The lone pair is not the missing term

The dipole is not a sum of bonds starts by refusing the additive model: adding bond dipoles as vectors gets the easy cases right and rests on a quantity — the bond dipole — with several incompatible definitions and no way to measure any of them in isolation.

The standard repair is offered in the same breath as the criticism. The lone pair contributes too. Ammonia’s moment is large because the lone pair’s contribution points the same way as the bonds’; nitrogen trifluoride’s is small because it points the other way.

That repair is testable, and testing it is this essay. A lone pair is one object, and what a lone pair is worth already found that its repulsive weight is not transferable between molecules either. If the correction is worth anything, the amount left over after the bonds are added up should be about the same for two molecules whose lone pair sits on the same atom in nearly the same geometry.

The arithmetic, which is one subtraction #

A pyramidal XY₃ has three equivalent bonds at a measured angle to each other. Elementary geometry turns that into the angle each bond makes with the threefold axis:

$$\cos\alpha = \sqrt{\frac{1 + 2\cos\theta}{3}}$$

so the vector sum of three equal bond moments is $3\mu_{\text{bond}}\cos\alpha$ along the axis and exactly nothing across it, by symmetry. The lone pair sits on the axis too. The two contributions are collinear, and the decomposition is a subtraction on one line.

At the tetrahedral angle that cosine is exactly $\tfrac13$, which is the check that the geometry half is right.

For the bond moment itself the model needs a number, and the one every additive treatment uses is Pauling’s own estimate of ionic character from an electronegativity difference, $1 - e^{-\Delta\chi^2/4}$, times the bond length and the elementary charge. That is a rough estimate and it is the rough estimate the additive model is always made with, which is the point: this section is testing the model as it is used.

The reversal, which the bond sum gets right #

Ammonia: hydrogen is less electronegative than nitrogen, so each bond moment points from the hydrogen towards the nitrogen — up towards the lone pair. Three of them sum to $+0.888$ D.

Nitrogen trifluoride: fluorine is more electronegative than nitrogen, so each bond moment points from the nitrogen down towards the fluorine, away from the lone pair. Three of them sum to $-1.713$ D.

Opposite directions, from one sign change in one electronegativity difference. And the individual bonds of the trifluoride are the more polar of the two — $1.305$ D against $0.786$ D — because fluorine’s electronegativity difference from nitrogen is larger and the bond is longer.

So a molecule whose bonds are half as polar again has a dipole six times smaller. That much the additive model explains completely, and it is a genuinely good explanation: it is qualitative, it follows from two numbers anybody can look up, and it gets the direction of a surprising fact right.

And a measured dipole does not carry a direction #

Before the leftover can be solved for, one thing has to be said that a table of dipole moments does not say.

A measured dipole is a magnitude. Which way a molecule’s moment points is a separate measurement — from a Stark effect in a known field, or from the isotope shifts in a rotational spectrum — and it is not in the number a data book prints. So the subtraction has two possible answers per molecule, and only where one of them would put the leftover on the wrong side of the central atom is the ambiguity resolved.

For ammonia it is resolved: the bond sum is $+0.888$ and the measured magnitude is $1.47$, so the leftover is either $+0.582$ or $-2.358$, and the second would have the lone pair’s contribution pointing away from the lone pair. $0.582$ D.

For nitrogen trifluoride it is not: the bond sum is $-1.713$ and the measured magnitude is $0.235$, so the leftover is either $+1.478$ or $+1.948$, and both point the right way. Between 1.48 and 1.95 D.

The repair fails, and it fails twice #

Take the kindest reading — the smallest possible trifluoride leftover, $1.478$, against ammonia’s $0.582$. That is a factor of $2.5$.

The same lone pair, on the same nitrogen, in molecules whose bond angles differ by four degrees, is required to contribute two and a half times as much in one as in the other. A correction that has to be refitted for every molecule it is applied to is a fitting parameter wearing an explanation’s clothes.

The obvious defence is that the two molecules are not quite the same shape — ammonia’s angle is $106.7°$ and the trifluoride’s $102.3°$ — and a more pyramidal molecule has its lone pair pointing more definitely along the axis. That defence can be sized rather than argued about. The geometrical factor $\cos\alpha$ is $0.376$ for ammonia and $0.437$ for the trifluoride: a difference of sixteen per cent, in the right direction, and an order of magnitude too small to cover a factor of two and a half.

The same is true of every other refinement available at this level. Changing the electronegativity scale moves the bond moments by tens of per cent, not by factors; using bond lengths rather than a formula for the charges moves them less. Nothing in the family of adjustments the additive model admits gets from $0.58$ to $1.48$.

And phosphorus trifluoride is worse. Its bond sum comes out at $-6.14$ D — phosphorus and fluorine differ by $1.79$ on the Pauling scale, which the exponential turns into 55 per cent ionic character over a $1.57$ Å bond — against a measured total of $1.03$ D. The leftover would have to be at least $5.1$ D, which is larger than the dipole moment of any small neutral molecule and enormously larger than anything a lone pair could supply.

So the additive model has already failed before the lone pair is reached. The bond moments themselves are wrong, by a factor of several, for a very polar bond — because a bond that ionic is not a point dipole at the bond midpoint, and because the charge it separates polarises everything else in the molecule.

What is actually being left out #

There are at least three things in the difference, and calling all of them “the lone pair” is what makes the repair unfalsifiable.

The lone pair, genuinely. A hybrid orbital pointing away from the bonds — with the $s$ character the angle does not fix the hybridisation computes from the measured angle — holds two electrons whose centroid is not at the nucleus, and that is a real contribution with a real sign.

Atomic polarisation. The core of the central atom distorts in the field of the bond charges, and its contribution opposes the bond sum. It is not a lone pair and it does not vanish for a molecule without one — which is why methane’s substituted derivatives are also badly described by bond sums, and why symmetry forbids a dipole is the only part of this subject that is exact.

The charges themselves being wrong. Pauling’s exponential is a correlation between ionic character and electronegativity difference fitted in the 1930s, and electronegativity is not one quantity is this site’s account of how much weight that kind of number will bear. For a difference of $1.79$ it is being extrapolated well past where it was fitted.

Only the first of those is what the repair claims to be adding, and the third is large enough to swamp it.

The two hydrides, where the bonds supply nothing #

Phosphine is the control case and it is worth a section of its own.

Phosphorus and hydrogen differ by $0.01$ on the Pauling scale, so the estimated ionic character of a P–H bond is $2.5 \times 10^{-5}$ and the bond sum is zero to three decimal places. Phosphine’s measured dipole is $0.574$ D.

So for this molecule the leftover is the whole moment, and no ambiguity of sign arises: everything the molecule has comes from something the additive model does not contain. That number, $0.574$ D, is as clean a measurement of “what a bond model leaves out for a pyramidal hydride” as this exercise can produce — and it does not agree with ammonia’s $0.582$ by accident so much as by the two molecules being similar in the ways that matter and the bond terms being negligible in one of them.

It also puts a scale on the argument. A lone-pair contribution of about half a debye is what these numbers support. The trifluorides need one and a half to five, which is the sense in which they refuse the repair rather than merely straining it.

The part that does survive #

None of this touches the qualitative argument, and it is worth separating carefully, because the qualitative argument is one of the better things in introductory chemistry.

The sign reversal is real and is computed correctly. Ammonia’s bonds and nitrogen trifluoride’s point in opposite directions along the axis; that follows from two electronegativity differences and one geometrical cosine; and it is the reason the trifluoride’s dipole is anomalously small. Any account of the two molecules that does not have that in it is missing the main fact.

The lone pair does contribute, and in the direction claimed. Every leftover computed here is positive — towards the lone pair — in all four molecules, which is what it would have to be for the story to be true at all.

What fails is the arithmetic: the contributions are not transferable, and the model cannot be turned into a calculation. It is a good account of a direction and a bad account of a magnitude, which is a distinction a ranking is not a difference makes for a different quantity and is the same distinction.

The molecule the arithmetic is about has its group recovered from its coordinates as C₃ᵥ, which permits a dipole along the threefold axis and forbids any component across it. That is the part of the problem symmetry settles, and it settles it without any reference to bonds or lone pairs at all.

What a real calculation does instead #

The additive model fails and something replaces it, and it is worth saying what — because the replacement is not a better set of bond dipoles.

A dipole moment is an integral over the whole charge distribution:

$$\boldsymbol\mu = \sum_A Z_A \mathbf{R}_A - \int \rho(\mathbf r)\, \mathbf r \, d^3r$$

nuclei minus electrons, with no decomposition into bonds anywhere in it. A calculation computes the density, does the integral, and returns three numbers. There are no bond dipoles in the answer and no lone-pair term either.

What the additive model was doing was partitioning a quantity that does not partition. The density is continuous; assigning parts of it to bonds and lone pairs is a convention, and the fact that different conventions give different partial charges for the same wavefunction is the standing statement of that.

So the honest position is that the bond-dipole picture is a mnemonic. It gets the sign reversal between ammonia and nitrogen trifluoride right, which is a real and useful thing to get right, and it gets it right because that reversal follows from the direction of two electronegativity differences rather than from any of the arithmetic. Everything quantitative in it is a fit.

That distinction — between what survives the model being wrong and what does not — is the one the dipole argument has drawn from the start, and it is why the next step has to compute a density rather than improve a decomposition.

Where the model stops #

A point dipole per bond is the whole assumption. A real bond’s charge distribution has a quadrupole and higher moments, and what a dipole cannot tell apart is about how little the lowest moment determines.

The geometry is measured and the charges are estimated. Bond angles and lengths here are experimental; the ionic characters are from a formula. Mixing measurement and estimate is unavoidable for this model and is worth flagging, since the errors quoted above are all in the estimated half.

No induction, no polarisation, no anisotropy. Each bond moment is treated as fixed and independent of the others, which is the definition of an additive model and is where the failure is.

Nothing here computes a dipole from a wavefunction. That would need a many-electron calculation, a separate and much harder job — a Gaussian is the wrong shape is one reason why.

There is one more comparison the four molecules support and it points the same way. Ammonia and phosphine have almost the same leftover — $0.582$ and $0.574$ D — and almost nothing else in common: their bond angles differ by thirteen degrees, their bonds by four tenths of an ångström, and their central atoms are in different rows.

That is either a coincidence or the one piece of transferability in the whole exercise, and there is no way to tell from four molecules. What can be said is that the two cases where the bond sum is small are the two where the leftover agrees, and the two where the bond sum is large are the two where it does not — which is what would happen if the bond-moment estimate, rather than the lone-pair term, were carrying the error.

What a transferable lone-pair term would have to survive #

There is a reason to expect the leftover to vary between these molecules that has nothing to do with the bonds, and it makes the near-agreement between ammonia’s and phosphine’s more awkward rather than less.

A lone pair’s contribution to a moment is a charge times how far its centroid sits from the nucleus, and that distance depends on the orbital’s shape. A lone pair in a pure s orbital is spherically symmetric about the nucleus and contributes exactly nothing; one in a pure p orbital has its density displaced to one side and contributes the most. Anything between contributes in proportion to how much p character it has.

The p character is not free to be anything: it is fixed by the bond angle, through the same arithmetic that relates hybridisation to geometry. A central atom with wide bond angles puts more s into its bonds and therefore more p into its lone pair; one with angles near ninety degrees uses nearly pure p orbitals for its bonds and leaves the lone pair nearly pure s.

Ammonia’s angle is 107° and phosphine’s is 93°. So phosphine’s lone pair should be much closer to pure s than ammonia’s, and its contribution to the moment should be substantially smaller — not slightly, but by most of it, since a pure s pair contributes zero.

The leftovers are 0.582 and 0.574 debye.

Two readings are available and neither helps the repair. Either the lone-pair term is not what the leftover is made of, in which case the agreement is a coincidence between two quantities that are not the same quantity; or it is, and it fails to behave in the one way it can be predicted to behave.

The second is the more damaging, because it means the term cannot be fixed by parameterising it better. A quantity that ought to vary by a factor of several between two molecules, and does not, is not measuring the thing its name says.

What testing the repair adds #

The additive model can be refused for a stated reason: the bond dipole is not a measurable quantity. A molecule’s group can change three times along one soft coordinate, and the lowest moment determines very little of a charge distribution.

This essay tests the standard repair to the additive model. Adding a lone-pair term is not a repair, because the term is not transferable between two molecules sharing a lone pair, and because for a very polar case it would have to be five debye. What the exercise leaves standing is the qualitative half — the sign reversal between ammonia and its trifluoride, which follows from two electronegativity differences and a cosine, and which the additive model gets exactly right.

The open question is the quantity nobody in this argument has computed: what a real charge distribution’s dipole is, which needs a density rather than a set of bond charges, and which is the first place the dipole argument needs a calculation rather than a construction.