Structure of Atom — Class 11 MCQs with Answers

Structure of Atom — important questions & MCQs with answers (Class 11 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Charge on a proton:

   A.+1 C

   B.−1.6×10⁻¹⁹ C

   C.0

   D.+1.6×10⁻¹⁹ C✓ Correct

   Answer: D. +1.6×10⁻¹⁹ C

   Explanation: A proton carries the elementary charge, +1.6×10⁻¹⁹ C — the same size as the electron's charge but opposite sign. Picking −1.6×10⁻¹⁹ C swaps it with the electron; 0 C is the neutron's charge.

2. Q2Easy

   Maximum number of electrons in a shell with principal quantum number n:

   A.2n²✓ Correct

   B.n²

   C.n

   D.2n

   Answer: A. 2n²

   Explanation: A shell of principal quantum number n holds n² orbitals, and each orbital takes 2 electrons of opposite spin, so capacity is 2n². For n=2 that is 8 electrons; quoting the bare n² = 4 counts the orbitals and forgets to pair them.

3. Q3Easy

   Discovered the atomic nucleus:

   A.Rutherford✓ Correct

   B.Thomson

   C.Bohr

   D.Dalton

   Answer: A. Rutherford

   Explanation: Rutherford's 1911 gold-foil α-scattering experiment showed most α-particles passed straight through but a few bounced back, revealing a small, dense, positively charged nucleus. Thomson found the electron, not the nucleus.

4. Q4Easy

   Maximum electrons in p subshell:

   A.2

   B.6✓ Correct

   C.10

   D.14

   Answer: B. 6

   Explanation: A p subshell has 3 orbitals (m_l = −1, 0, +1), each holding 2 electrons, so its maximum capacity is 3×2 = 6. Confusing this with the d subshell's 10 or f's 14 is the usual slip.

5. Q5Easy

   Discoverer of electron:

   A.Rutherford

   B.J.J. Thomson✓ Correct

   C.Chadwick

   D.Bohr

   Answer: B. J.J. Thomson

   Explanation: J.J. Thomson identified the electron in 1897 from cathode-ray deflection in electric and magnetic fields, measuring its charge-to-mass ratio e/m. Rutherford's later work found the nucleus, not the electron.

6. Q6Easy

   Maximum electrons in d subshell:

   A.2

   B.6

   C.14

   D.10✓ Correct

   Answer: D. 10

   Explanation: A d subshell has l = 2, so m_l runs −2…+2 — five orbitals — and each holds 2 electrons of opposite spin, giving 5×2 = 10. Picking 6 mistakes it for the p subshell's count.

7. Q7Medium

   Energy of nth Bohr orbit of H: En = −13.6/n² eV. E₂ =

   A.−13.6 eV

   B.−1.51 eV

   C.−6.8 eV

   D.−3.4 eV✓ Correct

   Answer: D. −3.4 eV

   Explanation: Bohr energy scales as En = −13.6/n² eV, so at n = 2, E₂ = −13.6/4 = −3.4 eV. Forgetting to square n before dividing gives the wrong −6.8 eV instead.

8. Q8Medium

   Principal quantum number (n) describes:

   A.Spin

   B.Energy level / shell✓ Correct

   C.Subshell shape

   D.Magnetic orientation

   Answer: B. Energy level / shell

   Explanation: The principal quantum number n specifies the shell or main energy level and roughly the electron's average distance from the nucleus, taking values 1, 2, 3, … Spin, subshell shape, and orientation are ms, l, m_l instead.

9. Q9Medium

   Radius of n=2 Bohr orbit of H atom (r₁ = 0.529 Å):

   A.0.529 Å

   B.2.116 Å✓ Correct

   C.1.058 Å

   D.4.232 Å

   Answer: B. 2.116 Å

   Explanation: Bohr radius scales as rₙ = n²r₁, so at n = 2, r₂ = 2² × 0.529 = 4 × 0.529 = 2.116 Å. Multiplying by n instead of n² gives the smaller, incorrect 1.058 Å.

10. Q10Medium

    For 3p orbital, l =

    A.1✓ Correct

    B.0

    C.2

    D.3

    Answer: A. 1

    Explanation: In the label 3p, the letter p fixes the azimuthal quantum number: p subshells always have l = 1, regardless of the principal number n = 3 in front. Only s (l=0), d (l=2), f (l=3) differ.

11. Q11Medium

    Wavenumber for H transition n₂ → n₁: ν̄ = R(1/n₁² − 1/n₂²). For Lyman first line:

    A.R(1/9 − 1/16)

    B.R(1/4 − 1/9)

    C.R(1 − 1/9)

    D.R(1 − 1/4)✓ Correct

    Answer: D. R(1 − 1/4)

    Explanation: The first Lyman line is the n = 2 → n = 1 transition, so ν̄ = R(1/1² − 1/2²) = R(1 − 1/4). Starting from n₁ = 3 or 4 instead of 1 would describe Balmer or Paschen lines, not Lyman.

12. Q12Hard

    Energy required for electron transition n=2 to n=4 in H atom:

    A.2.55 eV✓ Correct

    B.10.2 eV

    C.3.4 eV

    D.1.51 eV

    Answer: A. 2.55 eV

    Explanation: ΔE = 13.6(1/n₁² − 1/n₂²) = 13.6(1/4 − 1/16) = 13.6 × 3/16 = 2.55 eV. Using 1/16 − 1/4 instead (a negative result) or forgetting to subtract the fractions first are the usual arithmetic slips.

13. Q13Hard

    Pauli exclusion principle: no two electrons can have:

    A.All four quantum numbers same✓ Correct

    B.Same charge

    C.Same orbit

    D.Same spin always

    Answer: A. All four quantum numbers same

    Explanation: Pauli's exclusion principle forbids any two electrons in the same atom from sharing all four quantum numbers (n, l, m_l, ms) simultaneously — at minimum their spins must differ. Each orbital holds at most 2 electrons, of opposite spin.

14. Q14Hard

    Energy of n=1 of He⁺ (Z=2): E = −13.6 Z²/n². Equals:

    A.−13.6 eV

    B.−54.4 eV✓ Correct

    C.−27.2 eV

    D.−6.8 eV

    Answer: B. −54.4 eV

    Explanation: For a hydrogen-like ion, E = −13.6 Z²/n² eV; for He⁺ (Z=2, n=1), that is −13.6 × 4 = −54.4 eV. Using Z instead of Z² gives the smaller-magnitude −27.2 eV, a common slip.

Structure of Atom — FAQs

More Class 11 Chemistry chapters

• 01Some Basic Concepts of Chemistry
• 03Classification of Elements and Periodicity
• 04Chemical Bonding and Molecular Structure
• 05Thermodynamics
• 06Equilibrium
• 07Redox Reactions

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