Quantum Numbers and Orbitals: Rules, Counts and Shapes

Electrons as waves: the de Broglie wavelength

In 1924 de Broglie proposed that any moving particle has a wavelength λ = h/(mv), where h is the Planck constant, m the mass and v the speed. For everyday objects the wavelength is far too small to notice. A 0.145 kg baseball at 40.0 m/s has λ = 1.14 × 10⁻³⁴ m. For an electron, with a mass of 9.109 × 10⁻³¹ kg (NIST CODATA), the wavelength is comparable to the size of an atom.

Worked example 1. An electron moving at 2.188 × 10⁶ m/s has λ = h/(mv). With the CODATA values h = 6.62607 × 10⁻³⁴ J·s and m = 9.10938 × 10⁻³¹ kg, λ = (6.62607 × 10⁻³⁴)/[(9.10938 × 10⁻³¹)(2.188 × 10⁶)] = 3.324 × 10⁻¹⁰ m, or 0.3324 nm. This speed is the one Bohr's model assigns to the electron in hydrogen's lowest orbit, whose radius is the Bohr radius, a₀ = 5.29177 × 10⁻¹¹ m (NIST CODATA). The orbit's circumference is 2πa₀ = 3.325 × 10⁻¹⁰ m, so one electron wavelength fits around it, with agreement to better than 0.02%. Quantized energy levels are what you get when an electron wave must fit its space without cancelling itself, much as a guitar string can vibrate only at certain frequencies.

Uncertainty and the end of fixed orbits

The Heisenberg uncertainty principle sets a limit on how well position and momentum can be known together: Δx·Δp ≥ ħ/2, where ħ = h/2π = 1.0546 × 10⁻³⁴ J·s.

Worked example 2. Confine an electron to a region 1.0 × 10⁻¹⁰ m wide, about the size of an atom. Then Δp ≥ (1.0546 × 10⁻³⁴ J·s)/(2 × 1.0 × 10⁻¹⁰ m) = 5.3 × 10⁻²⁵ kg·m/s, and Δv = Δp/m ≥ 5.8 × 10⁵ m/s. The uncertainty in speed is a sizable fraction of the speed itself, so a definite path, like a planet's orbit, is meaningless for an electron in an atom.

Quantum mechanics replaces the path with a wavefunction, ψ. Its square, ψ², is the probability density: multiplied by the volume of a small region, it gives the probability of finding the electron there. An orbital is a wavefunction for one electron, and the familiar orbital pictures are surfaces that enclose most of that probability, often 90%.

Allowed values of n, l, mₗ and mₛ

Solving the wave equation for the hydrogen atom produces three quantum numbers, each restricted by the one before it. A fourth describes the electron's spin. The restrictions are the whole game.

• Principal quantum number n = 1, 2, 3, …. It sets the shell and, in hydrogen, the energy. Larger n means a larger, higher-energy orbital.
• Angular momentum quantum number l = 0, 1, …, n − 1. It sets the subshell and the shape: l = 0, 1, 2, 3 are s, p, d, f.
• Magnetic quantum number mₗ = −l, …, 0, …, +l, giving 2l + 1 values. It sets the orientation of the orbital in space.
• Spin quantum number mₛ = +½ or −½. It does not depend on the other three.

Worked example: testing six sets of quantum numbers

To test a set (n, l, mₗ, mₛ), check each number against the one before it, in order. Stop at the first violation.

(3, 3, 0, +½) is not allowed: l must be at most n − 1 = 2. (4, 2, −3, −½) is not allowed: with l = 2, mₗ runs only from −2 to +2. (3, 2, 1, 0) is not allowed: mₛ must be +½ or −½, never 0. (2, 1, −1, −½) is allowed and describes an electron in a 2p orbital. (5, 0, 0, +½) is allowed and describes a 5s electron. (4, 3, −3, +½) is allowed and describes a 4f electron.

To name the orbital from an allowed set, write n followed by the letter for l. The value of mₗ picks one orbital within the subshell but does not change its name in this notation.

Counting orbitals and electrons under a restriction

Counting questions become routine if you list the allowed l values, then the mₗ values for each, and only then apply the restriction. Each orbital, meaning each allowed (n, l, mₗ) combination, holds two electrons, one with each spin.

A shell with principal quantum number n contains n² orbitals and 2n² electrons. For n = 4, l can be 0, 1, 2 or 3, giving 1 + 3 + 5 + 7 = 16 orbitals and 32 electrons.

Worked example 3. How many electrons in an atom can have n = 3 and mₗ = +1? The l values for n = 3 are 0, 1 and 2. mₗ = +1 is possible only when l ≥ 1, so there is one such orbital in 3p and one in 3d: two orbitals, four electrons. How many can have n = 5 and mₛ = +½? Shell 5 has 5² = 25 orbitals, each with one +½ electron, so 25. How many can have n = 4 and l = 2? That is the 4d subshell: 5 orbitals and 10 electrons.

Orbital shapes and counting nodes

A node is a place where ψ, and so the probability of finding the electron, is zero. An orbital has n − 1 nodes in total. Of these, l are angular nodes, planes or cones through the nucleus, and the remaining n − l − 1 are radial nodes, spherical shells at fixed distances. OpenStax Chemistry 2e (Section 6.3) gives the radial-node rule.

s orbitals (l = 0) are spherical with no angular nodes, so a 2s orbital has one radial node and a 3s orbital has two. p orbitals (l = 1) have two lobes on opposite sides of the nucleus with one nodal plane between them; the three are labeled pₓ, p_y and p_z by the axis they point along. d orbitals (l = 2) have two angular nodes. Four of them have four lobes; the fifth, d_z², has two lobes along the z axis and a ring around the middle.

Worked example 4. A 3p orbital has 3 − 1 − 1 = 1 radial node and 1 angular node. A 4d orbital has 1 radial node and 2 angular nodes, 3 in total. A 3d orbital has no radial nodes, and a 6s orbital has 5.

When orbitals in the same shell differ in energy

In hydrogen, the energy depends only on n, so 3s, 3p and 3d have the same energy. In atoms with more than one electron, the other electrons shield the nucleus unevenly. An s electron penetrates closer to the nucleus than a p or d electron of the same shell, feels more of the nuclear charge and is lower in energy. The result is the splitting s < p < d < f within a shell, which sets the filling order covered in the electron configurations guide.

Orbitals in the same subshell, such as the three 2p orbitals, stay equal in energy in an isolated atom. They are called degenerate. Surrounding atoms in a molecule or crystal can break that degeneracy.

Quantum-number mistakes to check for

Nearly every error comes from skipping the listing step. When in doubt, write out every allowed value before counting.

• Letting l equal n. The largest allowed l is n − 1, so there is no 1p or 2d orbital.
• Counting mₗ values from 0 to l instead of from −l to +l, which undercounts orbitals.
• Counting orbitals when the question asks for electrons, or the reverse. Each orbital holds two electrons.
• Giving mₛ the value 0 or 1. Spin is +½ or −½ only.
• Using n − l − 1 as the total number of nodes. That formula counts radial nodes only; the total is n − 1.
• Treating the 90% boundary surface as a hard edge. The electron density fades smoothly and has no sharp boundary.