Say what it encloses

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.

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 drawn, 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 $|\psi|^2$ over that region. For a contour at level $c$, the region is where $|\psi| \geq c$, so the enclosed fraction is

$$P(c) = \frac{\int_{|\psi| \geq c} |\psi|^2 \, dV}{\int |\psi|^2 \, dV},$$

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

Making it exact #

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

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

The obvious way is different — sample a three-dimensional grid, sort by $|\psi|$, accumulate until the target fraction is reached. That puts the contour somewhere inside whichever shell the level happens to fall in, and on a practical grid it places 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 a stated fraction exists to fix precisely.

The separable version agrees with the analytic answer to about one part in two thousand, and every contour here is 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 here is checked for what it encloses, and the check 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 is refused rather than drawn.

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 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.

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 $|Y|$ 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.

A convention worth adopting #

If a single recommendation came out of this subject 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.

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 here that shows several orbitals together draws them at one enclosed fraction and one scale, so the sizes on the page are the sizes.

Four orbitals drawn at one stated fraction and one scale is the comparison this makes possible: the sizes on the page are the sizes, because every surface encloses the same amount of its own density. Without the bisection there is no such comparison to be made — the pictures would be four contours at four unrelated values.

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.

What it costs now #

Little enough that the excuse has expired, and the itemised bill is worth having because it is the argument for adopting the convention.

Finding the level costs about forty integrals. The enclosed fraction falls monotonically as the level rises, so solving $P(c) = 0.9$ is a bisection with a guaranteed unique answer, and forty steps take it well past the precision anybody could draw. Each step is one evaluation of the enclosed probability.

Each evaluation costs one pass over a radial profile. This is where the separability pays. Rather than integrating over a volume, the density is accumulated once as a function of radius and once as a function of direction, and the enclosed probability at any level is then read off those profiles. The profiles are computed once per orbital and cached, so the fortieth bisection step costs no more than the first.

Verifying the answer costs one more evaluation, and it is the one that matters. The level found by bisection is fed back through the integral, and the figure requires the result to be the fraction its caption claims to within half a per cent. That check is made for every figure.

The whole of it is a few milliseconds per orbital, done once. Against that, the cost of not doing it is a picture whose size carries no information, which is what a century of textbooks have printed.

There is one more line on the invoice, and it is the largest. The check as described sounds circular — the level is found by an integral and then verified by the same integral — and it is not quite, because the two runs are separated by everything in between. What it catches is not an error in the bisection but an error anywhere on the path from bisection to drawing: a level handed to the wrong orbital, a scale factor applied twice, a caption copied from a neighbouring figure. Those are the failures that actually happen, and each of them produces a picture that looks entirely correct.

Which is why the check is also fed something wrong on purpose: a contour captioned with a fraction it does not enclose, and a contour drawn at three times the level its caption implies, both of which must be refused. A check that has never rejected anything would pass unnoticed for ever and prove nothing.

Ninety is arbitrary, and it is not equally arbitrary #

The convention proposed here has two parts, and only one of them is doing work. State the fraction is the part that matters, and any fraction stated is better than none. Ninety is a choice, and it is worth asking whether it is a good one, because the answer is a number rather than a preference.

The test is how much the picture moves when the fraction does. Take hydrogen’s 1s, whose enclosed fraction has a closed form, and ask what a one-percentage-point uncertainty in the convention does to the radius drawn:

The last row is not a typographical accident. At ninety-nine per cent, one point in either direction takes the radius from 3.76 bohr to 22.18, because the upper end is approaching a fraction of one and the tail of an exponential has no outer edge to reach.

So the fractions are not interchangeable even as conventions. A picture drawn at a high fraction is a picture whose size is decided mostly by the convention, and the closer the fraction is to one the more of the drawing is an artefact of the choice rather than a property of the orbital. A picture drawn at a low fraction is stable and shows too little.

Ninety sits where the curve is still flat — five and a half per cent of movement for a point of ambiguity, against the low-fraction floor of three — and where most of the density is inside the surface. That is a defensible compromise rather than a good number, and the useful part is that its defensibility is checkable: anybody preferring a different fraction can run the same column and see what it costs.

It also explains an asymmetry in how the convention should be read. Two pictures at ninety per cent are comparable to within a few per cent even if one author’s integration was sloppy. Two at ninety-nine are not comparable at all.

Where the model stops #

Two limits, and the first applies to every orbital picture here.

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 to read on #

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.

One last remark about what the convention costs the reader rather than the author. Nothing. A caption stating an enclosed fraction is six words longer and gives back a number the reader can compare against any other picture drawn the same way — which is the whole difference between a library of drawings and a library of measurements.

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.