ICSE Chemistry 2026 Chemistry Answer Key ICSE Board Exam 2026 Science Paper 2 Chemistry Solutions ICSE Class 10 Board Exam Solutions Prediction Paper Bright Tutorials Nashik Road

ICSE Chemistry 2026 Board Exam — Complete Solutions & Answer Key

T

Tushar Parik

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Updated 14 March 2026
22 min read
Board Exam 2026 — Complete Solutions

ICSE Chemistry 2026 — Board Exam Complete Solutions & Answer Key

11 March 2026  |  Science Paper 2 (Chemistry)  |  Maximum Marks: 80

🌟 Bright Tutorials scored ~70–78% prediction accuracy!

🎯 Prediction Accuracy — How We Did

Bright Tutorials released two prediction papers before the ICSE Chemistry 2026 board exam. Here is how our predictions matched the actual paper:

Paper Accuracy Key Matches
Prediction Paper V1 63–69% Electrolysis of CuSO₄ with Cu electrodes, Haber Process, Contact Process, Bayer’s Process, IUPAC naming, organic conversions, mole concept numericals, salt analysis format
Prediction Paper V2 70–78% Electroplating with silver (MCQ + detailed question), molten PbBr₂ electrolysis, amphoteric oxide assertion-reason, acidulated water electrolysis diagram, ethene→ethane addition, Bayer’s Process with NaOH reasoning, pH-based identification, ethanol conversions flow chart, Fe₂O₃ + CO equation, stainless steel & alloys

Exact / Near-Exact Matches from Our Predictions:

  • Electrolysis of CuSO₄ with copper electrodes — appeared in both V1 and V2, and in the actual paper (Q2-i)
  • Electroplating with silver — V2 MCQ Q1(ii) matched the actual paper Q5(iii) almost exactly
  • Bayer’s Process with NaOH reasoning — both V1 (Q7-i) and V2 (Q7-i) predicted this, actual paper Q7(ii)
  • Contact Process (SO₃) — V1 predicted the catalyst MCQ, actual paper Q5(ii) asked about SO₃ collection
  • Haber Process — V1 and V2 both predicted it, actual paper Q2(i) asked about the full process with diagram
  • Electrolysis of acidulated water — V2 diagram-based MCQ closely matched actual paper Q1(xiv)
  • Molten PbBr₂ electrolysis — V2 Q8(ii) predicted this, actual paper Q2(iv-b) had the anode reaction
  • Ethene + H₂ → Ethane — V2 Q6(i-b) predicted this addition reaction, actual paper Q5(i)
  • Ethanol conversions (dehydration, esterification, combustion) — V2 Q6 predicted the flow chart, actual paper Q7(iv)
  • Fe₂O₃ + CO equation — both V1 (Q8-iii) and V2 (Q8-iii) predicted, though not in actual paper explicitly
  • Salt analysis with ZnO (yellow hot, white cold) — V2 Q4(iii) predicted Zn²⁺ identification, actual paper Q3(i) had Cu²⁺/Cl⁻
  • IUPAC naming & structural formulas — both papers predicted butan-2-ol, 2-methylpropane, butanal, pent-2-yne type questions
  • Assertion-Reason on amphoteric oxide — V2 Q1(xv) closely related to actual paper themes
  • Periodic table — elements with same config 2,8,2 — V1 Q3(iii) had element Z=12, actual paper Q1(iii) and Q1(viii) used same config
  • C + conc. H₂SO₄ — V1 Q3(ii-c) predicted this exact equation, actual paper Q1(i) tested it as MCQ

SECTION A — 40 Marks

Attempt all questions from this Section.

Question 1 — Multiple Choice Questions [15 Marks]

(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: (b) Carbon
C + 2H₂SO₄(conc.) → CO₂↑ + 2SO₂↑ + 2H₂O. Both CO₂ and SO₂ turn lime water milky. V1 Predicted

(ii) Which of the following element pairs will form an ionic bond?

P: Elements of Group 1 & Group 2    Q: Elements of Group 14 & Group 16
R: Elements of Group 2 & Group 17    S: Elements of Group 15 & Group 18

Answer: (c) R — Group 2 & Group 17
Group 2 (metals like Mg, Ca) + Group 17 (non-metals like Cl, Br) → ionic bond by transfer of electrons. The large electronegativity difference leads to complete electron transfer.

(iii) The electronic configuration of an element is 2,8,2. The hydroxide of this element can produce ___ hydroxyl ions per molecule.

(a) 3    (b) 2    (c) 1    (d) 0

Answer: (b) 2
Element is Magnesium (Z=12, config 2,8,2). Mg(OH)₂ produces 2 OH⁻ ions per molecule. V1 Predicted (Z=12)

(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: (a) Copper metal is deposited at the negative electrode
Cu²⁺ + 2e⁻ → Cu at cathode (negative electrode). The anode (positive electrode) dissolves, losing mass — not increasing. V1 + V2 Predicted

(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: (c) C₂H₅Cl + HCl
Substitution reaction in UV light: one H atom in ethane is replaced by a Cl atom, producing chloroethane (C₂H₅Cl) and hydrogen chloride (HCl).

(vi) A hydrocarbon X (butane) undergoes cracking → Y (propene) + Z (methane). Which compound(s) is/are unsaturated?

(a) X only    (b) Y only    (c) X and Z    (d) Y and Z

Answer: (b) Y only
Propene (CH₂=CHCH₃) has a C=C double bond, making it unsaturated. Butane (C₄H₁₀) and methane (CH₄) are both saturated hydrocarbons.

(vii) Rita added dilute HCl to four metals. Correct observations:

1. Copper → gas given off    2. Iron → gas given off
3. Magnesium → no gas    4. Zinc → gas given off

Answer: (d) Both 2 and 4
Iron and zinc are above hydrogen in the activity series — they react with dil. HCl to give H₂ gas. Copper is below hydrogen (does NOT react). Magnesium DOES react vigorously, so statement 3 is wrong.

(viii) An atom of X forms an ion X²⁺. The atomic number of X is:

(a) 16    (b) 10    (c) 12    (d) 14

Answer: (c) 12
X loses 2 electrons to form X²⁺. Element with valency 2 that forms a cation = Magnesium (Z=12, config 2,8,2). V1 Predicted (Z=12)

(ix) Salts can be prepared by action of dilute acid on: 1. bases, 2. carbonates, 3. metals, 4. sulphites. The method which cannot be used for copper salts is:

(a) 1    (b) 2    (c) 3    (d) 4

Answer: (c) 3 — Metals
Copper is below hydrogen in the activity series, so it does NOT react with dilute acids to form salts. Methods 1 (bases), 2 (carbonates), and 4 (sulphites) all work for copper salts.

(x) The compound with the highest melting point is:

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

Answer: (b) Sodium chloride
NaCl is an ionic compound with strong electrostatic forces between Na⁺ and Cl⁻ ions. Ionic compounds have very high melting points (~801°C for NaCl). Methane, ammonia, and ethanol are covalent with much weaker intermolecular forces.

(xi) Assertion (A): Dilute H₂SO₄ is a stronger electrolyte than concentrated H₂SO₄.
Reason (R): Dilute H₂SO₄ has a higher concentration of mobile ions.

Answer: (c) Both A and R are true and R is the correct explanation of A
Dilute H₂SO₄ is more ionized (more mobile H⁺ and SO₄²⁻ ions) than concentrated H₂SO₄, which is viscous and less dissociated. The higher concentration of mobile ions makes it a better electrolyte.

(xii) Volume of CO₂ from 5 litres propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

(a) 5 litres    (b) 15 litres    (c) 10 litres    (d) 20 litres

Answer: (b) 15 litres
By Gay-Lussac's Law: 1 vol C₃H₈ → 3 vol CO₂. So 5L propane produces 5 × 3 = 15L CO₂.

(xiii) An unsaturated hydrocarbon with 3 carbon atoms and 6 hydrogen atoms:

(a) propane    (b) propyne    (c) propene    (d) cyclopropane

Answer: (c) Propene
C₃H₆ = propene (CH₂=CH—CH₃), which has a C=C double bond making it unsaturated. Propane is C₃H₈ (saturated), propyne is C₃H₄ (triple bond).

(xiv) Assertion (A): In electrolysis of acidified water, volume of H₂ is twice the volume of O₂.
Reason (R): Water has H and O in ratio 1:2 by volume.

Answer: (a) A is true but R is false
A is correct: 2H₂O → 2H₂ + O₂, giving H₂:O₂ ratio of 2:1. R is false — water has H:O in ratio 2:1 by volume, NOT 1:2. V2 Predicted

(xv) In electrolysis of molten Al₂O₃: Which reaction occurs at the negative electrode?

1. Al³⁺ + 3e⁻ → Al    2. 2O²⁻ → O₂ + 4e⁻
3. C + O₂ → CO₂    4. Al₂O₃ ⇌ 2Al³⁺ + 3O²⁻

Answer: (a) 1 — Al³⁺ + 3e⁻ → Al
At cathode (negative electrode), Al³⁺ ions gain electrons to form aluminium metal. Reaction 2 occurs at the anode, reaction 3 is the burning of carbon anodes, and reaction 4 is the dissociation of alumina. V1 + V2 Predicted

Question 2 — Short Answers [25 Marks]

(i) Haber Process Diagram [5 Marks] V1 + V2 Predicted

(a) Name the process:

Haber Process

(b) Catalyst used:

Finely divided iron (Fe) with molybdenum as promoter.

(c) How is ammonia separated?

The gas mixture is cooled/liquefied. Ammonia has a higher boiling point than N₂ and H₂, so it condenses into liquid ammonia while unreacted gases are recycled back to the catalytic chamber.

(d) Two properties demonstrated by Fountain Experiment:

1. Ammonia is highly soluble in water.
2. Ammonia is lighter than air (less dense than air).

(ii) Identify Terms/Compounds [5 Marks]

(a) Compound + water → acetylene:

Calcium carbide (CaC₂)
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂↑

(b) Gas from sodium propanoate + soda lime:

Ethane (C₂H₆)
CH₃CH₂COONa + NaOH →(CaO) C₂H₆ + Na₂CO₃ (Decarboxylation)

(c) Reddish-brown precipitate from ferric chloride + alkali:

Ferric hydroxide / Iron(III) hydroxide — Fe(OH)₃ V1 + V2 Predicted

(d) Pair of electrons not shared during bonding:

Lone pair

(e) Relative molecular mass expressed in grams:

Gram molecular mass (Molar mass)

(iii) Match Column A with Column B [5 Marks]

Column A Answer Column B
(a) CuO + C → Cu + CO → 4 Redox (C is oxidized, CuO is reduced)
(b) Al³⁺ + 3e⁻ → Al → 3 Reduction (gain of electrons)
(c) PbBr₂ → Pb²⁺ + 2Br⁻ → 5 Electrolytic dissociation
(d) 2O²⁻ − 2e⁻ → O₂ → 1 Oxidation (loss of electrons)
(e) HCl → H⁺ + Cl⁻ → 2 Ionization

(iv) Fill in the Blanks [5 Marks]

(a) Oxide that dissolves in KOH:

ZnO (amphoteric — reacts with both acids and bases). ZnO + 2KOH → K₂ZnO₂ + H₂O

(b) Anode reaction in molten PbBr₂:

2Br⁻ − 2e⁻ → Br₂ V2 Predicted

(c) Volume of 8g O₂ at STP:

Moles = 8/32 = 0.25 mol. Volume = 0.25 × 22.4 = 5.6 litres

(d) Metal that does not give H₂ with cold dilute HNO₃:

Cu (Copper) — copper is below hydrogen in the activity series; it reacts with HNO₃ to give NO/NO₂, not H₂.

(e) A polar covalent compound:

HCl — electronegativity difference between H (2.1) and Cl (3.0) creates a dipole but is not large enough for complete ionic character.

(v) IUPAC Names & Structural Formulas [5 Marks] V1 + V2 Predicted

(a) HO—CH(CH₃)—CH₂—CH₃ → IUPAC name:

Butan-2-ol

(a2) CH₃—CH₂—CH(CH₃)—CH₂—CH₃ → IUPAC name:

3-Methylpentane

(b) Structural formulas:

1. Butanal: CH₃—CH₂—CH₂—CHO

2. Pent-2-yne: CH₃—C≡C—CH₂—CH₃

3. Isomer of n-butane: 2-Methylpropane — CH₃—CH(CH₃)—CH₃

SECTION B — 40 Marks

Attempt any four questions from this Section.

Question 3 — Salt Analysis, Equations & Bonding [10 Marks]

(i) Salt Analysis [2 Marks]

White precipitate with AgNO₃ soluble in NH₄OH → ?
Pale blue precipitate with NaOH → ?

Anion: Cl⁻ (Chloride) — White ppt of AgCl, soluble in NH₄OH (forms soluble [Ag(NH₃)₂]⁺)
Cation: Cu²⁺ (Copper) — Pale blue ppt is Cu(OH)₂
Salt Y = Copper(II) chloride — CuCl₂

(ii) NaNO₃ + conc. H₂SO₄ ≤200°C [2 Marks]

NaNO₃ + H₂SO₄(conc.) → NaHSO₄ + HNO₃

(a) Why temperature below 200°C?

Above 200°C, HNO₃ decomposes into NO₂, H₂O and O₂. Below 200°C, HNO₃ is obtained as a pure vapour without decomposition.

(b) Why is conc. H₂SO₄ used?

It is a non-volatile acid (high boiling point ~338°C) that can displace the more volatile HNO₃ (bp 83°C) from its salt.

(iii) Element A — Group 15, Period 2 [3 Marks]

(a) Element A = Nitrogen (N) (Z=7, config 2,5)

(b) Compound with hydrogen = NH₃ (Ammonia)

(c) Dot and cross structure of NH₃: Nitrogen (5 valence electrons) shares 3 electrons with 3 hydrogen atoms, forming 3 covalent bonds, with one lone pair remaining on nitrogen. The molecule has a pyramidal shape.

(iv) Complete and Balance Equations [3 Marks]

(a) C₂H₄Br₂ + 2KOH(alcoholic) →

C₂H₂ + 2KBr + 2H₂O (Dehydrohalogenation → Ethyne/Acetylene)

(b) NH₃(excess) + Cl₂ →

8NH₃(excess) + 3Cl₂ → N₂ + 6NH₄Cl

(c) C₁₂H₂₂O₁₁ + conc. H₂SO₄ →

12C + 11H₂O (Dehydration of sugar — charring by conc. H₂SO₄ acting as a dehydrating agent) V1 + V2 Predicted

Question 4 — Acids, Bases & Periodic Table [10 Marks]

(i) Differentiate [2 Marks]

(a) Acetic acid vs Sulphuric acid — replaceable hydrogen ions:

Acetic acid (CH₃COOH) has 1 replaceable hydrogen ion per molecule (monobasic).
Sulphuric acid (H₂SO₄) has 2 replaceable hydrogen ions per molecule (dibasic).

(b) Electrolyte vs Metallic conductor:

In an electrolyte, electricity is conducted by ions (cations and anions moving in solution).
In a metallic conductor, electricity is conducted by free/delocalized electrons.

(ii) Avogadro's Law Calculation [2 Marks]

Mass of NH₃ = 34g occupies same volume as Cl₂ at same T and P. Find the number of Cl₂ molecules.

Molar mass of NH₃ = 14 + 3 = 17 g/mol
Moles of NH₃ = 34/17 = 2 mol

By Avogadro’s Law: equal volumes at same T and P contain equal number of moles.
Moles of Cl₂ = 2 mol

Number of molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules

(iii) Elements P, Q, R in Same Period [3 Marks]

P = metal (loses electron, Group 1), Q = shares electrons (Group 14), R = high electronegativity (Group 17)

(a) Most difficult to reduce: R — Highest electron affinity, strongest tendency to gain electrons; already very stable, resists further reduction.

(b) Smallest atomic radius: R — Atomic radius decreases across a period from left to right due to increasing nuclear charge pulling electrons closer.

(c) Decreasing ionization potential: R > Q > P — Ionization energy increases across a period (more nuclear charge = harder to remove electrons).

(iv) Salt Preparation Equations [3 Marks]

(a) Lead(II) carbonate:

Pb(NO₃)₂ + Na₂CO₃ → PbCO₃↓ + 2NaNO₃
(Double decomposition / Precipitation — PbCO₃ is insoluble)

(b) Copper(II) chloride:

CuO + 2HCl → CuCl₂ + H₂O
(Base + Acid method, since Cu does not react with dilute acids directly)

(c) Iron(II) chloride:

Fe + 2HCl → FeCl₂ + H₂↑
(Metal + Acid method — iron is above hydrogen in the activity series)

Question 5 — Organic Chemistry & Electroplating [10 Marks]

(i) Ethene + H₂ → Ethane [3 Marks] V2 Predicted

(a) Type of reaction: Addition (Hydrogenation)

(b) Catalyst: Nickel (Ni) at 200°C [also Platinum (Pt) or Palladium (Pd)]

(c) Balanced equation:
C₂H₄ + H₂ →(Ni, 200°C) C₂H₆

(ii) Contact Process — SO₃ [3 Marks] V1 Predicted

(a) Gas collected: Sulphur trioxide (SO₃)

(b) If dissolved directly in water: Sulphuric acid (H₂SO₄). SO₃ + H₂O → H₂SO₄

(c) Why not dissolve directly in water: The reaction of SO₃ with water is extremely exothermic and produces a dense fog/mist of H₂SO₄ droplets that is difficult to condense and handle. Instead, SO₃ is dissolved in concentrated H₂SO₄ to form oleum (H₂S₂O₇), which is then carefully diluted with water to get the desired concentration.

(iii) Electroplating Spoon with Silver [4 Marks] V2 Predicted

(a) Cathode reaction:
Ag⁺ + e⁻ → Ag (silver deposited on the spoon)

(b) Why AgNO₃ is not preferred as electrolyte:
Silver nitrate solution produces a non-uniform, rough, and non-adherent deposit (crystalline/spongy). Sodium argentocyanide [Na[Ag(CN)₂]] gives a smooth, uniform, bright coating that adheres well.

(c) Why AC current is not used:
Alternating current reverses polarity continuously, so the silver would alternately deposit and dissolve. DC ensures continuous one-directional deposition of silver on the cathode.

(d) Observation at anode:
The silver anode gradually dissolves / decreases in mass, as silver atoms lose electrons and enter solution as Ag⁺ ions: Ag → Ag⁺ + e⁻

Question 6 — Reasons, True/False & Numericals [10 Marks]

(i) Give Reasons [2 Marks]

(a) HCl cannot form acid salt:

HCl is a monobasic acid (has only one replaceable hydrogen atom). Acid salts are formed by partial replacement of H⁺ ions, which requires at least two replaceable hydrogen atoms (dibasic or polybasic acid). Since HCl has only one H⁺, it can only form normal salts (e.g., NaCl).

(b) Electronegativity increases across a period:

Atomic radius decreases across a period (same shell, increasing nuclear charge). The smaller atom has a stronger pull on shared electrons, resulting in higher electronegativity.

(ii) True or False [2 Marks]

(a) “Organic compounds having different molecular formula but same structural formula are called isomers”

False — Isomers have the SAME molecular formula but DIFFERENT structural formulas.

(b) “A salt is a compound formed by partial or complete replacement of the hydrogen ion of an acid by a metal or electropositive ion”

True — This is the standard definition of a salt. Partial replacement gives acid salts; complete replacement gives normal salts.

(iii) Nitric Acid Storage [3 Marks] V2 Predicted

(a) Error: Anil stored HNO₃ in a transparent container exposed to sunlight. It should be stored in dark-coloured (amber) bottles away from light.

(b) Brown fumes: HNO₃ decomposes in sunlight:
4HNO₃ → 4NO₂↑ (brown) + 2H₂O + O₂
The brown fumes are nitrogen dioxide (NO₂).

(c) Remove yellowish tinge: Pass/bubble dry air through the acid, which removes dissolved NO₂. This oxidizes NO₂ and restores the colourless, pure HNO₃.

(iv) Zinc + Conc. HNO₃ — Numerical [3 Marks]

Zn + 4HNO₃(conc.) → Zn(NO₃)₂ + 2H₂O + 2NO₂    |    Given: 32.5g Zn, At. wt. Zn = 65

(a) Moles of Zn:
Moles = 32.5 / 65 = 0.5 moles

(b) Mass of HNO₃ required:
From equation: 1 mol Zn reacts with 4 mol HNO₃
Moles of HNO₃ = 0.5 × 4 = 2 mol
Molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol
Mass = 2 × 63 = 126 g

(c) Volume of NO₂ at STP:
1 mol Zn → 2 mol NO₂
Moles of NO₂ = 0.5 × 2 = 1 mol
Volume = 1 × 22.4 = 22.4 litres at STP

Question 7 — Reactions, Bayer’s Process & Conversions [10 Marks]

(i) Carbon + Conc. HNO₃ [2 Marks]

(a) Acid used: Concentrated nitric acid (HNO₃)

(b) Equation:
C + 4HNO₃(conc.) → CO₂↑ + 4NO₂↑ + 2H₂O
HNO₃ acts as an oxidizing agent, oxidizing carbon to CO₂.

(ii) Bayer’s Process [2 Marks] V1 + V2 Predicted

(a) Why NaOH is added:
It selectively dissolves aluminium oxide (Al₂O₃) which is amphoteric, while impurities like Fe₂O₃ and SiO₂ remain undissolved and can be filtered out as “red mud”.

(b) Equation:
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

(iii) Observations [3 Marks]

(a) NaOH + Ca(NO₃)₂:

A white precipitate of calcium hydroxide Ca(OH)₂ is formed.
Ca(NO₃)₂ + 2NaOH → Ca(OH)₂↓ + 2NaNO₃

(b) Dil. HCl + FeS:

A colourless gas with the smell of rotten eggs (H₂S) is evolved.
FeS + 2HCl → FeCl₂ + H₂S↑

(c) Amphoteric metal + hot conc. alkali:

The metal dissolves with evolution of hydrogen gas.
Example: Zn + 2NaOH(conc.) → Na₂ZnO₂ + H₂↑
Or: Al + NaOH + H₂O → NaAlO₂ + 3/2 H₂↑

(iv) Ethanol Conversions [3 Marks] V2 Predicted

Ethanol → Ethene (A):
A = Dehydration (excess conc. H₂SO₄ at 170°C, or Al₂O₃)
C₂H₅OH → C₂H₄ + H₂O

Ethanol → Ethyl acetate (B):
B = Acetic acid / Ethanoic acid (CH₃COOH) in presence of conc. H₂SO₄
C₂H₅OH + CH₃COOH →(conc. H₂SO₄) CH₃COOC₂H₅ + H₂O (Esterification)

Ethanol → CO₂ + H₂O (C):
C = Combustion (burning in air/oxygen)
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

Question 8 — Mole Concept, Periodic Table & Solutions [10 Marks]

(i) Gas Z Calculation [2 Marks]

5.6L at STP weighs 32g. Find molecular weight and vapour density.

At STP, 22.4L = 1 mole
Moles = 5.6 / 22.4 = 0.25 mol
Molecular weight = 32 / 0.25 = 128 g/mol

Vapour density = MW / 2 = 128 / 2 = 64

(ii) Name the Following [2 Marks]

(a) Most electronegative element of Period 2: Fluorine (F)
(F has the highest electronegativity of all elements — 4.0 on the Pauling scale)

(b) Largest atom of Period 3: Sodium (Na)
(Leftmost element in the period has the largest atomic radius — fewest protons pulling on the same shell)

(iii) Elements Li, K, Ca, F [3 Marks]

(a) Least atomic radius: Fluorine (F)
Period 2, Group 17 — smallest atom due to high nuclear charge relative to its size (only 2 shells with 9 protons).

(b) Two valence electrons: Calcium (Ca)
Z=20, config 2,8,8,2 — Group 2 element with 2 electrons in outermost shell.

(c) Most electropositive: Potassium (K)
Group 1, Period 4 — lowest ionization energy among the given elements, most readily loses its valence electron.

(iv) Three Beakers: Acetic Acid (K), NaCl Solution (L), Distilled Water (M) [3 Marks]

(a) Contains only ions: L (Sodium chloride solution)
NaCl is a strong electrolyte that fully dissociates into Na⁺ and Cl⁻ ions in solution.

(b) Contains only molecules: M (Distilled water)
Pure H₂O exists almost entirely as molecules with negligible self-ionization (Kw = 10⁻¹⁴).

(c) pH less than 7: K (Aqueous acetic acid)
Acetic acid is a weak acid that partially ionizes in water, producing H⁺ ions → pH < 7.

Prediction Accuracy — Final Summary

Topic Area V1 Prediction V2 Prediction Appeared in Actual Paper?
Electrolysis of CuSO₄ (Cu electrodes) YES (Q1-iv)
Haber Process (diagram + details) YES (Q2-i)
Contact Process (SO₃) YES (Q5-ii)
Bayer’s Process + NaOH reasoning YES (Q7-ii)
Electroplating with silver YES (Q5-iii)
Molten PbBr₂ electrolysis YES (Q2-iv-b)
Ethene + H₂ → Ethane (Addition) YES (Q5-i)
Ethanol conversions (dehydration, esterification) YES (Q7-iv)
Element Z=12 (Mg) properties YES (Q1-iii, viii)
C + conc. H₂SO₄ equation YES (Q1-i)
HNO₃ storage in dark bottles YES (Q6-iii)
IUPAC naming + structural formulas YES (Q2-v)
Sugar charring with conc. H₂SO₄ YES (Q3-iv-c)
Electrolysis of acidulated water (H₂:O₂ ratio) YES (Q1-xiv)
Al₂O₃ electrolysis (cathode reaction) YES (Q1-xv)
Overall Prediction Match Rate
V1: 63–69%  •  V2: 70–78%

ICSE Biology 2026 Prediction Paper — Coming Soon!

Our Biology prediction papers are being prepared based on the same deep analysis methodology that achieved 70–78% accuracy for Chemistry. Stay tuned!

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Important Notes for Students

  • Verify with your teachers: While we have prepared these solutions with great care, always cross-check answers with your school teachers and textbooks.
  • Marking scheme: Board marking may award partial marks for correct steps even if the final answer differs slightly. Always show your working clearly.
  • Chemical equations: Ensure you balance all equations and include state symbols and conditions where required.
  • Diagram questions: The actual exam included diagrams which cannot be fully reproduced here. Refer to your textbook for standard lab diagrams.
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