ICSE Class 10 Chemistry — 2026 Improvement

salt + water

Reason — A metal oxide is basic in nature. When it reacts with an acid, neutralisation takes place to give a salt and water only.

CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

water

Reason — In a water molecule (H₂O), oxygen forms two polar O—H covalent bonds and still carries two unshared (lone) pairs of electrons. The bent shape makes the molecule polar. Ammonia has only one lone pair, while chlorine (Cl₂) and oxygen (O₂) are non-polar molecules.

A pale blue precipitate is formed that dissolves in excess of ammonium hydroxide to form a deep blue solution.

Reason — On adding ammonium hydroxide dropwise, a pale blue precipitate of copper hydroxide is formed. In excess ammonium hydroxide, the precipitate dissolves to form a deep blue solution of tetraammine copper(II) sulphate.

CuSO₄ + 2NH₄OH ⟶ Cu(OH)₂ ↓ [pale blue] + (NH₄)₂SO₄

Both 3 and 4

Reason — Group 1 elements (alkali metals) are good conductors of heat and electricity, so statements 1 and 2 are true. They readily lose their single valence electron, so they are strong reducing agents (not oxidising agents); hence statement 3 is false. They form ionic (electrovalent) compounds, not covalent compounds, with non-metals; hence statement 4 is also false.

C₂H₆

Reason — C₂H₆ (ethane) is a saturated hydrocarbon (alkane) containing only single bonds, and saturated hydrocarbons undergo substitution reactions. C₂H₄, C₂H₂ and C₁₀H₁₈ are unsaturated and undergo addition reactions.

A greenish yellow gas is liberated that turns moist starch iodide paper blue-black.

Reason — Manganese dioxide oxidises concentrated hydrochloric acid on heating to liberate chlorine, a greenish-yellow gas. Chlorine liberates iodine from starch iodide paper, turning it blue-black.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

3 moles of hydrogen

Reason — At STP, 1 mole of any gas occupies 22.4 L.

• 3 moles of hydrogen = 3 × 22.4 = 67.2 L
• 1 mole of helium = 22.4 L
• 64 g of oxygen = $\frac{64}{32}$ = 2 moles = 44.8 L
• 44 g of carbon dioxide = $\frac{44}{44}$ = 1 mole = 22.4 L

Hence, 3 moles of hydrogen occupy the maximum volume (67.2 L).

X and Y

Reason — An electrovalent (ionic) compound is formed between a metal and a non-metal. X (atomic number 13, configuration 2, 8, 3) is a metal that loses 3 electrons to form X³⁺, and Y (atomic number 7, configuration 2, 5) is a non-metal that gains 3 electrons to form Y^3-. They combine to form an electrovalent compound (XY). Z (atomic number 10) is a noble gas and does not form compounds.

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

Reason — Noble gases have completely filled valence shells (a stable octet, or a duplet in the case of helium). Because of this stable configuration they have almost no tendency to gain or lose electrons, and so they are unreactive under standard conditions. Thus the reason correctly explains the assertion.

Froth flotation method

Reason — In the froth flotation method, the powdered ore is stirred with water and pine oil. The ore particles are preferentially wetted by the oil and rise up with the froth, while the gangue particles are wetted by water and settle down. This separation based on preferential wetting is the froth flotation method.

3 shells and 2 valence electrons

Reason — The period number gives the number of shells and, for Groups 1 and 2, the group number gives the number of valence electrons. Hence an element in Period 3 and Group 2 has 3 shells and 2 valence electrons (electronic configuration 2, 8, 2).

23

Reason — Vapour density = $\frac{\text{Molecular mass}}{2}$. Molecular mass of NO₂ = 14 + 2(16) = 46. So vapour density = $\frac{46}{2}$ = 23.

Hydrogen

Reason — Reduction (gain of electrons) takes place at the cathode. Hydrogen ions (H⁺) migrate to the cathode and gain electrons to be reduced to hydrogen gas. Bromine, chlorine and oxygen are formed by oxidation (loss of electrons) at the anode.

2H⁺ + 2e^- ⟶ H₂

Sodium atom loses an electron to form Na⁺ and chlorine atom gains an electron to form Cl^-

Reason — Sodium has one valence electron which it loses to attain a stable octet, forming Na⁺. Chlorine has seven valence electrons and gains one electron to complete its octet, forming Cl^-. These oppositely charged ions are held together by an ionic bond.

polar covalent compound

Reason — HCl is a polar covalent compound formed by the unequal sharing of electrons between hydrogen and the more electronegative chlorine. Being polar, it dissolves readily in water and ionises to form H₃O⁺ and Cl^- ions.

(a)

1. R — Concentrated sulphuric acid (H₂SO₄)
2. S — Sodium nitrate (NaNO₃) [potassium nitrate, KNO₃, is also acceptable]

(b) Only glass apparatus is used because nitric acid vapours attack rubber and cork.

(c) 4HNO₃ $\xrightarrow{\text{heat / light}}$ 4NO₂ + 2H₂O + O₂

(d) Dense reddish-brown fumes of nitrogen dioxide are evolved and a blue solution of copper(II) nitrate is formed.

Cu + 4HNO₃ [conc.] ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Lead(II) chloride (PbCl₂)

(b) Ammonium hydroxide (NH₄OH)

(c) Ferric ion / Iron(III) ion (Fe³⁺)

(d) Avogadro's Law

(e) Electrodes

(a) The IUPAC name for formaldehyde is methanal.

(b) A black solid, copper oxide reacts with dilute sulphuric acid to form a blue-coloured solution.

(c) The drying agent not used for drying ammonia is P₂O₅.

(d) The salt formed when iron reacts with dilute HCl is FeCl₂.

(e) Ammonia reacts with excess chlorine to form NCl₃.

(a) Structural diagrams:

1. ethanal (CH₃CHO)

2. acetic acid (CH₃COOH)

3. 2-bromopropane (CH₃CHBrCH₃)

(b) IUPAC names:

1. 2-methylbutane
2. Prop-1-ene (propene)

(a) Element D (configuration 2, 8, 1) is an alkali metal (sodium). It reacts with water to form an alkali (sodium hydroxide).

(b) D has a valency of 1 (forms D⁺) and C has a valency of 3 (forms C^3-). Therefore the formula of the compound is D₃C.

(a) X — It reacts with both an acid (HCl) and a base (NaOH), which is the property of an amphoteric oxide.

(b) Z — It reacts with a base (NaOH) but not with an acid, which is the property of an acidic oxide.

$$\begin{matrix} 2\text{KNO}_3 & \longrightarrow & 2\text{KNO}_2 & + & \text{O}_2 \\ 2 \text{ moles} && 2 \text{ moles} && 1 \text{ mole} \\ 2[39 + 14 + 3(16)] && 2[39 + 14 + 2(16)] && 2(16) \\ = 202 \text{ g} && = 170 \text{ g} && = 32 \text{ g} \\ \end{matrix}$$

(a) From the equation, 202 g of KNO₃ gives 22.4 L of O₂ at STP.

∴ 20.2 g of KNO₃ gives $\frac{22.4}{202}$ × 20.2 = 2.24 L of O₂ at STP.

(b) From the equation, 202 g of KNO₃ gives 170 g of KNO₂.

∴ 20.2 g of KNO₃ gives $\frac{170}{202}$ × 20.2 = 17 g of KNO₂.

Hence, 2.24 L of oxygen is obtained at STP and 17 g of KNO₂ is formed.

(a) Set-up Q — It uses sodium argentocyanide as the electrolyte, which releases silver ions slowly and gives a uniform, smooth and adherent coating of silver on the spoon (cathode).

(b) Set-up P — It uses silver nitrate solution, which ionises readily and releases a high concentration of silver ions, so silver is deposited too quickly, giving an uneven and rough coating.

(c) At the anode, the silver plate dissolves into the electrolyte:

Ag - e^- ⟶ Ag⁺

(a) O₂ — Oxygen is a colourless gas evolved during the thermal decomposition of metallic nitrates.

(b) NO — Nitric oxide (colourless) on oxidation gives the reddish-brown gas nitrogen dioxide.

2NO + O₂ ⟶ 2NO₂

(a)

Hydronium ion (H3O+)	Ammonium ion (NH4+)
It has one lone pair of electrons (on the oxygen atom).	It has no lone pair of electrons; all four electron pairs are bonded.

(b)

Calcium chloride	Calcium nitrate
On adding silver nitrate solution, a white precipitate of silver chloride is formed.	On adding silver nitrate solution, no precipitate is formed (no visible change).

CaCl₂ + 2AgNO₃ ⟶ 2AgCl ↓ [white ppt.] + Ca(NO₃)₂

(a) C₅H₈ and C₈H₁₄ — Both follow the general formula of alkynes, C_nH_2n-2, which contain a carbon–carbon triple bond.

(b) C₆H₆ — It represents benzene, a cyclic (aromatic) hydrocarbon.

(a) Aluminium oxide — Bauxite is hydrated aluminium oxide, Al₂O₃·2H₂O.

(b) Iron(II) chloride — Fe²⁺ ions form a dirty green precipitate of iron(II) hydroxide with NaOH.

(c) Sodium bisulphite — Sodium hydrogen sulphite (NaHSO₃) still has a replaceable hydrogen ion, so it is an acid salt.

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	26.7	12	$\frac{26.7}{12}$ = 2.225	$\frac{2.225}{2.2}$ = 1
Hydrogen	2.2	1	$\frac{2.2}{1}$ = 2.2	$\frac{2.2}{2.2}$ = 1
Oxygen	71.1	16	$\frac{71.1}{16}$ = 4.44	$\frac{4.44}{2.2}$ = 2

Simplest ratio C : H : O = 1 : 1 : 2

Hence, empirical formula is CHO₂.

Empirical formula weight = 12 + 1 + 2(16) = 45

n = $\frac{\text{Relative molecular mass}}{\text{Empirical formula weight}}$ = $\frac{90}{45}$ = 2

Molecular formula = 2 × (CHO₂) = C₂H₂O₄

(a) The solution is neutral, so its pH = 7.

(b) The solution is an acid (an acid reacts with ferrous sulphide to give hydrogen sulphide), so its pH is less than 7.

(c) The solution is an alkali (an alkali on heating with an ammonium salt gives pungent-smelling ammonia gas), so its pH is greater than 7.

(a) Electronic configuration of P (atomic number 19) = 2, 8, 8, 1

(b) Q has 6 valence electrons and gains 2 electrons to complete its octet, so its valency is 2.

(c) Element R has a tendency to gain one electron, so it is a non-metal.

(d) P has 4 electron shells (2, 8, 8, 1), so it belongs to Period 4.

(a) False — Brass is an alloy of copper and zinc, both of which are metals; it does not contain a non-metal.

(b) True — Duralumin is an alloy of aluminium, copper, magnesium and manganese, so it contains copper.

(c) False — Vapour density is half the molecular mass (Molecular mass = 2 × Vapour density), not twice it.

(a) Na₂CO₃ + 2HNO₃ ⟶ 2NaNO₃ + H₂O + CO₂

(b) Carbon dioxide (CO₂)

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

(a) Fluorspar (calcium fluoride, CaF₂) — it contains the divalent metal calcium.

(b)

1. Y (anode): 2O^2- - 4e^- ⟶ O₂
2. Z (cathode): Al³⁺ + 3e^- ⟶ Al

(c) The powdered coke sprinkled over the electrolyte prevents the carbon anodes from burning in air.

(a) Non-volatile nature — being a non-volatile (less volatile) acid, it displaces the more volatile hydrogen chloride from a metal chloride.

(b) Dehydrating property — it removes the water of crystallisation from blue copper sulphate crystals, leaving behind white anhydrous copper sulphate.

(a) Carbonate ion (CO₃^2-) — the gas evolved is carbon dioxide, which turns lime water milky but has no effect on potassium dichromate.

(b) Chloride ion (Cl^-) — the gas evolved is hydrogen chloride, which gives a white precipitate of silver chloride (insoluble in dilute nitric acid) with silver nitrate.

(a) Zinc sulphate — G: Simple displacement (zinc, being above hydrogen, displaces hydrogen from dilute sulphuric acid).

(b) Silver chloride — F: Precipitation (silver chloride is an insoluble salt).

(c) Ammonium nitrate — E: Neutralization by titration (a soluble salt of the alkali ammonium hydroxide with nitric acid).

(a) Metal X is lead.

(b) The bromide is in the molten state.

(c) The sulphate of metal X will have the formula XSO₄.

(a) P, Si, Al, Mg, Na

(b) I, Br, Cl, F

(c) He, Ne, Ar, Ca

(a) L belongs to Group 2 (valency 2, forms L²⁺) and P belongs to Group 17 (valency 1, forms P^-). One atom of L transfers one electron each to two atoms of P, forming the electrovalent compound LP₂.

(b) Q — It is a noble gas placed at the top of Group 18 (rightmost and highest in the table), where the atomic size is smallest and the outermost electrons are most tightly held, giving it the highest ionization potential.

(c) Both M and N belong to Group 1. Going down the group, the atomic size increases and the outermost electron is held less tightly by the nucleus, so it is lost more easily. Since N lies below M in the group, N loses its valence electron more readily and is therefore more reactive than M.

(a) P is ethyne (acetylene), C₂H₂.

(b) The general formula of the homologous series (alkynes) is C_nH_2n-2.

(c) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(d) It undergoes an addition reaction with hydrogen.