By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Mastering atomic structure unlocks 10-15% of IIT JEE Chemistry—that’s 30-45 marks in JEE Main and 50+ marks in JEE Advanced. It’s the foundation for periodic trends, chemical bonding, and spectroscopy, and it appears in every single JEE paper. If you can solve these problems fast and accurately, you’re already ahead of 80% of test-takers.
Before diving in, ensure you understand:1. Rutherford’s nuclear model – Electrons revolve around a dense nucleus.2. Electromagnetic spectrum basics – Wavelength, frequency, and energy relationships (E = hν).3. Basic algebra and logarithms – For solving Bohr’s formula and quantum number problems.
If any of these are shaky, stop now and review them first.
Formula: rₙ = (0.529 Å) × n² / Z - rₙ = Radius of the nth orbit (in Ångströms) - n = Principal quantum number (1, 2, 3, ...) - Z = Atomic number (number of protons) MEMORISE THIS – You’ll use it for hydrogen-like atoms (H, He⁺, Li²⁺, etc.).
rₙ = (0.529 Å) × n² / Z
rₙ
n
Z
Formula: Eₙ = – (13.6 eV) × Z² / n² - Eₙ = Energy of the electron in the nth orbit (in eV) - Z = Atomic number - n = Principal quantum number MEMORISE THIS – Negative sign means the electron is bound to the nucleus.
Eₙ = – (13.6 eV) × Z² / n²
Eₙ
Formula: 1/λ = R × Z² × (1/n₁² – 1/n₂²) - λ = Wavelength of light (in meters) - R = Rydberg constant = 1.097 × 10⁷ m⁻¹ (given on exam sheet) - n₁ = Lower energy level - n₂ = Higher energy level - Z = Atomic number Given on exam sheet – But you must know how to apply it.
1/λ = R × Z² × (1/n₁² – 1/n₂²)
λ
R
n₁
n₂
l
mₗ
mₛ
MEMORISE THIS TABLE – Examiners love asking for valid quantum number sets.
MEMORISE THIS – Questions on nodes and shapes appear every year.
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f...
If same (n + l), lower n = lower energy
Pauli Exclusion Principle – No two electrons can have the same four quantum numbers.
Each orbital holds max 2 electrons (opposite spins).
Hund’s Rule – Electrons fill degenerate orbitals (same energy) singly first, with parallel spins.
MEMORISE THESE RULES – Violating them is an instant wrong answer.
Ask yourself: - Is it about Bohr’s model (energy, radius, wavelength)? - Is it about quantum numbers (valid sets, max electrons)? - Is it about orbital shapes/nodes? - Is it about electronic configuration (ground state, exceptions)?
E
1 eV = 1.6 × 10⁻¹⁹ J
–l
+l
Question: Calculate the radius of the 2nd orbit of He⁺ (Z = 2).
Step 1: Identify problem type → Bohr’s radius. Step 2: Given: n = 2, Z = 2. Step 3: Use rₙ = (0.529 Å) × n² / Z. Step 4: Plug in values: r₂ = (0.529 Å) × (2)² / 2 = (0.529 × 4) / 2 = 1.058 Å. Step 5: Check units → Å is correct.
n = 2
Z = 2
r₂ = (0.529 Å) × (2)² / 2 = (0.529 × 4) / 2 = 1.058 Å
Answer: 1.058 Å
1.058 Å
What we did and why: We used Bohr’s radius formula because the question asked for the size of an orbit in a hydrogen-like ion (He⁺). The formula directly relates n and Z to the radius.
Question: Find the wavelength of light emitted when an electron in Li²⁺ (Z = 3) jumps from n = 3 to n = 2.
n = 3
Step 1: Identify problem type → Wavelength using Rydberg formula. Step 2: Given: Z = 3, n₁ = 2, n₂ = 3. Step 3: Use 1/λ = R × Z² × (1/n₁² – 1/n₂²). Step 4: Plug in values: 1/λ = (1.097 × 10⁷ m⁻¹) × 3² × (1/2² – 1/3²) = 1.097 × 10⁷ × 9 × (1/4 – 1/9) = 1.097 × 10⁷ × 9 × (5/36) = 1.097 × 10⁷ × 1.25 = 1.371 × 10⁷ m⁻¹ Step 5: Take reciprocal: λ = 1 / (1.371 × 10⁷) ≈ 7.29 × 10⁻⁸ m = 72.9 nm.
Z = 3
n₁ = 2
n₂ = 3
1/λ = (1.097 × 10⁷ m⁻¹) × 3² × (1/2² – 1/3²)
= 1.097 × 10⁷ × 9 × (1/4 – 1/9)
= 1.097 × 10⁷ × 9 × (5/36)
= 1.097 × 10⁷ × 1.25
= 1.371 × 10⁷ m⁻¹
λ = 1 / (1.371 × 10⁷) ≈ 7.29 × 10⁻⁸ m = 72.9 nm
Answer: 72.9 nm
72.9 nm
What we did and why: We used the Rydberg formula because the question involved an electron transition in a hydrogen-like ion (Li²⁺). The formula connects energy levels to wavelength, and we converted the final answer to nm for convenience.
Question: Which of the following is not a valid set of quantum numbers for an electron in a 3d orbital? (A) n = 3, l = 2, mₗ = –1, mₛ = +½ (B) n = 3, l = 2, mₗ = 3, mₛ = –½ (C) n = 3, l = 2, mₗ = 0, mₛ = +½ (D) n = 3, l = 2, mₗ = –2, mₛ = –½
n = 3, l = 2, mₗ = –1, mₛ = +½
n = 3, l = 2, mₗ = 3, mₛ = –½
n = 3, l = 2, mₗ = 0, mₛ = +½
n = 3, l = 2, mₗ = –2, mₛ = –½
Step 1: Identify problem type → Quantum number validity. Step 2: Recall rules: - l must be 0 to (n–1) → For n = 3, l can be 0, 1, 2. - mₗ must be –l to +l → For l = 2, mₗ can be –2, –1, 0, +1, +2. - mₛ can only be +½ or –½. Step 3: Check each option: - (A) Valid (mₗ = –1 is allowed). - (B) Invalid (mₗ = 3 is not allowed for l = 2). - (C) Valid (mₗ = 0 is allowed). - (D) Valid (mₗ = –2 is allowed). Step 4: Confirm answer → (B) is invalid.
0
(n–1)
0, 1, 2
l = 2
–2, –1, 0, +1, +2
+½
–½
mₗ = –1
mₗ = 3
mₗ = 0
mₗ = –2
Answer: (B)
What we did and why: We systematically checked each quantum number against the rules. The trap was in option (B), where mₗ exceeded the allowed range for l = 2. Always verify mₗ first—it’s the most common mistake.
Z = 1
l = n
2p³
↑ ↑ ↑
↑↓ ↑
3p
2 nodes
Z > 1
mₗ = ±(l+1)
s¹ d⁵
s¹ d¹⁰
s² d⁴
s² d⁹
[Ar] 4s¹ 3d⁵
[Ar] 4s¹ 3d¹⁰
Listen up—this is your last-minute cheat sheet.
rₙ = 0.529 × n² / Z
Eₙ = –13.6 × Z² / n²
Wavelength: 1/λ = R × Z² × (1/n₁² – 1/n₂²).
Quantum numbers:
mₛ = ±½.
Orbitals:
Nodes: n–1 total (n–l–1 radial, l angular).
n–1
n–l–1
Electronic config:
1s 2s 2p 3s 3p 4s 3d 4p...
Exceptions: Cr (4s¹ 3d⁵), Cu (4s¹ 3d¹⁰).
4s¹ 3d⁵
4s¹ 3d¹⁰
Exam traps:
Final tip: If stuck, draw the orbital diagram—it’s faster than memorising rules. Now go crush that exam!
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