JEE Main Chemistry · Chapter 2

Atomic Structure

Bohr model, hydrogen spectrum, de Broglie and Heisenberg, quantum numbers, orbitals and electronic configuration.

NCERT Class 11JEE Main: 1–2 Qs per shift 8 min read

1. Chapter at a Glance

Atomic structure explains what atoms are made of and how their electrons are arranged. We move from early models (Thomson, Rutherford) to Bohr's model of the hydrogen atom, then to the quantum mechanical model with orbitals and quantum numbers. The arrangement of electrons decides the chemical behaviour of every element. JEE Main usually asks one or two questions — often numericals on the Bohr model and spectral lines, and conceptual questions on quantum numbers and nodes.

What JEE asks most
  • Radius, energy and velocity of an electron in the n-th Bohr orbit of hydrogen-like species.
  • Wavelengths of spectral lines and the number of lines emitted.
  • de Broglie wavelength and Heisenberg's uncertainty principle.
  • Quantum numbers, nodes of orbitals, electronic configuration and unpaired electrons.

2. Sub-atomic Particles and Early Models

ParticleDiscovered byChargeMass
ElectronJ. J. Thomson (cathode rays)−1.602 × 10−19 C9.11 × 10−31 kg
ProtonGoldstein (anode rays), Rutherford+1.602 × 10−19 C1.673 × 10−27 kg
NeutronChadwick01.675 × 10−27 kg
  • Thomson's model ("plum pudding"): positive sphere with electrons embedded.
  • Rutherford's α-scattering experiment: most α-particles passed straight through gold foil, a few were deflected, and very few bounced back. Conclusion: a tiny, dense, positively charged nucleus with electrons around it. The model could not explain the stability of the atom or line spectra.
  • Atomic number Z = number of protons; mass number A = protons + neutrons. Isotopes have the same Z, different A; isobars have the same A; isotones have the same number of neutrons.

3. Bohr's Model of Hydrogen-like Species

Bohr proposed that the electron moves only in certain allowed orbits in which its angular momentum is a multiple of h/2π (mvr = nh/2π), and that energy is emitted or absorbed only when it jumps between orbits.

Quantity (n-th orbit, atomic number Z)FormulaDepends on
Radiusrn = 0.529 n²/Z Ån²/Z
EnergyEn = −13.6 Z²/n² eV = −2.18 × 10−18 Z²/n² JZ²/n²
Velocityvn = 2.18 × 10⁶ Z/n m s−1Z/n

Kinetic energy = −En and potential energy = 2En. The ionisation energy of H is 13.6 eV.

n = 1 n = 2 n = 3 n = 4 n = ∞ Lyman (UV) Balmer (visible) E₁ = −13.6 eVE₂ = −3.4 eV
Transitions to n = 1 form the Lyman series (ultraviolet); transitions to n = 2 form the Balmer series (visible). Energy levels crowd together as n increases.

Spectral lines

1/λ = RH Z² (1/n1² − 1/n2²),   RH = 1.097 × 10⁷ m−1
Seriesn1 (lower level)Region
Lyman1Ultraviolet
Balmer2Visible
Paschen3Infrared
Brackett4Infrared
Pfund5Infrared

Maximum number of lines when an electron falls from level n to the ground state: n(n − 1)/2. Within one series ending at n1: n − n1.

4. Dual Nature and Uncertainty

  • Photon energy: E = hν = hc/λ, h = 6.626 × 10−34 J s.
  • Photoelectric effect: hν = hν0 + ½mv² (work function hν0). Emission depends on frequency, not intensity; intensity decides the number of electrons.
  • de Broglie wavelength: λ = h/mv = h/p. For an electron accelerated through V volts, λ = 12.27/√V Å.
  • Heisenberg's uncertainty principle: Δx · Δp ≥ h/4π — position and momentum cannot both be known exactly.

5. Quantum Numbers and Orbitals

Quantum numberSymbolValuesTells us
Principaln1, 2, 3, …Shell, size and energy
Azimuthall0 to n − 1Subshell and shape (s, p, d, f)
Magneticml−l to +lOrientation of the orbital
Spinms+½, −½Spin of the electron
  • A subshell with azimuthal number l has 2l + 1 orbitals; a shell n has n² orbitals and at most 2n² electrons.
  • Nodes: radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.
  • Shapes: s is spherical; p is dumb-bell shaped; dz² has a doughnut ring, other d orbitals are four-lobed.

Filling electrons

  • Aufbau principle: fill orbitals in order of increasing (n + l); if equal, lower n first: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, …
  • Pauli exclusion principle: no two electrons in an atom have all four quantum numbers the same (so an orbital holds at most 2 electrons of opposite spin).
  • Hund's rule: electrons occupy degenerate orbitals singly with parallel spins before pairing.
  • Exceptions due to the extra stability of half-filled and fully filled d subshells: Cr = [Ar]3d⁵4s¹, Cu = [Ar]3d¹⁰4s¹.
  • Magnetic moment (spin-only): μ = √(n(n + 2)) BM, n = number of unpaired electrons.

6. Formula Master Table

QuantityFormula
Bohr radiusrn = 0.529 n²/Z Å
Bohr energyEn = −13.6 Z²/n² eV
Bohr velocityvn = 2.18 × 10⁶ Z/n m s−1
Angular momentummvr = nh/2π
Rydberg formula1/λ = R Z²(1/n1² − 1/n2²)
Number of spectral linesn(n − 1)/2
Photon energyE = hc/λ
de Broglieλ = h/mv; electron: 12.27/√V Å
UncertaintyΔx·Δp ≥ h/4π
Nodesradial n − l − 1, angular l
Magnetic momentμ = √(n(n + 2)) BM

7. Mind Map

Atomic Structure
Particles & models
  • e, p, n
  • Rutherford nucleus
  • Isotopes, isobars
Bohr model
  • r ∝ n²/Z, E ∝ Z²/n²
  • Spectral series
  • n(n − 1)/2 lines
Dual nature
  • Photoelectric effect
  • λ = h/mv
  • Uncertainty
Quantum model
  • n, l, m, s
  • Nodes
  • Aufbau, Pauli, Hund

8. Tips and Tricks

Shortcuts
  • For ratio questions between species or orbits, use only the proportionality (r ∝ n²/Z, E ∝ Z²/n², v ∝ Z/n) — constants cancel.
  • Shortest wavelength of a series: put n2 = ∞. Longest wavelength: n2 = n1 + 1.
  • For transition metal ions, remove electrons from 4s before 3d: Fe = [Ar]3d⁶4s², Fe³⁺ = [Ar]3d⁵ (5 unpaired).
  • Count nodes: 3p has 1 radial and 1 angular node; 4d has 1 radial and 2 angular nodes.
Mistakes students make
  • Using the Bohr model for multi-electron atoms — it works only for one-electron species (H, He⁺, Li²⁺ …).
  • Forgetting that energy is negative; "higher energy" means closer to zero.
  • Writing l = n for a subshell (l can be at most n − 1).

9. Important One-liners

  • The charge-to-mass ratio of the electron was measured by Thomson; the charge by Millikan's oil-drop experiment.
  • The radius of the nucleus is about 10−15 m, while the atom is about 10−10 m.
  • The ground-state energy of hydrogen is −13.6 eV; for He⁺ it is −54.4 eV.
  • The Balmer series is the only series of hydrogen in the visible region.
  • An orbital can hold a maximum of two electrons, with opposite spins.
  • Half-filled and fully filled subshells are extra stable due to symmetry and exchange energy.
  • The 4s orbital is filled before 3d, but in ions 4s electrons are removed first.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Numerical

An electron in a hydrogen atom falls from n = 5 to the ground state. What is the maximum number of spectral lines that can be emitted?

Show solution

Maximum lines = n(n − 1)/2 = 5 × 4/2 = 10.

Answer: 10

Q2Easy

The energy of an electron in the second orbit of He⁺ is:

  1. −3.4 eV
  2. −13.6 eV
  3. −54.4 eV
  4. −6.8 eV
Show solution

E = −13.6 Z²/n² = −13.6 × 4/4 = −13.6 eV (Z = 2, n = 2).

Answer: (B)

Q3Easy

The number of radial nodes in a 3p orbital is:

  1. 0
  2. 1
  3. 2
  4. 3
Show solution

Radial nodes = n − l − 1 = 3 − 1 − 1 = 1. (It also has l = 1 angular node.)

Answer: (B)

Q4Medium

The ratio of the radius of the second Bohr orbit of He⁺ to that of the third Bohr orbit of Li²⁺ is:

  1. 2 : 3
  2. 3 : 2
  3. 4 : 9
  4. 1 : 1
Show solution

r ∝ n²/Z. He⁺ (n = 2, Z = 2): 4/2 = 2. Li²⁺ (n = 3, Z = 3): 9/3 = 3. Ratio 2 : 3.

Answer: (A)

Q5Easy

Which set of quantum numbers is NOT possible?

  1. n = 2, l = 2, m = 0, s = +½
  2. n = 3, l = 2, m = −2, s = −½
  3. n = 1, l = 0, m = 0, s = +½
  4. n = 4, l = 3, m = +3, s = −½
Show solution

l can be at most n − 1. For n = 2, l can only be 0 or 1, so l = 2 is not allowed. All other sets obey the rules.

Answer: (A)

Q6Medium

The de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V is approximately:

  1. 12.27 Å
  2. 1.227 Å
  3. 0.1227 Å
  4. 122.7 Å
Show solution

λ = 12.27/√V Å = 12.27/10 = 1.227 Å.

Answer: (B)

Q7Medium

The spin-only magnetic moment of Fe³⁺ (Z of Fe = 26) is closest to:

  1. 4.90 BM
  2. 5.92 BM
  3. 3.87 BM
  4. 2.83 BM
Show solution

Fe = [Ar]3d⁶4s²; Fe³⁺ loses the two 4s electrons and one 3d electron: [Ar]3d⁵, with 5 unpaired electrons.

μ = √(5 × 7) = √35 ≈ 5.92 BM.

Answer: (B)

Q8Hard

The shortest wavelength in the Balmer series of hydrogen is approximately (RH = 1.097 × 10⁷ m−1):

  1. 656 nm
  2. 486 nm
  3. 365 nm
  4. 122 nm
Show solution

Shortest wavelength means the largest energy jump: n2 = ∞ to n1 = 2.

1/λ = R(1/4 − 0) ⇒ λ = 4/R = 4/(1.097 × 10⁷) m ≈ 3.65 × 10−7 m = 365 nm.

(656 nm is the longest Balmer line, and 122 nm is the first Lyman line.)

Answer: (C)

11. Quick Revision Summary

Remember before the exam
  • r ∝ n²/Z, E ∝ −Z²/n², v ∝ Z/n (one-electron species only).
  • 1/λ = RZ²(1/n1² − 1/n2²); Lyman UV, Balmer visible; lines = n(n − 1)/2.
  • λ = h/mv; Δx·Δp ≥ h/4π; E = hc/λ.
  • l = 0 to n − 1; orbitals 2l + 1; nodes: radial n − l − 1, angular l.
  • Aufbau (n + l), Pauli, Hund; Cr and Cu exceptions; μ = √(n(n + 2)) BM.
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