Comprehensive notes, solved MCQs, and Short Questions for Class 9 Chemistry Chapter 2 Atomic Structure. Covers Atomic Models, Electronic Configuration, and Isotopes.
Ancient Concept: Democritus suggested matter is made of indivisible particles called atoms.
Dalton's Atomic Theory (1808): John Dalton proposed that atoms are indivisible hard spheres. Elements are composed of atoms.
Modern View: Atom is divisible into subatomic particles: Electrons, Protons, Neutrons.
Experiment (1879): Sir William Crookes performed experiments in a discharge tube at very low pressure. When high voltage was applied, rays were emitted from the cathode.
Properties of Cathode Rays:
Experiment (1886): Goldstein observed positive rays in a discharge tube with a perforated cathode. These rays passed through the holes (canals) of the cathode.
Origin: Produced when cathode rays collide with residual gas molecules, ionizing them ($M + e^- \to M^+ + 2e^-$).
Properties:
Discovery (1932): Sir James Chadwick discovered neutrons by bombarding Beryllium with Alpha particles ($\alpha$).
Reaction: $^9_4Be + ^4_2He \to ^{12}_6C + ^1_0n$
Properties:
Experiment (1911): Rutherford bombarded a very thin Gold Foil ($0.00004$ cm) with Alpha particles produced from Polonium.
Observations:
Conclusions:
Concept: Based on Max Planck's Quantum Theory. Electrons revolve in fixed circular orbits (shells) with fixed energy (quantized).
Postulates:
Shells (n): Main energy levels. K (n=1, max 2e), L (n=2, max 8e), M (n=3, max 18e), N (n=4, max 32e). Max electrons = $2n^2$.
Subshells: Sub-division of shells. s (max 2e), p (max 6e), d (max 10e), f (max 14e).
Aufbau Principle/Rule: Electrons fill lowest energy subshells first.
Order: 1s < 2s < 2p < 3s < 3p...
Examples:
Definition: Atoms of the same element having same atomic number (same protons) but different mass number (different neutrons).
Properties: Same chemical properties (depend on electrons) but different physical properties (depend on mass).
Examples:
Definition: Average mass of atoms of an element as compared to 1/12th the mass of one atom of Carbon-12 ($1 amu = 1.66 \times 10^{-24}g$).
Calculation Formula: Relative Atomic Mass = $\frac{(\text{Mass}_1 \times \%\text{Abundance}_1) + (\text{Mass}_2 \times \%\text{Abundance}_2)}{100}$
Example (Chlorine): $\frac{(35 \times 75) + (37 \times 25)}{100} = 35.5$ amu.