Quantum Atomic Structure & Orbitals
CollegeThe electron as a wave
The Bohr model (fixed circular orbits) fails for anything beyond hydrogen. The rigorous treatment solves the time-independent Schrödinger equation, Ĥψ = Eψ, where the wavefunction ψ encodes everything knowable about an electron's state and |ψ|² gives the probability density of finding it at a point in space — an orbital is a three-dimensional region enclosing (conventionally) 90% of that probability.
Solving Ĥψ = Eψ for the hydrogen atom yields four quantum numbers that fully specify an electron's state:
| Symbol | Name | Allowed values | Determines |
|---|---|---|---|
| n | Principal | 1, 2, 3, … | Shell / energy level |
| ℓ | Azimuthal | 0 … n−1 | Subshell shape (s, p, d, f) |
| mℓ | Magnetic | −ℓ … +ℓ | Orbital orientation |
| ms | Spin | +½ or −½ | Electron spin direction |
The Pauli exclusion principle forbids any two electrons in one atom sharing all four quantum numbers — hence at most two electrons (opposite spin) per orbital. Orbital shapes follow directly from ℓ: s (ℓ=0) is spherical; p (ℓ=1) is dumbbell-shaped with three orientations; d (ℓ=2) has more complex, mostly four-lobed shapes.
Why multi-electron atoms need shielding and Zeff
The Schrödinger equation is exactly solvable only for one-electron systems (H, He⁺, Li²⁺). For any atom with more than one electron, electron–electron repulsion makes an exact analytic solution impossible; the equation must be solved approximately (e.g. the Hartree–Fock self-consistent field method), where each electron moves in an averaged field from the nucleus and all other electrons.
This gives rise to shielding: inner electrons partially screen outer electrons from the full nuclear charge, so an outer electron feels a reduced effective nuclear charge, Zeff = Z − S (S = shielding constant, estimable via Slater's rules). Shielding also explains orbital penetration: an electron in 4s has a small but nonzero probability density very close to the nucleus (penetrating through the 3d shielding), giving it slightly lower energy than 3d in neutral atoms at the point 4s first fills — the microscopic reason behind the Aufbau filling order and behind anomalies like Cr's [Ar]3d⁵4s¹ configuration (a half-filled 3d subshell is unusually stable, favoured over the "expected" 3d⁴4s²).
Worked example — quantum numbers and an anomalous configuration
(a) List all allowed (n, ℓ, mℓ) combinations for n = 2, and state the maximum electron capacity of the n = 2 shell.
- For n = 2, ℓ can be 0 or 1 (ℓ = 0 … n−1).
- ℓ = 0 (the 2s subshell): mℓ = 0 only → 1 orbital.
- ℓ = 1 (the 2p subshell): mℓ = −1, 0, +1 → 3 orbitals.
- Total orbitals in n = 2: 1 + 3 = 4; each holds 2 electrons (opposite spin) → capacity = 8 electrons, matching the n² × 2 rule (2² × 2 = 8).
(b) Explain why chromium (Z = 24) has configuration [Ar]3d⁵4s¹ instead of the "expected" [Ar]3d⁴4s².
- Aufbau would naively predict 3d⁴4s² by filling 4s fully before 3d.
- 3d and 4s are extremely close in energy once several 3d electrons are present, so exchange energy (a quantum stabilisation from electrons of parallel spin occupying separate degenerate orbitals) becomes decisive.
- A half-filled 3d⁵ subshell (five parallel-spin electrons, one per d orbital) maximises exchange stabilisation, more than compensating for promoting one 4s electron.
- Result: chromium's ground state is [Ar]3d⁵4s¹, not 3d⁴4s².
Why can't an electron have quantum numbers n = 2, ℓ = 2?
ℓ must satisfy 0 ≤ ℓ ≤ n−1. For n = 2, the maximum allowed ℓ is 1 (giving 2s and 2p). ℓ = 2 (a d subshell) does not exist until n = 3, so this combination violates the quantum number rules.
What does |ψ|² physically represent, and why do we speak of "90% probability boundaries" for orbitals?
|ψ|² is the probability density of locating the electron at a given point — it is nonzero (though vanishingly small) at essentially all distances from the nucleus, so an orbital has no sharp edge. Chemists draw a surface enclosing the volume where the electron is found with 90% probability purely as a practical convention for visualising orbital shape.
Explain, using Zeff and shielding, why ionisation energy increases across Period 2 despite added electron–electron repulsion.
Across a period, electrons are added to the same shell, where they shield each other very poorly (same-shell shielding is weak), while the nuclear charge Z increases by one each step. The poor shielding cannot offset the rising nuclear charge, so Zeff rises steadily left to right, pulling valence electrons in more tightly and raising ionisation energy overall (with small, explainable dips at half-filled and fully-filled subshells due to exchange energy and reduced repulsion).
Why does the Schrödinger equation have no exact analytic solution for helium?
Helium has two electrons, so the Hamiltonian includes an electron–electron repulsion term (1/r₁₂) that couples the two electrons' coordinates. This makes the equation a genuine three-body problem (nucleus + 2 electrons) with no closed-form solution, unlike hydrogen's clean two-body problem — hence the need for approximation methods like Hartree–Fock or configuration interaction.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1Write the ground-state electron configuration of copper (Z = 29).
- Expected filling would give [Ar] 3d9 4s2.
- A filled 3d10 is more stable, so one 4s electron moves to 3d.
- This gives the anomalous configuration.
Example 2How many electrons can the n = 2 shell hold?
- n = 2 contains the 2s and 2p subshells.
- 2s holds 2 electrons; 2p holds 6 electrons.
- Total = 2 + 6 = 8.
Now you try
Work each one out first, then tap to reveal the worked answer.