ICSE Class 10 Chemistry Board Exam Question Paper 2020 with Solutions

ICSE Grade 10 Chemistry Board Exam Question Paper with Solutions

SECTION I

Question 1

(a) Choose the correct answer from the options given below: [5 Marks]
(i) The element with highest ionization potential, is:
A. Hydrogen
B. Caesium
C. Radon
D. Helium

Answer: (D) Helium

Helium is a noble gas with a completely filled valence shell and a very small atomic size, resulting in the highest ionization potential among all elements.

Teacher's Note:
a) Ionization potential increases across a period from left to right and decreases down a group.
b) Noble gases have exceptionally high ionization potentials due to their stable electronic configurations.

(ii) The inert electrode used in the electrolysis of acidified water, is: [1 Mark]
A. Nickel
B. Platinum
C. Copper
D. Silver

Answer: (B) Platinum

Platinum or graphite is commonly used as an inert electrode because it does not participate in the chemical reaction during electrolysis.

Teacher's Note:
a) An inert electrode only provides a surface for the transfer of electrons without getting oxidized or reduced itself.
b) Active electrodes like copper take part in the reaction, whereas platinum and carbon (graphite) are typical examples of inert electrodes.

(iii) A compound with low boiling point, is: [1 Mark]
A. Sodium chloride
B. Calcium chloride
C. Potassium chloride
D. Carbon tetrachloride

Answer: (D) Carbon tetrachloride

Carbon tetrachloride is a covalent compound held together by weak intermolecular forces of attraction, giving it a low boiling point compared to electrovalent compounds.

Teacher's Note:
a) Electrovalent compounds like sodium chloride, calcium chloride, and potassium chloride have high melting and boiling points due to strong electrostatic forces.
b) Covalent compounds generally possess low melting and boiling points.

(iv) The acid which can produce carbon from cane sugar, is: [1 Mark]
A. Concentrated Hydrochloric acid
B. Concentrated Nitric acid
C. Concentrated Sulphuric acid
D. Concentrated Acetic acid

Answer: (C) Concentrated Sulphuric acid

Concentrated sulphuric acid acts as a dehydrating agent and removes water molecules from cane sugar (C₁₂H₂₂O₁₁), leaving behind a spongy black mass of carbon.

Teacher's Note:
a) This reaction demonstrates the strong dehydrating property of concentrated H₂SO₄.
b) Cane sugar is charred to carbon, also known as sugar charcoal.

(v) The organic compound having a triple carbon-carbon covalent bond, is: [1 Mark]
A. C₃H₄
B. C₃H₆
C. C₃H₈
D. C₄H₁₀

Answer: (A) C₃H₄

Propyne has the molecular formula C₃H₄, which corresponds to the general formula C_nH_2n-2 for alkynes containing a triple bond.

Teacher's Note:
a) C₃H₆ is an alkene (propene) with a double bond.
b) C₃H₈ and C₄H₁₀ are alkanes containing only single covalent bonds.

(b) State one relevant observation for each of the following reactions: [5 Marks]
(i) Action of concentrated nitric acid on copper.
(ii) Addition of excess ammonium hydroxide into copper sulphate solution.
(iii) A piece of sodium metal is put into ethanol at room temperature.
(iv) Zinc carbonate is heated strongly.
(v) Sulphide ore is added to a tank containing oil and water, and then stirred or agitated with air.

Answer:
(i) Dense reddish-brown fumes of nitrogen dioxide gas are evolved, and the solution turns bluish-green.
(ii) A pale blue precipitate is initially formed which dissolves in excess ammonium hydroxide to give a clear deep inky-blue solution.
(iii) Effervescence occurs and a colorless, odorless gas (hydrogen) is evolved which burns with a pop sound.
(iv) A yellowish-white residue is formed when hot, which turns white on cooling, along with the evolution of a colorless, odorless gas that turns lime water milky.
(v) A froth or foam is formed on the surface carrying the concentrated sulphide ore particles upwards.

Teacher's Note:
a) Students must mention specific color changes and gas characteristics precisely.
b) For copper with conc. HNO₃, mentioning reddish-brown fumes is essential for full credit.

(c) Write a balanced chemical equation for each of the following: [5 Marks]
(i) Reaction of carbon powder and concentrated nitric acid.
(ii) Reaction of excess ammonia with chlorine.
(iii) Reaction of lead nitrate solution with ammonium hydroxide.
(iv) Producing ethyne from bromo ethane using Zn/Cn couple in alcohol.
(v) Completed combustion of ethane

Answer:
(i) C + 4HNO₃ (conc.) → CO₂ + 2H₂O + 4NO₂
(ii) 8NH₃ (excess) + 3Cl₂ → N₂ + 6NH₄Cl
(iii) Pb(NO₃)₂ + 2NH₄OH → Pb(OH)₂↓ + 2NH₄NO₃
(iv) CH₃CH₂Br + Zn / Alc → CH ≡ CH + ZnBr₂ (or C₂H₅Br + Zn + HBr → C₂H₂ + ZnBr₂ + H₂)
(v) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Teacher's Note:
a) Ensure all chemical equations are fully balanced with physical states where necessary.
b) Note that part (iv) in the paper text has minor OCR numbering typos (listed as vi instead of v), but all five sub-parts are answered sequentially.

(d) (i) Draw the structural formula for each of the following: [5 Marks]
1. 2,2 dimethyl pentane
2. methanol
3. Iso propane
(ii) Write the IUPAC name for the following compounds:
1. Acetaldehyde
2. Acetylene

Answer:
(i) 1. 2,2-dimethyl pentane structure:
            CH₃
             | 
CH₃ - C - CH₂ - CH₂ - CH₃
             | 
           CH₃
2. Methanol structure:
H - C - O - H (with three H atoms bonded to the central carbon)
3. Iso propane (Note: Propane has no isomers; normal propane is meant):
CH₃ - CH₂ - CH₃
(ii) 1. Ethanal
2. Ethyne

Teacher's Note:
a) Isopropane does not exist chemically as propane has only 3 carbons, so structural representation for normal propane is accepted.
b) IUPAC names must follow standard nomenclature rules strictly.

(e) State one relevant reason for each of the following: [5 Marks]
(i) Graphite anode is preferred to platinum in the electrolysis of molten lead bromide.
(ii) Soda lime is preferred to sodium hydroxide in the laboratory preparation of methane.
(iii) Hydrated copper sulphate crystals turn white on heating.
(iv) Concentrated nitric acid appears yellow, when it is left for a while in a glass bottle.
(v) Hydrogen chloride gas fumes in moist air.

Answer:
(i) The vapors of bromine evolved at the anode react with platinum, destroying it, whereas graphite remains unaffected.
(ii) Sodium hydroxide is deliquescent and attacks glass, whereas soda lime does not attack glass and is not deliquescent.
(iii) On heating, water of crystallization is lost, leaving behind anhydrous copper sulphate which is white.
(iv) Concentrated nitric acid undergoes thermal decomposition in sunlight to form reddish-brown nitrogen dioxide gas which dissolves in the acid, giving it a yellow color.
(v) Hydrogen chloride gas dissolves in the moisture of the air to form tiny droplets of hydrochloric acid which appear as white fumes.

Teacher's Note:
a) Reasons should be scientifically concise and address the specific cause mentioned in the question.
b) Mentioning the decomposition of HNO₃ into NO₂ is crucial for part (iv).

(f) Calculate: [5 Marks]
(i) The amount of each reactant required to produce 750 ml of carbon dioxide, when two volumes of carbon monoxide combine with one volume of oxygen to produce two volumes of carbon dioxide. 2CO + O₂ → 2CO₂
(ii) The volume occupied by 80 g of carbon dioxide at STP.
(iii) Calculate the number of molecules in 4.4 gm of CO₂ [Atomic mass of C = 12, O = 16]
(iv) State the law associated in question no. (f)(i) above.

Answer:
(i) According to Gay-Lussac's Law of Combining Volumes, 2 volumes of CO produce 2 volumes of CO₂. Therefore, 750 ml of CO₂ requires 750 ml of CO. 2 volumes of CO combine with 1 volume of O₂, so 750 ml of CO requires 750 / 2 = 375 ml of O₂.
(ii) Molar mass of CO₂ = 12 + (16 × 2) = 44 g/mol. Number of moles = 80 / 44 = 1.818 moles. Volume at STP = 1.818 × 22.4 = 40.73 liters.
(iii) Moles of CO₂ = 4.4 / 44 = 0.1 moles. Number of molecules = 0.1 × 6.023 × 10²³ = 6.023 × 10²² molecules.
(iv) Gay-Lussac's Law of Combining Volumes.

Teacher's Note:
a) Note that part (f)(ii) has a typo in the question paper text ("RO g" read as 80 g).
b) Always write proper units in numerical calculations.

(g) Give one word or a passage following statement: [5 Marks]
(i) The chemical bond formed by a shared pair of electrons. Each bonding atom contributing one electron to the pair.
(ii) Electrode used as cathode in electrorefining of impure copper.
(iii) The substance prepared by adding other metals to a base metal in appropriate proportions to obtain certain desirable properties.
(iv) The tendency of an atom to attract electrons to itself when combined in a compound.
(v) The reaction in which carboxylic acid reacts with alcohol in the presence of conc. H₂SO₄ to form a substance having a fruity smell.

Answer:
(i) Single covalent bond
(ii) Pure copper strip
(iii) Alloy
(iv) Electronegativity
(v) Esterification

Teacher's Note:
a) Terms must be exact scientific definitions.
b) Part (ii) expects "thin strip of pure copper".

(h) Fill in the blanks from the choices given in brackets: [5 Marks]
(i) The polar covalent compound in gaseous state that does not conduct electricity is __________ (carbon tetra chloride, ammonia, methane)
(ii) A salt prepared by displacement reaction is __________ (ferric chloride, ferrous chloride, silver chloride)
(iii) The number of moles in 11 gm of nitrogen gas is __________ (0.39, 0.49, 0.29) [atomic mass of N = 14]
(iv) An alkali which completely dissociates into ions is __________ (ammonium hydroxide, calcium hydroxide, lithium hydroxide)
(v) An alloy used to make statues is __________ (bronze, brass, fuse metal)

Answer:
(i) ammonia
(ii) ferrous chloride
(iii) 0.39 (11 / 28 = 0.392 moles)
(iv) lithium hydroxide
(v) bronze

Teacher's Note:
a) Ammonia is polar covalent, but dry ammonia gas does not conduct electricity as it has no free ions.
b) Moles of nitrogen gas (N₂, molar mass 28 g/mol) in 11g = 11/28 = 0.39 moles.

SECTION II

Attempt any four questions from this Section

Question 2

(a) The following table represent the elements and the atomic number. With reference to this, answer the following using only the alphabets given in the table. [3 Marks]

(i) Which element combines with hydrogen to form a basic gas?
(ii) Which element has an electron affinity zero?
(iii) Name the element, which forms an ionic compound with chlorine.

Answer:
(i) Q (Nitrogen, atomic number 7, forms ammonia NH₃ which is a basic gas)
(ii) R (Neon, atomic number 10, is a noble gas with zero electron affinity)
(iii) P (Aluminium, atomic number 13, forms ionic aluminium chloride)

Teacher's Note:
a) Students must answer using only the alphabets provided in the table.
b) Noble gases have stable octets/duplets, so their electron affinity is zero.

(b) Draw the electron dot diagram for the compounds given below. Represent the electrons by (.) and (x) in the diagram. [Atomic No.: Ca = 20, O = 8, Cl = 17, H = 1] [3 Marks]
(i) Calcium oxide
(ii) Chlorine molecule
(iii) Water molecule

Answer:
(i) Calcium oxide: Ca²⁺ [ : O : ]^2- (Calcium transfers 2 electrons to oxygen).
(ii) Chlorine molecule: Cl - Cl single covalent bond with 6 non-bonding electrons around each chlorine atom.
(iii) Water molecule: Central oxygen atom bonded with two hydrogen atoms through single covalent bonds, with two lone pairs on oxygen.

Teacher's Note:
a) Use dots (.) and crosses (x) to distinguish electrons of combining atoms clearly.
b) Show charges on ions for electrovalent compounds.

(c) Choose the correct word which refers to the process of electrolysis from A to E to match the description (i) to (iv). A: Oxidation B: Cathode C: Anode D: An electrolyte E: Reduction [4 Marks]
(i) Conducts electricity in aqueous or in molten state.
(ii) Loss of electron takes place at anode.
(iii) A reducing electrode.
(iv) Electrode connected to the positive end or terminal of the battery.

Answer:
(i) D: An electrolyte
(ii) A: Oxidation
(iii) B: Cathode
(iv) C: Anode

Teacher's Note:
a) Cathode is the negative electrode where reduction (gain of electrons) takes place, hence it acts as a reducing electrode.
b) Anode is connected to the positive terminal of the battery where oxidation occurs.

Question 3

(a) Baeyer's process is used to concentrate bauxite ore to alumina. Give balanced chemical equations for the reaction taking place for its conversion from bauxite to alumina. [3 Marks]

Answer:
1. Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
2. NaAlO₂ + 2H₂O → NaOH + Al(OH)₃↓
3. 2Al(OH)₃ → Al₂O₃ + 3H₂O (on strong heating)

Teacher's Note:
a) Baeyer's process involves leaching powdered bauxite with hot caustic soda solution.
b) All three steps with proper balancing are necessary for full marks.

(b) Complete the following by selecting the correct option from the choices given: [3 Marks]
(i) pH of acetic acid is greater than dilute Sulphuric acid. So acetic acid contains __________ concentration of H⁺ ions. (greater, same, low)
(ii) The indicator which does not change colour on passage of HCl gas is __________ . (methyl orange, moist blue litmus, phenolphthalein)
(iii) The acid which cannot act as an oxidizing agent is (conc. H₂SO₄, conc. HNO₃, conc. HCl)

Answer:
(i) low
(ii) phenolphthalein
(iii) conc. HCl

Teacher's Note:
a) Higher pH means lower concentration of hydrogen ions.
b) Phenolphthalein is colorless in acidic medium and remains unaffected by dry or gaseous HCl.

(c) Match the gases given in column I to the identification of the gases mentioned in column II: [4 Marks]

Answer:
(i) - D
(ii) - C
(iii) - B
(iv) - A

Teacher's Note:
a) Hydrogen sulphide gas turns moist lead acetate paper silvery black due to the formation of lead sulphide.
b) Nitric oxide is colorless, but on reacting with oxygen, it forms nitrogen dioxide which is reddish-brown.

Question 4

(a) Differentiate between the following pairs based on the information given in the brackets: [3 Marks]
(i) Conductor and electrolyte (conducting particles)
(ii) Cations and anions (formation from an atom)
(iii) Acid and Alkali (formation of type of ions)

Answer:

Teacher's Note:
a) Use a tabular format for differentiation questions for clarity.
b) Ensure the specific parameter in brackets is addressed directly.

(b) Draw the structures of isomers of pentane. [3 Marks]

Answer:
Pentane has three isomers:
1. Normal pentane (n-pentane):
CH₃ - CH₂ - CH₂ - CH₂ - CH₃
2. Isopentane (2-methylbutane):
            CH₃
             | 
CH₃ - CH - CH₂ - CH₃
3. Neopentane (2,2-dimethylpropane):
            CH₃
             | 
CH₃ - C - CH₃
             | 
           CH₃

Teacher's Note:
a) Draw full structural or condensed structural formulas showing all carbon chains clearly.
b) State both common names and IUPAC names for precision.

(c) Hydrogen chloride gas is prepared in the laboratory using concentrated sulphuric acid and sodium chloride. Answer the questions that follow based on this reaction: [4 Marks]
(i) Give the balanced chemical equation for the reaction with suitable condition (s) if any.
(ii) Why is concentrated sulphuric acid used instead of concentrated nitric acid?
(iii) How is the gas collected?
(iv) Name the drying agent not used for drying the gas.

Answer:
(i) NaCl + H₂SO₄ (conc.) → NaHSO₄ + HCl↑ (Temperature below 200°C)
(ii) Concentrated nitric acid is volatile, whereas concentrated sulphuric acid is non-volatile and has a high boiling point.
(iii) By upward displacement of air (downward delivery).
(iv) Quicklime (Calcium oxide, CaO)

Teacher's Note:
a) Quicklime is basic and reacts with hydrogen chloride gas, hence it cannot be used as a drying agent.
b) Concentrated H₂SO₄ is the preferred drying agent.

Question 5

(a) Distinguish between the following pairs of compounds using a reagent as a chemical test: [3 Marks]
(i) Calcium nitrate and Zinc nitrate solution.
(ii) Ammonium sulphate crystals and Sodium sulphate crystals.
(iii) Magnesium chloride and Magnesium nitrate solution.

Answer:
(i) Add sodium hydroxide solution dropwise and then in excess. Zinc nitrate gives a gelatinous white precipitate soluble in excess NaOH, while calcium nitrate gives a white precipitate sparingly soluble or insoluble in excess NaOH.
(ii) Add sodium hydroxide and warm. Ammonium sulphate evolves ammonia gas with a pungent smell which turns moist red litmus blue, whereas sodium sulphate does not give ammonia gas.
(iii) Add silver nitrate solution followed by dilute nitric acid. Magnesium chloride gives a curdy white precipitate of silver chloride, whereas magnesium nitrate does not react.

Teacher's Note:
a) Reagent must be specified clearly along with distinct observations for both compounds.
b) Confirmatory tests for ions form the basis of analytical chemistry questions.

(b) Calculate the percentage of: [3 Marks]
(i) Fluorine
(ii) Sodium and
(iii) Aluminium
in sodium aluminium fluoride [Na₃AlF₆], to the nearest whole number. [Atomic Mass: Na = 23, Al = 27, F = 19]

Answer:
Molar mass of Na₃AlF₆ = (23 × 3) + 27 + (19 × 6) = 69 + 27 + 114 = 210 g/mol.
(i) Percentage of Fluorine = (114 / 210) × 100 = 54.28% = 54%
(ii) Percentage of Sodium = (69 / 210) × 100 = 32.85% = 33%
(iii) Percentage of Aluminium = (27 / 210) × 100 = 12.85% = 13%

Teacher's Note:
a) Always calculate total molecular mass first before finding individual percentages.
b) Round off to the nearest whole number as specified in the question.

(c) (i) State the volume occupied by 40 gm of methane at STP, if its vapour density (V.D.) is 8. [4 Marks]
(ii) Calculate the number of moles present in 160 gm of NaOH. [Atomic Mass: Na = 23, H = 1, O = 16]

Answer:
(i) Molecular mass = 2 × Vapour Density = 2 × 8 = 16 g/mol.
Number of moles = Mass / Molar mass = 40 / 16 = 2.5 moles.
Volume at STP = 2.5 × 22.4 = 56 liters.
(ii) Molar mass of NaOH = 23 + 1 + 16 = 40 g/mol.
Number of moles = 160 / 40 = 4 moles.

Teacher's Note:
a) Vapour density is half of molecular mass.
b) One mole of any gas at STP occupies 22.4 liters.

Question 6

(a) Identify the salts P, Q, R from the following observations: [3 Marks]
(i) Salt P has light bluish green colour. On heating, it produces a black coloured residue. Salt P produces brisk effervescence with dil. HCl and the gas evolved turns lime water milky, but no action with acidified potassium dichromate solution.
(ii) Salt Q is white in colour. On strong heating, it produces buff yellow residue and liberates reddish brown gas. Solution of salt Q produces chalky white insoluble precipitate with excess of ammonium hydroxide.
(iii) Salt R is black in colour. On reacting with concentrated HCl, it liberates a pungent greenish yellow gas which turns moist starch iodide paper blue black.

Answer:
(i) Salt P: Copper carbonate (CuCO₃)
(ii) Salt Q: Lead nitrate (Pb(NO₃)₂)
(iii) Salt R: Manganese dioxide (MnO₂)

Teacher's Note:
a) Bluish-green salt yielding black CuO on heating indicates copper carbonate.
b) Greenish-yellow chlorine gas liberated with MnO₂ and conc. HCl confirms manganese dioxide.

(b) Identify the substance underlined in each of the following: [3 Marks]
(i) The electrode that increases in mass during the electro-refining of silver.
(ii) The acid that is a dehydrating as well as a drying agent.
(iii) The catalyst used to oxidize ammonia into nitric oxide.

Answer:
(i) Cathode (Pure silver strip)
(ii) Concentrated sulphuric acid (Concentrated H₂SO₄)
(iii) Platinum (Pt)

Teacher's Note:
a) During electro-refining, pure metal deposits on the cathode, increasing its mass.
b) Platinum gauze acts as a catalyst in Ostwald's process for catalytic oxidation of ammonia.

(c) Copy and complete the following paragraph using the options given in brackets: [4 Marks]
Alkenes are a homologous series of (i) __________ (saturated / unsaturated) hydrocarbons characterized by the general formula (ii) (C_nH_2n+2 / C_nH_2n). Alkenes undergo (iii) __________ (addition / substitution) reactions and also undergo (iv) __________ (hydrogenation / dehydrogenation) to form alkanes.

Answer:
(i) unsaturated
(ii) C_nH_2n
(iii) addition
(iv) hydrogenation

Teacher's Note:
a) Alkenes contain a carbon-carbon double bond, making them unsaturated hydrocarbons.
b) Addition of hydrogen to alkenes (hydrogenation) yields saturated alkanes.

Question 7

(a) Write balanced chemical equations, for the preparation of the given salts (i) to (iii) by using the methods A to C respectively: [3 Marks]
A: Neutralization B: Precipitation C: Titration
(i) Copper sulphate
(ii) Zinc carbonate (Note: paper text has minor numbering glitch reading i, ii, ii)
(iii) Ammonium sulphate

Answer:
(i) CuO + H₂SO₄ → CuSO₄ + H₂O (Neutralization)
(ii) ZnSO₄ + Na₂CO₃ → ZnCO₃↓ + Na₂SO₄ (Precipitation)
(iii) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O (Titration)

Teacher's Note:
a) Different salts require specific preparation methods depending on their solubility.
b) Insoluble salts are prepared by precipitation (ionic synthesis).

(b) Name the following elements: [3 Marks]
(i) An alkaline earth metal present in group 2 and period 3.
(ii) A trivalent metal used to make light tools.
(iii) A monovalent non-metal present in fluorspar.

Answer:
(i) Magnesium (Mg)
(ii) Aluminium (Al)
(iii) Fluorine (F)

Teacher's Note:
a) Magnesium is in group 2 with atomic number 12, belonging to period 3.
b) Fluorspar is calcium fluoride (CaF₂), containing fluorine.

(c) An aqueous solution of nickel (II) sulphate was electrolyzed using nickel electrodes. Observe the diagram and answer the questions that follow: [4 Marks]

[Figure: Electrolytic cell showing power supply connected to nickel cathode and nickel anode immersed in an aqueous solution of nickel (II) sulphate]

(i) What do you observe at the cathode and anode respectively?
(ii) Name the cation that remains as a spectator ion in the solution.
(iii) Which equation for the reaction at the anode is correct?
1. Ni → Ni²⁺ + 2e^-
2. Ni + 2e^- → Ni²⁺
3. Ni²⁺ + 2e^- → 2e^-
4. Ni²⁺ + 2e^- → Ni

Answer:
(i) Cathode: Nickel gets deposited (increases in size/mass). Anode: Nickel anode dissolves (decreases in mass).
(ii) Hydrogen ion (H⁺)
(iii) 1. Ni → Ni²⁺ + 2e^-

Teacher's Note:
a) In electrorefining/electroplating with active metal electrodes, the anode dissolves while the cathode accumulates the metal.
b) Hydrogen ions remain in solution as spectator ions since nickel ions preferentially discharge at the cathode due to their position in the electrochemical series.