⚛ Atomic Structure
Complete Study Notes with Solved Examples, MCQs & Practice Questions
About the Author: This comprehensive notes guide was researched and compiled from standard university reference texts by Sudhir Nama, an IIT Bombay alumnus, author, and competitive chemistry educator.
📋 What's Inside
- Historical Development of Atomic Structure Theory
- Atomic Spectra & Bohr Model
- Schrödinger Equation & Wave Functions
- Quantum Numbers — Deep Dive
- Aufbau, Pauli & Hund's Rule
- Shielding & Slater's Rules
- Periodic Properties
- Solved Examples
- Chapter MCQs (with Answers)
- 25 Practice Questions
- Exam Tips & Tricks
1. Historical Development of Atomic Theory
The story of the atom didn't begin in a lab — it started in ancient Greece. But it took centuries of real experiments before we got a theory that actually holds up. Let's walk through the landmarks.
The Big Names Scientist Timeline
- ~600 BCE — Maharishi Kanada an Indian philosopher. He proposed that matter can be subdivided, but this division cannot go on forever. The ultimate, indivisible, and indestructible particle of matter is the parmanu (atom)
- ~460 BCE — Democritus proposes that matter is made of tiny indivisible particles called "atomos"
- 1808 — John Dalton publishes A New System of Chemical Philosophy — atoms of same element are identical, combine in simple ratios
- ~1811 — Avogadro proposes equal volumes of gases at same T and P contain equal number of molecules
- 1860 — Cannizzaro standardises atomic weights at Karlsruhe conference
- 1869 — Mendeleev & Meyer independently propose periodic tables
- 1896 — Becquerel discovers radioactivity of uranium
- 1897 — J. J. Thomson identifies the electron; charge/mass = 1.76 × 10¹¹ C/kg
- 1909 — Millikan's oil-drop experiment measures electronic charge: 1.60 × 10⁻¹⁹ C; mass of electron = 9.11 × 10⁻³¹ kg
- 1911 — Rutherford's gold foil experiment — nucleus is tiny, heavy, positively charged; rest is empty space
- 1913 — Moseley determines nuclear charges (atomic number Z) via X-ray emission; Z is more fundamental than atomic mass
- 1913 — Niels Bohr publishes quantum theory of the atom
- 1920s — de Broglie, Heisenberg, Schrödinger — wave mechanics era begins
⚠ Dalton's Limitation: Dalton assumed H₂O was "HO" (monatomic H and O). He didn't know about diatomic molecules, so his particle ratio for hydrogen and oxygen reacting to form water was off. The correct equation is 2H₂ + O₂ → 2H₂O.
Periodic Table — Key Points
- A horizontal row = Period; a vertical column = Group
- IUPAC numbering: Groups 1 to 18 (this is what modern exams follow)
- Mendeleev predicted properties of undiscovered elements — gallium, scandium, germanium, polonium — and was proven right
- Groups 1–2: filling s-orbitals | Groups 13–18: filling p-orbitals | Groups 3–12 (transition metals): filling d-orbitals | Lanthanides & Actinides (58–71, 90–103): filling f-orbitals
2. Atomic Spectra & The Bohr Model
When atoms are excited (by heat or electric discharge), they emit light at specific energies — not a rainbow, but discrete lines. This is atomic emission spectrum, and it gave us the keys to understanding electron energy levels.
Balmer's Equation (1885)
Balmer Series Formula
E = RH × (1/nl² – 1/nh²)
where RH = 2.179 × 10⁻¹⁸ J (Rydberg constant for hydrogen)
nl = lower quantum number | nh = higher quantum number (nh > nl)
where RH = 2.179 × 10⁻¹⁸ J (Rydberg constant for hydrogen)
nl = lower quantum number | nh = higher quantum number (nh > nl)
Energy of Emitted/Absorbed Light
E = hν = hc/λ = hcν̃
h = 6.626 × 10⁻³⁴ J·s (Planck's constant)
c = 2.998 × 10⁸ m/s (speed of light)
λ = wavelength (usually in nm)
ν̃ = wavenumber (usually in cm⁻¹)
h = 6.626 × 10⁻³⁴ J·s (Planck's constant)
c = 2.998 × 10⁸ m/s (speed of light)
λ = wavelength (usually in nm)
ν̃ = wavenumber (usually in cm⁻¹)
Spectral Series of Hydrogen
| Series Name | nl (lower level) | Region | Visible? |
|---|---|---|---|
| Lyman | 1 | Ultraviolet (UV) | No |
| Balmer | 2 | Visible + UV | Yes (partially) |
| Paschen | 3 | Infrared (IR) | No |
| Brackett | 4 | Far IR | No |
| Pfund | 5 | Far IR | No |
Exam Trick: For Balmer series, visible lines correspond to nh = 3, 4, 5, 6 → nl = 2. Red line: nh = 3 (longest wavelength, lowest energy). Violet line: nh = 6 (shortest wavelength, highest energy).
Bohr's Quantum Theory of Hydrogen Atom
In 1913, Niels Bohr proposed that electrons move in stable circular orbits with no energy loss. They can jump between levels by absorbing or emitting photons of specific energy.
Bohr Energy Level Formula
E = R × (1/nl² – 1/nh²)
R = 2π²mZ²e⁴ / [(4πε₀)²h²]
For hydrogen (Z=1): R = RH = 2.179 × 10⁻¹⁸ J = 13.61 eV
For other one-electron species: multiply RH by Z²
R = 2π²mZ²e⁴ / [(4πε₀)²h²]
For hydrogen (Z=1): R = RH = 2.179 × 10⁻¹⁸ J = 13.61 eV
For other one-electron species: multiply RH by Z²
- Works perfectly for H, He⁺, Li²⁺, Be³⁺ (one-electron species only)
- Fails for multi-electron atoms — can't account for electron-electron repulsions
- As n → ∞, energy → 0 (electron escapes the atom)
- Energy levels are closer together at high n and far apart at low n
de Broglie's Wave-Particle Duality (1920s)
Louis de Broglie proposed something wild: every moving particle has wave properties. Massive objects have wavelengths too small to measure, but electrons — being tiny — show real observable wave behaviour.
de Broglie Equation
λ = h / (mu)
λ = wavelength of the particle | h = Planck's constant
m = mass of particle | u = velocity of particle
λ = wavelength of the particle | h = Planck's constant
m = mass of particle | u = velocity of particle
Heisenberg's Uncertainty Principle
Heisenberg Uncertainty Principle
Δx · Δpₓ ≥ h/(4π)
Δx = uncertainty in position | Δpₓ = uncertainty in momentum
Δx = uncertainty in position | Δpₓ = uncertainty in momentum
This is fundamental: the more precisely we know an electron's momentum (energy), the less precisely we can know its location. So instead of describing precise orbits (like Bohr), we describe orbitals — regions of space where the probability of finding an electron is high.
Key Distinction: Bohr gave us "orbits" (defined circular paths). Modern quantum mechanics gives us "orbitals" (probability clouds). These are NOT the same thing. Exams test this distinction often.
3. The Schrödinger Equation & Wave Functions
In 1926, Schrödinger expressed the wave nature of electrons mathematically. His equation is the foundation of modern quantum chemistry.
Schrödinger Equation (simplified notation)
Ĥψ = Eψ
Ĥ = Hamiltonian operator (kinetic + potential energy)
ψ = wave function (describes the orbital)
E = energy of the electron
Ĥ = Hamiltonian operator (kinetic + potential energy)
ψ = wave function (describes the orbital)
E = energy of the electron
What Does ψ Mean?
- ψ itself has no direct physical meaning
- ψ² = probability density = probability of finding the electron at a given point in space
- Larger |ψ|² at a point → electron more likely to be found there
Conditions for a Valid Wave Function
- Single-valued: Only one probability value at each point in space
- Continuous: ψ and its first derivatives must be continuous everywhere
- Approaches zero: ψ → 0 as r → ∞ (atom must be finite)
- Normalised: ∫|ψ|² dτ = 1 over all space (total probability = 1)
- Orthogonal: ∫ψ_A · ψ_B* dτ = 0 for different orbitals A and B
Particle in a Box — The Simplest Case
This is a model problem that shows how quantum mechanics gives discrete (quantised) energy levels. Imagine a particle trapped in a 1D box of length 'a' where potential energy is zero inside and infinite outside.
Energy of Particle in 1D Box
En = n²h² / (8ma²)
n = 1, 2, 3, ... (quantum number, cannot be zero)
m = mass of particle | a = length of the box
n = 1, 2, 3, ... (quantum number, cannot be zero)
m = mass of particle | a = length of the box
Wave Function of Particle in Box
ψn = √(2/a) × sin(nπx/a)
0 ≤ x ≤ a
0 ≤ x ≤ a
Why n = 0 is not allowed: If n = 0, then ψ = 0 everywhere — meaning there is no particle. So n must start at 1, not 0.
The Hamiltonian Operator for Hydrogen
Full Hamiltonian (1-electron atom)
Ĥ = –(h²/8π²m)(∂²/∂x² + ∂²/∂y² + ∂²/∂z²) – Ze²/(4πε₀r)
First term = kinetic energy operator
Second term = potential energy V (electrostatic attraction between nucleus and electron)
First term = kinetic energy operator
Second term = potential energy V (electrostatic attraction between nucleus and electron)
Attractive interactions (positive nucleus, negative electron) have negative potential energy. Electron near nucleus: large negative V. At infinite distance: V = 0. This is shown clearly in hydrogen's energy level diagram.
Separating ψ into Radial and Angular Parts
Wave Function Factorisation
ψ(r, θ, φ) = R(r) · Θ(θ) · Φ(φ) = R(r) · Y(θ, φ)
R(r) = radial function (depends on n and l)
Y(θ, φ) = angular function (depends on l and ml; gives orbital shape)
R(r) = radial function (depends on n and l)
Y(θ, φ) = angular function (depends on l and ml; gives orbital shape)
Radial Probability Function
Radial Probability Function
P(r) = 4πr² R²(r)
- Gives the probability of finding the electron at distance r from nucleus, summed over all angles
- All orbitals including s-orbitals have zero probability at the nucleus (r = 0), because 4πr² = 0 at r = 0
- The Bohr radius a₀ = 52.9 pm corresponds to the maximum of ψ² for the hydrogen 1s orbital
- Electron density falls off rapidly beyond its maximum as r increases
Nodal Surfaces
A node is a surface where ψ = 0 (and therefore ψ² = 0 — zero probability of finding the electron). Nodes arise naturally from the wave nature of electrons.
| Type of Node | Condition | Shape | Count |
|---|---|---|---|
| Angular Node | Y(θ, φ) = 0 | Plane or cone | = l |
| Radial Node (Spherical) | R(r) = 0 | Sphere | = n – l – 1 |
| Total Nodes | Both types | — | = n – 1 |
Memory shortcut for nodes:
Angular nodes = l
Radial nodes = n – l – 1
Total nodes = n – 1
Example: 3p orbital → angular = 1, radial = 3–1–1 = 1, total = 2
Angular nodes = l
Radial nodes = n – l – 1
Total nodes = n – 1
Example: 3p orbital → angular = 1, radial = 3–1–1 = 1, total = 2
| Orbital | n | l | Angular Nodes | Radial Nodes | Total |
|---|---|---|---|---|---|
| 1s | 1 | 0 | 0 | 0 | 0 |
| 2s | 2 | 0 | 0 | 1 | 1 |
| 2p | 2 | 1 | 1 | 0 | 1 |
| 3s | 3 | 0 | 0 | 2 | 2 |
| 3p | 3 | 1 | 1 | 1 | 2 |
| 3d | 3 | 2 | 2 | 0 | 2 |
| 4s | 4 | 0 | 0 | 3 | 3 |
| 4d | 4 | 2 | 2 | 1 | 3 |
4. Quantum Numbers — The Complete Picture
Four quantum numbers completely describe any electron in any atom. Think of them as the "address" of an electron.
| Symbol | Name | Allowed Values | What It Tells Us |
|---|---|---|---|
| n | Principal | 1, 2, 3, 4, ... | Major energy level; shell size; overall energy |
| l | Angular Momentum (Azimuthal) | 0, 1, 2, ..., n–1 | Shape of orbital; angular momentum magnitude |
| ml | Magnetic | –l, ..., 0, ..., +l | Orientation of orbital in space |
| ms | Spin | +½ or –½ | Direction of electron's magnetic moment (spin up or down) |
Orbital Labels by l Value
| l value | Orbital type | Number of orbitals | Max electrons |
|---|---|---|---|
| 0 | s | 1 | 2 |
| 1 | p | 3 | 6 |
| 2 | d | 5 | 10 |
| 3 | f | 7 | 14 |
| 4 | g | 9 | 18 |
Number of orbitals in a subshell = 2l + 1
Max electrons in a subshell = 2(2l + 1)
Max electrons in a shell n = 2n²
Max electrons in a subshell = 2(2l + 1)
Max electrons in a shell n = 2n²
Shapes of Orbitals
- s orbitals: Spherically symmetric. Each successive s orbital (2s, 3s...) has more radial nodes and is larger
- p orbitals: Dumbbell-shaped with one nodal plane through nucleus. Three orientations: pₓ, p_y, pz along x, y, z axes
- d orbitals: Five d orbitals — dxy, dxz, dyz (cloverleaf in planes), dx²-y² (cloverleaf in xy plane), dz² (dumbbell along z with torus ring)
- Orbital lobes with different shading = different signs of the wave function ψ (important for bonding)
- Outer boundary shown in diagrams = 90% of total electron density
5. Aufbau Principle, Pauli & Hund's Rule
Aufbau Principle
Electrons fill orbitals starting from the lowest energy. The word "Aufbau" is German for "building up." Klechkowsky's rule gives the filling order:
Klechkowsky's Rule: Fill in increasing order of (n + l). When two orbitals have the same (n + l), fill the one with lower n first.
Example: 4s has n+l = 4+0 = 4; 3d has n+l = 3+2 = 5 → so 4s fills before 3d ✓
Example: 4s has n+l = 4+0 = 4; 3d has n+l = 3+2 = 5 → so 4s fills before 3d ✓
Standard Filling Order
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p
Pauli Exclusion Principle
No two electrons in the same atom can have all four quantum numbers identical. At least one quantum number must differ. In practice: each orbital holds at most 2 electrons, and they must have opposite spins (+½ and –½).
Hund's Rule of Maximum Multiplicity
When electrons fill orbitals of the same energy (degenerate), they go into separate orbitals with parallel spins first, before pairing up. This minimizes electrostatic repulsion.
- Electrons in same orbital repel each other more than electrons in different orbitals
- Parallel-spin electrons in separate orbitals are stabilized by exchange energy (ε)
- Paired electrons in the same orbital carry extra Coulombic repulsion energy (c)
Coulombic vs Exchange Energy — The p² Example
For carbon (1s²2s²2p²), three arrangements of 2p electrons are possible:
| State | Arrangement | Coulombic (c) | Exchange (ε) | Total Energy |
|---|---|---|---|---|
| (1) Paired in same orbital | ↑↓ _ _ | 1c (repulsion) | 0ε | Highest energy |
| (2) Unpaired, opposite spin | ↑ ↓ _ | 0c | 0ε | Intermediate |
| (3) Unpaired, same spin ✓ | ↑ ↑ _ | 0c | –1ε (stable) | Lowest energy ✓ |
Key formula for Exchange Energy:
Number of exchange pairs for n parallel-spin electrons = n(n–1)/2
Each exchange lowers energy by ε. More parallel-spin electrons = more stable.
Number of exchange pairs for n parallel-spin electrons = n(n–1)/2
Each exchange lowers energy by ε. More parallel-spin electrons = more stable.
p⁴ Configuration (e.g., Oxygen)
Oxygen has four electrons in the 2p subshell. The most stable arrangement has 2 unpaired electrons: ↑↓ ↑ ↑ (one pair forced, two unpaired). This gives: 3 exchange interactions (from the 3 electrons with ↑ spin: pairs 1–2, 1–3, 2–3) and 1 Coulombic term. Total = –3ε + c.
Why does oxygen have lower IE than nitrogen? In nitrogen (↑ ↑ ↑), no pairing needed. In oxygen (↑↓ ↑ ↑), the fourth electron must share an orbital, increasing repulsion. So it's easier to remove an electron from oxygen — lower ionisation energy.
6. Shielding & Slater's Rules
In multi-electron atoms, inner electrons partially block (shield) the nuclear charge from outer electrons. The effective nuclear charge Z* is what an electron actually "feels."
Effective Nuclear Charge
Z* = Z – S
Z = actual nuclear charge (atomic number)
S = shielding constant (calculated using Slater's rules)
Z = actual nuclear charge (atomic number)
S = shielding constant (calculated using Slater's rules)
Slater's Rules — Step by Step
Step 1: Write electron configuration in Slater groups:
(1s) | (2s, 2p) | (3s, 3p) | (3d) | (4s, 4p) | (4d) | (4f) | (5s, 5p) | ...
Step 2: Rules for ns or np valence electrons:
- Each electron in the same group contributes 0.35 to S (exception: 1s electrons contribute 0.30 to each other)
- Each electron in the (n–1) shell contributes 0.85 to S
- Each electron in (n–2) or lower shells contributes 1.00 to S
- Electrons to the right (higher groups) contribute 0 to S
Step 3: Rules for nd or nf valence electrons:
- Each electron in the same (nd) or (nf) group contributes 0.35 to S
- Each electron in all groups to the left contributes 1.00 to S
SOLVED EXAMPLE — Slater's Rules
Find Z* for a 2p electron in oxygen (Z = 8)
Slater grouping: (1s²)(2s², 2p⁴) — we want Z* for one of the 2p electrons
Same group contribution (2s, 2p): 5 other electrons × 0.35 = 1.75
Next inner shell (1s): 2 electrons × 0.85 = 1.70
Total S = 1.75 + 1.70 = 3.45
Z* = 8 – 3.45 = 4.55
Same group contribution (2s, 2p): 5 other electrons × 0.35 = 1.75
Next inner shell (1s): 2 electrons × 0.85 = 1.70
Total S = 1.75 + 1.70 = 3.45
Z* = 8 – 3.45 = 4.55
✓ The 2p electron "feels" ~57% of the full +8 nuclear charge
SOLVED EXAMPLE — 3d vs 4s in Nickel
Calculate Z* for 3d and 4s electrons in Nickel (Z = 28, config: [Ar]4s²3d⁸)
Slater grouping: (1s²)(2s²,2p⁶)(3s²,3p⁶)(3d⁸)(4s²)
For 3d electron:
Same group (3d): 7 × 0.35 = 2.45
All groups to left: 18 × 1.00 = 18.00
S = 20.45 → Z* = 28 – 20.45 = 7.55
For 4s electron:
Same group (4s): 1 × 0.35 = 0.35
n–1 groups (3s,3p,3d = 16 electrons): 16 × 0.85 = 13.60
n–2 and lower (1s,2s,2p = 10 electrons): 10 × 1.00 = 10.00
S = 23.95 → Z* = 28 – 23.95 = 4.05
For 3d electron:
Same group (3d): 7 × 0.35 = 2.45
All groups to left: 18 × 1.00 = 18.00
S = 20.45 → Z* = 28 – 20.45 = 7.55
For 4s electron:
Same group (4s): 1 × 0.35 = 0.35
n–1 groups (3s,3p,3d = 16 electrons): 16 × 0.85 = 13.60
n–2 and lower (1s,2s,2p = 10 electrons): 10 × 1.00 = 10.00
S = 23.95 → Z* = 28 – 23.95 = 4.05
✓ Z*(3d) = 7.55 >> Z*(4s) = 4.05 → 3d electrons are held much more tightly → 4s is removed first in ionisation (Ni²⁺ has config [Ar]3d⁸, not [Ar]3d⁶4s²)
Why 4s Fills Before 3d But 3d Electrons Are Removed First
- In neutral atoms: 4s has lower energy than 3d (that's why it fills first)
- When an ion forms, removing electrons increases Z* for remaining electrons
- 3d orbital has shorter most-probable distance from nucleus → stabilises more as Z increases
- In transition metal cations: 3d is always lower in energy than 4s → s electrons always lost first
Exceptions in Transition Metals
Certain atoms don't follow the simple filling order because half-filled and fully-filled d subshells have extra stability (from high exchange energy):
| Element | Expected Configuration | Actual Configuration | Reason |
|---|---|---|---|
| Cr (Z=24) | [Ar] 4s² 3d⁴ | [Ar] 4s¹ 3d⁵ | Half-filled 3d (d⁵) extra stable |
| Cu (Z=29) | [Ar] 4s² 3d⁹ | [Ar] 4s¹ 3d¹⁰ | Fully-filled 3d (d¹⁰) extra stable |
| Mo (Z=42) | [Kr] 5s² 4d⁴ | [Kr] 5s¹ 4d⁵ | Half-filled 4d extra stable |
| Pd (Z=46) | [Kr] 5s² 4d⁸ | [Kr] 5s⁰ 4d¹⁰ | Fully-filled 4d |
7. Periodic Properties of Atoms
7.1 Ionisation Energy (IE)
Energy needed to remove an electron from a gaseous atom or ion: A^n⁺(g) → A^(n+1)⁺(g) + e⁻
- Across a period (left → right): IE generally increases (more protons, electrons added to same shell, higher Z*)
- Down a group (top → bottom): IE decreases (outer electrons farther from nucleus, more shielded)
- Exception — B vs Be: B has lower IE than Be because B's 2p electron is shielded by the 2s electrons and is in a higher energy subshell
- Exception — O vs N: O has lower IE than N because O's fourth 2p electron must pair up in an orbital, increasing repulsion (Coulombic energy)
- Noble gases have the highest IE in each period
- Alkali metals have the lowest IE in each period
7.2 Electron Affinity (EA)
Energy needed to remove an electron from a negative ion: A⁻(g) → A(g) + e⁻ (this definition avoids sign confusion)
- Pattern of EA across Z is similar to IE, but shifted by one element (one more electron in each species)
- Noble gases and alkaline earth metals have very low or negative EA (stable configurations)
- Halogens have high EA (close to noble gas configuration)
- All EA values are much smaller than corresponding IE because removing an electron from a negative ion is easier
7.3 Covalent and Ionic Radii
- Across a period (left → right): atomic radius decreases (more protons pull electrons inward)
- Down a group (top → bottom): atomic radius increases (more electron shells)
- Cations are smaller than parent atoms (electrons removed, nuclear charge same)
- Anions are larger than parent atoms (electrons added, increased repulsion)
- Isoelectronic series (same electron count): radius decreases as nuclear charge increases
| Ion | Protons | Electrons | Crystal Radius (pm) |
|---|---|---|---|
| O²⁻ | 8 | 10 | 126 |
| F⁻ | 9 | 10 | 119 |
| Na⁺ | 11 | 10 | 116 |
| Mg²⁺ | 12 | 10 | 86 |
| Al³⁺ | 13 | 10 | 68 |
Isoelectronic Trend: All ions above have 10 electrons. As proton count goes up (8 → 13), radius drops (126 → 68 pm). More protons pull the same number of electrons closer.
8. Solved Examples
Example 1 — Balmer Series Energy Calculation
Calculate the energy (in joules) of the transition from nh = 3 to nl = 2 in hydrogen.
E = RH × (1/nl² – 1/nh²)
E = 2.179 × 10⁻¹⁸ × (1/4 – 1/9)
E = 2.179 × 10⁻¹⁸ × (0.25 – 0.1111)
E = 2.179 × 10⁻¹⁸ × 0.1389
E = 3.025 × 10⁻¹⁹ J
E = 2.179 × 10⁻¹⁸ × (1/4 – 1/9)
E = 2.179 × 10⁻¹⁸ × (0.25 – 0.1111)
E = 2.179 × 10⁻¹⁸ × 0.1389
E = 3.025 × 10⁻¹⁹ J
✓ This is the red line (λ ≈ 656 nm) in the visible Balmer series
Example 2 — de Broglie Wavelength
Calculate the de Broglie wavelength of an electron moving at 1/10 the speed of light.
u = c/10 = 2.998 × 10⁷ m/s
m = 9.11 × 10⁻³¹ kg
h = 6.626 × 10⁻³⁴ J·s
λ = h/(mu) = 6.626 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.998 × 10⁷)
λ = 6.626 × 10⁻³⁴ / (2.731 × 10⁻²³)
λ = 2.43 × 10⁻¹¹ m = 24.3 pm
m = 9.11 × 10⁻³¹ kg
h = 6.626 × 10⁻³⁴ J·s
λ = h/(mu) = 6.626 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.998 × 10⁷)
λ = 6.626 × 10⁻³⁴ / (2.731 × 10⁻²³)
λ = 2.43 × 10⁻¹¹ m = 24.3 pm
✓ This is on the order of atomic dimensions — electrons really do show wave properties!
Example 3 — Nodal Analysis of dx²–y²
Identify the angular nodal surfaces of the dx²–y² orbital. Angular function Y = (constant)(x² – y²)/r²
For angular nodes, set Y = 0:
(x² – y²) = 0
→ x² = y²
→ x = y AND x = –y
These are two planes, each containing the z-axis and making 45° angles with the x and y axes.
(x² – y²) = 0
→ x² = y²
→ x = y AND x = –y
These are two planes, each containing the z-axis and making 45° angles with the x and y axes.
✓ Two angular nodal planes (as expected for l = 2). Wave function is positive where x > y, negative where x < y.
Example 4 — Coulombic and Exchange Energies (Oxygen, p⁴)
For oxygen's 2p⁴, determine whether ↑↓ ↑ ↑ or ↑↓ ↑↓ _ is the ground state.
State A: ↑↓ ↑ ↑ (two unpaired)
Coulombic: 1c (one pair)
Exchange interactions for ↑ electrons (3 of them): 3 pairs (1–2, 1–3, 2–3) → 3ε
Total energy contribution: c – 3ε
State B: ↑↓ ↑↓ _ (zero unpaired)
Coulombic: 2c (two pairs)
Exchange for ↑ electrons (2): 1 pair → 1ε; for ↓ (2): 1 pair → 1ε → total 2ε
Total energy contribution: 2c – 2ε
Coulombic: 1c (one pair)
Exchange interactions for ↑ electrons (3 of them): 3 pairs (1–2, 1–3, 2–3) → 3ε
Total energy contribution: c – 3ε
State B: ↑↓ ↑↓ _ (zero unpaired)
Coulombic: 2c (two pairs)
Exchange for ↑ electrons (2): 1 pair → 1ε; for ↓ (2): 1 pair → 1ε → total 2ε
Total energy contribution: 2c – 2ε
✓ State A has lower Coulombic energy (1c vs 2c) AND more exchange stabilisation (3ε vs 2ε). State A is the ground state. This confirms Hund's rule.
9. Chapter MCQs (with Explanations)
MCQ 01 — JEE Level
According to Heisenberg's uncertainty principle, if an electron's exact momentum is known, what can we say about its position?
(A) It is precisely known
(B) It is completely unknown
(C) It is also precisely known
(D) It cannot be determined at all in principle
✓ Correct Answer: (D)
If Δp → 0 (exact momentum), then Δx → ∞ (completely indeterminate position). The position cannot be known in principle — this is a fundamental quantum mechanical limitation, not a measurement limitation.
MCQ 02 — JEE/IIT-JAM Level
The number of radial nodes in a 4f orbital is:
(A) 4
(B) 3
(C) 0
(D) 1
✓ Correct Answer: (C)
Radial nodes = n – l – 1 = 4 – 3 – 1 = 0. The 4f orbital has 3 angular nodes (l=3) and 0 radial nodes. Total nodes = n–1 = 3, all angular.
MCQ 03 — CSIR-NET Level
Bohr's model correctly predicts spectral lines for which of the following?
(A) He atom
(B) Li⁺ ion
(C) He⁺ ion
(D) Li atom
✓ Correct Answer: (C)
Bohr's model works only for one-electron species: H, He⁺ (one electron), Li²⁺, Be³⁺. He has two electrons, Li has three, Li⁺ has two — all fail with Bohr. He⁺ has only one electron and works perfectly.
MCQ 04 — JEE Level
The ground state electron configuration of Cr (Z=24) is:
(A) [Ar] 4s² 3d⁴
(B) [Ar] 4s¹ 3d⁵
(C) [Ar] 4s² 3d³ 4p¹
(D) [Ar] 4s² 3d⁴ 4p⁰
✓ Correct Answer: (B)
Cr gets extra stability from half-filled 3d⁵ configuration. One electron moves from 4s to 3d, giving [Ar] 4s¹ 3d⁵. This maximises exchange energy (10 exchange pairs for 5 parallel-spin d electrons).
MCQ 05 — GATE Level
Using Slater's rules, the effective nuclear charge Z* experienced by a 3p electron in Cl (Z=17, config: [Ne]3s²3p⁵) is:
(A) 6.10
(B) 6.45
(C) 5.75
(D) 7.00
✓ Correct Answer: (A)
For 3p electron: same group (3s²,3p⁴) = 6 × 0.35 = 2.10; n–1 group (2s²,2p⁶) = 8 × 0.85 = 6.80; n–2 group (1s²) = 2 × 1.00 = 2.00. S = 10.90. Z* = 17 – 10.90 = 6.10
MCQ 06 — JEE Level
Which series of hydrogen atom emission lies in the ultraviolet region?
(A) Balmer series
(B) Paschen series
(C) Lyman series
(D) Brackett series
✓ Correct Answer: (C)
Lyman series (nl=1) involves transitions to the ground state — these are the highest energy transitions in hydrogen, corresponding to UV radiation. Balmer (nl=2) includes visible light. Paschen and Brackett are IR.
MCQ 07 — JEE Advanced Level
The pz orbital has Y = (constant) × z/r. Its angular nodal surface is the:
(A) yz-plane (x = 0)
(B) xz-plane (y = 0)
(C) xy-plane (z = 0)
(D) No nodal surface
✓ Correct Answer: (C)
For angular node, set Y = 0 → z = 0 → this is the xy-plane. The pz orbital is positive above the xy-plane (z > 0) and negative below it (z < 0), with a nodal plane at z = 0.
MCQ 08 — JEE/BITSAT Level
The ionisation energy of oxygen is lower than that of nitrogen. This is because:
(A) Oxygen has more protons
(B) Oxygen's 2p electrons are more shielded
(C) A paired 2p electron in oxygen experiences extra repulsion
(D) Nitrogen is in a higher period
✓ Correct Answer: (C)
Nitrogen has 2p³ (↑ ↑ ↑, all unpaired). Oxygen has 2p⁴ (↑↓ ↑ ↑), where one electron must pair up. The Coulombic repulsion (c) in the paired orbital makes that electron easier to remove.
MCQ 09 — IIT-JAM Level
In a particle-in-a-box model, what happens to the energy levels if the box length 'a' is doubled?
(A) Energy doubles
(B) Energy halves
(C) Energy becomes one-quarter
(D) Energy quadruples
✓ Correct Answer: (C)
E = n²h²/(8ma²). If a → 2a, then E → n²h²/(8m×4a²) = E/4. Energy becomes one-quarter of original. This is why larger conjugated systems absorb at longer wavelengths (lower energy).
MCQ 10 — CSIR-NET Level
The total number of orbitals in the n = 4 shell is:
(A) 8
(B) 16
(C) 32
(D) 4
✓ Correct Answer: (B)
Total orbitals in nth shell = n². For n=4: 4² = 16 orbitals (1 from 4s + 3 from 4p + 5 from 4d + 7 from 4f = 16). Max electrons = 2n² = 32.
📝 25 Practice Questions
Based on the complete chapter — JEE, GATE, CSIR-NET level. Answers at the bottom.
Q.1
Which of the following has the largest number of unpaired electrons?
(A) Fe²⁺
(B) Cr³⁺
(C) Mn²⁺
(D) V²⁺
Q.2
The de Broglie wavelength of a particle is inversely proportional to its:
(A) Energy
(B) Mass × velocity
(C) Temperature
(D) Charge
Q.3
How many angular nodes does a 4d orbital have?
(A) 0
(B) 1
(C) 2
(D) 3
Q.4
Among Na⁺, Mg²⁺, Al³⁺, which is the largest ion and why?
(A) Al³⁺ — highest charge
(B) Na⁺ — fewest protons pulling 10 electrons
(C) Mg²⁺ — intermediate
(D) All same size (isoelectronic)
Q.5
Which quantum number determines the shape of an orbital?
(A) n
(B) l
(C) ml
(D) ms
Q.6
The Pauli exclusion principle states that:
(A) Electrons fill lowest energy orbitals first
(B) Electrons in same subshell have parallel spins first
(C) No two electrons can have the same set of all four quantum numbers
(D) Electrons in degenerate orbitals pair up first
Q.7
What is the electron configuration of Cu²⁺ (Z=29)?
(A) [Ar] 4s¹ 3d⁸
(B) [Ar] 3d⁹
(C) [Ar] 4s² 3d⁷
(D) [Ar] 4s¹ 3d¹⁰
Q.8
The radial probability function is zero at the nucleus for all orbitals because:
(A) R(r) = 0 at r = 0
(B) 4πr² = 0 at r = 0
(C) Electrons cannot be near nuclei
(D) ψ is undefined at origin
Q.9
Ionisation energy of Be is greater than B. This is because:
(A) Be has a higher atomic mass
(B) B has a 2p electron shielded by 2s, making it easier to remove
(C) Be has more electrons
(D) B is farther right in the period
Q.10
Which transition in hydrogen gives the longest wavelength in the Paschen series?
(A) n = 4 → n = 3
(B) n = 5 → n = 3
(C) n = 6 → n = 3
(D) n = ∞ → n = 3
Q.11
The wave function ψ for the particle in a box at n=2 has how many nodes between x=0 and x=a?
(A) 0
(B) 1
(C) 2
(D) 3
Q.12
In the Schrödinger equation, the Hamiltonian operator includes which two energy contributions?
(A) Nuclear and gravitational
(B) Kinetic and potential
(C) Rotational and vibrational
(D) Spin and orbital
Q.13
Mendeleev predicted properties of which element that was later discovered?
(A) Bromine
(B) Gallium
(C) Zinc
(D) Copper
Q.14
For a 5f orbital, the number of radial nodes is:
(A) 1
(B) 2
(C) 3
(D) 0
Q.15
The exchange energy ε arises from:
(A) Electrostatic repulsion between electrons
(B) Quantum mechanical exchanges between electrons of same spin in separate orbitals
(C) Attraction between electron and nucleus
(D) Spin-orbit coupling
Q.16
The effective nuclear charge Z* on a valence electron increases as you move across a period from left to right. The primary reason is:
(A) More electrons are added to inner shells
(B) Nuclear charge increases while electrons are added to the same shell (poor shielding)
(C) Atomic radius increases
(D) More protons increase shielding
Q.17
Which of the following sets of quantum numbers is NOT valid?
(A) n=2, l=1, ml=0, ms=+½
(B) n=3, l=2, ml=–2, ms=–½
(C) n=2, l=2, ml=1, ms=+½
(D) n=4, l=3, ml=+3, ms=–½
Q.18
Millikan's oil-drop experiment determined the:
(A) Mass-to-charge ratio of the electron
(B) Charge of the electron
(C) Radius of the nucleus
(D) Number of protons in an atom
Q.19
Which rule states that electrons in degenerate orbitals occupy them singly before pairing?
(A) Aufbau principle
(B) Pauli exclusion principle
(C) Hund's rule of maximum multiplicity
(D) Klechkowsky's rule
Q.20
In the Bohr model, what happens to the energy as n → ∞?
(A) Energy → –∞
(B) Energy → 0
(C) Energy → RH
(D) Energy stays constant
Q.21
The 2s orbital of hydrogen has a nodal sphere at r = 2a₀. This is an example of a:
(A) Angular node
(B) Conical node
(C) Radial node
(D) Planar node
Q.22
Transition metal cations (like Fe²⁺) have no s electrons in their outer levels because:
(A) s electrons are always most stable
(B) After electron removal, Z* increases, stabilising d orbitals more than ns
(C) d electrons are always removed first
(D) The 4s subshell doesn't exist in ions
Q.23
For an electron in a hydrogen 1s orbital, the most probable distance from the nucleus is:
(A) Infinite
(B) Zero
(C) a₀ = 52.9 pm (Bohr radius)
(D) 2a₀
Q.24
Which of the following statements about crystal radii is correct?
(A) Cations are larger than their parent atoms
(B) For isoelectronic ions, radius increases as nuclear charge increases
(C) Anions are larger than their parent atoms
(D) All ions in a period have the same radius
Q.25
Moseley's contribution to atomic theory established that atomic number Z is:
(A) Equal to atomic mass
(B) The number of neutrons
(C) More fundamental than atomic mass for classifying elements
(D) Related to isotope number
Answer Key:
Q1-C | Q2-B | Q3-C | Q4-B | Q5-B | Q6-C | Q7-B | Q8-B | Q9-B | Q10-A
Q11-B | Q12-B | Q13-B | Q14-A | Q15-B | Q16-B | Q17-C | Q18-B | Q19-C | Q20-B
Q21-C | Q22-B | Q23-C | Q24-C | Q25-C
Q1-C | Q2-B | Q3-C | Q4-B | Q5-B | Q6-C | Q7-B | Q8-B | Q9-B | Q10-A
Q11-B | Q12-B | Q13-B | Q14-A | Q15-B | Q16-B | Q17-C | Q18-B | Q19-C | Q20-B
Q21-C | Q22-B | Q23-C | Q24-C | Q25-C
10. Exam Tips & Tricks
🎯 High-Yield Topics by Exam
| Exam | Most Tested Topics from This Chapter |
|---|---|
| JEE | Bohr model, quantum numbers, electronic configurations, periodic trends (IE, EA, radius) |
| JEE Mains | de Broglie, Heisenberg, Bohr energy levels, aufbau/Hund/Pauli, exceptions (Cr, Cu) |
| JEE Advanced | Nodal surfaces, Slater's rules, Coulombic/exchange energies, particle-in-a-box |
| CSIR-NET | Schrödinger equation, radial/angular functions, shielding theory, periodic properties |
| IIT-JAM | Quantum numbers, wave functions, particle-in-a-box, Z* calculations |
| GATE | Slater's rules, effective nuclear charge, transition metal configurations |
| TGT/PGT | Historical development, spectral series, quantum numbers, aufbau principle |
⚡ Quick Memory Tricks
- Spectral series in order of energy (highest to lowest): Lyman > Balmer > Paschen > Brackett > Pfund. Mnemonic: "Let Boys Play Big Pianos"
- Nodes at a glance: For any orbital nl: angular = l, radial = n–l–1. Fast check: 3d → l=2, radial=3–2–1=0 ✓
- Filling order trick: Write 1s, then diagonals from top-right to bottom-left: 1s/2s,2p/3s,3p,3d/4s,4p,4d,4f/... fill each diagonal going down-right
- Transition metal ions: ALWAYS remove s electrons first to write ion configs. Fe²⁺: start from Fe ([Ar]4s²3d⁶), remove 4s² → [Ar]3d⁶
- Cr and Cu exceptions: Only Cr (d⁵) and Cu (d¹⁰) in first row. In second row: Mo (d⁵), Pd (d¹⁰). In third row: Pt (d¹⁰)
- IE zigzag: In Period 2, IE dips at B (from Be) and at O (from N). Same pattern repeats in all periods at +2, +3 positions
- Z* increases left → right across period because electrons in the same shell shield each other poorly (only 0.35 each)
- Slater's shortcut: For d electrons, ALL electrons to the left contribute 1.00 to S (full shielding). For s/p, only (n–2)+ contribute 1.00; (n–1) contributes 0.85
🚫 Common Mistakes to Avoid
- Don't confuse "orbits" (Bohr — circular paths) with "orbitals" (Schrödinger — probability clouds)
- Don't say ψ gives probability — it's ψ² (probability density) that gives probability
- Don't forget: in Slater's rules for d/f electrons, ALL groups to the left contribute 1.00 (not 0.85)
- Don't write Cu²⁺ as [Ar]4s¹3d⁸ — ions don't have s electrons (Cu²⁺ = [Ar]3d⁹)
- Don't mix up Coulombic energy (bad — pairing in same orbital) and Exchange energy (good — parallel spins in different orbitals)
- For Bohr model, remember: energy levels are negative (bound electrons). The more negative, the more stable (lower energy)
- The 1s orbital has the maximum ψ² at the nucleus (ψ² is highest there), but the radial probability function 4πr²ψ² = 0 at r=0 because 4πr²=0
📌 Important Constants to Memorise
| Constant | Symbol | Value |
|---|---|---|
| Planck's constant | h | 6.626 × 10⁻³⁴ J·s |
| Speed of light | c | 2.998 × 10⁸ m/s |
| Electron mass | me | 9.11 × 10⁻³¹ kg |
| Electronic charge | e | 1.60 × 10⁻¹⁹ C |
| Rydberg constant (H) | RH | 2.179 × 10⁻¹⁸ J = 13.61 eV |
| Bohr radius | a₀ | 52.9 pm = 0.529 Å |
Final Tip for JEE: This chapter carries questions every year — especially quantum numbers, electronic configurations (including exceptions), periodic trends, and spectral lines. Spend extra time on Slater's rules calculations and the Coulombic/exchange energy concept if you're aiming for GATE/CSIR-NET. The nodal surface problems (angular vs radial) are almost guaranteed in JEE Advanced and IIT-JAM.
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