CBSE Class 11 Chemistry MCQs, chapter by chapter

895 practice questions for CBSE Class 11 Chemistry, organised by NCERT chapter (9 chapters). Every question includes a written explanation. Pick a chapter below, or try the samples first.

CBSE Class 11 Chemistry mock tests and papers

Practice by chapter

Some Basic Concepts of Chemistry

81 questions in timed practice

Structure of Atom

115 questions · 2 topics

Classification of Elements and Periodicity in Properties

75 questions in timed practice

Chemical Bonding and Molecular Structure

108 questions · 1 topic

Chemical Thermodynamics

93 questions in timed practice

Equilibrium

105 questions

Redox Reactions

75 questions in timed practice

Organic Chemistry - Some Basic Principles and Techniques

118 questions · 2 topics

Hydrocarbons

125 questions · 4 topics

Sample questions

Q1
medium
Assertion (A): Aniline undergoes electrophilic substitution much faster than benzene. Reason (R): The lone pair on the nitrogen of the $\text{-NH}_2$ group is delocalised into the ring, raising the electron density at the ortho and para positions.
  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: A - Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason explains the assertion. The nitrogen lone pair is donated into the ring by a $+M$ effect, which raises the ring's electron density and makes it a much better target for an electrophile. The extra density gathers at the ortho and para carbons, which is also why aniline yields mainly those products.
Q2
medium
Which is the correctly balanced half-reaction, in basic medium, for the reduction of permanganate ion to manganese dioxide?
  1. A $\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$
  2. B $\text{MnO}_4^- + 2\text{H}_2\text{O} + 5e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$
  3. C $\text{MnO}_4^- + 4\text{OH}^- + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$
  4. D $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$
Show answer and explanation

Correct answer: D - $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

Manganese falls from $+7$ to $+4$, so exactly three electrons are needed, which rules out the five-electron version. In the key, both sides carry six oxygens and four hydrogens, and the charge is $-1 - 3 = -4$ on the left against $-4$ on the right. The version with $\text{H}^+$ is the acidic-medium form, and moving hydroxide to the reactant side leaves the charge at $-8$ against $0$.
Q3
medium
Assertion (A): Bohr's model cannot account for the Zeeman effect, the splitting of spectral lines in a magnetic field. Reason (R): Bohr's model assumes that the angular momentum of the orbiting electron is quantised in units of $h/2\pi$.
  1. A (a) Both A and R are true, and R is the correct explanation of A.
  2. B (b) Both A and R are true, but R is not the correct explanation of A.
  3. C (c) A is true, but R is false.
  4. D (d) A is false, but R is true.
Show answer and explanation

Correct answer: B - (b) Both A and R are true, but R is not the correct explanation of A.

Both statements are true. Splitting in a magnetic field needs states that differ in their orientation in space, an idea the Bohr model has no room for, since it allows only flat circular orbits labelled by a single number. R states one of Bohr's postulates correctly, but that postulate fixes the allowed radii and energies and says nothing about behaviour in a magnetic field, so R does not explain A.
Q4
medium
Assertion (A): Cyclopentane is classified as an alicyclic compound. Reason (R): Alicyclic compounds are cyclic compounds that contain a benzene ring.
  1. A Both A and R are true, and R explains A.
  2. B Both A and R are true, but R does not explain A.
  3. C A is true, but R is false.
  4. D A is false, but R is true.
Show answer and explanation

Correct answer: C - A is true, but R is false.

A is true: cyclopentane is a ring of carbon atoms with no aromatic character, which is what alicyclic means. R is false and describes aromatic compounds instead. Alicyclic compounds are rings that behave much like open chain aliphatic compounds, whereas a ring system built round benzene, such as toluene or naphthalene, is aromatic and not alicyclic.
Q5
hard
Which of these substances would be expected to have the smallest standard molar entropy at $298\ \text{K}$?
  1. A Diamond
  2. B Liquid water
  3. C Solid iodine
  4. D Water vapour
Show answer and explanation

Correct answer: A - Diamond

Absolute entropies are measured upwards from the perfect crystal at zero kelvin, and they grow with the freedom particles have to move. Diamond is a rigid covalent network of light carbon atoms locked in fixed positions, so it has almost no such freedom and its standard molar entropy is among the smallest known. Water vapour is the highest of the four, and solid iodine, though a solid, has heavy weakly held molecules with far more accessible arrangements than diamond.
Q6
easy
In Thomson's model of the atom, how are the electrons and the positive charge arranged?
  1. A All the positive charge sits in a tiny central core while electrons fill the rest of the volume
  2. B Electrons and positive charge occupy separate hemispheres of the atom
  3. C Electrons revolve in fixed circular orbits around a dense central positive nucleus
  4. D Electrons are embedded in a uniform sphere of positive charge, like seeds in a watermelon
Show answer and explanation

Correct answer: D - Electrons are embedded in a uniform sphere of positive charge, like seeds in a watermelon

Thomson pictured the atom as a sphere of positive charge of atomic size with the electrons embedded in it, so that the atom as a whole is neutral. Electrons revolving around a central nucleus, and a tiny core holding all the positive charge, are both features of Rutherford's later model rather than Thomson's. Separate hemispheres of charge was never proposed and would not give a neutral, stable arrangement.

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