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Class 11 · CBSE

Structure of Atom — NCERT Class 11 Chemistry

Complete NCERT Solutions for CBSE Class 11 Chemistry Chapter 2 — all intext and chapter-end exercise questions on subatomic particles, atomic models, quantum numbers, orbitals, and electronic configuration answered clearly.

Updated 11 June 202610 min read

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This page provides complete, step-by-step NCERT solutions for Class 11 Chemistry Chapter 2 — Structure of Atom. Every intext question and chapter-end exercise question is answered with clear reasoning. Use the section headings to quickly find the question you need help with.

Key Concepts (Quick Recap)

  • Subatomic particles: Electron (−1, ~0 u), Proton (+1, ~1 u), Neutron (0, ~1 u)
  • Thomson's model: Atom is a positive sphere with embedded electrons (plum pudding)
  • Rutherford's model: Nucleus at centre, electrons revolve around it; could not explain stability or line spectra
  • Bohr's model: Electrons revolve in fixed orbits (stationary states); E = −13.6/n² eV for H atom; explains hydrogen spectrum
  • Electromagnetic spectrum: Radio → microwave → IR → visible → UV → X-rays → gamma rays
  • Dual nature of matter: de Broglie wavelength λ = h/mv; matter has both particle and wave nature
  • Heisenberg uncertainty principle: Δx × Δp ≥ h/4π; cannot simultaneously determine exact position and momentum
  • Quantum numbers: n (principal), l (azimuthal), mₗ (magnetic), mₛ (spin)
  • Orbitals: s (spherical, l=0), p (dumbbell, l=1, 3 orientations), d (l=2, 5 orientations), f (l=3, 7 orientations)
  • Electronic configuration: Aufbau principle, Hund's rule, Pauli exclusion principle
  • Hydrogen spectrum: Lyman (UV), Balmer (visible), Paschen, Brackett, Pfund (IR) series

Intext Questions

Page 16–17 (Discovery of Electron, Cathode Rays)

Q2.1: What are cathode rays? How are they produced?

AnswerCathode rays are streams of fast-moving electrons emitted from the cathode in a discharge tube when a high voltage (10,000–15,000 V) is applied across electrodes in a gas at very low pressure (~0.01 mm Hg). They were discovered by William Crookes and their particle nature was established by J.J. Thomson.

Q2.2: What is the charge and mass of an electron?

AnswerCharge of an electron = −1.6 × 10⁻¹⁹ C (≈ −1 unit). Mass of an electron = 9.1 × 10⁻³¹ kg ≈ 0.00055 u ≈ 1/1837 of a proton's mass.

Q2.3: How was the neutron discovered?

AnswerChadwick (1932) bombarded beryllium with alpha particles and observed a neutral radiation that could knock out protons from paraffin. By measuring the recoil energies, he calculated that the particle had mass ≈ 1 u and zero charge — the neutron.

Page 22–23 (Rutherford Model, Atomic Number, Mass Number)

Q2.4: Describe Rutherford's alpha-particle scattering experiment.

AnswerA beam of alpha particles (He²⁺) from ²¹⁰Po was directed at a thin gold foil. Most alpha particles passed straight through, some were deflected at small angles, and a few (about 1 in 8000) bounced back. This led Rutherford to conclude that the atom has a tiny, dense, positively charged nucleus with most of the atom being empty space.

Q2.5: What are the limitations of Rutherford's model?

Answer(1) It could not explain the stability of the atom — an electron in circular motion should radiate energy and spiral into the nucleus. (2) It could not explain the line spectra of hydrogen — a continuously losing electron should produce a continuous spectrum, not line spectra.

Q2.6: Define atomic number and mass number.

AnswerAtomic number (Z) = number of protons in the nucleus = number of electrons in a neutral atom. Mass number (A) = number of protons + number of neutrons in the nucleus. Isotopes have same Z but different A; isobars have same A but different Z.

Page 33–34 (Bohr Model, Hydrogen Spectrum)

Q2.7: What is Bohr's model of the atom?

AnswerBohr postulated: (1) Electrons revolve in certain fixed circular orbits (stationary states) without radiating energy. (2) The angular momentum of an electron is quantised: mvr = nh/2π, where n = 1, 2, 3… (3) Energy is emitted or absorbed only when an electron jumps from one orbit to another, with ΔE = hν.

Q2.8: Calculate the energy of an electron in the first Bohr orbit of hydrogen.

AnswerE₁ = −13.6/n² eV = −13.6/1² = −13.6 eV. The negative sign indicates the electron is bound to the nucleus.

Q2.9: What is the radius of the first Bohr orbit?

Answerr₁ = 0.529 Å (or 52.9 pm). For orbit n, rₙ = 0.529 × n² Å.

Q2.10: What are the different series in the hydrogen spectrum?

Answer
  • Lyman series: n₁ = 1 (UV region)
  • Balmer series: n₁ = 2 (visible region)
  • Paschen series: n₁ = 3 (IR region)
  • Brackett series: n₁ = 4 (IR region)
  • Pfund series: n₁ = 5 (IR region)

Page 44–46 (Quantum Numbers, Orbitals)

Q2.11: Define the four quantum numbers.

Answer
  • n (principal): Determines size and energy; n = 1, 2, 3…
  • l (azimuthal): Determines shape; l = 0, 1, 2…(n−1); s=0, p=1, d=2, f=3
  • mₗ (magnetic): Determines orientation; mₗ = −l to +l (2l+1 values)
  • mₛ (spin): Spin orientation; mₛ = +½ or −½

Q2.12: How many orbitals are present in the L shell?

AnswerL shell: n = 2. l = 0 (one 2s orbital) and l = 1 (three 2p orbitals). Total = 4 orbitals.

Q2.13: Write the electronic configuration of Cr (Z = 24).

AnswerExpected: [Ar] 3d⁴4s². Actual: [Ar] 3d⁵4s¹. Half-filled d-subshell (3d⁵) provides extra stability.

Chapter-End Exercise

Q1. What is the minimum number of electrons that must be removed from an atom to give it a charge of +3?

AnswerThree electrons must be removed, since each electron carries −1 charge. Removing 3 electrons gives a net +3 charge.

Q2. Calculate the number of electrons which will together weigh one gram.

AnswerMass of one electron = 9.1 × 10⁻²⁸ g. Number = 1/(9.1 × 10⁻²⁸) = 1.1 × 10²⁷ electrons.

Q3. Calculate the mass and charge of one mole of electrons.

AnswerMass = 6.022 × 10²³ × 9.1 × 10⁻³¹ kg = 5.48 × 10⁻⁷ kg. Charge = 6.022 × 10²³ × 1.6 × 10⁻¹⁹ C = 9.635 × 10⁴ C ≈ 96350 C (1 Faraday).

Q4. Calculate the energy required to remove an electron from the first Bohr orbit of hydrogen.

AnswerEnergy required = ionisation energy = E∞ − E₁ = 0 − (−13.6) = 13.6 eV.

Q5. What is the longest wavelength of light that can cause the transition from ground state to first excited state in H atom?

AnswerΔE = E₂ − E₁ = (−13.6/4) − (−13.6) = 13.6 × 3/4 = 10.2 eV = 10.2 × 1.6 × 10⁻¹⁹ J = 1.632 × 10⁻¹⁸ J. λ = hc/ΔE = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1.632 × 10⁻¹⁸) = 1.218 × 10⁻⁷ m = 121.8 nm.

Q6. Calculate the de Broglie wavelength of an electron moving with 1% of the speed of light.

Answerv = 0.01 × 3 × 10⁸ = 3 × 10⁶ m/s. λ = h/mv = (6.626 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 3 × 10⁶) = 2.427 × 10⁻¹⁰ m = 2.427 Å.

Q7. The kinetic energy of an electron is 4.55 × 10⁻²⁵ J. Calculate its de Broglie wavelength.

AnswerKE = ½mv², so v = √(2KE/m) = √(2 × 4.55 × 10⁻²⁵ / 9.1 × 10⁻³¹) = √(10⁶) = 10³ m/s. λ = h/mv = (6.626 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 10³) = 7.28 × 10⁻⁷ m = 728 nm.

Q8. Which of the following are not possible? (i) n = 1, l = 1; (ii) n = 2, l = 0; (iii) n = 3, l = 3.

Answerl can range from 0 to n−1.
  • (i) n = 1, l = 1: Not possible (l max = 0 for n = 1)
  • (ii) n = 2, l = 0: Possible (2s orbital)
  • (iii) n = 3, l = 3: Not possible (l max = 2 for n = 3)

Q9. Using s, p, d, f notations, describe the orbital with: (i) n = 2, l = 1; (ii) n = 4, l = 0; (iii) n = 5, l = 3.

Answer
  • (i) n = 2, l = 1 → 2p
  • (ii) n = 4, l = 0 → 4s
  • (iii) n = 5, l = 3 → 5f

Q10. Explain why the following set of quantum numbers is not allowed: n = 3, l = 2, mₗ = −3.

Answer: For l = 2, mₗ can be −2, −1, 0, +1, +2 only. mₗ = −3 is not allowed becausemₗcannot exceed l.

Q11. How many electrons in an atom may have the following quantum numbers: (i) n = 3; (ii) n = 4, l = 1?

Answer
  • (i) n = 3: Maximum electrons = 2n² = 2(9) = 18 electrons
  • (ii) n = 4, l = 1 (4p): Number of orbitals = 2l + 1 = 3. Electrons = 3 × 2 = 6 electrons

Q12. Write the electronic configurations of: (i) Cr (Z = 24), (ii) Cu (Z = 29), (iii) Fe²⁺ (Z = 26).

Answer
  • (i) Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (half-filled d for stability)
  • (ii) Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (fully-filled d for stability)
  • (iii) Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ (two 4s electrons removed first)

Q13. Calculate the wavelength, frequency, and wavenumber of a light wave whose period is 2.0 × 10⁻¹⁰ s.

Answerν = 1/T = 1/(2.0 × 10⁻¹⁰) = 5.0 × 10⁹ Hz. λ = c/ν = (3.0 × 10⁸)/(5.0 × 10⁹) = 0.06 m = 6 cm. Wavenumber = 1/λ = 1/0.06 = 16.67 m⁻¹.

Q14. What is the number of photons of light with wavelength 4000 pm that provide 1 J of energy?

AnswerE = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4000 × 10⁻¹²) = 4.97 × 10⁻¹⁹ J. Number of photons = 1/(4.97 × 10⁻¹⁹) = 2.012 × 10¹⁸ photons.

Q15. The threshold frequency for a metal is 5 × 10¹⁴ s⁻¹. Will a photon of wavelength 6000 Å cause photoemission?

AnswerFrequency of photon ν = c/λ = (3 × 10⁸)/(6000 × 10⁻¹⁰) = 5 × 10¹⁴ s⁻¹. This equals the threshold frequency exactly. Yes, photoemission will just begin (E = hν₀, so kinetic energy of emitted electron = 0).

Q16. An electron is in one of the 3d orbitals. Give the possible values of n, l, and mₗ.

AnswerFor a 3d orbital: n = 3, l = 2, mₗ = −2, −1, 0, +1, +2 (any one of these five values).

Q17. An atom of an element contains 29 electrons and 35 neutrons. Deduce: (i) the electronic configuration, (ii) the number of unpaired electrons.

AnswerZ = 29 (Cu). (i) Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹. (ii) Only the 4s¹ electron is unpaired: 1 unpaired electron.

Q18. Show that the circumference of the nth Bohr orbit is an integral multiple of the de Broglie wavelength of the electron.

AnswerCircumference = 2πrₙ. de Broglie wavelength λ = h/mv. Bohr's condition: mvrₙ = nh/2π, so 2πrₙ = nh/mv = nλ. Therefore, 2πrₙ = nλ — the circumference is n times the de Broglie wavelength, an integral multiple.

Q19. The velocity of electron in the nth orbit is vₙ = v₁/n. Calculate the velocity of the electron in the 3rd orbit.

Answerv₁ = 2.18 × 10⁶ m/s (Bohr velocity). v₃ = v₁/3 = 2.18 × 10⁶/3 = 7.27 × 10⁵ m/s.

Q20. Which of the following are iso-electronic species? Na⁺, K⁺, Mg²⁺, Ca²⁺, S²⁻, Ar.

AnswerCount total electrons:
  • Na⁺ (10 e⁻), Mg²⁺ (10 e⁻), Ne (10 e⁻) — iso-electronic group
  • K⁺ (18 e⁻), Ca²⁺ (18 e⁻), Ar (18 e⁻), S²⁻ (18 e⁻), Cl⁻ (18 e⁻) — iso-electronic group

Na⁺ and Mg²⁺ are iso-electronic; K⁺, Ca²⁺, Ar, and S²⁻ are iso-electronic.

Internal Links

  • [Chapter 2 Guide — Structure of Atom](/school/cbse/class/11/chemistry/cbse-structure-of-atom)
  • [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
  • [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)

Frequently Asked Questions

The four quantum numbers are: (1) Principal quantum number (n) — determines the size and energy of the orbital, (2) Azimuthal quantum number (l) — determines the shape of the orbital (l = 0 to n−1), (3) Magnetic quantum number (mₗ) — determines the orientation of the orbital (mₗ = −l to +l), (4) Spin quantum number (mₛ) — describes the spin of the electron (+½ or −½).
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