Atomic Structure

Concept Core

From Bohr's orbits to quantum numbers — the complete picture.

Bohr's Model — Energy & Radius

For hydrogen-like atoms (H, He⁺, Li²⁺ ...) with atomic number Z:

Eₙ = −13.6 × Z²/n² eV (energy of nth orbit) rₙ = 0.529 × n²/Z Å (radius of nth orbit) vₙ = 2.18×10⁶ × Z/n m/s (velocity in nth orbit)

Ground state of H: n=1, E₁ = −13.6 eV, r₁ = 0.529 Å (Bohr radius a₀).

Spectral Lines — Rydberg Formula

When electron transitions from n₂ → n₁ (n₂ > n₁):

1/λ = R∞ × Z² (1/n₁² − 1/n₂²)

R∞ = 1.097×10⁷ m⁻¹ (Rydberg constant)

Series	n₁	Region
Lyman	1	UV
Balmer	2	Visible
Paschen	3	IR

Quantum Numbers — The Address of an Electron

n (principal): 1, 2, 3... Shell. Determines energy (mainly) and size.

ℓ (azimuthal): 0 to n−1. Subshell. 0=s, 1=p, 2=d, 3=f. Determines shape.

mₗ (magnetic): −ℓ to +ℓ. Orientation. (2ℓ+1) values total.

mₛ (spin): +½ or −½ only. Up or down spin.

Rules for Electron Configuration

Aufbau Principle: Fill lower energy orbitals first. Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p...

Pauli Exclusion: No two electrons in an atom can have all 4 quantum numbers identical. Maximum 2 electrons per orbital (opposite spins).

Hund's Rule: In degenerate orbitals (same n, ℓ), electrons fill singly before pairing. All single electrons have same spin.

de Broglie Wavelength

All matter has wave character. Wavelength associated with particle of mass m and velocity v:

λ = h/mv = h/p

h = 6.626×10⁻³⁴ J·s (Planck's constant). Smaller objects (electrons) have observable wavelengths. Large objects (cricket ball) have negligibly small λ.

Heisenberg Uncertainty Principle

Cannot simultaneously measure position and momentum with perfect precision:

Δx · Δp ≥ h/4π

The more precisely we know position (Δx small), the less precisely we can know momentum (Δp large), and vice versa. This is a fundamental quantum limit, not a measurement limitation.

Electronic Configurations — Exceptions (EAPCET Favourites)

Two key exceptions due to stability of half-filled and fully-filled subshells:

Chromium (Cr, Z=24): Expected [Ar]3d⁴4s² → Actual [Ar]3d⁵4s¹ (half-filled d is more stable)

Copper (Cu, Z=29): Expected [Ar]3d⁹4s² → Actual [Ar]3d¹⁰4s¹ (fully-filled d is most stable)

Extra stability comes from symmetrical charge distribution and exchange energy.

Formula Vault

Every atomic structure formula for EAPCET.

Bohr's Energy

Eₙ = −13.6 Z²/n² eV

n = shell number, Z = atomic number

Bohr's Radius

rₙ = 0.529 n²/Z Å

r₁(H) = 0.529 Å = a₀

Orbital Velocity

vₙ = 2.18×10⁶ Z/n m/s

Decreases as n increases

Rydberg Formula

1/λ = R∞Z²(1/n₁²−1/n₂²)

R∞ = 1.097×10⁷ m⁻¹

Energy of Photon

E = hν = hc/λ

h = 6.626×10⁻³⁴ J·s

de Broglie Wavelength

λ = h/mv = h/p

All matter has wave nature

Heisenberg Uncertainty

Δx·Δp ≥ h/4π

Also ΔE·Δt ≥ h/4π

Max electrons in shell

2n²

n=1: 2; n=2: 8; n=3: 18...

Orbitals in subshell

2ℓ + 1

s:1, p:3, d:5, f:7

Number of spectral lines

n(n−1)/2

From nth orbit to ground state

Worked Examples

5 questions from Bohr model to quantum numbers — all EAPCET patterns.

EasyFind energy of electron in 3rd orbit of H atom▾

Calculate the energy of the electron in the 3rd Bohr orbit of hydrogen atom.

Eₙ = −13.6 × Z²/n² eV. For H: Z = 1, n = 3

E₃ = −13.6 × 1/9 = −1.51 eV

✓ E₃ = −1.51 eV

EasyHow many spectral lines when electron falls from n=4 to n=1?▾

An electron transitions from n=4 orbit to n=1 in hydrogen. How many spectral lines are emitted?

Number of lines = n(n−1)/2 where n = number of levels involved = 4 (levels 1, 2, 3, 4)

= 4(4−1)/2 = 4×3/2 = 6 spectral lines

Transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 ✓ (6 total)

✓ 6 spectral lines

MediumWrite quantum numbers for the 19th electron of potassium▾

Potassium has Z = 19. Write all four quantum numbers for its last (19th) electron.

K configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The 19th electron goes into 4s orbital.

n = 4 (4th shell), ℓ = 0 (s subshell), mₗ = 0 (only one s orbital)

mₛ = +½ (first electron in orbital)

✓ n=4, ℓ=0, mₗ=0, mₛ=+½

EAPCET Levelde Broglie wavelength of electron accelerated through 100 V▾

Find the de Broglie wavelength of an electron (m = 9.1×10⁻³¹ kg) accelerated through 100 V. (h = 6.626×10⁻³⁴ J·s, e = 1.6×10⁻¹⁹ C)

KE gained = eV = 1.6×10⁻¹⁹ × 100 = 1.6×10⁻¹⁷ J

½mv² = 1.6×10⁻¹⁷ → mv² = 3.2×10⁻¹⁷ → mv = √(m × mv²) = √(9.1×10⁻³¹ × 3.2×10⁻¹⁷)

p = mv = √(2meV) = √(2 × 9.1×10⁻³¹ × 1.6×10⁻¹⁷) = 5.41×10⁻²⁴ kg·m/s

λ = h/p = 6.626×10⁻³⁴ / 5.41×10⁻²⁴ ≈ 1.22×10⁻¹⁰ m = 1.22 Å

✓ λ = 1.22 Å (shortcut: λ(Å) = 12.27/√V for electrons)

Trap QuestionElectronic config of Cr — and why Cu is [Ar]3d¹⁰4s¹ not [Ar]3d⁹4s²▾

Write the electronic configuration of (a) Cr (Z=24) and (b) Cu (Z=29). ⚠️ Most students write incorrect configs.

Expected for Cr: [Ar]3d⁴4s² (filling in sequence). Actual: [Ar]3d⁵4s¹

Reason: Half-filled 3d⁵ (5 electrons, one per orbital) has extra stability from maximum exchange energy and symmetric arrangement.

Expected for Cu: [Ar]3d⁹4s². Actual: [Ar]3d¹⁰4s¹

Reason: Fully-filled 3d¹⁰ is more stable than 3d⁹. One electron "promotes" from 4s to 3d to complete it.

✓ Cr: [Ar]3d⁵4s¹ | Cu: [Ar]3d¹⁰4s¹ — half-filled and fully-filled d are extra stable

Mistake DNA

5 errors from distractor analysis — this chapter is a classic mistake minefield.

🌀

Wrong Electronic Configuration for Cr and Cu

The most common error in this chapter — writing Cr as [Ar]3d⁴4s² instead of [Ar]3d⁵4s¹.

❌ Wrong

Cr: [Ar]3d⁴4s² ✗
Cu: [Ar]3d⁹4s² ✗

✓ Correct

Cr: [Ar]3d⁵4s¹ ✓
Cu: [Ar]3d¹⁰4s¹ ✓
Exceptions to Aufbau

Memorise Cr and Cu as the two Aufbau exceptions in Period 4. Half-filled (3d⁵) and fully-filled (3d¹⁰) d subshells are extra stable.

🔢

Counting Spectral Lines Incorrectly

Students add transitions instead of using n(n-1)/2 formula, and often overcount or undercount.

❌ Wrong

From n=4 to n=1:
"4 transitions" (only
counting 4→3→2→1) ✗

✓ Correct

All possible: 4→3,4→2,4→1
3→2,3→1, 2→1 = 6 ✓
Formula: 4(3)/2 = 6 ✓

Every possible transition from any level to any lower level produces a spectral line. Use n(n−1)/2 where n is the number of energy levels involved.

🎯

Confusing Principal Quantum Number with Shell Count

Students say "3rd shell has 3 electrons max" when it can hold 2(3)² = 18.

❌ Wrong

n=3 shell holds max
3 or 9 electrons ✗

✓ Correct

n=3 holds max 2n² = 18
electrons ✓
(3s²+3p⁶+3d¹⁰ = 18)

Formula: max electrons in nth shell = 2n². n=1:2, n=2:8, n=3:18, n=4:32. This is a direct EAPCET question every few years.

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Applying Bohr Model to Multi-Electron Atoms

Bohr's model is only valid for hydrogen-like species (H, He⁺, Li²⁺). Applying En = −13.6Z²/n² to Na or Fe gives wrong answers.

❌ Wrong

Energy of electron in
Na (Z=11): E = −13.6×121/n²
Applying Bohr to Na ✗

✓ Correct

Bohr works only for
H, He⁺, Li²⁺, Be³⁺ etc.
(one electron systems) ✓

Multi-electron atoms have electron-electron repulsion which Bohr didn't account for. Use quantum mechanical model (orbitals, not orbits) for multi-electron systems.

⬆️

Hund's Rule Violation — Pairing Before Half-Fill

Students pair electrons in p or d orbitals before all are singly occupied.

❌ Wrong

Carbon (2p²):
↑↓ _ _ (both in one
p orbital) ✗

✓ Correct

Carbon (2p²):
↑ ↑ _ (one each in
separate p orbitals) ✓

Hund's Rule: maximum multiplicity. Fill all 2ℓ+1 orbitals singly with same spin before any pairing. Nitrogen is 2p³ with all three p orbitals half-filled.

Chapter Intelligence

Atomic structure is the foundation of all chemistry. Master it once, benefit everywhere.

EAPCET Topic Weightage (2019–2024)

Electronic configuration

Quantum numbers

Bohr model calculations

Spectral lines / series

de Broglie / Heisenberg

High-Yield PYQ Patterns

Config of Cr, Cu exceptions Quantum numbers of nth electron Number of spectral lines Energy of Bohr orbit de Broglie wavelength Max e⁻ in given shell/subshell

Exam Strategy

Electronic configuration questions: write the full config, then check for Cr/Cu exceptions. These are tested every year.
Quantum number questions: given an electron (e.g., "21st electron of Sc"), write the full config, identify the last orbital, and read off all four quantum numbers.
Bohr model shortcut: E₂/E₁ = n₁²/n₂². Don't recompute from scratch — use the ratio to find any orbit's energy if you know one.
Spectral lines: use n(n-1)/2. For "minimum lines in visible" → Balmer series (n₁=2), for UV → Lyman (n₁=1).
de Broglie for electrons accelerated through V volts: λ = 12.27/√V Å (shortcut). Memorise this for speed in the exam.

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

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