Atomic Structure

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Atomic Structure is the foundation of chemistry — it deals with the model of the atom, quantum numbers, electronic configuration, and atomic properties. In JEE, this chapter contributes about 5–8% of the chemistry paper and frequently appears as direct numerical problems or as a prerequisite in other chapters.

Key Facts for JEE Quick Recall:

• Rutherford’s Gold Foil Experiment established the nuclear model: a dense, positively charged nucleus surrounded by electrons. The α-particle scattering angle and impact parameter relationship confirms nuclear radius ≈ 10⁻¹⁵ m.
• Bohr’s Model (for hydrogen and hydrogen-like species): Electron orbits with quantized angular momentum L = mvr = nh/2π. The energy of the nth level is Eₙ = −13.6 Z²/n² eV. This gives the Rydberg formula for spectral lines: ν = R(1/n₁² − 1/n₂²) where R = 3.29 × 10¹⁵ Hz.
• de Broglie’s Equation: λ = h/mv. JEE frequently asks: “Calculate the de Broglie wavelength of an electron accelerated through 150 V.” Answer: λ = h/√(2meV) = 1.227/√V nm ≈ 0.1 nm.
• Heisenberg’s Uncertainty Principle: Δx · Δp ≥ h/4π. If Δv = 0.1 v for an electron, can we locate it inside an atom? Solve to see that confining an electron within an atom (Δx ≈ 10⁻¹⁰ m) violates the principle — meaning Bohr’s precise orbits are fundamentally incompatible with quantum mechanics.
• Quantum Numbers Summary:

| n | 1, 2, 3… | Shell | Azimuthal l = 0 to n−1 | | m | −l to +l | Orbital orientation | Magnetic spin ms = ±½ | | Spin (s) | +½, −½ | Paired in orbitals | Hund’s rule: max multiplicity |

• Electronic Configuration: Aufbau principle (n+l rule), Pauli exclusion principle (no two electrons can have all four quantum numbers identical), Hund’s rule of maximum multiplicity. The anomalous configuration of Cr ([Ar] 4s¹ 3d⁵) and Cu ([Ar] 4s¹ 3d¹⁰) is a JEE favourite.
• Atomic Properties: Atomic radius (increases down a group, decreases across a period), ionization enthalpy (peaks at noble gases and high ionization energy elements like N, Be), electron affinity (most negative at halogens), electronegativity (Pauling scale, F = 4.0 highest).

⚡ Exam tip: When asked about the number of subshells or orbitals, use the formulas: subshells in shell n = n, orbitals = n². For a 3p³ configuration, write all four quantum numbers for any one electron. Don’t forget m = +1 for one of them, not just 0.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Atomic Structure — JEE Chemistry Study Guide

1. Development of Atomic Models

Rutherford’s nuclear model couldn’t explain why electrons don’t spiral into the nucleus despite continuous radiation. Bohr addressed this by introducing quantized stationary states, but his model had limitations — it failed for multi-electron atoms and couldn’t explain fine spectral lines or the Zeeman effect.

The wave mechanical model replaced orbits with orbitals — three-dimensional probability distributions ψ² for finding an electron. The Schrödinger wave equation ∂²ψ/∂x² + ∂²ψ/∂y² + ∂²ψ/∂z² + (8π²m/h²)(E − V)ψ = 0 gives wave functions that describe atomic orbitals.

2. Quantum Numbers — Deep Dive

• Principal quantum number (n): Determines energy level and average distance from nucleus. Eₙ ∝ −1/n². For hydrogen, Eₙ = −13.6/n² eV.
• Azimuthal quantum number (l): Determines subshell shape and orbital angular momentum. Values: 0 (s), 1 (p), 2 (d), 3 (f). Also contributes to fine structure via spin-orbit coupling.
• Magnetic quantum number (mₗ): For a given l, mₗ = −l, −(l−1), …, 0, …, +(l−1), +l. Total of (2l+1) values.
• Spin quantum number (mₛ): Only ±½, related to intrinsic angular momentum S = √(s(s+1))·h/2π where s = ½.

Problem-Solving Example: How many electrons in Fe (Z = 26) can have n = 3, l = 2 simultaneously? n = 3, l = 2 means 3d subshell. For d: l = 2, so mₗ = −2, −1, 0, +1, +2 (5 orbitals). Each holds 2 electrons with opposite spin = 10 electrons total.

3. Electronic Configuration — Writing and Exceptions

The (n + l) rule determines filling order: lower (n+l) fills first; if equal, lower n fills first. Order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p.

Exceptions in d-block: Cr and Cu prefer half-filled and fully-filled d subshells respectively. Similarly, Mo, Pt, Au show anomalies. For ions: remove from outermost shell first (4s before 3d).

4. Atomic Properties and Periodic Trends

• Ionization Energy (IE): Energy to remove the most loosely held electron. General trend: increases across a period (excluding IE dips at B, Al, Ga and noble gas peak). First IE decreases down a group (atomic size increases). ΔH = IE (always endothermic, positive).
• Electron Affinity (EA): Energy released when an electron is added. Most negative for halogens (−Cl < −F due to small size/repulsion). Chalcogens also show negative EA. O₂ has lower EA than S due to inter-electron repulsion in the small 2p subshell.
• Electronegativity: Allred-Rochow (covalent radius based) and Mulliken (average of IE and EA) scales. Fluorine = 4.0 on Pauling scale.
• Effective Nuclear Charge (Zₑff): Zₑff = Z − S (screening constant). Slater’s rules: for ns, np electrons: electrons in same group shield 0.35 (except 1s: 0.30); electrons in (n−1) shell: 0.85; electrons in (n−2) or lower: 1.00. Zₑff increases across a period, explaining decreasing atomic radius.

⚡ Exam tip: Questions on the “photoelectric effect” type — given wavelength/frequency of incident radiation and work function, find kinetic energy of ejected electron: KE = hν − hν₀ = hc/λ − φ. Also note that de Broglie wavelength of a photon is λ = hc/E.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Atomic Structure — Comprehensive JEE Advanced Notes

1. Wave-Particle Duality and the Failure of Classical Physics

Classical physics assumed electrons behaved as particles in defined orbits. Three experimental facts demolished this:

• Black-body radiation (Planck): Energy is quantized — E = nhν
• Photoelectric effect (Einstein): Light of frequency ν can eject electrons if ν > ν₀ (threshold). KEₘₐₓ = h(ν − ν₀)
• Compton scattering: Photon transfers momentum to electron; Δλ = h/mₑc (1 − cos θ). For θ = 180° (backscatter), maximum wavelength shift = 2h/mₑc = 0.0048 nm (Compton wavelength).

de Broglie’s hypothesis: λ = h/p = h/mv. This was confirmed by Davisson-Germer (electron diffraction by Ni crystal) and Thomson (electron beam diffraction through thin gold foil showing ring pattern).

2. Heisenberg’s Uncertainty Principle — Mathematical Derivation

Starting from Fourier analysis of wave packets: a wave packet of finite length Δx must contain a range of wavelengths Δλ ≈ λ²/Δx. Multiplying by momentum p = h/λ:

Δp ≈ h · (Δλ/λ²) = h · (λ/Δx · 1/λ²) = h/Δx

Refined to: Δx · Δp ≥ ℏ/2 = h/4π (the standard form).

Consequence for atomic physics: If an electron is confined to an atom (Δx ≈ 10⁻¹⁰ m), then Δv ≥ h/(4πmₑΔx) ≈ 10⁶ m/s — so velocity is fundamentally uncertain at atomic scales. This rules out Bohr’s precise circular orbits.

3. The Schrödinger Equation for Hydrogen Atom

Time-independent Schrödinger equation: Ĥψ = Eψ

For hydrogen: Ĥ = −(h²/8π²mₑ)(∂²/∂r² + (2/r)(∂/∂r)) + (l(l+1)h²/8π²mₑr²) − (e²/4πε₀r)

Solutions exist only when three quantum numbers are integers. Energy depends only on n:

Eₙ = −(mₑe⁴/8ε₀²h²) · (1/n²) = −13.6/n² eV

The Rydberg constant R = mₑe⁴/8ε₀²h³c = 1.097 × 10⁷ m⁻¹

4. Hydrogen Wave Functions (Normalized)

Radial part Rₙₗ(r) and angular part Yₗₘ(θ, φ). Key results:

Orbital	Radial nodes	Angular nodes	Shape
1s	0	0	Spherical
2s	1	0	Spherical with a node
2p	0	1	Dumbbell
3d	0	2	Double dumbbell / cloverleaf

The radial probability distribution P(r) = r²|Rₙₗ(r)|². For 1s, the maximum occurs at r = a₀ (Bohr radius = 0.529 Å).

5. Quantum Numbers from Schrödinger Solutions

n = 1, 2, 3,… (positive integers) l = 0, 1, 2, …, n−1 mₗ = −l, …, 0, …, +l mₛ = +½ or −½ (Dirac’s relativistic equation later refined this)

Degeneracy: For hydrogen, all orbitals with the same n are degenerate (same energy). This degeneracy is broken in multi-electron atoms by electron-electron repulsion and spin-orbit coupling.

6. Many-Electron Atoms and Spectroscopic Terms

For multi-electron atoms, LS coupling (Russell-Saunders) gives term symbols: ²S₊₁ where S = sum of spin quantum numbers, L = total orbital angular momentum (S, P, D, F for L = 0, 1, 2, 3), J = L + S for less than half-filled, J = |L − S| for more than half-filled.

Example: Carbon ground state — 1s² 2s² 2p². Two p electrons. Following Hund’s rule: maximum multiplicity = 3 (parallel spins). Term symbol: ³P. J = 0, 1, 2 for ³P₀, ³P₁, ³P₂.

7. Zeeman Effect and Stark Effect

• Zeeman effect: Splitting of spectral lines in a magnetic field. Normal Zeeman (Δm = 0, ±1 transitions) gives 3 lines. Anomalous Zeeman involves electron spin.
• Stark effect: Splitting in an electric field (linear Stark for hydrogen, quadratic for other atoms).

8. Atomic Properties — Quantitative Treatment

Slater’s Rules (detailed): For 1s electron: each other electron in 1s = 0.30 For ns, np electron:

• Same group: 0.35 each
• (n−1) shell: 0.85 each
• (n−2) or lower: 1.00 each

Calculate Zₑff for Na (1s² 2s² 2p⁶ 3s¹): Zₑff(3s) = 11 − [(0.35×0) + (0.85×8) + (1.00×2)] = 11 − 8.8 = 2.2

Ionization Energy Trends: IE₁(N) > IE₁(O) — though both are in period 2, O has lower IE because removing an electron breaks the half-filled 2p subshell stability of N.

9. Photoelectron Spectroscopy (PES)

XPS (X-ray Photoelectron Spectroscopy) measures binding energy directly. Core electrons have higher binding energy. Chemical shift indicates oxidation state and electronegativity of neighbours.

10. JEE-Advanced Specific Problem Types

1. Given wavelength → find velocity of photoelectron: Use KE = h(c/λ) − φ. Watch units: λ in nm, φ in eV.
2. Uncertainty principle in JEE: Often combined with orbit radius — if uncertainty in position is given as the radius, find minimum uncertainty in momentum.
3. Spectral transitions: Balmer series (visible, n₁ = 2), Lyman (UV, n₁ = 1), Paschen (IR, n₁ = 3). Find wavelength, frequency, or energy of emitted photon.
4. Quantum numbers counting: “How many electrons in Mn have n = 3, l = 2, mₗ = 0?” — apply Pauli principle to count allowed spin states.

⚡ Exam tip: When asked about “maximum number of electrons” in a subshell with a given n and l, remember: mₗ has (2l+1) values and mₛ has 2 values, so total = 2(2l+1). For n = 4, l = 2 (4d): 2(2×2+1) = 10 electrons.

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