreases and the effective pull of the nucleus on the bonding electrons increases. Hence, the tendency of an atom to attract the shared pair of electrons (electronegativity) increases across a period.

(a) False

Explanation – Isomers are organic compounds having the same molecular formula but different structural formulae.

(b) True

Explanation – A salt is indeed formed when the hydrogen ion of an acid is partially or completely replaced by a metal or electropositive radical (such as the ammonium ion).

(a) The error made by Anil was that he stored the nitric acid in a transparent container exposed to sunlight. Nitric acid should be stored in a dark-coloured (brown) reagent bottle kept in a cool and dark place to prevent decomposition.

(b) Nitric acid is unstable and decomposes on exposure to light and heat to release reddish-brown nitrogen dioxide (NO₂) gas, water and oxygen:

\mathrm{4HNO_3} \xrightarrow{light/heat} \mathrm{4NO_2} \uparrow + \mathrm{2H_2O} + \mathrm{O_2} \uparrow

The brown fumes seen in the container are due to NO₂ gas.

(c) The yellowish tinge in nitric acid (caused by dissolved NO₂) can be removed by bubbling air or carbon dioxide gas through the acid. This drives out the dissolved nitrogen dioxide and the acid becomes colourless again.

Zn + 4HNO₃ ⟶ Zn(NO₃)₂ + 2H₂O + 2NO₂

(a) Number of moles of Zn = \frac{\mathrm{Mass}}{\mathrm{Molar\;mass}}=\frac{32.5}{65}=0.5 mole

(b) From the equation, 65 g of Zn reacts with 252 g of HNO₃.

∴ 32.5 g of Zn reacts with \frac{252}{65}\times 32.5=126 g of HNO₃.

(c) From the equation, 65 g of Zn produces 2 × 22.4 = 44.8 L of NO₂ at STP.

∴ 32.5 g of Zn produces \frac{44.8}{65}\times 32.5=22.4 L of NO₂ at STP.

Hence, 0.5 mole of zinc is required, 126 g of nitric acid is needed, and 22.4 L of NO₂ is liberated at STP.

(a) Concentrated Nitric Acid (HNO₃)

(b) C + 4HNO₃ [conc.] ⟶ CO₂ + 2H₂O + 4NO₂

(a) Bauxite contains alumina along with impurities such as ferric oxide and sand. Alumina is amphoteric and dissolves in hot concentrated caustic soda to form soluble sodium meta-aluminate, while the insoluble impurities remain as red mud and are removed by filtration.

(b) Al₂O₃·2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

(a) A white precipitate of calcium hydroxide is formed, which is insoluble in excess sodium hydroxide.

Ca(NO₃)₂ + 2NaOH ⟶ Ca(OH)₂ ↓ [white ppt.] + 2NaNO₃

(b) A colourless gas with the smell of rotten eggs, hydrogen sulphide (H₂S), is evolved with effervescence.

FeS + 2HCl ⟶ FeCl₂ + H₂S ↑

(c) The amphoteric metal (e.g., zinc or aluminium) dissolves in the hot concentrated alkali showing vigorous effervescence (bubbling) due to the evolution of a colourless, odourless gas. This gas burns with a “pop” sound when a burning splinter is brought near the mouth of the test tube, confirming it is hydrogen.

Zn + 2NaOH ⟶ Na₂ZnO₂ (sodium zincate) + H₂ ↑

A → Concentrated Sulphuric Acid (conc. H₂SO₄)

B → Acetic acid (CH₃COOH)

C → Oxygen (O₂)

Explanation –

Ethanol ⟶ Ethene (with reactant A): This is dehydration of ethanol. A is concentrated sulphuric acid (conc. H₂SO₄) at 170°C.

\mathrm{C_2H_5OH} \xrightarrow[\;170^\circ\mathrm{C}\;]{\mathrm{Conc.\ H_2SO_4}} \mathrm{C_2H_4} + \mathrm{H_2O}

Ethanol ⟶ Ethyl acetate (with reactant B): This is esterification. B is acetic acid (CH₃COOH) in the presence of conc. H₂SO₄ (acts as a catalyst).

\mathrm{C_2H_5OH} + \mathrm{CH_3COOH} \xrightarrow{\mathrm{Conc.\ H_2SO_4}} \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}

Ethanol ⟶ Carbon dioxide + Water (with reactant C): This is combustion. C is oxygen (O₂).

C₂H₅OH + 3O₂ ⟶ 2CO₂ + 3H₂O

Hence, A = Concentrated H₂SO₄, B = Acetic acid (CH₃COOH), and C = Oxygen (O₂).

At STP, 22.4 L of any gas = 1 mole

∴ Number of moles of Z = \frac{5.6}{22.4}=0.25 mole

Molecular weight of Z = \frac{\mathrm{Mass}}{\mathrm{Number\;of\;moles}}=\frac{32}{0.25}=128 g/mol

Vapour density = \frac{\mathrm{Molecular\;weight}}{2}=\frac{128}{2}=64

Hence, the molecular weight of Z is 128 and its vapour density is 64.

(a) Fluorine (F)

(b) Sodium (Na)

(a) Fluorine (F)

(b) Calcium (Ca)

(c) Potassium (K)

Explanation –

(a) Fluorine (F) has the least atomic radius. It is on the rightmost side of period 2, where atomic radius is minimum.

(b) Calcium (Ca) has two valence electrons (electronic configuration: 2, 8, 8, 2). It belongs to Group 2 of the periodic table.

(c) Potassium (K) is the most electropositive. Within Group 1, electropositive character increases down the group; potassium is below lithium and hence is more electropositive.

(a) L — Sodium chloride is a strong electrolyte and ionises completely in water, dissociating fully into Na⁺ and Cl⁻ ions.

NaCl ⟶ Na⁺ + Cl⁻

(b) M — Distilled water consists almost entirely of H₂O molecules and contains no ions (its self-ionisation is negligible).

(c) K — Aqueous acetic acid is acidic in nature (it is a weak acid which partially ionises to give H⁺ ions), and hence has a pH less than 7.

CH₃COOH ⇌ CH₃COO⁻ + H⁺