Molecular Geometry & Bonding
what the shape decides
Orbitals

Nodes

An orbital with quantum numbers n and l has exactly n−l−1 radial nodes and l angular ones. That is a count, it is exact, and it is the fastest way to catch a drawing that is wrong.
11 min read 6 figures Counted, not quotedSay what it encloses

An orbital is a function, and a node is a surface on which it is exactly zero — and stays zero, rather than passing through zero on the way somewhere. The number of them is fixed by the quantum numbers and by nothing else.

The radial function of 3sThe radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.node 1.90node 7.10most probable radius 13.10 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function2 radial nodes · one electron
Fig. 1 The radial function of a 3s orbital, which crosses zero twice. Those crossings are its two radial nodes, and their positions were found by walking the computed function rather than looked up.

An orbital with quantum numbers nn and ll has exactly n−l−1n - l - 1 radial nodes and exactly ll angular ones. Both counts are exact, both are computable, and together they are the sharpest check available on whether an orbital has been drawn correctly.

Two kinds

A radial node is a sphere. The wavefunction changes sign as the radius crosses it, so the orbital has an inner region of one sign and an outer region of the other. A 2s orbital has one; a 3s has two.

An angular node is a surface through the nucleus — a plane, or a pair of cones. The wavefunction changes sign as a direction crosses it. A p orbital has one, which is the plane between its lobes; a d orbital has two.

The two kinds arise from the two factors of the wavefunction. The radial part R(r)R(r) contributes the spheres and the angular part Y(Ω)Y(\Omega) contributes the surfaces through the origin, and because the function is a product, it vanishes wherever either factor does.

Why the counts are what they are

The radial function for given nn and ll is an exponential times a polynomial in rr of degree n−l−1n - l - 1, and a polynomial of that degree has that many positive roots here. The angular function for given ll is a spherical harmonic, whose nodal structure is ll surfaces.

Adding them gives n−1n - 1 nodes in total for every orbital of principal quantum number nn, however the total is split between the two kinds. A 3s has two radial and none angular; a 3p has one and one; a 3d has none and two. All three have two.

Orbitals at the 90 per cent contourSeveral orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density.one scale across the plate3s2 radial · 0 angular3pz1 radial · 1 angular3dz20 radial · 2 angulareach level solved for separately90% · one electron
Fig. 2 Three orbitals with the same principal quantum number and the same total number of nodes, distributed differently between radial and angular. The 3s surface is three nested shells, the 3p is two shells of lobes, and the 3d is a single set.

That is a genuinely useful pattern. It says the total node count is a property of the shell and the split is a property of the shape, and it explains why higher-ll orbitals of the same shell are the ones that look simple.

What a node is not

Three misreadings, and the third is the one that produces real confusion.

Not a place the electron avoids. The density is zero on the node, which is a set of measure zero — a surface has no volume, so the probability of finding the electron exactly there is zero for the same reason it is zero for any exact position.

Not a barrier. Nothing prevents the electron from being on either side. The question “how does it get from one lobe to the other” presupposes a trajectory, and there is none.

Not the same as a small value. A node is exactly zero, and the sign changes across it. A wavefunction that becomes very small without changing sign has no node there, and the distinction is what makes the count exact rather than approximate.

Counting them, and the bug that found

The counts are checked on this site by walking the computed function, and doing it properly was harder than it looks.

Radial nodes are easy: step outward and count sign changes, ignoring the far tail where the function has decayed into arithmetic noise and a spurious flip means nothing.

Angular nodes are not. The first version walked one great circle across the sphere and counted sign changes, which works for a p orbital and fails badly for others. A circle chosen in the plane containing both nodal planes of a dxyd_{xy} orbital touches them only tangentially — the function is proportional to cos⁡2t\cos^2 t along it and never goes negative — so the count came out zero for an orbital with two.

The fix is to walk many randomly oriented circles and take the largest count, since a generic circle cuts each nodal surface exactly twice. That version gets every orbital right.

It is worth dwelling on what that error looked like. The drawing was unchanged, the shape was correct, and the reported node count was wrong by two. Nothing about the picture would have revealed it, which is exactly the situation the checks on this site exist for.

What the node does to the picture

A radial node has a consequence for the drawn surface that almost every published picture leaves out.

If the contour level is low enough, the isosurface has a component inside the node and another outside it: a shell inside a shell. That is the correct picture of a 2s orbital at any sensible level, and the single ball usually drawn is what the surface looks like only when the contour is high enough to have lost the inner region entirely.

The 2s orbitalThe 2s orbital at the contour enclosing 90 per cent of its density — a level solved for by integration rather than chosen. The two colours are the two signs of the wavefunction, which is what distinguishes a bonding interaction from an antibonding one.2sencloses 90% of the densitycontour at |ψ| = 7.40e-31 radial node0 angular nodescontour solved for by integrating the density2 shells · one electron
Fig. 3 The 2s orbital at its ninety-per-cent contour: two nested shells. The inner one is the density inside the radial node, and it is a real part of the surface rather than an artefact.

The site checks this directly: the number of shells in the drawn surface must be one more than the number of radial nodes. A 2s surface claimed to be a single shell throws.

Nodes and energy

There is a pattern worth knowing that connects nodes to something measurable.

More nodes means higher energy. A function with more sign changes has more curvature, curvature is kinetic energy, and so within a given atom the orbitals sort by total node count. For hydrogen the energy depends on nn alone, which is the same statement, since the total node count is n−1n-1.

For a many-electron atom the degeneracy breaks and the s orbital of a shell lies below the p, which lies below the d. The usual explanation is penetration — the s orbital has density close to the nucleus where the shielding by inner electrons is incomplete — and the radial distribution is where that becomes visible.

The radial function of 2sThe radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.node 2.00most probable radius 5.24 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function1 radial node · one electron
Fig. 4 The 2s radial distribution, showing the small inner peak inside the node. That peak is the penetration: density close to the nucleus, where the other electrons screen it least, and it is why 2s lies below 2p in every atom but hydrogen.

Nodes in a molecule

The idea transfers directly and it is where it earns its keep.

When two atomic orbitals combine, the antibonding combination has a node between the nuclei that neither parent had. That node is the whole reason the combination is higher in energy: density has been expelled from the region between the nuclei, where it would have been attracted to both.

So counting nodes in a molecular orbital is counting antibonding character, and the ordering of a molecular orbital diagram is very largely a node count. Overlap decides which combinations form; the nodes decide which of them is which.

What the count cannot tell

Two limits.

It does not fix the positions. The count is exact; where the radial nodes sit depends on ZZ and on the details of the radial function, and for a many-electron atom on the approximation used.

It is a property of the one-electron function, and the same caution applies to every quantity here. In a many-electron atom there are no exact orbitals, so the node count is a property of whichever approximate orbitals have been chosen. It happens to be robust — every reasonable approximation gives the same counts — but it is not a measured quantity.

Where the idea came from

Nodes are older than quantum mechanics by a long way. They belong to the physics of standing waves, and Chladni was scattering sand on vibrating plates to make them visible in the 1780s.

That lineage is not decorative. The hydrogen wavefunctions are the standing waves of a particular three-dimensional problem, the spherical harmonics were developed by Laplace and Legendre for gravitation, and the mathematics arrived a century before the application. When Schrödinger solved the hydrogen atom in 1926 the functions were already known objects; what was new was what they were for.

The node count also carries a historical irony. Spectroscopists had labelled series of lines sharp, principal, diffuse and fundamental decades before anybody knew what distinguished them — and those labels became s, p, d, f, which is to say the letters now encoding the angular node count were chosen to describe the appearance of spectral lines.

The whole set, counted

A summary table is worth having, because the counts are simple and the pattern is not obvious until it is written out.

Orbital n l Radial Angular Total
1s 1 0 0 0 0
2s 2 0 1 0 1
2p 2 1 0 1 1
3s 3 0 2 0 2
3p 3 1 1 1 2
3d 3 2 0 2 2

Every entry was produced by walking the computed function rather than copied, and the totals are what make the pattern visible: every orbital of principal quantum number n has n − 1 nodes, however they are split.

Orbitals at the 90 per cent contourSeveral orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density.one scale across the plate3s2 radial · 0 angular3pz1 radial · 1 angular3dz20 radial · 2 angulareach level solved for separately90% · one electron
Fig. 5 The third shell, all three at one enclosed fraction and one scale. Three nested shells, two shells of lobes, and one set of lobes — the same total node count arranged three different ways.
The 3s orbitalThe 3s orbital at the contour enclosing 90 per cent of its density — a level solved for by integration rather than chosen. The two colours are the two signs of the wavefunction, which is what distinguishes a bonding interaction from an antibonding one.3sencloses 90% of the densitycontour at |ψ| = 2.64e-32 radial nodes0 angular nodescontour solved for by integrating the density3 shells · one electron
Fig. 6 The 3s surface at ninety per cent: three concentric shells, because two radial nodes cut the density into three regions. Drawing this as a single sphere would lose both nodes and most of what distinguishes 3s from 1s.

That total is the same as the number of nodes of a one-dimensional oscillator’s nth state, and it is not a coincidence: both are counting the sign changes of the nth solution of a Sturm–Liouville problem, and the theorem that the nth eigenfunction has n − 1 nodes is general. The hydrogen atom is a standing-wave problem, and its node counting is the standing-wave node counting.

Where the ladder goes next

The consequence for the drawn surface is what the contour encloses, where the level decides whether the inner shell of a 2s appears at all.

The related question the node count does not answer is where the electron actually is.

And the place nodes do physical work is overlap, where a node between two nuclei is what makes a combination antibonding.

What the pictures here cannot show. A node is a surface of exactly zero, and a drawing shows a contour of some non-zero level — so the node appears as a gap between shells rather than as itself. The claim that the function is exactly zero there is arithmetic, not something visible in any figure.

Where this goes next

2pzencloses 90% of the densitycontour at |ψ| = 9.48e-30 radial nodes1 angular nodecontour solved for by integrating the density1 shell · one electron What an orbital is Not a region the electron occupies, not a path it follows, and for any atom but hydrogen not an exact anything. An orbital is a one-electron wavefunction, and almost every difficulty in this subject comes from forgetting that. 50% of the density|ψ| = 1.48e-190% of the density|ψ| = 3.94e-299% of the density|ψ| = 8.44e-31slevels solved for by integration, drawn at one scale Say what it encloses An orbital picture is a contour at a level somebody chose, and almost no source says which. Two textbooks can draw the same orbital at visibly different sizes with the same caption, and both be printed in good faith. most probable radius 1.00 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function0 radial nodes · one electron Where the electron is The wavefunction is largest at the nucleus, the electron is most likely to be found a bohr out, and the ninety-per-cent contour is at 2.66. Three numbers, all correct, all answering different questions.