Radial Probability Distributions and Orbital Nodes

The radial probability distribution, 4πr²R(r)², gives the probability of finding an electron at a distance r from the nucleus, and orbital nodes are the places where the wave function is zero: an orbital has n − l − 1 radial nodes (spherical surfaces) and l angular nodes (planes or cones), a total of n − 1. In brief: the 1s electron of hydrogen is most likely to be found at r = a₀ = 52.9 pm, the 2s orbital has one radial node at r = 2a₀ (106 pm), and the 3s orbital has two nodes.

The wave function of an electron has a value at every point in space, but chemists need to know how far from the nucleus an electron is typically found, and where it is never found. These questions explain the sizes of atoms, why s electrons feel the nucleus more strongly than p electrons and how orbitals are drawn. This article defines the radial distribution function, gives a table of nodes and typical radii, explains penetration and shielding and works through calculations for hydrogen.

What is the difference between density and radial distribution?

The probability density |ψ|² is the probability per unit volume at a point. For an s orbital, it is largest at the nucleus (r = 0) and decreases with distance. But there are very few points close to the nucleus, and many points further out: the volume of a thin spherical shell at distance r is 4πr²dr, which grows as r². The probability of finding the electron in a shell of thickness dr at a distance r is therefore:

P(r)dr = 4πr2 R(r)2 dr

where R(r) is the radial part of the wave function. The function 4πr²R² is the radial probability distribution. It is zero at the nucleus (because of the factor r²), rises to a maximum, and falls off. For the 1s orbital of hydrogen the maximum is at the Bohr radius, a₀ = 52.9 pm. The total probability of finding the electron somewhere is 1, which is the area under the curve.

What are nodes?

A node is a region where the wave function changes sign and has zero value, so the probability of finding the electron there is zero.

• Radial nodes are spherical surfaces at certain distances from the nucleus. The number is n − l − 1.
• Angular nodes are planes (or cones) through the nucleus. The number is l. A p orbital has one nodal plane, which passes through the nucleus, a d orbital has two nodal surfaces.
• The total number of nodes is n − 1.

The wave function has opposite signs (phases) on either side of a node. This is the reason for the plus and minus signs, or shading in two colors, used in orbital diagrams: they give the sign of ψ and are important for how orbitals overlap in bond formation.

Nodes of common orbitals

At what distance are electrons found?

For orbitals of the same n, the radial distribution of an s orbital has several maxima, and the outermost one is the highest. An s orbital thus has a small probability close to the nucleus (the inner peaks) in addition to its main maximum far away. Orbitals with the same n but higher l have a single main maximum that lies somewhat closer to the nucleus. The average distance of the electron in a hydrogen-like atom is given by ⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)].

Radii of hydrogen orbitals

What are the shapes of orbitals?

• s orbitals are spherical, without angular nodes.
• p orbitals have two lobes on opposite sides of the nucleus (“dumbbell”), separated by a nodal plane through the nucleus, with opposite signs of ψ in the two lobes. The three p orbitals point along the x, y and z axes.
• d orbitals: four of the five d orbitals have four lobes (“cloverleaf”) with two perpendicular nodal planes; the fifth (d_z²) has two lobes along the z axis and a ring in the xy plane.
• f orbitals have more complex shapes, with three angular nodes.

The boundary surface of an orbital is drawn so that it encloses a large fraction of the probability (usually 90 %). The orbital does not have a sharp edge: the probability falls off gradually.

What is penetration?

Because s orbitals have a nonzero probability near the nucleus, an electron in an s orbital “penetrates” through the inner electrons and feels more of the nuclear charge than a p electron in the same shell, which has zero probability at the nucleus. Penetration decreases in the order s > p > d > f for the same n. The energy of the orbitals in a many-electron atom thus rises in the same order: E(ns) < E(np) < E(nd) < E(nf). This is the reason for the order of filling in which 4s fills before 3d, and the reason for the lifting of the degeneracy of the subshells in atoms with more than one electron.

Worked examples

Example 1: number of nodes

How many radial and angular nodes does a 4p orbital have? n = 4, l = 1. Radial nodes = 4 − 1 − 1 = 2, angular nodes = 1, a total of 3 = n − 1.

Example 2: the radial node of 2s

The radial function of the 2s orbital of hydrogen is proportional to (2 − r/a₀) e^−r/2a₀. It is zero when r = 2a₀ = 2 × 52.9 = 106 pm. The wave function is positive inside this sphere and negative outside.

Example 3: nodes of 3s

The radial function of the 3s orbital is proportional to (27 − 18ρ + 2ρ²) e^−ρ/3, with ρ = r/a₀. Setting the polynomial to zero: 2ρ² − 18ρ + 27 = 0, so ρ = [18 ± √(324 − 216)] / 4 = (18 ± 10.39) / 4, which gives ρ = 1.90 and 7.10. The nodes are at r = 1.90 a₀ = 100 pm and r = 7.10 a₀ = 376 pm.

Example 4: average radius

For the 2p orbital of hydrogen, ⟨r⟩ = (a₀/2)[3 × 4 − 1 × 2] = (a₀/2)(10) = 5.0 a₀ = 265 pm. For a hydrogen-like ion with Z = 2 (He⁺), the radius is divided by Z: 1s ⟨r⟩ = 1.5 a₀ / 2 = 0.75 a₀ = 40 pm.

Example 5: comparing orbitals

Which has a higher probability of being found close to the nucleus, the 2s or the 2p electron? The 2s, since the 2p wave function is zero at the nucleus. In spite of this, the average distance of the 2s electron (6.0 a₀) is larger than that of the 2p (5.0 a₀) in hydrogen, because the outer maximum of 2s lies further out. In many-electron atoms, the penetration still makes the 2s orbital lower in energy than the 2p.

Which mistakes are common?

• Confusing density and radial distribution: the density is highest at the nucleus, but the radial distribution is zero there.
• Forgetting that p orbitals have a nodal plane through the nucleus: the probability of finding the electron at the nucleus is zero.
• Counting nodes wrongly: radial nodes are n − l − 1, and not n − 1.
• Thinking that the electron moves between the lobes through the node: the node is not a barrier or a path.
• Treating the orbital boundary as a hard wall: it only encloses most of the probability.