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Chemistry.
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Grade 11 · Lahore Board · 2024 · Group I
Complete objective paper · English questions

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17 questions. 17 marks.

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QUESTION 1 · 1 MARKEqual masses of methane and oxygen gases are mixed in an empty container at 25 °C. The fraction of the total pressure exerted by oxygen is:

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Correct answer: A) 1/3

Dalton’s law of partial pressures: the fraction of total pressure contributed by a gas equals its mole fraction.

  1. Let equal masses of CH4 and O2 be m each.
  2. Moles: n(CH4) = m/16 and n(O2) = m/32.
  3. Mole fraction of O2 = (m/32) / (m/16 + m/32) = (1/32) / (3/32) = 1/3, so oxygen exerts one-third of the total pressure.
QUESTION 2 · 1 MARKOrbitals having same energy are called:

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Correct answer: C) Degenerate orbitals

Orbitals with identical energy are called degenerate orbitals.

  1. The term 'degenerate' in atomic structure means equal in energy.
  2. Orbitals of the same subshell in a free atom often have this property.
  3. Therefore the correct name is degenerate orbitals.
QUESTION 3 · 1 MARKFor which system does the equilibrium constant Kc has units of (concentration)⁻¹:

Source note: Option C is printed as “2NO ⇌ N₂O₄”, an unbalanced equation. The intended standard reaction is 2NO₂ ⇌ N₂O₄; its concentration exponents still identify C.

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Correct answer: C) 2NO ⇌ N2O4

For a reaction, units of Kc depend on the difference between total products and reactants stoichiometric coefficients.

  1. For the intended dimerization 2NO₂ ⇌ N₂O₄, write Kc = [N₂O₄]/[NO₂]².
  2. Subtract the total reactant exponent from the product exponent: 1 − 2 = −1. Hence the school-level units are (mol L⁻¹)⁻¹.
  3. Option A has exponent 2 − 4 = −2, while B and D have exponent 0. Therefore C is the intended choice. The printed “NO” is a source error, not a balanced chemical equation.
QUESTION 4 · 1 MARKAll the photochemical reactions are usually:

Source note: The printed word “All” is too broad. C is the usual textbook answer for constant illumination with essentially complete absorption; reaction order depends on conditions.

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Correct answer: C — Zero order reactions (under the usual textbook light-limited assumption)

When the absorbed photon supply is fixed, a photochemical rate can be independent of reactant concentration.

  1. Light supplies the photons that initiate a photochemical reaction.
  2. In the textbook limiting case, the reactant absorbs essentially all available light and the illumination remains constant. The number of absorbed photons per second then stays approximately constant.
  3. The rate is approximately k[A]⁰ = k, so the expected choice is zero order, C.
  4. This is not a universal rule: under other absorption conditions or mechanisms, photochemical reactions can have other reaction orders.
QUESTION 5 · 1 MARKThe largest number of molecules are present in:

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Correct answer: A) 3.6 g of H2O

For equal comparison of molecules, compare number of moles: larger moles means more molecules.

  1. Moles in each sample are mass ÷ molar mass.
  2. H2O: 3.6/18 = 0.20 mol; C2H5OH: 4.8/46 ≈ 0.104 mol; CO: 2.8/28 = 0.10 mol; N2O5: 5.4/108 = 0.05 mol.
  3. The 3.6 g water sample has the greatest amount, 0.20 mol. Since number of molecules = moles × Avogadro’s constant, it contains the most molecules.
QUESTION 6 · 1 MARKWhich of the following hydrogen halide has the highest percentage of ionic character:

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Correct answer: C) HF

Ionic character increases with electronegativity difference; HF has the largest H–X electronegativity difference among hydrogen halides.

  1. Compare electronegativity of halogens: F is the most electronegative.
  2. Therefore the H–F bond is the most polar among HCl, HBr, HF and HI.
  3. Greater polarity means greater ionic character, so HF is correct.
QUESTION 7 · 1 MARKThe colour of iodine in organic layer is:

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Correct answer: C) Purple

Iodine dissolves in many organic solvents to give a violet/purple colour.

  1. Iodine is much more soluble in non-polar organic solvents than in water.
  2. In the organic layer it shows a characteristic violet to purple colour.
  3. Hence the correct option is purple.
QUESTION 8 · 1 MARKThe bond order of N2 molecule is:

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Correct answer: B) Three

Bond order = (number of bonding electrons − number of antibonding electrons) / 2.

  1. Use bond order = (number of bonding electrons − number of antibonding electrons)/2.
  2. For N₂, there are 10 bonding electrons and 4 antibonding electrons in the molecular-orbital description.
  3. Bond order = (10 − 4)/2 = 3, corresponding to a triple bond. Choose B.
QUESTION 9 · 1 MARKThe cathodic reaction in the electrolysis of dil. H2SO4 with pt electrode is:

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Correct answer: B) Reduction

Cathode is the electrode where reduction takes place.

  1. At the cathode, cations gain electrons.
  2. In dilute H2SO4, H+ ions are reduced to H2 gas.
  3. Therefore the cathodic reaction is reduction.
QUESTION 10 · 1 MARKΔHv value of C6H14 should be --- than that of C2H6:

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Correct answer: A) Greater

Heat of vaporization increases with molecular size and stronger intermolecular forces in a homologous series.

  1. C6H14 is larger than C2H6 and has stronger London dispersion forces.
  2. More energy is needed to separate its molecules into vapour.
  3. So ΔHv for C6H14 is greater than that for C2H6.
QUESTION 11 · 1 MARKThe phenomenon of isotopy was introduced by:

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Correct answer: A) Soddy

Frederick Soddy introduced the concept of isotopes.

  1. Isotopes are atoms of the same element with different mass numbers.
  2. This concept was introduced by Frederick Soddy.
  3. Hence the correct option is Soddy.
QUESTION 12 · 1 MARKDiamond is bad conductor because:

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Correct answer: C) There are no free electrons present in the crystal of diamond to conduct electricity

Diamond is a giant covalent solid with all valence electrons tied up in σ-bonds, so it lacks mobile charge carriers.

  1. Each carbon atom in diamond is sp3 hybridized and bonded to four others.
  2. All valence electrons are used in strong covalent bonds.
  3. Because no free electrons are available, diamond does not conduct electricity well.
QUESTION 13 · 1 MARKDuring liquefaction of gases the intermolecular spaces:

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Correct answer: A) Decreases

Liquefaction brings particles closer together, so intermolecular spaces become smaller.

  1. Gas molecules are far apart in the gaseous state.
  2. On liquefaction, attractive forces pull molecules nearer.
  3. Therefore intermolecular space decreases.
QUESTION 14 · 1 MARKRutherford model of atom failed because:

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Correct answer: C) It did not account for the stability of atom

According to classical physics, orbiting electrons should lose energy and fall into the nucleus, so Rutherford’s model could not explain atomic stability.

  1. Rutherford proposed a nuclear model with electrons revolving around the nucleus.
  2. Classical theory predicts that accelerated electrons should radiate energy continuously.
  3. Thus the model could not explain why atoms remain stable.
QUESTION 15 · 1 MARKPaper chromatography is known as:

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Correct answer: B) Partition chromatography

In paper chromatography, components distribute between water held in paper fibers and the moving solvent, so it is partition chromatography.

  1. The stationary phase is water adsorbed on cellulose fibers of the paper.
  2. The mobile phase is the solvent that moves up the paper.
  3. Separation occurs by partitioning between these two phases.
QUESTION 16 · 1 MARKWhich one is not a state function:

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Correct answer: C) Work

State functions depend only on the state of the system, while work depends on the path taken.

  1. Temperature, internal energy and volume are state functions.
  2. Work changes with the path of the process, so it is not a state function.
  3. Therefore work is the correct answer.
QUESTION 17 · 1 MARKMolarity of pure water is:

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Correct answer: C) 55.5

Pure water contains about 55.5 moles per liter because 1 L of water has a mass close to 1000 g and molar mass 18 g mol⁻¹.

  1. Moles in 1 L water ≈ 1000/18.
  2. 1000/18 ≈ 55.5 mol.
  3. Therefore the molarity of pure water is about 55.5 M.

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Original exam paper sourced from Ilmkidunya ↗. Pastpaper’s solutions are study support, not an official board marking key. Where an option is unclear in the scan, the source note is retained.