NeetCrackers Chemistry Notes
Complete Notes

Atomic Structure

Every topic explained simply, with practice questions and answers after each one.

📑 Topics Covered

  1. Discovery of Sub-atomic Particles
  2. Atomic Number, Mass Number, Isotopes & Isobars
  3. Early Atomic Models
  4. Nature of Electromagnetic Radiation
  5. Planck's Quantum Theory & the Photoelectric Effect
  6. Atomic Spectra & Bohr's Model
  7. Dual Nature of Matter & Heisenberg's Principle
  8. Quantum Numbers
  9. Shapes of Atomic Orbitals
  10. Filling Electrons: Aufbau, Pauli & Hund's Rule
  11. Electronic Configuration & Stability

Discovery of Sub-atomic Particles

For a long time, atoms were considered the smallest indivisible unit of matter. Experiments in the late 19th and early 20th centuries revealed that atoms are themselves made of smaller particles — electrons, protons, and neutrons.

Discovery of the Electron

J.J. Thomson studied cathode ray discharge tubes — a gas at low pressure sealed inside a tube with electrodes, subjected to high voltage. The rays travelling from the cathode to the anode were found to bend towards a positive plate in an electric field, showing they carry a negative charge. Thomson calculated the charge-to-mass ratio ($e/m$) of these particles, called electrons, as $1.76 \times 10^{8}\ \text{C/g}$.

Later, R.A. Millikan's oil drop experiment determined the actual charge on an electron: $e = 1.6 \times 10^{-19}\ \text{C}$. Combining this with Thomson's ratio gave the electron's mass: $9.1 \times 10^{-28}\ \text{g}$ — about 1/1837th the mass of a hydrogen atom.

Discovery of the Proton

Using a modified cathode ray tube with a perforated cathode, Goldstein observed rays travelling in the direction opposite to cathode rays — called canal rays or anode rays. These carried a positive charge, and the lightest such particle (obtained using hydrogen gas) was named the proton, with a mass of $1.672 \times 10^{-24}\ \text{g}$ (about 1837 times that of an electron) and a charge of $+1.6 \times 10^{-19}\ \text{C}$.

Discovery of the Neutron

James Chadwick discovered the neutron by bombarding a beryllium sheet with alpha particles, producing a neutral particle with mass almost equal to that of a proton ($1.675 \times 10^{-24}\ \text{g}$) and no charge. This explained the mass discrepancy that protons and electrons alone couldn't account for.

ParticleChargeMassDiscovered by
Electron−1 (−1.6×10⁻¹⁹ C)9.1×10⁻²⁸ gJ.J. Thomson
Proton+1 (+1.6×10⁻¹⁹ C)1.672×10⁻²⁴ gGoldstein
Neutron01.675×10⁻²⁴ gChadwick
Q. What key observation from the cathode ray experiment showed that electrons carry a negative charge?
Ans: The cathode rays were deflected towards the positively charged plate in an external electric field — a negative charge is attracted towards a positive plate, confirming the rays (electrons) carry negative charge.

Atomic Number, Mass Number, Isotopes & Isobars

Key Definitions $$Z\ (\text{Atomic number}) = \text{Number of protons}$$ $$A\ (\text{Mass number}) = \text{Number of protons} + \text{Number of neutrons}$$ $$\text{Number of neutrons} = A - Z$$

A neutral atom also has electrons equal in number to its protons.

Isotopes, Isobars & Isotones

Definitions

Isotopes — same atomic number (Z), different mass number (A); e.g., $^1_1\text{H}$, $^2_1\text{H}$, $^3_1\text{H}$. Isobars — same mass number, different atomic number; e.g., $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$. Isotones — same number of neutrons, different atomic and mass numbers; e.g., $^{14}_{6}\text{C}$ and $^{16}_{8}\text{O}$ (both have 8 neutrons).

Q. An atom has a mass number of 23 and 12 neutrons. Find its atomic number and the number of electrons in the neutral atom.
Ans: Z = A − neutrons = 23 − 12 = 11. Number of electrons in a neutral atom = number of protons = 11.

Early Atomic Models

Thomson's Model

J.J. Thomson proposed that an atom is a sphere of uniform positive charge with electrons embedded in it, like seeds in a watermelon (or "plum pudding"). It successfully explained the overall neutrality of the atom but failed to explain later experimental results like the scattering of alpha particles.

Rutherford's Nuclear Model

In the famous gold foil (alpha-particle scattering) experiment, a beam of alpha particles was directed at a thin gold foil. Most particles passed straight through, some deflected at small angles, and a very small fraction bounced almost straight back.

Conclusions Drawn

Most of the atom is empty space (since most particles passed through undeflected). A small, dense, positively charged centre — the nucleus — repelled the few particles that came close to it, causing large-angle deflections. Electrons revolve around this nucleus in circular orbits, held by electrostatic attraction.

Where This Model Fails

Rutherford's model couldn't explain why revolving electrons — which are accelerating charges and should continuously emit energy according to classical electromagnetic theory — don't spiral into the nucleus. It also failed to explain the line spectrum of hydrogen.

Q. Why did most alpha particles pass straight through the gold foil without any deflection?
Ans: Because most of the volume of an atom is empty space, so the majority of alpha particles didn't encounter any charged region dense enough to deflect them.

Nature of Electromagnetic Radiation

Light and other forms of radiant energy travel as electromagnetic waves — oscillating electric and magnetic fields perpendicular to each other and to the direction of travel. All electromagnetic radiation travels through vacuum at the same speed, $c = 3 \times 10^8\ \text{m/s}$.

Wave Relations $$c = \nu \lambda$$ $$\bar{\nu} = \dfrac{1}{\lambda}$$

where $\nu$ = frequency (Hz or s⁻¹), $\lambda$ = wavelength (m), and $\bar\nu$ = wavenumber (m⁻¹).

The complete range of electromagnetic radiation — from long-wavelength radio waves to short-wavelength gamma rays — is called the electromagnetic spectrum, and includes radio, microwave, infrared, visible, ultraviolet, X-ray, and gamma regions in increasing order of frequency (and energy).

Q. Calculate the frequency of light with a wavelength of 500 nm.
Ans: $\nu = c/\lambda = (3\times10^8)/(500\times10^{-9}) = 6\times10^{14}\ \text{Hz}$.

Planck's Quantum Theory & the Photoelectric Effect

Classical physics couldn't explain why heated bodies emit radiation of certain frequencies (black-body radiation) or why light striking a metal knocks out electrons only above a certain frequency. Max Planck resolved this by proposing that energy is emitted or absorbed in discrete packets called quanta (photons, in the case of light), not continuously.

Planck's Equation $$E = h\nu = \dfrac{hc}{\lambda}$$

where $h$ = Planck's constant $= 6.626 \times 10^{-34}\ \text{J s}$.

Photoelectric Effect

Observation

When light of sufficient frequency strikes a metal surface, electrons are ejected instantly — but only if the frequency of light exceeds a minimum value called the threshold frequency ($\nu_0$), regardless of the light's intensity. Increasing intensity beyond that threshold increases the number of electrons ejected, not their individual energy; increasing frequency increases the kinetic energy of each ejected electron.

Einstein's Photoelectric Equation $$h\nu = h\nu_0 + \tfrac{1}{2}m_ev^2$$

Energy of incident photon = work function ($h\nu_0$, the minimum energy to eject an electron) + kinetic energy of the ejected electron.

Q. Calculate the energy of one photon of light with a frequency of $5 \times 10^{14}$ Hz.
Ans: $E = h\nu = (6.626\times10^{-34})(5\times10^{14}) = 3.31\times10^{-19}\ \text{J}$.

Atomic Spectra & Bohr's Model

When hydrogen gas is excited electrically and the emitted light is passed through a prism, it produces not a continuous rainbow but a series of sharp, separate lines — a line (atomic) spectrum. This was the key evidence Bohr's model needed to explain, which Rutherford's model could not.

Postulates of Bohr's Model

  1. Electrons revolve around the nucleus in fixed circular paths called orbits or shells, each with a definite energy — these are called stationary states.
  2. Electrons don't radiate energy while in a stationary state, regardless of classical expectations.
  3. Energy is absorbed when an electron jumps to a higher orbit, and emitted (as a photon) when it falls to a lower orbit — with the energy of the photon equal to the energy difference between the two orbits: $\Delta E = h\nu$.
  4. The angular momentum of an electron in an orbit is quantised: $mvr = \dfrac{nh}{2\pi}$, where $n = 1, 2, 3...$ is the orbit number.
Bohr's Formulas (for hydrogen-like species) $$r_n = 0.529 \times \dfrac{n^2}{Z}\ \overset{\circ}{\text{A}}$$ $$E_n = -2.18 \times 10^{-18} \times \dfrac{Z^2}{n^2}\ \text{J per atom}$$

where $n$ = orbit number, $Z$ = atomic number. The negative sign shows the electron is bound to the nucleus; energy becomes less negative (higher) as $n$ increases.

Solved Example

Calculate the radius of the second Bohr orbit of hydrogen.

$r_2 = 0.529 \times \dfrac{2^2}{1} = 0.529 \times 4 = 2.116\ \overset{\circ}{\text{A}}$

Where Bohr's Model Falls Short

It works well only for single-electron species (H, He⁺, Li²⁺) and fails to explain the spectra of multi-electron atoms, the splitting of spectral lines in a magnetic field (Zeeman effect), or the wave nature of electrons — limitations resolved later by the quantum mechanical model.

Q. What causes the line spectrum of hydrogen, according to Bohr's model?
Ans: Electrons falling from a higher-energy orbit to a lower-energy orbit emit a photon with energy exactly equal to the energy gap between the two orbits — since orbit energies are fixed (quantised), only specific photon energies (and hence specific wavelengths/lines) are possible, producing a line spectrum rather than a continuous one.

Dual Nature of Matter & Heisenberg's Uncertainty Principle

de Broglie's Hypothesis

Louis de Broglie proposed that just as light shows both wave and particle behaviour, matter (electrons, protons, and even larger objects) also has an associated wavelength.

de Broglie Equation $$\lambda = \dfrac{h}{mv} = \dfrac{h}{p}$$

where $p = mv$ is momentum. This wavelength is only significant (measurable) for very small masses, like electrons — for everyday objects it's far too small to detect.

Heisenberg's Uncertainty Principle

Statement

It is impossible to simultaneously determine, with perfect precision, both the exact position and exact momentum of a microscopic particle like an electron. This isn't a limitation of measuring instruments — it's a fundamental property of nature at that scale, and directly follows from the wave nature of matter.

Mathematical Form $$\Delta x \cdot \Delta p \ge \dfrac{h}{4\pi}$$

where $\Delta x$ = uncertainty in position, $\Delta p$ = uncertainty in momentum. This principle led directly to the idea of orbitals (regions of probability) replacing Bohr's fixed, well-defined orbits.

Q. Why is the wave nature of a moving car not noticeable in everyday life, even though de Broglie's equation applies to all matter?
Ans: Because the car's mass is enormous compared to an electron's, and de Broglie wavelength (λ = h/mv) is inversely proportional to mass — for a large mass, the resulting wavelength is so extremely small that it has no observable/measurable effect.

Quantum Numbers

Since Bohr's fixed orbits gave way to the quantum mechanical model, an electron's position and energy are now described using four quantum numbers.

Quantum NumberSymbolDescribesAllowed Values
PrincipalnShell / energy level & size of orbital1, 2, 3...
Azimuthal (angular)lSub-shell & shape of orbital0 to (n−1)
MagneticmₗOrientation of orbital in space−l to +l (including 0)
SpinmₛDirection of electron spin+½ or −½

Sub-shells corresponding to $l = 0, 1, 2, 3$ are named s, p, d, f respectively. For a given $n$, the number of orbitals in a sub-shell is $(2l+1)$, and the total number of orbitals in a shell is $n^2$.

Solved Example

For n = 3, list the possible values of l and the sub-shells present.

$l$ can be $0, 1, 2$ → sub-shells present are $3s, 3p, 3d$.

Q. What are the possible values of the magnetic quantum number for l = 2?
Ans: mₗ can range from −l to +l, so for l = 2: −2, −1, 0, +1, +2 — a total of 5 values, corresponding to the 5 orbitals of a d sub-shell.

Shapes of Atomic Orbitals

Nodes

A node is a region where the probability of finding an electron is zero. Radial nodes $= n - l - 1$; angular nodes $= l$; total nodes $= n - 1$.

Q. Calculate the number of radial and angular nodes for a 3p orbital.
Ans: Radial nodes = n − l − 1 = 3 − 1 − 1 = 1. Angular nodes = l = 1. Total nodes = n − 1 = 2.

Filling Electrons: Aufbau, Pauli & Hund's Rule

Aufbau Principle

Electrons occupy orbitals in order of increasing energy — lower-energy orbitals fill first. The order generally follows the $(n+l)$ rule: orbitals with lower $(n+l)$ values fill first; if two orbitals have the same $(n+l)$ value, the one with lower $n$ fills first.

Standard Filling Order $$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s$$

Pauli Exclusion Principle

Statement

No two electrons in the same atom can have all four quantum numbers identical. This means a single orbital can hold a maximum of 2 electrons, and they must have opposite spins.

Hund's Rule of Maximum Multiplicity

Statement

Electrons are distributed among orbitals of the same sub-shell (equal energy) singly first, with parallel spins, before any orbital receives a second electron. Pairing begins only once every orbital in that sub-shell has one electron.

Q. Write the electron distribution across the three 2p orbitals for nitrogen (Z = 7), and state which rule this follows.
Ans: Nitrogen's configuration is 1s² 2s² 2p³ — all three 2p orbitals get one electron each (2pₓ¹ 2p_y¹ 2p_z¹) with parallel spins, following Hund's rule, rather than pairing up in fewer orbitals.

Electronic Configuration & Stability

The electronic configuration of an atom shows how its electrons are distributed across shells and sub-shells, written using the notation $n l^{x}$ (e.g., $1s^2\, 2s^2\, 2p^6$ for neon).

Stability of Half-filled & Fully-filled Configurations

Why They're Extra Stable

Half-filled ($p^3, d^5, f^7$) and fully-filled ($p^6, d^{10}, f^{14}$) sub-shells have extra stability due to two effects: greater symmetry of electron distribution, and greater exchange energy — the stabilising energy released when electrons with the same spin in degenerate orbitals exchange positions, which is maximum when the sub-shell is exactly half or fully filled.

Common Exceptions

Chromium (Z = 24) is $[\text{Ar}]\, 3d^5\, 4s^1$, not $3d^4\, 4s^2$ — one electron shifts from 4s to 3d to achieve a stable half-filled d sub-shell. Similarly, copper (Z = 29) is $[\text{Ar}]\, 3d^{10}\, 4s^1$, not $3d^9\, 4s^2$, to achieve a stable fully-filled d sub-shell.

Q. Write the electronic configuration of chromium (Z = 24) and explain why it's an exception to the normal Aufbau order.
Ans: $[\text{Ar}]\, 3d^5\, 4s^1$. The expected configuration by strict Aufbau order would be $3d^4\, 4s^2$, but one 4s electron shifts into the 3d sub-shell so that 3d becomes exactly half-filled (extra stable due to symmetry and exchange energy), making $3d^5\, 4s^1$ the actual, more stable configuration.

⚡ Formula Cheat-Sheet

Mass numberA = protons + neutrons
Wave-frequencyc = νλ
Photon energyE = hν = hc/λ
Bohr radiusr_n = 0.529 n²/Z Å
Bohr energyE_n = −2.18×10⁻¹⁸ Z²/n² J
de Broglie wavelengthλ = h/mv
Heisenberg principleΔx·Δp ≥ h/4π
Orbitals per shelln²
Max electrons per shell2n²