JEE Chemistry Structure Of Atom Notes Set 01

Fundamental Particles

Electron: discovered by J.J. Thomson from cathode rays; charge −1.602 × 10⁻¹⁹ C, mass 9.11 × 10⁻³¹ kg, e/m = 1.76 × 10¹¹ C/kg. Proton: positive charge of equal size, mass 1.673 × 10⁻²⁷ kg (about 1837 times the electron); found from anode (canal) rays. Neutron: no charge, mass 1.675 × 10⁻²⁷ kg; discovered by Chadwick in 1932 by bombarding beryllium with α-particles.
Cathode rays travel in straight lines, are deflected by electric and magnetic fields towards the positive side, and are the same whatever the gas or cathode. Anode rays are positive ions whose nature depends on the gas in the tube.
Protons and neutrons together are called nucleons. The nuclear radius is about 1.3 × 10⁻¹⁵ A^1/3 m (a few fermi), while an atom is about 10⁻¹⁰ m across.

Atomic Number, Mass Number and Atomic Species

Atomic number Z = number of protons = number of electrons in a neutral atom. Mass number A = protons + neutrons. Neutrons = A − Z.
Isotopes: same Z, different A (¹H, ²H, ³H). Isobars: same A, different Z (⁴⁰Ar and ⁴⁰Ca). Isotones: same number of neutrons. Isoelectronic species: same number of electrons (N³⁻, O²⁻, F⁻, Ne, Na⁺, Mg²⁺, Al³⁺). Isosters: same number of atoms and electrons (CO₂ and N₂O).

Atomic Models

Thomson's model: a positive sphere with electrons embedded in it (plum pudding). Rutherford's α-particle scattering: most particles went straight through, a few were deflected strongly and very few came back, so an atom is mostly empty space with a tiny, dense, positively charged nucleus; electrons move around it.
Failures of Rutherford's model: a revolving electron should radiate energy and spiral into the nucleus, and it cannot explain line spectra.

Electromagnetic Radiation and Planck's Theory

A wave is described by wavelength λ, frequency ν, wave number \( \bar\nu=1/\lambda \) and speed c = νλ = 3 × 10⁸ m/s.
Planck's quantum theory: energy is emitted or absorbed in packets called quanta (photons), E = hν = hc/λ, with h = 6.626 × 10⁻³⁴ J s. The photoelectric effect shows that light has particle character: electrons are ejected only if ν exceeds the threshold frequency ν0, and the maximum kinetic energy is KE = hν − hν0.

Bohr's Model of the Hydrogen Atom

Postulates: electrons move in certain stationary circular orbits without radiating; the angular momentum in an allowed orbit is mvr = nh/2π (n = 1, 2, 3, ...); energy is emitted or absorbed only when the electron jumps between orbits, ΔE = hν.
Results for hydrogen-like species (nuclear charge Z): r_n = 0.529 n²/Z Å; E_n = −13.6 Z²/n² eV = −2.18 × 10⁻¹⁸ Z²/n² J = −1312 Z²/n² kJ/mol; v_n = 2.18 × 10⁶ Z/n m/s. Kinetic energy = −E, potential energy = 2E.
The model works only for one-electron species (H, He⁺, Li²⁺) and cannot explain fine structure, the Zeeman and Stark effects, or the spectra of multi-electron atoms.

Hydrogen Spectrum

Lines arise from jumps between orbits: \( \bar\nu=\frac{1}{\lambda}=R_H Z^{2}\left(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\right) \), R_H = 1.097 × 10⁷ m⁻¹.
Series: Lyman n₁ = 1 (ultraviolet), Balmer n₁ = 2 (visible), Paschen n₁ = 3, Brackett n₁ = 4, Pfund n₁ = 5 (all infrared). The first line of a series is 'n₂ = n₁ + 1', the series limit is n₂ = ∞. The number of lines produced when an electron falls from level n is n(n − 1)/2.

Dual Nature and Uncertainty

de Broglie: every moving particle has a wavelength λ = h/mv = h/p. It is measurable only for very light particles such as electrons; a cricket ball has λ of about 10⁻³⁵ m. Bohr's condition mvr = nh/2π is the same as 2πr = nλ. Davisson and Germer confirmed the wave nature of electrons by diffraction.
Heisenberg's uncertainty principle: Δx · Δp ≥ h/4π, so the exact position and momentum of a microscopic particle cannot be known together. This rules out fixed orbits and leads to the idea of orbitals, regions of space where the probability of finding the electron is high.
Schrödinger's wave equation gives the wave function ψ; ψ² is the probability density. Quantum numbers appear naturally as the solutions.

Quantum Numbers

Principal n = 1, 2, 3, ...: size and energy of the shell, maximum 2n² electrons. Azimuthal l = 0 to n − 1: shape of the subshell (s, p, d, f for l = 0, 1, 2, 3); orbital angular momentum = √(l(l + 1)) h/2π. Magnetic m = −l to +l: orientation, 2l + 1 orbitals in a subshell. Spin s = +½ or −½.
Number of orbitals in a shell = n². Radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.

Shapes of Orbitals

s orbitals are spherical; p orbitals are dumb-bell shaped along x, y and z with one nodal plane; four d orbitals (d_xy, d_yz, d_zx, d_{x²−y²}) are double dumb-bell with two nodal planes, while d_{z²} has a dumb-bell with a ring (collar) around its middle. The 3p orbital has one spherical and one planar node.

Rules for Filling Electrons

Aufbau principle: orbitals are filled in order of increasing (n + l); if (n + l) is equal, the orbital with lower n is filled first: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d.
Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers the same, so an orbital holds at most two electrons of opposite spin. Hund's rule: electrons occupy degenerate orbitals singly with parallel spins before pairing.
Extra stability of half-filled and completely filled subshells explains Cr = [Ar]3d⁵ 4s¹ and Cu = [Ar]3d¹⁰ 4s¹ (and similarly Mo, Ag, Au). Cations lose electrons first from the outermost shell: Fe²⁺ = [Ar]3d⁶.

Magnetic Behaviour

Species with unpaired electrons are paramagnetic and those with all electrons paired are diamagnetic. The spin-only magnetic moment is \( \mu=\sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons. Unpaired electrons: Cr³⁺ 3, Mn²⁺ 5, Fe³⁺ 5, Ni²⁺ 2, Cu⁺ 0.

Solved Examples

Example 1. Find the wavelength of the radiation emitted when the electron of a hydrogen atom falls from n = 3 to n = 2.
Answer:
1/λ = R(1/2² − 1/3²) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹.
λ = 6.56 × 10⁻⁷ m = 656 nm (red line of the Balmer series).

Example 2. Calculate the de Broglie wavelength of an electron moving at 1.0 × 10⁶ m/s.
Answer:
λ = h/mv = 6.626 × 10⁻³⁴/(9.11 × 10⁻³¹ × 1.0 × 10⁶).
λ = 7.27 × 10⁻¹⁰ m.

Example 3. Write the ground-state configuration of Fe³⁺ and find its number of unpaired electrons.
Answer:
Fe (Z = 26) = [Ar]3d⁶ 4s². Fe³⁺ loses two 4s electrons and one 3d electron.
Fe³⁺ = [Ar]3d⁵, with five unpaired electrons (n = 5).

Example 4. Find the radius and energy of the electron in the second orbit of He⁺.
Answer:
r = 0.529 × n²/Z = 0.529 × 4/2 = 1.058 Å.
E = −13.6 Z²/n² = −13.6 × 4/4 = −13.6 eV.