ICSE Class 10 Chemistry Board Exam Question Paper 2016 with Solutions

ICSE Class 10 Chemistry Board Exam Question Paper with Solutions

SECTION-I

Question 1

(a) Fill in the blanks with the choices given in brackets. [5]
1. Metals are good .......... (oxidizing agents/reducing agents) because they are electron .......... (acceptors/donors).
2. Electrovalent compounds have .......... (high/low) melting points.
3. Higher the pH value of a solution, the more .......... (acidic/alkaline) it is.
4. .......... (AgCl/PbCl₂) white precipitate is soluble in excess NH₄OH.
5. Conversion of ethene to ethane is an example of .......... (hydration/hydrogenation).

Answer:
1. Reducing agents, donors
2. High
3. Alkaline
4. AgCl
5. Hydrogenation

Teacher's Note:
a) Metals readily lose electrons to form positive ions, which makes them good reducing agents.
b) Electrovalent compounds possess strong electrostatic forces of attraction between oppositely charged ions, requiring high thermal energy to break.

(b) Choose the correct answer from the options given below: [5]
1. An element with the atomic number 19 will most likely combine chemically with the element whose atomic number is:
(A) 17
(B) 11
(C) 18
(D) 20
2. The ratio between the number of molecules in 2g of hydrogen and 32g of oxygen is:
(A) 1 : 2
(B) 1 : 0.01
(C) 1 : 1
(D) 0.01 : 1
[Given that H = 1, O = 16]
3. The two main metals in Bronze are:
(A) Copper and zinc
(B) Copper and lead
(C) Copper and nickel
(D) Copper and tin
4. The particles present in strong electrolytes are:
(A) Only molecules
(B) Mainly ions
(C) Ions and molecules
(D) Only atoms
5. The aim of the Fountain Experiment is to prove that:
(A) HCl turns blue litmus red
(B) HCl is denser than air
(C) HCl is highly soluble in water
(D) HCl fumes in moist air.

Answer: (A) 17

Element with atomic number 19 is Potassium (2, 8, 8, 1), which forms an ionic bond with Chlorine (atomic number 17, electronic configuration 2, 8, 7).

Teacher's Note:
a) Element 19 needs to lose 1 electron, whereas element 17 needs to gain 1 electron to achieve stability.
b) Metals react with non-metals to form electrovalent compounds.

Answer: (C) 1 : 1

Moles of H₂ = 2 / 2 = 1 mole. Moles of O₂ = 32 / 32 = 1 mole. Ratio of molecules is equal to ratio of moles.

Teacher's Note:
a) Equal moles of any gas contain the same number of molecules under Avogadro's hypothesis.
b) Always convert mass to moles before finding molecular ratios.

Answer: (D) Copper and tin

Bronze is an alloy primarily composed of copper and tin.

Teacher's Note:
a) Memorize common alloys and their constituent metals for objective questions.
b) Brass is composed of copper and zinc, while bronze consists of copper and tin.

Answer: (B) Mainly ions

Strong electrolytes undergo almost complete dissociation in aqueous solution yielding mainly ions.

Teacher's Note:
a) Strong electrolytes conduct electricity well due to a high concentration of free mobile ions.
b) Weak electrolytes exist predominantly as molecules with very few ions.

Answer: (C) HCl is highly soluble in water

The fountain experiment demonstrates the high solubility of hydrogen chloride gas in water.

Teacher's Note:
a) Rapid dissolution creates a partial vacuum, drawing water up through the jet to form the fountain.
b) This property is specific to gases like HCl and ammonia (NH₃).

(c) Write balanced chemical equations for each of the following: [5]
1. Action of warm water on AlN.
2. Action of hot and concentrated Nitric acid on copper.
3. Action of Hydrochloric acid on sodium bicarbonate.
4. Action of dilute Sulphuric acid on Sodium Sulphite.
5. Preparation of ethanol from Ethyl Chloride.

Answer:
1. AlN + 3H₂O → Al(OH)₃ + NH₃↑
2. Cu + 4HNO₃ (conc.) → Cu(NO₃)₂ + 2H₂O + 2NO₂↑
3. NaHCO₃ + HCl → NaCl + H₂O + CO₂↑
4. Na₂SO₃ + H₂SO₄ (dil.) → Na₂SO₄ + H₂O + SO₂↑
5. C₂H₅Cl + KOH (aq) → C₂H₅OH + KCl

Teacher's Note:
a) Ensure all state symbols and gas/precipitate arrows are clearly marked where required.
b) Differentiate between reactions of dilute and concentrated acids on metals and salts.

(d) State your observations when: [5]
1. Dilute Hydrochloric acid is added to Lead nitrate solution and the mixture is heated.
2. Barium chloride solution is mixed with Sodium Sulphate Solution.
3. Concentrated Sulphuric acid is added to Sugar Crystals.
4. Dilute Hydrochloric acid is added to Copper carbonate.
5. Dilute Hydrochloric acid is added to Sodium thiosulphate.

Answer:
1. A white precipitate of lead chloride appears which dissolves on heating and reappears on cooling.
2. A white insoluble precipitate of Barium sulphate is obtained.
3. White crystals of sugar turn to a black spongy mass known as sugar charcoal.
4. A brisk effervescence with the release of a colourless, odourless, acidic gas that extinguishes a glowing splint (carbon dioxide) is evolved.
5. A colourless gas with the smell of burning sulphur (sulphur dioxide) is released, and a yellow precipitate/crystal of sulphur settles at the bottom.

Teacher's Note:
a) Observation questions require exact descriptions of color changes, precipitates, or gas evolutions.
b) Mention the solubility of precipitates and the specific characteristic smell or test of gases.

(e) Identify the term/substance in each of the following: [5]
1. The tendency of an atom to attract electrons to itself when combined in a compound.
2. The method used to separate ore from gangue by preferential wetting.
3. The catalyst used in the conversion of ethyne to ethane.
4. The type of reactions alkenes undergo.
5. The electrons present in the outermost shell of an atom.

Answer:
1. Electron affinity (or Electronegativity)
2. Froth Floatation Method
3. Nickel (Ni) [or Platinum/Palladium]
4. Addition reaction
5. Valence electron

Teacher's Note:
a) Read the definitions carefully to distinguish between electron affinity and electronegativity.
b) Remember that alkenes possess double bonds, making them prone to addition reactions.

(f) (i) A gas of mass 32gms has volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.
(ii) How much Calcium oxide is formed when 82g of calcium nitrate is heated? Also find the volume of nitrogendioxide evolved:
2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂
(Ca = 40, N = 14, O = 16) [5]

Answer:
(i) Mass of 20 litres of gas at S.T.P. = 32 g
Volume of 1 mole of gas at S.T.P. = 22.4 litres
Gram molecular weight = (32 / 20) × 22.4 = 35.84 g ≈ 36 g.

(ii) Molecular mass of Ca(NO₃)₂ = 40 + 2 × (14 + 48) = 40 + 2 × 62 = 328 g.
From the equation:
2 moles [2 × 328 = 656 g] of Ca(NO₃)₂ give 2 moles [2 × 56 = 112 g] of CaO and 4 moles [4 × 22.4 = 89.6 litres] of NO₂ at S.T.P.
1. Mass of CaO formed from 82 g of Ca(NO₃)₂:
= (112 / 328) × 82 = 28 g of CaO.
2. Volume of NO₂ evolved at S.T.P. from 82 g of Ca(NO₃)₂:
= (89.6 / 328) × 82 = 22.4 litres of nitrogen dioxide.

Teacher's Note:
a) Apply unitary method or mole concept carefully, ensuring atomic masses are added correctly.
b) State units clearly at every step of numerical calculations.

(g) Match the salts given in Column I with their method of preparation given in Column II: [5]

Answer:
(i) - (C) Neutralisation
(ii) - (A) Simple displacement
(iii) - (E) Combination
(iv) - (B) Titration
(v) - (D) Precipitation

Teacher's Note:
a) Match each salt based on the chemical reactions involved in laboratory preparations.
b) Double check the preparation routes for soluble and insoluble salts.

(h) (i) Write the IUPAC names of each of the following: [5]
1. H - C = C - C - H (with H atoms attached: Prop-1-ene)
2. H - C - C ≡ C - C - H (with H atoms attached: But-2-yne)
3. H - C = O (with C, H atoms attached: Ethanal)
(ii) Rewrite the following sentences by using the correct symbol > (greater than) or < (less than) in the blanks given:
1. The ionization potential of Potassium is .......... that of Sodium.
2. The electronegativity of Iodine is .......... that of Chlorine.

Answer:
(i) 1. Prop-1-ene
2. But-2-yne (Note: OCR prints Butr-2-yne, correct IUPAC name is But-2-yne)
3. Ethanal (Ethanol/Ethanal as per structure, printed as ethan-1-al)

(ii) 1. Potassium ionization potential < Sodium ionization potential
2. Iodine electronegativity < Chlorine electronegativity

Teacher's Note:
a) Ionization potential and electronegativity both decrease down a group as atomic size increases.
b) Number carbon chains correctly to assign lowest locants for double or triple bonds.

SECTION-II

Question 2:

(a) Use the letters only written in the Periodic Table given below to answer the questions that follow: [4]
[Figure: Partial periodic table showing groups I, II, III, IV, V, VI, VII, 0 and periods 1 to 5. Element L in group 0 period 1; Q in group II period 2; E, G, J, Z, M in period 3; R in group II period 3; T in group II period 4]
1. State the number of valence electrons in atom J.
2. Which element shown forms ions with a single negative charge?
3. Which metallic element is more reactive than R?
4. Which element has its electrons arranged in four shells?

Answer:
1. 5 valence electrons
2. Element M
3. Element T
4. Element T

Teacher's Note:
a) Group number corresponds to the number of valence electrons for representative elements.
b) Metallic character and reactivity increase down a group.

(b) Fill in the blanks by selecting the correct word from the brackets: [2]
1. If an element has a low ionization energy then it is likely to be .......... (metallic/non metallic).
2. If an element has seven electrons in its outermost shell then it is likely to have the .......... (largest/smallest) atomic size among all the elements in the same period.

Answer:
1. Metallic
2. Smallest

Teacher's Note:
a) Metals lose electrons easily, which corresponds to low ionization energy.
b) Across a period from left to right, atomic size decreases while nuclear charge increases, making halogens the smallest in their respective periods.

(c) The following table shows the electronic configuration of the elements W, X, Y, Z:

Answer the following questions based on the table above:
1. What type of Bond is formed between: [2]
1. W and X    2. Y and Z
2. What is the formula of the compound formed between: [2]
1. X and Z    2. W and X

Answer:
1. 1. Ionic bond
2. Covalent bond
2. 1. XZ (or ZX)
2. WX

Teacher's Note:
a) Metal-nonmetal combinations form ionic bonds, whereas nonmetal-nonmetal combinations form covalent bonds.
b) Use valencies to cross-multiply and determine chemical formulas.

Question 3:

(a) Write a balanced chemical equation for each the following: [3]
1. Burning of ethane in plentiful supply of air.
2. Action of water on Calcium carbide.
3. Heating of Ethanol at 170°C in the presence of conc. Sulphuric acid.

Answer:
1. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
2. CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂↑
3. C₂H₅OH [conc. H₂SO₄ / 170°C] → C₂H₄↑ + H₂O

Teacher's Note:
a) Concentrated sulfuric acid acts as a dehydrating agent in the preparation of ethene from ethanol.
b) Ensure complete balancing of hydrocarbon combustion equations.

(b) Give the structural formulae of each of the following: [3]
1. 2-methyl propane
2. Ethanoic acid
3. Butan-2-ol

Answer:
1. 2-Methylpropane: Central carbon bonded to three methyl groups and one hydrogen atom.
2. Ethanoic acid: CH₃ - COOH (with C=O and C-OH group on the second carbon).
3. Butan-2-ol: CH₃ - CH(OH) - CH₂ - CH₃.

Teacher's Note:
a) Draw all carbon, hydrogen, and functional group bonds clearly in structural formulae.
b) Verify that each carbon atom forms exactly four covalent bonds.

(c) Equation for the reaction when compound A is bubbled through bromine dissolved in carbon tetrachloride is as follows:
[Figure: A + Br₂/CCl₄ → CH₂Br - CH₂Br]
1. Draw the structure of A.
2. State your observation during this reaction. [2]

Answer:
1. Compound A is Ethene (C₂H₄). Structure: H₂C=CH₂ (Double bond between two carbon atoms, each attached to two hydrogen atoms).
2. The reddish-brown colour of bromine in carbon tetrachloride gets decolourised.

Teacher's Note:
a) Decolourisation of bromine water/solution is a standard confirmatory test for unsaturation (alkenes and alkynes).
b) Addition reactions destroy the multiple bond.

(d) Fill in the blanks using the appropriate words given below: [2]
(Sulphur dioxide, Nitrogen dioxide, Nitric oxide, Sulphuric acid)
1. Cold, dilute nitric acid reacts with copper to give ..........
2. Hot, concentrated nitric acid reacts with sulphur to form ..........

Answer:
1. Nitric oxide
2. Sulphuric acid

Teacher's Note:
a) Dilute nitric acid typically yields nitric oxide (NO) upon reaction with metals like copper.
b) Concentrated oxidizing acids like nitric acid oxidize non-metals such as sulfur to sulfuric acid.

Question 4:

(a) Identify the gas evolved and give the chemical test in each of the following case: [2]
1. Dilute hydrochloric acid reacts with sodium sulphite.
2. Dilute hydrochloric acid reacts with iron (II) sulphide.

Answer:
1. Sulphur dioxide (SO₂) gas is evolved.
Chemical test: It turns acidified potassium dichromate solution from orange to green.
2. Hydrogen sulphide (H₂S) gas is evolved.
Chemical test: It turns lead acetate paper silvery black.

Teacher's Note:
a) Always state both the gas name and its specific confirmatory test.
b) H₂S has a characteristic smell of rotten eggs.

(b) State your observations when ammonium hydroxide solution is added drop by drop and then in excess to each of the following solutions: [2]
1. copper sulphate solution.
2. zinc sulphate solution.

Answer:
1. A pale blue/bluish-white precipitate is formed which dissolves in excess ammonium hydroxide to give an inky/deep blue solution.
2. A gelatinous white precipitate is formed which dissolves in excess ammonium hydroxide to give a clear, colourless solution.

Teacher's Note:
a) Mention both the precipitate formation stage and its behaviour in excess reagent.
b) Deep blue solution indicates formation of tetraamminecopper(II) complex.

(c) Write equations for the reactions taking place at the two electrodes (mentioning clearly the name of the electrode) during the electrolysis of: [4]
1. Acidified copper sulphate solution with copper electrodes.
2. Molten lead bromide with inert electrodes.

Answer:
1. Acidified CuSO₄ with Cu electrodes:
At Anode (Copper): Cu - 2e^- → Cu²⁺
At Cathode (Copper): Cu²⁺ + 2e^- → Cu
2. Molten lead bromide with inert (graphite) electrodes:
At Anode (Graphite): Br^- - e^- → Br, then Br + Br → Br₂
At Cathode (Graphite): Pb²⁺ + 2e^- → Pb

Teacher's Note:
a) Note that active copper electrodes participate in the reaction during copper refining/electrolysis.
b) Inert electrodes do not take part in the chemical reaction.

(d) 1. Name the product formed at the anode during the electrolysis of acidified water using platinum electrodes.
2. Name the metallic ions that should be present in the electrolyte when an article made of copper is to be electroplated with silver. [2]

Answer:
1. Oxygen gas (O₂)
2. Silver ions (Ag⁺)

Teacher's Note:
a) During electroplating, the electrolyte must contain the ions of the metal being plated (coating metal).
b) Hydroxide ions or water molecules discharge at the anode during acidified water electrolysis, yielding oxygen.

Question 5:

(a) A gas cylinder contains 12 × 10²⁴ molecules of oxygen gas. If Avogadro's number is 6 × 10²³; Calculate:
(i) the mass of oxygen present in the cylinder.
(ii) the volume of oxygen at S.T.P. present in the cylinder. [O = 16] [2]

Answer:
(i) Number of moles = (12 × 10²⁴) / (6 × 10²³) = 20 moles.
Mass of oxygen = Moles × Molar mass = 20 × 32 g = 640 g.
(ii) Volume of oxygen at S.T.P. = Moles × 22.4 litres = 20 × 22.4 = 448 litres.

Teacher's Note:
a) Use the relation: Number of moles = Given molecules / Avogadro number.
b) Multiply moles by molar mass for mass, and by 22.4 L for volume at S.T.P.

(b) A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. [C = 12, H = 1] [3]

Answer:

Empirical formula = C₂H₅.
Empirical formula mass = (2 × 12) + (5 × 1) = 29.
Molecular mass = 2 × Vapour Density = 2 × 29 = 58.
n = Molecular mass / Empirical formula mass = 58 / 29 = 2.
Molecular formula = (Empirical formula)_n = (C₂H₅)₂ = C₄H₁₀.

Teacher's Note:
a) Always determine the empirical formula first using percentage composition and atomic masses.
b) Multiply the empirical formula by n (Molecular Mass / Empirical Formula Mass) to get the molecular formula.

(c) The equation 4NH₃ + 5O₂ → 4NO + 6H₂O, represents the catalytic oxidation of ammonia. If 100 cm³ of ammonia is used calculate the volume of oxygen required to oxidise the ammonia completely. [3]

Answer:
According to Gay-Lussac's Law of Combining Volumes:
4 volumes of NH₃ require 5 volumes of O₂.
Therefore, 1 volume of NH₃ requires (5 / 4) volumes of O₂.
Volume of O₂ required for 100 cm³ of NH₃ = (5 / 4) × 100 cm³ = 125 cm³.

Teacher's Note:
a) Gas stoichiometric calculations depend directly on the molar coefficients from the balanced equation.
b) State the gas law applied clearly in the solution steps.

(d) By drawing an electron dot diagram show the formation of Ammonium Ion [Atomic No.: N = 7 and H = 1] [2]

Answer:
Ammonia (NH₃) molecule has a lone pair on nitrogen and three covalent bonds with hydrogen atoms. When it reacts with a hydrogen ion (proton, H⁺), the lone pair is donated to form a coordinate (dative) bond, resulting in the ammonium ion [NH₄]⁺.
[Figure: Electron dot diagram showing ammonia molecule with a lone pair on N, combining with H⁺ to form [H-N(H)-H]⁺ with one dative bond]

Teacher's Note:
a) Clearly show the lone pair on the nitrogen atom of ammonia.
b) Indicate the coordinate bond and enclose the entire structure in square brackets with a positive charge.

Question 6:

(a) Name the gas evolved when the following mixtures are heated: [2]
1. Calcium hydroxide and Ammonium Chloride.
2. Sodium Nitrite and Ammonium Chloride.

Answer:
1. Ammonia gas (NH₃)
2. Nitrogen gas (N₂)

Teacher's Note:
a) Heating an ammonium salt with a fixed alkali liberates ammonia gas.
b) Heating a mixture of sodium nitrite and ammonium chloride produces nitrogen gas.

(b) Write balanced chemical equations for each of the following: [2]
(i) When excess of ammonia is treated with chlorine.
(ii) An equation to illustrate the reducing nature of ammonia.

Answer:
(i) 8NH₃ (excess) + 3Cl₂ → 6NH₄Cl + N₂↑
(ii) 3CuO + 2NH₃ [Δ] → 3Cu + 3H₂O + N₂↑

Teacher's Note:
a) Note the difference between excess ammonia and excess chlorine reactions with chlorine.
b) Ammonia reduces metallic oxides like copper oxide to metal while being oxidized to nitrogen gas.

(c) A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.
A = Typical acid property
B = Non volatile acid
C = Oxidizing agent
D = Dehydrating agent
Choose the property (A, B, C or D) depending on which is relevant to each of the following: [3]
1. Preparation of Hydrogen chloride gas.
2. Preparation of Copper sulphate from copper oxide.
3. Action of conc. Sulphuric acid on Sulphur.

Answer:
1. B - Non volatile acid
2. A - Typical acid property
3. C - Oxidizing agent

Teacher's Note:
a) Non-volatile sulfuric acid displaces more volatile acids like HCl from their salts.
b) Concentrated sulfuric acid acts as a strong oxidizing agent when reacting with non-metals like sulfur.

(d) Give reasons why: [3]
1. Sodium Chloride will conduct electricity only in fused or aqueous solution state.
2. In the electroplating of an article with silver, the electrolyte sodium argento-cyanide solution is preferred over silver nitrate solution.
3. Although copper is a good conductor of electricity, it is a non-electrolyte.

Answer:
1. In solid state, sodium chloride has strong electrostatic forces holding its ions in a crystal lattice with no free mobile ions. In fused or aqueous state, ions become free to conduct electricity.
2. Silver nitrate causes very rapid deposition leading to a non-uniform coating, whereas sodium argento-cyanide undergoes slow dissociation, ensuring a smooth and even coating.
3. Copper conducts electricity through the flow of free electrons (metallic conduction) and does not undergo chemical decomposition or form ions in solution, hence it is a non-electrolyte.

Teacher's Note:
a) Electrolytes must contain free mobile ions either in molten state or in solution.
b) Complex salts provide a slow and steady supply of metal ions for uniform electroplating.

Question 7:

(a) (i) Name the solution used to react with Bauxite as a first step in obtaining pure aluminium oxide, in the Baeyer's process.
(ii) Write the equation for the reaction where the aluminium oxide for the electrolytic extraction of aluminium is obtained by heating aluminium hydroxide.
(iii) Name the compound added to pure alumina to lower the fusion temperature during the electrolytic reduction of alumina.
(iv) Write the equation for the reaction that occurs at the cathode during the extraction of aluminium by electrolysis.
(v) Explain why it is preferable to use a number of graphite electrodes as anode instead of a single electrode, during the above electrolysis. [5]

Answer:
(i) Concentrated sodium hydroxide solution (Caustic soda, NaOH)
(ii) 2Al(OH)₃ [1000°C] → Al₂O₃ + 3H₂O
(iii) Cryolite (Na₃AlF₆) [or Fluorspar / Alumina-cryolite-fluorspar mixture]
(iv) At Cathode: Al³⁺ + 3e^- → Al
(v) Oxygen gas evolved at the anode reacts with the hot graphite rods to form carbon dioxide, causing them to burn away gradually. Using multiple graphite electrodes allows easy replacement without stopping the process.

Teacher's Note:
a) Baeyer's process uses caustic soda to dissolve bauxite and separate impurities like iron oxide.
b) Cryolite lowers the melting point of alumina and increases its electrical conductivity.

(b) State what would you observe when:
1. Washing Soda Crystals are exposed to the atmosphere.
2. The salt ferric chloride is exposed to the atmosphere. [2]

Answer:
1. Washing soda crystals (sodium carbonate decahydrate) lose their water of crystallization and crumble to form an amorphous white powder (efflorescence).
2. Ferric chloride absorbs moisture from the atmosphere and dissolves in it to form a saturated solution (deliquescence).

Teacher's Note:
a) Efflorescence is the loss of water of crystallization when exposed to dry air.
b) Deliquescence is the absorption of atmospheric moisture by a substance to form a solution.

(c) Identify the cations in each of the following case:
1. NaOH solution when added to the Solution (A) gives a reddish brown precipitate.
2. NH₄OH Solution when added to the Solution (B) gives white ppt which does not dissolve in excess.
3. NaOH Solution when added to Solution (C) gives white ppt which is insoluble in excess. [3]

Answer:
1. Ferric ion (Fe³⁺)
2. Lead ion (Pb²⁺)
3. Calcium ion (Ca²⁺)

Teacher's Note:
a) Analytical chemistry identification tables based on alkali precipitates are crucial for these questions.
b) Calcium hydroxide is sparingly soluble, leading to an insoluble white precipitate with excess NaOH.