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Structure of Atom – Class 11 Chemistry Chapter 2 Notes & NCERT PDF

Class 11ChemistryChapter 2NCERT book: Chemistry Part I - Class 11

Notes and a simple summary of Chapter 2 of the NCERT Class 11 Chemistry book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 2: Structure of Atom

Introduction

Dalton pictured atoms as indivisible, but experiments from the late 19th century showed atoms contain smaller particles. This unit traces the journey from the discovery of the electron to the quantum mechanical model, and ends with the rules for writing electronic configurations — the foundation of chemical bonding and the periodic table.

Key Concepts

2.1 Discovery of Sub-atomic Particles

  • Cathode ray discharge tube: cathode rays travel from cathode to anode, move in straight lines, are deflected by electric and magnetic fields as negatively charged particles, and their nature does not depend on the gas or electrode material — they are electrons.
  • J.J. Thomson measured the charge-to-mass ratio: e/mₑ = 1.758820 × 10¹¹ C kg⁻¹.
  • R.A. Millikan's oil drop experiment gave the electron charge: −1.602176 × 10⁻¹⁹ C, so mₑ = 9.1094 × 10⁻³¹ kg.
  • Protons came from canal rays (anode rays); neutrons (neutral, slightly heavier than protons) were discovered by Chadwick (1932).
  • 2.2 Atomic Models

  • Thomson's model ("plum pudding"/watermelon): positive charge spread uniformly with electrons embedded.
  • Rutherford's α-particle scattering (gold foil): most α-particles passed straight through, a few were deflected, and very few (about 1 in 20,000) bounced back. Conclusions: most of the atom is empty space; positive charge and nearly all the mass are concentrated in a tiny nucleus; electrons move around it.
  • Atomic number (Z) = number of protons = number of electrons in a neutral atom. Mass number (A) = protons + neutrons. Notation: ᴬ_Z X.
  • Isotopes have the same Z but different A (¹H, ²H, ³H); isobars have the same A but different Z (¹⁴C and ¹⁴N).
  • Drawbacks of Rutherford's model: an accelerating (orbiting) charge should radiate energy and spiral into the nucleus, so the model cannot explain the stability of atoms or line spectra.
  • 2.3 Developments Leading to the Bohr Model

  • Wave nature of electromagnetic radiation: c = νλ (c = 3.0 × 10⁸ m s⁻¹); wavenumber ν̄ = 1/λ. Visible light spans about 400 nm (violet) to 750 nm (red).
  • Particle nature: Planck's quantum theory — energy is emitted or absorbed in packets, E = hν (h = 6.626 × 10⁻³⁴ J s). This explains black-body radiation.
  • Photoelectric effect: electrons are ejected only above a threshold frequency ν₀; the number of electrons depends on intensity but their kinetic energy depends on frequency: hν = hν₀ + ½mₑv². hν₀ is the work function.
  • Line spectra: hydrogen gives series named Lyman (n₁ = 1), Balmer (n₁ = 2, visible), Paschen (3), Brackett (4) and Pfund (5). Rydberg: ν̄ = 109,677 (1/n₁² − 1/n₂²) cm⁻¹.
  • 2.4 Bohr's Model for the Hydrogen Atom

  • Electrons move in fixed circular orbits (stationary states) of definite energy; energy changes only when an electron jumps between orbits: ΔE = hν.
  • Angular momentum is quantised: mₑvr = nh/2π.
  • Radius: rₙ = 52.9 n²/Z pm; energy: Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J. The negative sign means the bound electron has lower energy than a free electron.
  • Limitations: fails for multi-electron atoms, cannot explain splitting of lines in magnetic (Zeeman) and electric (Stark) fields, and contradicts the dual nature and uncertainty principle.
  • 2.5 Towards the Quantum Mechanical Model

  • de Broglie: matter has wave character, λ = h/mv. Noticeable only for tiny particles like electrons (basis of the electron microscope).
  • Heisenberg's uncertainty principle (1927): Δx × Δp ≥ h/4π — position and momentum of an electron cannot both be known exactly, so fixed Bohr orbits are meaningless.
  • 2.6 Quantum Mechanical Model of Atom

  • Schrödinger's equation gives wave functions ψ (orbitals); |ψ|² gives the probability density of finding the electron.
  • Quantum numbers: principal n (shell, size, energy: 1, 2, 3…); azimuthal l (subshell and shape: 0 to n − 1 → s, p, d, f); magnetic mₗ (orientation: −l to +l, giving 2l + 1 orbitals); spin mₛ (+½ or −½). A shell has n² orbitals and holds 2n² electrons.
  • Shapes: s orbitals are spherical; p orbitals are dumb-bell shaped (pₓ, p_y, p_z); d orbitals are mostly double dumb-bell. Number of nodes: radial = n − l − 1; angular = l; total = n − 1.
  • Energies: in hydrogen, energy depends only on n; in multi-electron atoms it depends on n + l — the lower the (n + l) value, the lower the energy; for equal (n + l), lower n comes first. Order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s…
  • Filling rules: Aufbau principle (lowest energy first), Pauli exclusion principle (no two electrons have all four quantum numbers the same, so an orbital holds at most two electrons with opposite spin), Hund's rule (orbitals of a subshell are singly occupied with parallel spins before pairing).
  • Stability of half-filled and completely filled subshells explains exceptions: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹, due to symmetry and higher exchange energy.
  • Worked Examples

    1. Sub-atomic particles in ⁸⁰₃₅Br. Protons = electrons = 35; neutrons = 80 − 35 = 45.

    2. Species with 18 electrons, 16 protons, 16 neutrons. Z = 16 (sulphur), A = 32, two extra electrons → ³²₁₆S²⁻.

    3. Wavelength of a radio station at 1368 kHz. λ = c/ν = (3.0 × 10⁸) / (1.368 × 10⁶) = 219.3 m (radio waves).

    4. Energy of one mole of photons of frequency 5 × 10¹⁴ Hz. E(one photon) = hν = 6.626 × 10⁻³⁴ × 5 × 10¹⁴ = 3.313 × 10⁻¹⁹ J; × 6.022 × 10²³ = 199.5 kJ mol⁻¹.

    5. Photoelectric effect. Threshold ν₀ = 7.0 × 10¹⁴ s⁻¹, incident ν = 1.0 × 10¹⁵ s⁻¹: KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × 3.0 × 10¹⁴ = 1.988 × 10⁻¹⁹ J.

    6. Transition n = 5 → n = 2 in hydrogen. ΔE = 2.18 × 10⁻¹⁸ (1/4 − 1/25) = 4.58 × 10⁻¹⁹ J (emitted); ν = ΔE/h = 6.91 × 10¹⁴ Hz; λ = c/ν ≈ 434 nm (Balmer, visible).

    7. First orbit of He⁺ (Z = 2). E₁ = −2.18 × 10⁻¹⁸ × 4 = −8.72 × 10⁻¹⁸ J; r₁ = 52.9/2 = 26.45 pm.

    8. de Broglie and uncertainty. A 0.1 kg ball at 10 m s⁻¹ has λ = 6.626 × 10⁻³⁴ m (undetectable). Locating an electron within 0.1 Å gives Δv ≥ h/(4πmΔx) ≈ 5.79 × 10⁶ m s⁻¹ — enormous, which is why fixed orbits fail. A photon of λ = 3.6 Å has mass m = h/λc = 6.135 × 10⁻³³ kg (the book's worked answer prints the power as 10⁻²⁹; recompute to see that 10⁻³³ is correct).

    Formula / Key-Facts Sheet

  • c = νλ; ν̄ = 1/λ; E = hν = hc/λ; h = 6.626 × 10⁻³⁴ J s
  • hν = hν₀ + ½mₑv² (photoelectric equation)
  • ν̄ = 109,677 (1/n₁² − 1/n₂²) cm⁻¹
  • rₙ = 52.9 n²/Z pm; Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J; mₑvr = nh/2π
  • λ = h/mv; Δx·Δp ≥ h/4π
  • Orbitals in shell n = n²; electrons = 2n²; orbitals in subshell = 2l + 1
  • Radial nodes = n − l − 1; angular nodes = l
  • Exam Focus

  • Numericals (3 marks): Bohr energy/radius, wavelength of a spectral line, photoelectric kinetic energy, de Broglie wavelength, uncertainty.
  • Quantum numbers (2 marks): write the orbital for given n and l (n = 4, l = 0 → 4s; n = 5, l = 3 → 5f); list all four quantum numbers for an electron.
  • Configuration (2-3 marks): write configurations up to Z = 30, explaining Cr and Cu.
  • Answer outline — "Why did Rutherford's model fail?" An orbiting electron is accelerating and must radiate energy (Maxwell's theory), so it would spiral into the nucleus; the model also cannot explain line spectra or the distribution of electrons.
  • Key Terms

    Electron (इलेक्ट्रॉन), Nucleus (नाभिक), Atomic number (परमाणु क्रमांक), Mass number (द्रव्यमान संख्या), Isotope (समस्थानिक), Isobar (समभारिक), Frequency (आवृत्ति), Wavelength (तरंगदैर्ध्य), Photon (फोटॉन), Threshold frequency (देहली आवृत्ति), Orbit (कक्षा), Orbital (कक्षक), Quantum number (क्वांटम संख्या), Uncertainty principle (अनिश्चितता सिद्धांत), Electronic configuration (इलेक्ट्रॉनिक विन्यास)

    Common Mistakes

  • Confusing orbit (Bohr's fixed path) with orbital (a region of probability).
  • Filling 3d before 4s — 4s has lower (n + l) and fills first.
  • Writing Cr as [Ar] 3d⁴ 4s² or Cu as [Ar] 3d⁹ 4s².
  • Dropping the negative sign in Eₙ or forgetting Z² for hydrogen-like ions such as He⁺.
  • Thinking intensity controls photoelectron energy — it controls only their number.
  • 💡 Key Learning Points

    • ✓Describe the discovery of the electron, proton and neutron and their charge and mass
    • ✓Explain Thomson's and Rutherford's models, and why Rutherford's model failed
    • ✓Apply the wave and particle nature of radiation: c = νλ, E = hν and the photoelectric effect
    • ✓Use Bohr's model to calculate energies, radii and spectral lines of hydrogen-like species
    • ✓Use quantum numbers, orbital shapes and the Aufbau, Pauli and Hund rules to write electronic configurations

    👨‍🏫 Teaching Tips

    • →Recreate the gold foil experiment with marbles rolled at a hidden object to show 'mostly empty space'
    • →Use a ladder analogy for quantised energy levels versus a ramp for continuous energy
    • →Show a prism or a CD spectrum, then a hydrogen line spectrum image, to contrast continuous and line spectra
    • →Solve one numerical for every formula on the board before assigning practice
    • →Teach the (n + l) rule with a 'diagonal arrows' chart drawn step by step by students
    • →Use orbital-box diagrams with arrows to practise Hund's rule and Pauli's principle
    • →Discuss Cr and Cu exceptions as a debate: why would a half-filled d subshell be preferred?

    📋 Assessment Questions

    1. Calculate protons, neutrons and electrons for a given ion or isotope
    2. Compute the energy, frequency and wavelength for a hydrogen electronic transition
    3. Find the kinetic energy of a photoelectron from threshold and incident frequencies
    4. Write quantum numbers and electronic configurations, including exceptions like Cr and Cu
    5. Explain the limitations of Rutherford's and Bohr's models

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