Class 11 Chemistry Notes

Structure of Atom Class 11 Notes

Complete, exam-ready notes on the structure of atom: discovery of the electron, proton and neutron, atomic models from Thomson to Bohr, quantum numbers, the Aufbau–Pauli–Hund rules for electronic configuration, and de Broglie's wave-particle duality — written for CBSE, JEE and NEET revision.

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Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the structure of an atom in one line?

An atom is a dense positive nucleus of protons and neutrons surrounded by electrons arranged in shells and orbitals, whose exact organisation is described by quantum numbers.

Subatomic Particles and Atomic Numbers

Isotopes and isobars

The atomic number Z is the number of protons; the mass number A is protons + neutrons. Isotopes share Z with different A (same element); isobars share A with different Z (different elements); isotones share the same neutron number.

Electron — charge −1.6 × 10⁻¹⁹ C, mass ≈ 9.1 × 10⁻³¹ kg (discovered by Thomson).
Proton — charge +1.6 × 10⁻¹⁹ C, mass ≈ 1.67 × 10⁻²⁷ kg (Rutherford).
Neutron — no charge, mass slightly more than the proton (Chadwick).

Atomic Models — Thomson and Rutherford

Rutherford's gold foil experiment

Most alpha particles passed straight through the foil, some deflected, a few bounced back. Conclusion: most of the atom is empty space with a tiny, dense, positive nucleus at the centre. The model could not explain why electrons orbiting the nucleus don't spiral in.

Failed classical model

Classical physics predicted an accelerating electron should radiate energy and collapse into the nucleus — which atoms obviously don't do. This instability forced Bohr's quantised model.

Bohr's Model of the Atom

Electrons move in fixed, allowed circular orbits (shells n = 1, 2, 3...) where angular momentum is quantised.
Energy is absorbed or emitted only when an electron jumps between orbits — as photons with exact energy.
Works beautifully for hydrogen (one electron) but fails for multi-electron atoms.

E_n = -\frac{2.18\times10^{-18}\,Z^2}{n^2}\,\text{J},\qquad r_n = 0.529\,\frac{n^2}{Z}\,\text{Å},\qquad \frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Bohr energy, radius and Rydberg formula

Quantum Numbers and Orbitals

Quantum numbers

Four numbers specify an electron: n (principal — shell, energy), l (azimuthal — 0 to n−1, s/p/d/f subshell), m (magnetic — −l to +l, orbital orientation), and s (spin — ±½). A shell with quantum number n has n subshells, n² orbitals and a capacity of 2n² electrons.

\text{orbitals in a shell} = n^2,\qquad \text{electrons in a shell} = 2n^2

Shell capacity

Electronic Configuration — Aufbau, Pauli and Hund

Aufbau principle — electrons fill orbitals in order of increasing energy: 1s < 2s < 2p < 3s < 3p < 4s < 3d...
Pauli exclusion principle — no two electrons in an atom can have all four quantum numbers identical, so each orbital holds at most two electrons with opposite spins.
Hund's rule — within a subshell, electrons first occupy each orbital singly with parallel spins before pairing.

n + l rule

Fill the subshell with the lower (n + l) value first; if (n + l) ties, the lower n wins. This is why 4s fills before 3d — (4+0) vs (3+2).

de Broglie and Heisenberg Uncertainty

Wave-particle duality

Matter has a wavelength too: $\lambda = h/mv$ . For macroscopic objects it is undetectably small. Heisenberg's principle says position and momentum cannot both be known precisely: the product of their uncertainties is at least h/4π.

\lambda = \frac{h}{mv},\qquad \Delta x \cdot \Delta p \geq \frac{h}{4\pi}

de Broglie wavelength and uncertainty principle

Solved Examples

Example: Write the electronic configuration of $_{26}\text{Fe}$ (Z = 26) and state its total unpaired electrons.

Solution: Configuration is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$ . By Hund's rule the 3d⁶ subshell has 4 unpaired — filling the five d-orbitals singly first gives four unpaired electrons and one pair.

Example: What is the de Broglie wavelength of an electron moving at $2\times10^6\,\text{m/s}$ ? (h = 6.626 × 10⁻³⁴ J·s, mₑ = 9.1 × 10⁻³¹ kg)

Solution: $\lambda = \frac{6.626\times10^{-34}}{9.1\times10^{-31} \times 2\times10^6} \approx 3.64\times10^{-10}\,\text{m}$ (≈ 3.6 Å, in the same range as atomic sizes).

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

de Broglie wavelength

\lambda = \frac{h}{mv}

Heisenberg uncertainty

\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

Bohr energy

E_n = -2.18\times10^{-18}\,\frac{Z^2}{n^2}\,\text{J}

Bohr radius

r_n = 0.529\,\frac{n^2}{Z}\,\text{Å}

Rydberg formula

\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Photon energy

E = h\nu = \frac{hc}{\lambda}

Shell capacity

2n^2

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

n = 1,2,3...; l = 0 to n−1; m = −l to +l; s = ±½.
Pauli: two electrons per orbital with opposite spins.
Hund: fill orbitals singly with parallel spins before pairing.
Shell n holds n² orbitals and 2n² electrons.
4s fills before 3d (n + l rule).
de Broglie wavelength matters only for tiny particles like electrons.

FAQ

Common questions

What is the difference between Rutherford's and Bohr's atomic model?

Rutherford's model had electrons orbiting a dense nucleus with no explanation of stability; Bohr quantised the orbits so electrons can only occupy fixed energy levels and radiate only on jumping between them.

What are the four quantum numbers?

n (principal — shell/energy), l (azimuthal — subshell type), m (magnetic — orbital orientation) and spin quantum number s (±½). Together they uniquely identify an electron.

What is the difference between Aufbau, Pauli and Hund?

Aufbau says fill orbitals in order of increasing energy; Pauli says no two electrons share all four quantum numbers (max two per orbital, opposite spins); Hund says occupy each orbital of a subshell singly first, with parallel spins.

What is the Heisenberg uncertainty principle?

The position and momentum of a particle cannot both be determined exactly at the same time — the product of the uncertainties Δx·Δp is at least h/4π. It is significant only for subatomic particles.

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