Solved 2026 Improvement Paper ICSE Class 10 Chemistry

Acid + Metal oxide on reaction gives:

1. base + water
2. salt + water
3. base + salt
4. metal + salt

Answer

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

A polar covalent compound with two lone pairs of electrons is:

1. ammonia
2. water
3. chlorine
4. oxygen

Answer

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.

What do you observe when ammonium hydroxide solution is added first dropwise and then in excess to a solution of copper sulphate?

1. A pale blue precipitate that is insoluble in excess of ammonium hydroxide is seen.
2. A curdy white precipitate that is soluble in excess of ammonium hydroxide is seen.
3. A pale blue precipitate is formed that dissolves in excess of ammonium hydroxide to form a deep blue solution.
4. A curdy white precipitate that is insoluble in excess of ammonium hydroxide is seen.

Answer

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

Identify the statement(s) that is/are NOT true for Group 1 elements:

1. They are good conductors of heat.
2. They are good conductors of electricity.
3. They are strong oxidising agents.
4. They form covalent compounds with non-metals.

1. Only 1
2. Only 3
3. Both 3 and 4
4. Both 2 and 3

Answer

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.

The organic compound which undergoes substitution reaction is:

1. C₂H₆
2. C₂H₄
3. C₂H₂
4. C₁₀H₁₈

Answer

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.

What do you observe when concentrated hydrochloric acid is added to manganese dioxide and then heated?

1. A colourless gas is liberated that puts out a burning splinter with a 'pop' sound.
2. A reddish-brown gas is liberated that turns moist potassium iodide paper brown.
3. A colourless gas is liberated that turns lime water milky.
4. A greenish yellow gas is liberated that turns moist starch iodide paper blue-black.

Answer

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

Which of the following occupies the maximum volume at STP? [At. wt. C=12, O=16]

1. 3 moles of hydrogen
2. 1 mole of helium
3. 64 grams of oxygen
4. 44 grams of carbon dioxide

Answer

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 = $\dfrac{64}{32}$ = 2 moles = 44.8 L
• 44 g of carbon dioxide = $\dfrac{44}{44}$ = 1 mole = 22.4 L

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

The table below shows the atomic numbers of elements W, X, Y and Z.

Which of the above elements will combine to form an electrovalent compound?

1. Y and Z
2. C and Z
3. X and Y
4. X and W

Answer

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.

Assertion (A): Noble gases are unreactive under standard conditions.

Reason (R): Noble gases have complete valence shells making them stable and less likely to gain or lose electrons.

1. (A) is true but (R) is false.
2. (A) is false but (R) is true.
3. Both (A) and (R) are true and (R) is the correct explanation of (A).
4. Both (A) and (R) are true but (R) is not the correct explanation of (A).

Answer

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.

The process of dressing (purification) of the ore which involves separation of ore from gangue due to preferential wetting is:

1. Magnetic Separation
2. Hydrolytic method
3. Froth flotation method
4. Chemical method

Answer

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.

If an element A belongs to Period 3 and Group 2, then it will have:

1. 3 shells and 2 valence electrons
2. 2 shells and 3 valence electrons
3. 3 shells and 3 valence electrons
4. 2 shells and 2 valence electrons

Answer

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).

The vapour density of nitrogen dioxide [At. wt. N = 14, O = 16] is:

1. 15
2. 30
3. 46
4. 23

Answer

23

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

Metallic and non-metallic elements can both be extracted by electrolysis. Which of the following will undergo reduction at the electrode?

1. Bromine
2. Chlorine
3. Hydrogen
4. Oxygen

Answer

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 is in Group I of the Periodic Table and chlorine is in Group VII. Which statement correctly describes what happens when sodium bonds ionically with chlorine?

1. Sodium atom gains an electron to form Na^- and chlorine atom loses an electron to form Cl⁺
2. Sodium atom gains an electron to form Na⁺ and chlorine atom loses an electron to form Cl^-
3. Sodium atom loses an electron to form Na^- and chlorine atom gains an electron to form Cl⁺
4. Sodium atom loses an electron to form Na⁺ and chlorine atom gains an electron to form Cl^-

Answer

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.

Hydrogen chloride is soluble in water because it is a / an:

1. base
2. ionic compound
3. polar covalent compound
4. non-polar covalent compound

Answer

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.

The figure given below illustrates the apparatus used in the laboratory preparation of nitric acid.

(a) Name:

1. R (a liquid) and
2. S (a solid)

(b) Why should we use only glass apparatus in the laboratory preparation of nitric acid?

(c) Write an equation to show how nitric acid undergoes decomposition.

(d) What do you observe when copper is oxidized by concentrated nitric acid?

Answer

(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

Match Column 1 with Column 2 without repeating the options from Column 2.

Answer

Name the following:

(a) A salt which is soluble in hot water, but insoluble in cold water.

(b) An alkali in which both copper hydroxide and zinc hydroxide can be dissolved.

(c) The cation that forms a reddish brown precipitate when its salt solution reacts with a strong alkali.

(d) The law which states that 'equal volumes of all gases, at the same temperature and pressure, have the same number of molecules'.

(e) Two metal wires/plates through which electricity enters and leaves the electrolytic cell during electrolysis.

Answer

(a) Lead(II) chloride (PbCl₂)

(b) Ammonium hydroxide (NH₄OH)

(c) Ferric ion / Iron(III) ion (Fe³⁺)

(d) Avogadro's Law

(e) Electrodes

Complete the following sentences by choosing the correct word(s) from the brackets.

(a) The IUPAC name for formaldehyde is ............... (methanal / ethanal)

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

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

(d) The salt formed when iron reacts with dilute HCl is ............... (FeCl₃ / FeCl₂)

(e) Ammonia reacts with excess chlorine to form ............... (NCl₃ / NH₄Cl)

Answer

(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) Draw the structural diagram for the following organic compounds.

1. ethanal
2. acetic acid
3. 2-bromo propane

(b) Give IUPAC name for the following organic compounds.

Answer

(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)

Elements A, B, C & D have the following electronic configurations:

[A = 2, 8, 7    B = 1    C = 2, 5    D = 2, 8, 1]

Answer the following questions:

(a) Identify the element that will form an alkali when reacted with water.

(b) State the formula of the compound formed when D reacts with C.

Answer

(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.

The reactions of four different oxides W, X, Y and Z are given below:

• W reacts with hydrochloric acid, but not with sodium hydroxide.
• X reacts with both hydrochloric acid and sodium hydroxide.
• Y does not react with either hydrochloric acid or sodium hydroxide.
• Z reacts with sodium hydroxide, but not with hydrochloric acid.

From the information provided above, identify the oxide which is:

(a) amphoteric

(b) acidic

Answer

(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.

20.2 g of potassium nitrate is decomposed by heating according to the equation given below:

2KNO₃ ⟶ 2KNO₂ + O₂

Calculate the following:

(a) Volume of oxygen gas obtained at STP.

(b) Mass of KNO₂ formed.

[At. wt. of K = 39, N=14, O=16]

Answer

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

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

∴ 20.2 g of KNO₃ gives $\dfrac{22.4}{202}$ × 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 $\dfrac{170}{202}$ × 20.2 = 17 g of KNO₂.

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

A student wanted to electroplate a spoon with silver. He tried four different set-ups to electroplate the spoon as depicted in the diagrams given below.

Answer the following questions based on the above diagrams:

(a) Identify the set-up made by the student which would result in the uniform and smooth electroplating of the spoon with silver.

(b) In which set up will the coating on the spoon be uneven?

(c) Write the reaction taking place at the anode for the correct set-up.

Answer

(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⁺

Select the compound / element from the box given below for each of the following statements:

(a) The colourless gas obtained when a metallic nitrate undergoes thermal decomposition.

(b) The gas which on oxidation produces a coloured gas.

Answer

(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

Distinguish between the following:

(a) Hydronium ion and ammonium ion based on their lone pair of electrons.

(b) Solutions of calcium chloride and calcium nitrate using aqueous silver nitrate.

Answer

(a)

(b)

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

You are given the following molecular formulae of some hydrocarbons:

C₅H₈   C₅H₁₀   C₈H₁₄   C₇H₁₄   C₆H₆   C₆H₁₂

(a) Which two compounds represent unsaturated hydrocarbons having triple bonds?

(b) Identify the formula which represents a cyclic compound.

Answer

(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.

Choose the answer from the list of compounds given below:

sodium bisulphite, ferric oxide, iron (II) chloride, sodium sulphate, aluminium oxide

(a) A compound which is a major constituent of bauxite.

(b) The compound which will form a dirty green precipitate with NaOH.

(c) An acid salt.

Answer

(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.

The compound ‘A’ has the following percentage composition by mass, carbon 26.7%, oxygen 71.1%, hydrogen 2.2%. Determine the empirical formula of ‘A’. If the relative molecular mass of ‘A’ is 90, what is the molecular formula of ‘A’?

[At. wt. C=12, H=1, O=16]

Answer

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

Hence, empirical formula is CHO₂.

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

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

Molecular formula = 2 × (CHO₂) = C₂H₂O₄

What would you say about the pH of a solution-

(a) in which the concentration of H⁺ ions is equal to concentration of OH^- ions?

(b) which evolves hydrogen sulphide when added to ferrous sulphide?

(c) which produces a pungent smelling gas when heated with an ammonium salt?

Answer

(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.

Answer the following questions based on the information provided for the elements P, Q and R.

• P : atomic number = 19
• Q : electronic configuration = 2, 6
• R : has a tendency to gain one electron

(a) Write the electronic configuration of element P.

(b) What is the valency of element Q?

(c) Is element R a metal or a non-metal?

(d) Which period does P belong to?

Answer

(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.

State true or false for the following statements:

(a) Brass is an alloy which contains a non-metal as one of its constituents.

(b) Duralumin contains copper.

(c) Vapour density of gas is twice its molecular mass.

Answer

(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.

Deepak spilled some concentrated nitric acid accidentally in the laboratory. To neutralize it, he sprinkled sodium carbonate powder on it.

(a) Write the equation for the reaction that took place when he sprinkled sodium carbonate on nitric acid.

(b) Name the gas evolved in the reaction above.

(c) Write the equation for the reaction that would take place if Deepak had sprinkled carbon over concentrated nitric acid.

Answer

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

(b) Carbon dioxide (CO₂)

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

The following sketch illustrates the process of conversion of Alumina to Aluminum. Study the diagram and answer the following:

(a) Name the constituent of the electrolyte mixture which has a divalent metal in it.

(b) Write the reactions taking place at the electrodes:

1. Y (anode) and
2. Z (cathode) respectively.

(c) Give one use of sprinkling coke on the electrolyte.

Answer

(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.

Which property of concentrated H₂SO₄ is used in each of the following cases?

(a) For the production of HCl gas when it reacts with a metal chloride.

(b) For the conversion of blue copper sulphate crystals to white anhydrous copper sulphate.

Answer

(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.

Identify the anion present in the following compounds:

(a) Compound ‘C’ on heating with dilute hydrochloric acid liberates a gas which turns lime water milky but has no effect on potassium dichromate solution.

(b) Compound ‘D’ on heating with concentrated sulphuric acid liberates a gas which when passed through silver nitrate solution gave a white precipitate, insoluble in dilute nitric acid.

Answer

(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.

From the list given, choose the method which is most suitable for the preparation of the following salts?

• E - Neutralization by titration
• F - Precipitation
• G - Simple displacement
• H - Direct combination

(a) Zinc sulphate

(b) Silver chloride

(c) Ammonium nitrate

Answer

(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).

Metal X is low in the reactivity series and it is obtained by the electrolysis of its bromide.

Complete the following blanks by choosing the correct word from the options given in the brackets:

(a) Metal X is ............... (lead / sodium)

(b) The bromide is in the ............... state. (solution / molten)

(c) The sulphate of metal X will have the formula ............... (X₂SO₄ / XSO₄)

Answer

(a) Metal X is lead.

(b) The bromide is in the molten state.

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

Arrange the following as per the instructions given in the bracket.

(a) Na, Al, P, Si, Mg (increasing order of metallic character)

(b) Cl, F, Br, I (increasing order of electronegativity)

(c) He, Ar, Ne, Ca (increasing order of the number of shells)

Answer

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

(b) I, Br, Cl, F

(c) He, Ne, Ar, Ca

Study the part of the Periodic Table given below and answer the questions that follow: (Do not identify the elements)

(a) Draw the dot (.) and cross (×) structure of the compound formed between elements L and P.

(b) Which element has the highest ionization potential?

(c) Element N is more reactive than M. Justify.

Answer

(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.

The following diagram shows the laboratory preparation of a gaseous hydrocarbon (P) containing two carbon atoms.

(a) Identify the gaseous product P.

(b) Give the general formula of the homologous series which P belongs to.

(c) Write a balanced equation for the reaction that takes place.

(d) State the type of reaction that P undergoes when it reacts with hydrogen.

Answer

(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.