BSc Second Year (Sem 3 & 4) Chemistry – Daily Revision MCQ (English Medium)
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Physical adsorption (physisorption) of a gas on a solid surface is mainly due to:
Chapter 1: Adsorption and Surface Chemistry
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Answer: B. Weak van der Waals forces
Physisorption involves weak van der Waals attraction, so it has low enthalpy and is reversible; chemisorption involves chemical bonds.A Carnot engine works between a source at 500 K and a sink at 300 K. Its efficiency is:
Chapter 4: Second and Third Law of Thermodynamics
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Answer: A. 40%
η = 1 − T₂/T₁ = 1 − 300/500 = 0.4, i.e. 40%.Kohlrausch's law of independent migration of ions states that, at infinite dilution, the molar conductivity of an electrolyte equals:
Chapter 5: Electrochemistry
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Answer: C. The sum of the limiting molar conductivities of its cation and anion
At infinite dilution each ion moves independently, so Λ°m = ν₊λ°₊ + ν₋λ°₋ (a sum of ionic contributions).Which of the following ions is colourless in aqueous solution?
Chapter 7: Transition Elements
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Answer: D. Zn²⁺
Zn²⁺ has a completely filled 3d¹⁰ configuration, so no d–d transition is possible and it is colourless.Reduction of nitrobenzene with tin and hydrochloric acid (Sn/HCl) gives:
Chapter 8: Nitro Compounds
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Answer: B. Aniline
In acidic medium (Sn/HCl) the –NO₂ group is fully reduced to –NH₂, giving aniline (freed from its salt by alkali).What is the coordination number of cobalt in the complex ion [Co(en)₃]³⁺ (en = ethylenediamine)?
Chapter 10: Coordination Compounds
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Answer: B. 6
Each en is bidentate (two N donor atoms), so three en ligands give 3 × 2 = 6 donor atoms.The oxidation number of manganese in KMnO₄ is:
Chapter 11: Redox Reactions
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Answer: D. +7
K is +1 and four O give −8, so Mn = 0 − 1 + 8 = +7.In titrations using potassium permanganate (KMnO₄) in acidic medium, the end point is detected by:
Chapter 12: Titrimetric (Volumetric) Analysis
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Answer: C. KMnO₄ itself acting as a self-indicator
MnO₄⁻ is deep purple and its reduced form Mn²⁺ is nearly colourless, so the first lasting pink colour marks the end point.The carbylamine (isocyanide) test, with chloroform and alcoholic KOH, is given by:
Chapter 14: Amines
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Answer: A. Primary amines only
Only primary amines (R–NH₂) react with CHCl₃ and alcoholic KOH to form foul-smelling isocyanides.The Beckmann rearrangement converts a ketoxime, in the presence of acid, into:
Chapter 17: Molecular Rearrangements
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Answer: D. An N-substituted amide
In the Beckmann rearrangement the group anti to the –OH migrates to nitrogen, and the ketoxime becomes an N-substituted amide.
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All questions are written by PadhaiNotes in our own words from the common syllabus of Maharashtra universities (NEP). Your own university's syllabus may differ a little.