From Hydrogen to Many-Electron Atoms

Introduction

Hydrogen gave us exact orbitals, energies and quantum numbers. Every other atom has at least two electrons, and those electrons repel each other. That single extra ingredient changes the picture profoundly: the 2s and 2p levels no longer have the same energy, electrons shield one another from the nucleus, and the Pauli principle limits how orbitals can be filled. This page builds the bridge from the exact hydrogen solution to the approximate but powerful orbital picture of the whole periodic table.

Core explanation

The many-electron Hamiltonian. For an atom with nuclear charge Z and N electrons, in atomic units,

Ĥ = Σᵢ [−½∇ᵢ² − Z/rᵢ] + Σᵢ<ⱼ 1/rᵢⱼ.

The first sum is simply a hydrogen-like Hamiltonian for each electron. The second sum, over every pair of electrons, is the Coulomb repulsion 1/rᵢⱼ, which depends on the positions of two electrons at once. It couples the motions of all the electrons, so the Schrödinger equation can no longer be separated into independent one-electron problems.

The orbital approximation. The simplest workable approach replaces the exact wavefunction by a product of one-electron functions (orbitals):

Ψ(1, 2, …, N) ≈ φₐ(1) φ b(2) … φ N(N).

Each electron is imagined to move in the attraction of the nucleus plus the average repulsion of all the others. This averaged field is roughly spherical, so each orbital can still be labelled by n, l and mₗ, and its angular part is still a spherical harmonic. The radial parts, however, differ from those of hydrogen.

Shielding and effective nuclear charge. An electron far from the nucleus is partly screened by the electrons closer in. It behaves roughly as if it were attracted by a reduced charge,

Z eff = Z − σ,

where σ is the shielding constant. In helium, each electron shields the other only partly, and variational calculations give Z eff ≈ 1.69 rather than 2. For the outer 2s electron of lithium, the two 1s electrons shield most of the nuclear charge of 3, leaving Z eff ≈ 1.3.

Penetration lifts degeneracy. In hydrogen, energy depends only on n. In a many-electron atom, an s electron has a small but significant probability of being close to the nucleus, inside the inner shells, where it feels nearly the full nuclear charge. A p electron with the same n penetrates less, and a d electron less still. Hence for a given n the energies order s < p < d < f. This is why 2s fills before 2p and why 4s fills before 3d in potassium and calcium.

The Pauli principle. Electrons are identical fermions. The total wavefunction, including spin, must change sign when the labels of any two electrons are swapped. A simple product of orbitals does not satisfy this, but a Slater determinant of spin orbitals does. If two electrons are placed in the same spin orbital, two columns of the determinant become identical and it vanishes. This is the familiar rule that no two electrons can have the same four quantum numbers, so each spatial orbital holds at most two electrons of opposite spin.

Step-by-step reasoning

To build the ground-state configuration of a light atom:

1. Start from hydrogen-like orbitals labelled by n and l. 2. Order the subshells using penetration: 1s < 2s < 2p < 3s < 3p < 4s < 3d, and so on. 3. Place electrons in the lowest subshells, at most two per orbital with paired spins (Pauli). 4. Within a partly filled subshell, spread electrons over separate orbitals with parallel spins where possible (Hund's first rule).

Visual explanation

Plot radial distribution functions for 2s and 2p in lithium on the same axes. The 2p curve has one broad peak. The 2s curve has a similar outer peak but also a small inner hump lying inside the 1s region. That inner hump is penetration made visible: it is why 2s feels more nuclear charge and lies lower in energy.

Real-world analogy

Imagine standing at the back of a crowd listening to a speaker. People in front block and muffle the voice, so you hear a weaker version of it. If you occasionally squeeze through to the front, you hear the full voice for a moment. Outer electrons that penetrate the inner shells get these moments of stronger attraction.

Real-world example

The first ionisation energy of lithium is only 5.39 eV, far smaller than the 122 eV that would be needed to remove the electron from Li²⁺. The low value, which makes lithium a reactive metal used in batteries, is a direct consequence of shielding of the 2s electron by the filled 1s shell.

Why?

Why can't we just add up hydrogen energies for each electron? Because the repulsion between electrons is large, not a small correction. For helium, ignoring it predicts a total energy of −108.8 eV, whereas experiment gives −79.0 eV. An error of nearly 30 eV is larger than most chemical bond energies, so repulsion must be built into any useful model.

Common misconception

"The 4s orbital is always lower in energy than 3d." In potassium and calcium the 4s level fills first, but once the 3d orbitals are occupied in transition metals, 3d usually lies below 4s. That is why transition-metal atoms lose their 4s electrons first when forming cations such as Fe²⁺ ([Ar]3d⁶).

Worked example

Question: The ionisation energy of lithium is 5.39 eV. Treating the 2s electron as hydrogen-like with n = 2, estimate its effective nuclear charge.

Reasoning: For a hydrogen-like orbital, IE = 13.6 × Z eff²/n² eV. So Z eff² = 5.39 × 4/13.6 = 1.585, giving Z eff = 1.26.

Answer: Z eff ≈ 1.26, so the 1s pair shields about 1.74 of the three units of nuclear charge.

Quick check

1. Why is 2s lower in energy than 2p in a lithium atom although they are degenerate in hydrogen? Answer: The 2s electron penetrates the 1s shell more, so it experiences a larger effective nuclear charge and is more strongly bound.

Exam focus

Write the many-electron Hamiltonian and point out the 1/rᵢⱼ terms as the source of difficulty. Explain penetration and shielding with reference to radial distribution functions, and state the Pauli principle in its antisymmetry form, not only as "two electrons per orbital".

Advanced insight

The orbital approximation can be made quantitative through the Hartree–Fock method, which finds the best single Slater determinant variationally. Each electron then moves in the self-consistent average field of the others. Even this best orbital picture misses the instantaneous avoidance of electrons; the missing energy, called the correlation energy , is about 1.1 eV for helium and must be recovered by more advanced methods.

Summary

A many-electron atom has repulsion terms 1/rᵢⱼ that prevent exact separation. The orbital approximation assigns each electron a hydrogen-like orbital moving in an averaged field. Shielding reduces the effective nuclear charge, and penetration makes s < p < d for a given n. The Pauli principle requires an antisymmetric wavefunction, conveniently expressed as a Slater determinant, which limits each orbital to two electrons of opposite spin.

Practice questions

1. Identify the term in the helium Hamiltonian that prevents an exact solution and explain why. Answer: The repulsion 1/r₁₂. It depends on the coordinates of both electrons, so the equation cannot be separated into two independent one-electron problems. 2. Estimate Z eff for the 3s electron of sodium, given that its ionisation energy is 5.14 eV. Answer: Z eff² = 5.14 × 9/13.6 = 3.40, so Z eff ≈ 1.84. 3. Explain, using the Slater determinant, why two electrons cannot occupy the same spin orbital. Answer: Two identical columns (or rows) make a determinant zero, so the wavefunction would vanish everywhere, which is not an allowed state. 4. Arrange the 3s, 3p and 3d subshells of argon in order of increasing energy and justify the order. Answer: 3s < 3p < 3d, because s electrons penetrate the inner shells most and d electrons least, so s experiences the largest effective nuclear charge.