Atomic structure for JEE Main

Atomic structure is the chapter that the rest of chemistry is built on. Periodic trends, bonding, magnetic moments and colour all come back to where the electrons are. In JEE Main it is also a reliable source of numerical questions: a Bohr-model energy, a spectral wavelength or a count of electrons with a given set of quantum numbers can each be solved in a minute if the relations are clear.

The official 2026 syllabus covers electromagnetic radiation and the photoelectric effect, the hydrogen spectrum, the Bohr model with derivations of energy and radius and its limitations, the dual nature of matter and de Broglie's relation, the Heisenberg uncertainty principle, the quantum mechanical model, ψ and ψ² against r for 1s and 2s, quantum numbers, shapes of s, p and d orbitals, electron spin, the Aufbau, Pauli and Hund rules, and the extra stability of half-filled and filled subshells.

Radiation and the photoelectric effect

Light of frequency ν and wavelength λ travels at c = νλ and comes in photons of energy E = hν = hc/λ. A handy form: E (in eV) = 1240/λ (in nm).

When light falls on a metal, electrons are emitted only if hν exceeds the work function φ:

• KE_max = hν − φ, and the threshold frequency is ν₀ = φ/h
• increasing intensity increases the number of electrons, not their maximum energy

The same ideas appear in physics; see modern physics for stopping potential and graphs.

The Bohr model and the hydrogen spectrum

Bohr assumed that the electron moves in circular orbits in which its angular momentum is mvr = nh/2π. For a hydrogen-like species of atomic number Z:

A transition from n₂ down to n₁ emits a photon with

1/λ = R_H Z²(1/n₁² − 1/n₂²), where R_H = 1.097 × 10⁷ m⁻¹.

Electrons falling from level n to the ground state in a sample of many atoms can produce n(n − 1)/2 different lines.

Limitations of the Bohr model: it works only for one-electron species, cannot explain the fine structure of lines or the splitting in electric (Stark) and magnetic (Zeeman) fields, and conflicts with the uncertainty principle by giving the electron a definite path.

Dual nature and uncertainty

• de Broglie: λ = h/mv = h/√(2mK). For an electron accelerated through V volts, λ ≈ 1.227/√V nm.
• Bohr's condition is the same as fitting a whole number of wavelengths round the orbit: 2πr = nλ.
• Heisenberg: Δx × Δp ≥ h/4π. The position and momentum of an electron cannot both be known exactly. The effect matters for electrons and is negligible for everyday objects.

The quantum mechanical model

An orbital is a one-electron wave function ψ; ψ² gives the probability density of finding the electron at a point. For 1s, ψ² is largest at the nucleus and falls steadily. For 2s, ψ passes through zero once, giving a spherical node.

Counts that follow:

• orbitals in shell n: n²; electrons in shell n: 2n²
• orbitals in a subshell: 2l + 1; electrons: 2(2l + 1)
• radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1

Shapes: s is spherical, p is dumb-bell shaped (three orientations), and d has four clover-leaf orbitals plus d_z², which has two lobes and a ring.

Filling electrons

• Aufbau: fill in order of increasing n + l; for equal n + l, lower n first. So 4s (4 + 0) fills before 3d (3 + 2).
• Pauli: no two electrons in an atom have the same four quantum numbers, so an orbital holds at most two, with opposite spins.
• Hund: in a subshell, electrons occupy orbitals singly with parallel spins before pairing.

Half-filled and completely filled subshells have extra stability (symmetry and exchange energy). This produces the two exceptions every JEE aspirant must know: Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹.

Unpaired electrons make a species paramagnetic, with spin-only moment μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. When forming cations of transition metals, remove the 4s electrons first: Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s².

Worked problems

Problem 1: a Balmer wavelength. Find the wavelength of the line emitted when a hydrogen electron falls from n = 3 to n = 2.

1/λ = R_H(1/4 − 1/9) = R_H × 5/36.
λ = 36/(5 × 1.097 × 10⁷) = 6.56 × 10⁻⁷ m = 656 nm (red light, the first Balmer line).

Problem 2: matching energies. Which orbit of He⁺ has the same energy as the ground state of hydrogen?

E(H, n = 1) = −13.6 eV. For He⁺, E = −13.6 × 4/n² eV.
Setting 4/n² = 1 gives n = 2. Its radius is 52.9 × 4/2 = 105.8 pm, twice the hydrogen ground-state radius.

Problem 3: de Broglie wavelength. An electron is accelerated from rest through 100 V. Find its wavelength.

λ = 1.227/√100 nm = 0.123 nm (1.23 Å).
In full: λ = h/√(2meV) = 6.626 × 10⁻³⁴/√(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 100) ≈ 1.23 × 10⁻¹⁰ m.

Problem 4 (numerical answer): counting by quantum numbers. How many electrons in a ground-state chromium atom (Z = 24) have m_l = 0?

Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹.
Every s electron has m_l = 0: 2 + 2 + 2 + 1 = 7.
Each full p subshell has one orbital with m_l = 0, holding 2: 2p gives 2 and 3p gives 2.
3d⁵ has one electron in each d orbital, so 1 has m_l = 0.
Total = 7 + 2 + 2 + 1 = 12.

Traps to watch

• Using the Bohr formulas for many-electron atoms. They hold only for H, He⁺, Li²⁺ and similar species.
• Number of lines. n(n − 1)/2 is for many atoms. A single atom falling from n = 4 can give at most 3 lines.
• Energy sign. Bohr energies are negative; the energy to remove an electron from level n is +13.6 Z²/n² eV.
• Cation configurations. Remove 4s before 3d.
• Nodes. Angular nodes equal l, not n − 1.

Practice set

1. Find the radial and angular nodes of a 3d orbital and of a 4f orbital.
2. How many spectral lines can a sample of hydrogen atoms emit when electrons fall from n = 6 to n = 1?
3. Find the longest wavelength in the Lyman series of hydrogen.
4. The position of an electron is known to within 1 × 10⁻¹⁰ m. Find the minimum uncertainty in its velocity (h = 6.626 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg).
5. Which set is not allowed: (a) n = 3, l = 2, m_l = −2 (b) n = 2, l = 2, m_l = 1 (c) n = 4, l = 0, m_l = 0 (d) n = 3, l = 1, m_l = −1?
6. Find the spin-only magnetic moment of Ni²⁺ (Z = 28).
7. Convert the ionisation energy of hydrogen, 2.18 × 10⁻¹⁸ J per atom, to kJ/mol.
8. Light of wavelength 248 nm falls on a metal of work function 2.0 eV. Find the maximum kinetic energy of the emitted electrons.

Answers

1. 3d: radial 3 − 2 − 1 = 0, angular 2. 4f: radial 4 − 3 − 1 = 0, angular 3.
2. 6 × 5/2 = 15.
3. The longest wavelength is the 2 → 1 transition: 1/λ = R_H × 3/4, so λ = 4/(3 × 1.097 × 10⁷) ≈ 122 nm.
4. Δv ≥ h/(4πmΔx) = 6.626 × 10⁻³⁴/(4π × 9.11 × 10⁻³¹ × 10⁻¹⁰) ≈ 5.8 × 10⁵ m/s.
5. (b): l cannot equal n.
6. Ni²⁺ is 3d⁸ with 2 unpaired electrons: μ = √8 ≈ 2.83 BM.
7. 2.18 × 10⁻¹⁸ × 6.022 × 10²³ = 1.313 × 10⁶ J/mol ≈ 1,312 kJ/mol.
8. E = 1240/248 = 5.0 eV, so KE_max = 5.0 − 2.0 = 3.0 eV.

What to do next

• Derive rₙ and Eₙ once from mvr = nh/2π and the Coulomb force, so the scalings stick.
• Write the configurations of elements 1 to 30 and of their common ions from memory.
• Practise quantum-number counting questions with our guide to quantum numbers and configuration.
• Move on to periodic properties, which apply these configurations.