Dual behaviour of matter: de Broglie
Light behaves as a wave and as a particle. In 1924 de Broglie said matter should too: any moving particle has a wavelength
λ = h/mv = h/p (p = momentum).
For a cricket ball λ is so tiny (about 10−34 m) that no wave effect can be seen. For an electron, with its very small mass, λ is about the size of an atom, so wave effects matter. Electron diffraction proved it, and the electron microscope uses it.
Link to Bohr: if a whole number of electron waves fits on an orbit, 2πr = nλ = nh/mv, which gives mvr = nh/2π: exactly Bohr's rule.
Heisenberg's uncertainty principle
It is impossible to know the exact position and the exact momentum (or velocity) of an electron at the same time:
Δx × Δp ≥ h/4π, or Δx × Δv ≥ h/(4πm).
Why: to "see" an electron you must hit it with light of wavelength smaller than the region you want to measure. Such light has high energy and knocks the electron, changing its momentum. The sharper the position, the fuzzier the momentum.
What it means: the idea of a fixed path (orbit) for an electron is meaningless, so Bohr's model must go. For big objects (a ball) the uncertainty is far too tiny to notice.
Quantum mechanical model and orbitals
Schrödinger (1926) wrote an equation for the electron as a wave. Its solutions are wave functions ψ. ψ itself has no simple meaning, but ψ² gives the probability density: how likely the electron is to be found at a point.
- An orbital is the region of space around the nucleus where the chance of finding the electron is highest (usually drawn to hold 90 % of the chance).
- An orbit (Bohr) is a sharp circular path; an orbital is a 3D probability cloud. An orbital holds at most 2 electrons.
- The energy of an electron is quantised; the allowed energies come naturally from the equation.
- Nodes: places where ψ² = 0. Radial nodes = n − l − 1; angular nodes = l; total = n − 1.
Quantum numbers
| Quantum number | Values | Tells us |
|---|---|---|
| Principal n | 1, 2, 3 … (K, L, M …) | Shell, size and energy. Orbitals in shell n = n²; electrons = 2n². |
| Azimuthal l | 0 to n − 1 (s, p, d, f) | Subshell and shape. Orbitals in a subshell = 2l + 1. |
| Magnetic mₗ | −l … 0 … +l | Orientation of the orbital in space. |
| Spin mₛ | +½ or −½ | Spin direction of the electron (↑ or ↓). |
Example: for n = 3, l can be 0, 1, 2 (3s, 3p, 3d); number of orbitals = 1 + 3 + 5 = 9 = 3²; electrons = 18.
Orbital angular momentum = √(l(l+1)) · h/2π.
Shapes of s, p and d orbitals
- s (l = 0): spherical. Size grows 1s < 2s < 3s. 2s has 1 radial node, 3s has 2.
- p (l = 1): three dumbbells, pₓ, py, pz, along the three axes, with a nodal plane through the nucleus. Same energy (degenerate) in an atom. Starts from n = 2.
- d (l = 2): five orbitals from n = 3. dxy, dyz, dxz have four lobes between the axes; dx²−y² has four lobes along x and y; dz² is a dumbbell along z with a ring (doughnut) round the middle.
Energy of orbitals and the (n + l) rule
In hydrogen, energy depends only on n: 2s = 2p. In atoms with many electrons, energy also depends on l, because inner electrons shield outer ones, and s electrons get closer to the nucleus (penetrate) more than p, then d.
(n + l) rule: lower n + l means lower energy. If equal, lower n comes first. So 4s (4+0 = 4) fills before 3d (3+2 = 5).
Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s …
Filling rules: Aufbau, Pauli and Hund
- Aufbau principle: electrons go into the lowest-energy orbital first ("aufbau" = building up).
- Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers the same. So one orbital holds at most two electrons, with opposite spins (↑↓).
- Hund's rule of maximum multiplicity: in orbitals of equal energy (p, d, f), electrons first go in singly, with parallel spins; pairing starts only after each orbital has one. So N (2p³) is ↑ ↑ ↑, not ↑↓ ↑.
Electronic configuration and stability
Write subshells with electron counts as superscripts: O (8) = 1s² 2s² 2p⁴; K (19) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ = [Ar] 4s¹. The outermost electrons are valence electrons; the rest are core electrons.
Special stability of half-filled and fully filled subshells:
- Symmetry: an evenly filled subshell is more symmetric and lower in energy.
- Exchange energy: electrons with parallel spin in equal-energy orbitals can swap places; each swap releases energy. More parallel spins = more exchange = more stability. d⁵ has the most.
So Cr (24) = [Ar] 3d⁵ 4s¹, not 3d⁴4s², and Cu (29) = [Ar] 3d¹⁰ 4s¹, not 3d⁹4s².
Ions: remove electrons from the highest n first. Fe (26) = [Ar]3d⁶4s²; Fe²⁺ = [Ar]3d⁶; Fe³⁺ = [Ar]3d⁵ (half-filled, which is why Fe³⁺ is more stable).
Try it: fill and check
In the last story step, predict the configuration of nitrogen (Z = 7) and write it in your notebook with boxes and arrows. Then set Z = 7 and check that the three 2p electrons are single and parallel. Then compare Z = 23, 24, 25: watch the 4s electron move into 3d at Cr.
Board exam corner
Common questions: de Broglie wavelength and uncertainty numericals (2–3 marks); possible/impossible quantum number sets; number of orbitals/electrons in a shell; nodes; shapes of orbitals (diagram); state and explain Aufbau, Pauli, Hund; configurations of Z = 1–30 with Cr and Cu exceptions and why.
Key formulas and definitions
- λ = h/mv = h/p (de Broglie)
- Δx · Δp ≥ h/4π; Δx · Δv ≥ h/4πm
- l = 0 to n − 1; mₗ = −l to +l; mₛ = ±½
- Orbitals in shell = n²; electrons in shell = 2n²
- Orbitals in subshell = 2l + 1; electrons = 2(2l + 1)
- Radial nodes = n − l − 1; angular nodes = l; total = n − 1
- Orbital angular momentum = √(l(l+1)) h/2π
- (n + l) rule for filling order
Worked examples
1. Find the de Broglie wavelength of a 0.1 kg ball moving at 10 m/s.
λ = h/mv = 6.626 × 10⁻³⁴ ÷ (0.1 × 10) = 6.626 × 10⁻³⁴ m. Far too small to notice.
2. Find the de Broglie wavelength of an electron (m = 9.1 × 10⁻³¹ kg) moving at 2.19 × 10⁶ m/s.
λ = 6.626 × 10⁻³⁴ ÷ (9.1 × 10⁻³¹ × 2.19 × 10⁶) = 6.626 × 10⁻³⁴ ÷ 1.993 × 10⁻²⁴ = 3.32 × 10⁻¹⁰ m = 332 pm, about the size of an atom. (This is 2πr for H's first orbit: one whole wave fits.)
3. A 25 g ball's position is known within 10⁻⁵ m. Find the minimum uncertainty in its velocity.
Δv = h/(4πmΔx) = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 0.025 × 10⁻⁵) = 6.626 × 10⁻³⁴ ÷ 3.14 × 10⁻⁶ = 2.1 × 10⁻²⁸ m/s. Negligible.
4. An electron's position is known within 0.1 Å (10⁻¹¹ m). Find the minimum uncertainty in its velocity.
Δv = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 9.1 × 10⁻³¹ × 10⁻¹¹) = 6.626 × 10⁻³⁴ ÷ 1.143 × 10⁻⁴⁰ = 5.8 × 10⁶ m/s. Bigger than its own speed in an atom, so a sharp orbit makes no sense.
5. Write the quantum numbers of all orbitals in the 3d subshell and the number of radial and angular nodes.
n = 3, l = 2, mₗ = −2, −1, 0, +1, +2 (five orbitals). Radial nodes = 3 − 2 − 1 = 0; angular nodes = 2.
6. Which sets are not possible? (a) n = 2, l = 2, mₗ = 0 (b) n = 3, l = 1, mₗ = −1 (c) n = 1, l = 0, mₗ = +1 (d) n = 4, l = 3, mₗ = −3
(a) impossible: l must be ≤ n − 1 = 1. (c) impossible: for l = 0, mₗ can only be 0. (b) 3p and (d) 4f are fine.
7. How many electrons in an atom can have n = 3 and mₛ = −½?
Shell n = 3 has n² = 9 orbitals. Each has one electron with mₛ = −½. Answer 9.
8. Arrange 4s, 3d, 4p, 5s by energy using the (n + l) rule.
4s: 4, 3d: 5, 4p: 5, 5s: 5. So 4s is lowest. Among n + l = 5, lower n first: 3d < 4p < 5s. Order: 4s < 3d < 4p < 5s.
9. Write configurations of Cr (24) and Cu (29) and explain.
Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹. Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹. Moving one 4s electron makes 3d half-filled (Cr) or full (Cu). These are more symmetric and have more exchange energy, so they are more stable.
10. How many unpaired electrons are in Fe³⁺ (Fe, Z = 26)?
Fe = [Ar] 3d⁶ 4s². Remove 4s² then one 3d: Fe³⁺ = [Ar] 3d⁵. By Hund's rule all five are single: 5 unpaired electrons.
Common mistakes
- Filling 3d before 4s in neutral atoms, or removing 3d electrons before 4s in ions. Fill 4s first; remove 4s first.
- Writing 2d or 1p. l must be less than n: 1s only; 2s, 2p; d starts at n = 3.
- Confusing orbit and orbital. An orbit is a sharp path (Bohr); an orbital is a probability cloud.
- Pairing electrons in a p or d subshell before every orbital has one (breaking Hund's rule).