Class 10 Chemistry Answer Key | ICSE Exam 2025-26 Board Term 1

Question 1(i). A non-metal which reacts with concentrated sulphuric acid to form two gases which turn lime water milky is:

• (a) Sulphur
• (b) Carbon
• (c) Oxygen
• (d) Nitrogen

Answer is as follow: (b) Carbon

Solution:

Step 1: Understanding the Chemical Reaction:
Concentrated sulphuric acid (H₂SO₄) acts as a powerful oxidizing agent. When it reacts with elemental carbon, it oxidizes the carbon to carbon dioxide while being reduced to sulphur dioxide.

Step 2: Identifying the Product Gases:
Balanced chemical equation: C + 2H₂SO₄(conc.) → CO₂ + 2SO₂ + 2H₂O
The gaseous products are Carbon dioxide (CO₂) and Sulphur dioxide (SO₂).

Step 3: Lime Water Test:
Both CO₂ and SO₂ react with lime water [Ca(OH)₂] to form white precipitates (CaCO₃ and CaSO₃), making the solution milky.

Step 4: Final Answer:
Carbon is the non-metal in this list that produces both gases simultaneously upon reaction with concentrated sulphuric acid.

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Question 1(iv). With respect to the electrolysis of copper(II) sulphate solution using copper electrodes, which statement is correct?

• (a) Copper metal is deposited at the negative electrode.
• (b) Oxygen gas is produced at the positive electrode.
• (c) The positive electrode increases in mass.
• (d) The negative electrode decreases in mass.

Answer is as follow: (a) Copper metal is deposited at the negative electrode.

Solution:

Step 1: Understanding the Electrolysis Setup:
With active copper electrodes, the anode (positive) dissolves and the cathode (negative) receives deposits.

Step 2: Reaction at the Cathode (Negative):
Copper ions (Cu²⁺) move toward the cathode, gain electrons, and deposit as solid copper:
Cu²⁺ + 2e⁻ → Cu(s)
This increases the mass of the negative electrode.

Step 3: Reaction at the Anode (Positive):
The copper anode loses electrons and dissolves into the solution:
Cu(s) → Cu²⁺ + 2e⁻
This decreases the mass of the positive electrode.

Step 4: Final Answer:
Statement (a) is correct for this electrolytic process.

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Question 1(iii). The electronic configuration of an element is 2,8,2. The hydroxide of this element can produce how many hydroxyl ions per molecule?

• (a) 3
• (b) 2
• (c) 1
• (d) 4

Answer is as follow: (b) 2

Solution:

Step 1: Identifying the Element:
Total electrons = 2 + 8 + 2 = 12. Element with atomic number 12 is Magnesium (Mg).

Step 2: Determining Valency:
Magnesium has 2 electrons in its outermost shell. It loses these to form Mg²⁺, a divalent cation.

Step 3: Writing the Hydroxide Formula:
Hydroxide (OH⁻) has valency 1. To balance Mg²⁺, two OH⁻ ions are needed. The chemical formula is Mg(OH)₂.

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Question 1(v). Identify the equation that shows the reaction of ethane with chlorine in the presence of ultraviolet light:

• (a) C₂H₆ + Cl₂ → C₂H₆Cl₂
• (b) C₂H₆ + Cl₂ → C₂H₄Cl₂ + H₂
• (c) C₂H₆ + Cl₂ → C₂H₅Cl + HCl
• (d) C₂H₆ + Cl₂ → 2CH₃Cl

Answer is as follow: (c) C₂H₆ + Cl₂ → C₂H₅Cl + HCl

Solution:

Step 1: Understanding the Reaction Type:
Ethane is a saturated hydrocarbon (alkane). Alkanes undergo substitution reactions with halogens where hydrogen atoms are replaced one by one.

Step 2: The Role of UV Light:
UV light provides energy to break the Cl−Cl bond, initiating the free-radical chain reaction.

Step 3: Writing the Products:
One hydrogen atom from ethane (C₂H₆) is replaced by one chlorine atom to form chloroethane (C₂H₅Cl). The displaced hydrogen bonds with the second chlorine atom to form HCl.

Step 4: Final Answer:
The balanced equation is C₂H₆ + Cl₂ → C₂H₅Cl + HCl.

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Question 1(viii). An atom of X forms an ion according to the equation X + 2e⁻ → X²⁻. The atomic number of the atom X is:

• (a) 16
• (b) 10
• (c) 12
• (d) 14

Answer is as follow: (a) 16

Solution:

Step 1: Understanding Ion Formation:
X → X²⁻ implies the atom X accepts 2 electrons to form an anion.

Step 2: Applying the Octet Rule:
Non-metals gain electrons to complete their valence shell (usually 8 electrons). An atom gaining 2 electrons must have had 6 valence electrons.

Step 3: Checking the Atomic Numbers:
• 16 (Sulphur): 2, 8, 6 → gains 2 electrons to become 2, 8, 8 → matches.
• 10 (Neon): 2, 8 → stable noble gas.
• 12 (Magnesium): 2, 8, 2 → usually loses 2 electrons to form Mg²⁺.
• 14 (Silicon): 2, 8, 4 → usually forms covalent bonds.

Step 4: Final Answer:
The atomic number is 16.

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Question 1(ix). The method which cannot be used for the preparation of copper salts is:

• (a) 1 (Action of acid on bases)
• (b) 2 (Action of acid on carbonates)
• (c) 3 (Action of acid on metals)
• (d) 4 (Action of acid on sulphites)

Answer is as follow: (c) 3

Solution:

Step 1: Understanding Salt Preparation:
Soluble salts can be made by reacting an acid with a metal, base, or carbonate. However, the “Metal + Acid” method only works for metals more reactive than hydrogen.

Step 2: Checking Copper’s Reactivity:
Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute acids like HCl or H₂SO₄.

Step 3: Evaluating Alternatives:
Copper salts can be prepared by reacting copper oxide (a base) or copper carbonate with dilute acids, as these are neutralization/decomposition reactions not requiring hydrogen displacement.

Step 4: Final Answer:
Method 3 (Action of dilute acid on metals) cannot be used for copper.

Question 1(x). The compound that has the highest melting point amongst the following is:

• (a) Methane
• (b) Sodium chloride
• (c) Ammonia
• (d) Ethanol

Answer is as follow: (b) Sodium chloride

Solution:

Step 1: Comparing Bond Types:
The melting point of a substance depends on the strength of the forces holding its particles together.

Step 2: Analyzing the Options:
• Methane, Ammonia, Ethanol: Covalent compounds with weak intermolecular forces (Van der Waals or hydrogen bonding), usually gases or liquids at room temperature with very low melting points.
• Sodium chloride (NaCl): Ionic compound forming a giant 3D lattice held together by strong electrostatic forces between ions.

Step 3: Comparing Strengths:
Energy required to break ionic bonds is much higher than that needed to overcome intermolecular forces in covalent molecules.

Step 4: Final Answer:
Sodium chloride has the highest melting point.

Question 1(xi). Assertion (A): Dilute sulphuric acid is a stronger electrolyte than concentrated sulphuric acid.
Reason (R): Dilute sulphuric acid has a higher concentration of mobile ions than concentrated sulphuric acid.

• (a) (A) is true but (R) is false.
• (b) (A) is false but (R) is true.
• (c) Both (A) and (R) are true and (R) is the correct explanation of (A).
• (d) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer is as follow: (c) Both (A) and (R) are true and (R) is the correct explanation of (A).

Solution:

Step 1: Understanding Electrolytes:
The strength of an electrolyte depends on its ability to conduct electricity, which is determined by the presence of free, mobile ions.

Step 2: Comparing Dilute and Concentrated Acid:
Concentrated H₂SO₄ is largely molecular. Dilution with water causes the acid to ionize into H⁺ and SO₄²⁻ ions.

Step 3: Linking Ionization to Conductivity:
Dilute acid has a higher number of dissociated, mobile ions than concentrated acid, so it conducts electricity more effectively.

Step 4: Final Answer:
Both statements are true; the higher concentration of ions in dilute H₂SO₄ explains why it is a stronger electrolyte.

Question 1(xii). What volume of carbon dioxide is produced at STP when 5 litres of propane is burnt completely according to the equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

• (a) 10 litres
• (b) 15 litres
• (c) 20 litres
• (d) 25 litres

Answer is as follow: (b) 15 litres

Solution:

Step 1: Understanding Gay-Lussac’s Law:
According to Gay-Lussac’s Law of Combining Volumes, gases react in simple whole-number ratios by volume under the same temperature and pressure.

Step 2: Analyzing the Stoichiometry:
From the balanced equation: 1 volume of C₃H₈ reacts with 5 volumes of O₂ to produce 3 volumes of CO₂ and 4 volumes of H₂O. Thus, 1 litre of propane produces 3 litres of CO₂.

Step 3: Calculating for 5 Litres:
Volume of CO₂ = 5 litres × 3 = 15 litres

Step 4: Final Answer:
The volume of CO₂ produced is 15 litres.

Question 1(xiii). An unsaturated hydrocarbon with three carbon atoms and six hydrogen atoms is:

• (a) propyne
• (b) propane
• (c) propene
• (d) propanol

Answer is as follow: (c) propene

Solution:

Step 1: Analyzing the Molecular Formula:
The compound has 3 carbon atoms and 6 hydrogen atoms → molecular formula C₃H₆.

Step 2: Matching with General Formulas:
• Alkanes: CₙH₂ₙ₊₂ → C₃H₈ → Propane
• Alkenes: CₙH₂ₙ → C₃H₆ → Propene
• Alkynes: CₙH₂ₙ₋₂ → C₃H₄ → Propyne

Step 3: Verifying Unsaturation:
Unsaturated hydrocarbons contain double or triple bonds. C₃H₆ (propene) has one C=C double bond, making it unsaturated.

Step 4: Final Answer:
The compound is propene.

Question 1(xiv). Assertion (A): In the electrolysis of acidified water, the volume of hydrogen liberated is twice the volume of oxygen formed.
Reason (R): Water has hydrogen and oxygen in the ratio of 1:2 by volume.

• (a) (A) is true but (R) is false.
• (b) (A) is false but (R) is true.
• (c) Both (A) and (R) are true and (R) is the correct explanation of (A).
• (d) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer is as follow: (a) (A) is true but (R) is false.

Solution:

Step 1: Understanding Water Electrolysis:
2H₂O(l) → 2H₂(g) + O₂(g)

Step 2: Evaluating the Assertion:
From the equation, 2 volumes of hydrogen are produced for every 1 volume of oxygen. Assertion (A) is true.

Step 3: Evaluating the Reason:
Water (H₂O) has hydrogen and oxygen in a volume ratio of 2:1 (and by mass 2:16 = 1:8). Reason (R) states 1:2, which is false.

Question 1(xv). The reaction that takes place at the negative electrode (cathode) in the electrolysis of molten aluminium oxide is:

• (1) Al³⁺ + 3e⁻ → Al
• (2) 2O²⁻ → O₂ + 4e⁻
• (3) C + O₂ → CO₂
• (4) Al₂O₃ → 2Al³⁺ + 3O²⁻

• (a) 1
• (b) 2
• (c) 3
• (d) 4

Answer is as follow: (a) 1

Solution:

Step 1: Understanding Electrode Polarity:
The negative electrode is the cathode. Cations (positively charged ions) are attracted to the cathode for reduction.

Step 2: Identifying the Cation:
In molten Al₂O₃, the cation is Al³⁺.

Step 3: Describing the Cathode Reaction:
Al³⁺ ions gain 3 electrons each to form neutral aluminium atoms:
Al³⁺ + 3e⁻ → Al

Step 4: Final Answer:
Reaction 1 is the correct cathode reaction.

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Question 1(iv). Complete the following sentences by choosing the correct option from the brackets:

• (a) The oxide that dissolves in caustic potash (KOH) is [ZnO / CuO]
• (b) The reaction that takes place at the anode during the electrolysis of molten lead bromide is [2Br⁻ → Br₂ + 2e⁻ / 2Br⁻ + 2e⁻ → 2Br⁻]
• (c) The volume occupied by 8 grams of oxygen gas at STP is ______ litres. [15 / 8.96] [Atomic weight of O = 16]
• (d) ______ does not give hydrogen gas when it reacts with cold and very dilute nitric acid. [Cu / Mn]
• (e) ______ is a polar covalent compound [HCl / CCl₄]

Answer is as follow:

(a) ZnO

(b) 2Br⁻ → Br₂ + 2e⁻

(c) 5.6 litres (calculation: Molecular weight O₂ = 32 g, 32 g occupies 22.4 L → 8 g occupies 22.4 × 8 / 32 = 5.6 L)

(d) Cu

(e) HCl

Solution:

(a) ZnO: Zinc oxide is amphoteric, reacts with acids and bases like KOH. CuO is basic and does not dissolve in KOH.

(b) 2Br⁻ → Br₂ + 2e⁻: At the anode, oxidation occurs; bromide ions lose electrons to form bromine gas.

(c) 5.6 L: Using molar volume at STP (22.4 L for 32 g of O₂), 8 g occupies 22.4 × 8 / 32 = 5.6 L.

(d) Cu: Copper is below hydrogen in the reactivity series, does not liberate H₂ with dilute HNO₃.

(e) HCl: HCl is polar due to large electronegativity difference; CCl₄ is non-polar due to symmetry canceling bond dipoles.

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Question 3(i). Alex was given a solution of an unknown salt Y for analysis. He performed the following tests:

• Added silver nitrate solution → white precipitate, soluble in ammonium hydroxide.
• Added sodium hydroxide solution → pale blue precipitate.

Identify:

• (a) the anion
• (b) the cation

Answer:

(a) Anion: Cl⁻ (Chloride ion)

(b) Cation: Cu²⁺ (Copper(II) ion)

Solution:

(a) Anion: Chloride ion reacts with AgNO₃ to form a white precipitate of AgCl, which dissolves in excess NH₄OH forming [Ag(NH₃)₂]⁺, confirming Cl⁻.

(b) Cation: Copper(II) ions react with NaOH to give pale blue Cu(OH)₂ precipitate, which is a characteristic test for Cu²⁺.

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Question 3(ii). Study the reaction and answer the questions:

N aNO₃ + H₂SO₄ <200°C → NaHSO₄ + HNO₃

• (a) Give one reason for maintaining the temperature below 200°C.
• (b) Why is concentrated sulphuric acid used in the above reaction?

Answer:

(a) Reason for temperature < 200°C:

• Prevents formation of a hard crust of Na₂SO₄ on the glass, which is difficult to remove.
• Prevents thermal decomposition of HNO₃.
• Saves fuel and avoids cracking the glass apparatus.

(b) Use of concentrated H₂SO₄:

• Concentrated H₂SO₄ is non-volatile and can displace the more volatile HNO₃ from sodium nitrate when heated.

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Question 3(iii). Element A belongs to group 15 and period 2 of the Periodic Table.

• (a) Identify the element A.
• (b) Write the formula of the compound formed when element A combines with hydrogen.
• (c) Draw the dot (•) and cross (×) structure of the compound formed in (b).

Answer:

(a) Element A: Nitrogen (N)

(b) Formula of compound with hydrogen: NH₃ (Ammonia)

(c) Electron Dot Structure:

Nitrogen atom (center) shares three of its five valence electrons with three Hydrogen atoms, leaving one lone pair.

    H
    |
H — N ••
    |
    H

Here, •• represents the lone pair on nitrogen, and each H shares one electron with N.

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Question 3(iv). Complete and balance the following equations:

• (a) C₂H₄Br₂ + 2KOH (alcoholic) → ?
• (b) 8NH₃ (excess) + 3Cl₂ → ?
• (c) C₁₂H₂₂O₁₁ + conc. H₂SO₄ → ?

Answer:

(a) Dehydrohalogenation of ethylene dibromide:

C₂H₄Br₂ + 2KOH(alc.) → C₂H₂ + 2KBr + 2H₂O
Product: Ethyne (Acetylene)

(b) Reaction of excess ammonia with chlorine:

8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
Products: Ammonium chloride (white fumes) and Nitrogen gas

(c) Dehydration of cane sugar by concentrated H₂SO₄:

C₁₂H₂₂O₁₁ + conc. H₂SO₄ → 12C + 11H₂O
Concentrated H₂SO₄ removes water, leaving a black spongy mass of sugar charcoal

Question 4(i). Differentiate between the following pairs based on the criteria given:

• (a) Acetic acid and Sulphuric acid (number of replaceable hydrogen ions per molecule)
• (b) Electrolyte and Metallic conductor (the particles conducting electricity)

Answer:

(a) Acetic acid vs. Sulphuric acid:

• Acetic acid (CH₃COOH): Monobasic → 1 replaceable hydrogen ion per molecule.
• Sulphuric acid (H₂SO₄): Dibasic → 2 replaceable hydrogen ions per molecule.

(b) Electrolyte vs. Metallic conductor:

• Electrolyte: Conducts electricity through mobile ions (cations and anions).
• Metallic conductor: Conducts electricity through free electrons.

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Question 4(iii). Elements P, Q, and R are in the same period of the modern periodic table:

• P readily loses its one valence electron to form a stable ion.
• Q shares its electrons in bonding but does not form ions easily.
• R has high electronegativity.

Solution:

Based on the descriptions:

• P: Alkali Metal (Group 1)
• Q: Group 14 element (e.g., Carbon or Silicon)
• R: Halogen (Group 17)

(a) Which element would be most difficult to reduce?
Element P. Being highly electropositive, it readily loses electrons (oxidizes), so gaining electrons (reduction) is very difficult.

(b) Smallest atomic radius?
Element R. Atomic radius decreases across a period from left to right due to increasing nuclear charge, making R the smallest.

(c) Decreasing order of ionization potential:
Ionization potential increases across a period. Therefore, decreasing order: R > Q > P

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ICSE 10th Result 2021



Last Updated on : March 26, 2026