Molecular Geometry & Bonding
what the shape decides
Orbitals

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.
13 min read 7 figures Say what it enclosesCounted, not quotedOne electron only

Open any two chemistry textbooks at the page with the p orbitals on it. The two pictures will not be the same size relative to the atom, they will not be the same shape — one will be fatter, one more elongated — and both will be captioned the same way.

Neither book is careless. They are drawing different contours of the same function, and neither says which.

One orbital, three contoursThe same orbital drawn at three enclosed fractions, all at one scale. The pictures differ substantially, and a caption that says only "the orbital" does not distinguish them — which is why every figure on this site states its level.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
Fig. 1 One orbital, three contours, all at the same scale. The pictures differ substantially and a caption reading only “the 1s orbital” does not distinguish them. Every level here was solved for rather than chosen.

What the picture is a picture of

A wavefunction is a function of position, and an orbital is one of them. It has a value everywhere, falling off exponentially with distance and never quite reaching zero, so it has no edge and no surface of its own — which is already a departure from the way an orbital is usually first met.

To draw it, somebody has to pick a value and draw the surface where the function takes it. That surface is an isosurface, and it is the only thing an orbital picture ever is.

The choice of value is therefore not an implementation detail. It is the entire content of the drawing’s size and much of its shape, and it is made by the person drawing rather than by the physics. The same is true of every contour on this site, which is why every figure here states its level — and of the surfaces used to argue about overlap, where a badly chosen level would make a bonding interaction look larger or smaller than it is.

The stock caption, and why it is usually not true

The phrase almost every source reaches for is “the ninety per cent probability surface” — the contour inside which the electron will be found nine times in ten.

That is a well-defined object. Finding it requires integrating the density over the region inside a candidate surface, and adjusting the surface until the integral comes to nine tenths. It is not hard, and it is very rarely done.

What is usually done instead is to pick a level that gives a picture of a convenient size, and then to write the stock caption underneath. The result is a drawing that is not wrong about anything in particular and is not right about the thing its caption claims.

Doing the integral

The calculation is worth setting out because it is short.

The probability of finding the electron inside a region is the integral of ∣ψ∣2|\psi|^2 over that region. For a contour at level cc, the region is where ∣ψ∣≥c|\psi| \geq c, so the enclosed fraction is

P(c)=∫∣ψ∣≥c∣ψ∣2 dV∫∣ψ∣2 dV,P(c) = \frac{\int_{|\psi| \geq c} |\psi|^2 \, dV}{\int |\psi|^2 \, dV},

and the denominator is one for a normalised orbital. PP falls monotonically as cc rises — a higher contour encloses less — so solving P(c)=0.9P(c) = 0.9 is a one-dimensional root-find with a guaranteed unique answer.

Choosing a contour for 1sThe fraction of the density enclosed by a contour, against the contour's level. Picking a level is picking a point on this curve, and the usual practice of picking one that looks right is picking a point without knowing which.0%25%50%75%100%50% enclosed|ψ| = 1.48e-190% enclosed|ψ| = 3.94e-299% enclosed|ψ| = 8.44e-3contour level |ψ|fraction of the density enclosedevery level here was solved for, not chosen1s
Fig. 2 The enclosed fraction against the contour level. Choosing a level is choosing a point on this curve, and the curve is steep enough over the useful range that the choice matters a great deal.

Making it exact

For a hydrogenic orbital the integral separates, which makes the whole thing exact rather than approximate — and getting that right turned out to matter.

The wavefunction factors as ψ=R(r) Y(Ω)\psi = R(r)\,Y(\Omega). So along any direction, the region inside the contour is the set of radii where ∣R(r)∣≥c/∣Y(Ω)∣|R(r)| \geq c/|Y(\Omega)|: a one-dimensional threshold problem, solved on a fine radial grid, direction by direction.

The first version of this site’s code did it the obvious way instead — sample a three-dimensional grid, sort by ∣ψ∣|\psi|, accumulate until the target fraction is reached. That puts the contour somewhere inside whichever shell the level happens to fall in, and it placed the 1s ninety-per-cent radius at 2.58 bohr where the closed form gives 2.661. Three per cent out, in the one number the site exists to state precisely.

The separable version agrees with the analytic answer to about one part in two thousand, and every contour on this site is now solved that way.

What the numbers are

Having done the integral, the numbers are worth having, because they are not what most people expect.

For a 1s orbital the fifty-per-cent contour is at 1.337 bohr, the ninety at 2.661, the ninety-five at 3.148 and the ninety-nine at 4.203. So the “size” of a hydrogen atom depends by more than a factor of three on how much of the electron one insists on including.

That spread is the reason the convention matters. A picture drawn at fifty per cent and a picture drawn at ninety-nine are pictures of the same orbital and would be described by a reader as different sizes of atom.

The check the figures make

Every orbital figure on this site asserts what it encloses, and the assertion runs the other way from the construction.

The level is found by bisecting on the enclosed probability until the fraction is the one claimed. The figure then computes the enclosed probability of the level it is about to draw, and requires it to be the claimed fraction to within half a per cent. A picture captioned ninety per cent that encloses eighty-five throws, and the build stops.

That sounds circular and is not quite: the level is found by one route and verified by re-running the integral on the specific value that will be drawn. What it catches is not an error in the bisection but an error anywhere between the bisection and the drawing — a level passed to the wrong orbital, a scale factor applied twice, a caption copied from a neighbouring figure.

The 2pz orbitalThe 2pz 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.2pzencloses 90% of the densitycontour at |ψ| = 9.48e-30 radial nodes1 angular nodecontour solved for by integrating the density1 shell · one electron
Fig. 3 A 2p orbital at the ninety-per-cent contour, with the level printed beside it. The number is not decoration: it is what makes the picture a statement rather than an impression.

The same problem, seen in the radial function

The contour question has a one-dimensional shadow that is easier to see, and looking at it first makes the three-dimensional case obvious.

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 radial part of a 2s wavefunction and its radial distribution. Picking a contour level is picking a horizontal line across the first curve; the region inside the contour is where the curve lies above that line, and for this orbital there are two such regions because the function changes sign in between.

Read that way, three things become plain. A contour is a horizontal cut. Where the function has a node, a single cut produces several disconnected regions. And moving the cut up or down changes the enclosed area a great deal, because the function is steep where the density is.

The three-dimensional case adds only the angular factor, which scales the threshold direction by direction. That is why the surface bulges where ∣Y∣|Y| is large and pinches where it is small, and it is why a p orbital’s contour is two lobes rather than a sphere — the same level cuts the radial function at different radii depending on which way one looks.

What else the level decides

Two things beyond size, and the second is the interesting one.

Shape. A p orbital drawn at a low contour is a fat pair of lobes nearly touching at the nucleus; at a high contour it is two slim spindles well separated. The familiar dumbbell is somewhere in between, and which dumbbell it is depends on the level.

Whether the nodes show at all. A 2s orbital has a radial node, so its ninety-per-cent contour is a shell inside a shell. Almost every published 2s picture draws a single ball, which is what the surface looks like only if the contour is high enough to lose the inner region entirely — and drawing it that way hides the node that is the whole difference between 1s and 2s.

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. 5 The 2s orbital at its ninety-per-cent contour: two nested shells rather than one ball. The inner shell is the density inside the radial node, and it is there in every honest drawing of this surface.

A convention worth adopting

If a single recommendation came out of this site it would be this one, and it costs almost nothing.

State the enclosed fraction in the caption. Not “the 2p orbital” but “the 2p orbital at the contour enclosing ninety per cent of its density”. Six extra words, and they turn a picture into a claim.

Two benefits follow immediately. Pictures become comparable — two sources drawing the same orbital at the same stated fraction are drawing the same surface, and if they differ, one of them is wrong about something. And a reader who wants a different level knows what they are changing from.

There is a third benefit that matters more for teaching. A student who is told the level is a choice has been told something true about the picture, which is that it is a representation with a parameter in it. A student who is shown the picture without the parameter learns that an orbital has a shape and a size, which is not quite true and is very hard to unlearn.

Choosing a contour for 2pzThe fraction of the density enclosed by a contour, against the contour's level. Picking a level is picking a point on this curve, and the usual practice of picking one that looks right is picking a point without knowing which.0%25%50%75%100%60% enclosed|ψ| = 2.61e-290% enclosed|ψ| = 9.48e-395% enclosed|ψ| = 5.99e-3contour level |ψ|fraction of the density enclosedevery level here was solved for, not chosen2pz
Fig. 6 The choice, for a p orbital. Nothing about the physics prefers any point on this curve, and a caption that does not say which point was taken has left out the only free parameter in the drawing.

Where this leaves the pictures

None of this makes conventional orbital pictures wrong. They are contours of the right function, drawn at levels chosen for legibility, and they convey the shape and the symmetry correctly.

What they do not convey is scale, and the caption they usually carry implies a precision they do not have. Two consequences follow for a reader.

A picture cannot be used to compare sizes unless both were drawn at the same level, and almost no source says whether they were. The plate on this site that shows several orbitals together draws them at one enclosed fraction and one scale, so the sizes on the page are the sizes.

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 plate1s0 radial · 0 angular2s1 radial · 0 angular2pz0 radial · 1 angular3dz20 radial · 2 angulareach level solved for separately90% · one electron
Fig. 7 Four orbitals at the same enclosed fraction and the same scale. Drawn this way a 3d orbital is a very much larger object than a 1s, which is true and which panel-by-panel fitting conceals.

And a picture cannot be used to reason about where the electron is. That is a different question with a different answer, and it has its own essay: the most probable radius for a 1s electron is one bohr, and the ninety-per-cent contour is at 2.661.

Why nobody did this

It is worth being fair about the history rather than treating a century of chemists as careless.

Computing the enclosed probability of an arbitrary contour of a three-dimensional function is a numerical integration, and before computers it was genuinely laborious for anything but the spherical cases. For 1s and 2s the region is a sphere or a pair of nested spheres and the integral is analytic; for a p or a d orbital it is not, and the honest answer in 1935 was to draw something of about the right size.

The convention then outlived the constraint, as conventions do. The pictures were copied from book to book, the caption travelled with them, and by the time the integral was trivial nobody was checking it because everybody had seen the picture before.

Where the model stops

Two limits, and the first is the standing one on this site.

These are one-electron orbitals. Every wavefunction here is a hydrogen-like solution. A many-electron atom has no exact orbitals at all — the picture is a basis for an approximation — and the contour of a hydrogenic 2p is not the contour of carbon’s 2p, which is contracted by the nuclear charge and the other electrons.

A contour is not a boundary, and the node structure it may hide is part of what a level choice conceals. The electron is not inside the surface. There is density everywhere, and the ninety-per-cent surface is a statement about a proportion rather than about a region the electron occupies. The remaining tenth extends indefinitely.

Where the ladder goes next

The companion question is where the electron actually is, which the radial distribution answers and the contour does not.

The structural feature the contour choice can hide is the node, counted rather than recalled.

And the standing caution behind all of it is what an orbital is — a one-electron function, and a basis rather than a photograph.

What the pictures here cannot show. Every figure on this page is a surface, and a surface is a level set of a function that has no edge. The density outside the contour is not zero and is not drawn; the picture shows where a chosen proportion is, and cannot show where the rest goes.

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. 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. node 1.90node 7.10most probable radius 13.10 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function2 radial nodes · one electron 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.