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| From: |
<zoellner ^at^ cacao.issecc.fi.cnr.it> |
| Date: |
Fri Jan 13 08:51:38 1995 |
| Subject: |
SUMMARY charged species dipole moments |
To the CCL Subscribers:
On 23 December 1994, I posted some inter-related questions
concerning the calculation of dipole moments of charged species.
Even so near to a holiday season for many people, I received a
number of extremely helpful (some duplication, and I am reminded that
I must brush up on my Physics!) responses, and I have summarized those
responses (after slight editing for length in this message. To all of
you who sent responses, thank you very much!
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My original questions were as follows:
1) Can a species already carrying a charge have a dipole moment
as well?
1a) If so, can such a moment (multipole, perhaps?) be calculated
in any satisfactory way?
1b) If not, why not?
2) When calculations are carried out on charged species, the result
often includes a value for "dipole" moment, yet these values are
unreliable. If the "origin" of the ion is changed, with no other
changes in bonds, angles, or torsions, the value for the "dipole"
can also change. If the origin for charged species was defined
as always being at the center of mass, would the resulting value
for dipole moment have any validity?
3) Can charge be treated as an "additive" property so that a dipole
moment could be calculated for the uncharged species as the
structure determined for the charged species, and then adding
the extra charge after that? (In other words, making the dipole
moment of the charged species the same as the dipole moment for
the hypothetical uncharged species?)
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Response 1/17
From: carlos' at \`extreme.bio.cornell.edu
Carlos Faerman
1) Yes.
1a) It can be calculated but it is dependent upon the origin
of the coordinates that you choose. In other words, when
the net charge of the molecule is different from zero
the dipole moment is no longer invariant upon translations.
2) I don't have a definite answer on this issue, but see above.
3) This is not correct. You have to recalculate the dipole moment
once your molecule becomes charged. It will certainly change.
Think about the following:
Charge (q) times distance (r) has the right units for dipole.
Well, if you sum over i (qi times ri) for both the negative and
positive charges in you molecule (say water) then this will give
you a vector called the dipole moment. If you alter either the
negative charge or the positive charge, then you will
naturally alter the dipole.
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Response 2/17
From: aldert <-at-> rulglj.leidenuniv.nl
Aldert Westra Hoekzema
I think most of your questions are answered by A. D. Buckingham in his
article Quart. Rev. Chem. Soc. (London), 13 (1959), 183-?.
In short, it can be stated that only the lowest nonvanishing molecular
moment is independent of place. So if a molecule carries a charge
(zeroth moment), the dipole (first) moment, or any higher moment, is
place dependent. This does not mean that its value is unreliable.
For instance, if the dipole moment is calculated with respect to (any)
origin and next the molecule is moved, the new dipole vector is simply
the sum of the old one and the product of the molecular charge and the
diplacement vector (in fact this can be viewed as the movement of the
"center of charge").
As to question 3, if I understand it correctly, this would only be the
case if the charge is added to the hypothetical uncharged species in
"its" origin (in a undistributed way!).
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Response 3/17
From: bate()at()xray.chem.ruu.nl
Loes Kroon-Batenburg
1) Yes. The only nasty thing is that the value of the dipole moment
then depends on the choice of the origin. It is nevertheless
well defined and can be calculated, but the origin used has
to be reported.
3) The dipole moment of a molecule is, of course, U = sum (q.r).
In other words, the dipole of the charged molecule is different
from that of the uncharged (and this is what one would measure
experimentally). I have the feeling though, that is is not
exactly what you mean, but that you would like to represent the
charge distribution of a charged or non-charged molecule just
by a monopole plus a dipole moment. In that case your additivity
could be a first approximation.
Quadrupole moments are ALWAYS origin dependent.
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Response 4/17
From: vkitzing $#at#$ sunny.mpimf-heidelberg.mpg.de
Eberhard von Kitzing
For details, please see the book:
J.D. Jackson (1962). Classical Electrodynamics. New York, Wiley.
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Response 5/17
From: jstewart(-(at)-)fujitsu.com
Dr. James Stewart
Yes, the topic of a dipole for an ion is interesting. Purists shrug
and say that it is meaningless. In which case, they cannot protest
when the rest of us come up with a working definition.
I agree with your comment (2) - use the center of mass. In the MOPAC
manual, I write:
"Formally, the dipole moment for an ion is undefined, however it
is convenient to set up a `working definition'. Consider a
heteronuclear diatomic ion in a uniform electric field. The ion
will accelerate. To compensate for this, it is convenient to
consider the ion in an accelerating frame of reference. The ion
will experience a torque which acts about the center of mass,
in a manner similar to that of a polar molecule. This allows us to
define the dipole of an ion as the dipole the system would exhibit
while accelerating in a uniform electric field.
[then a bit of maths, but it's in LaTeX, so it doesn't read easily]
This general expression will work for all discrete species, charged
and uncharged, and is rotation and position invariant."
The best journal reference for this that I've seen is by A. D.
Buckingham, "Molecular Quadrupole Moments", in Quarterly Reviews.
Anyhow, in that article, Buckingham gives a closely reasoned argument,
one that I find quite convincing.
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Response 6/17
From: cmao771 $#at#$ charon.cc.utexas.edu
Isaac B.Bersuker
The fundamental notion in physics and chemistry is CHARGE and CHARGE
DISTRIBUTION (CD), while the expansion of this distribution in dipole,
quadrupole, ..., multipole terms is an APPROXIMATE yet often very
useful presentation. In the case of charged species, the use of
multipole expansion for the description of CD is useless or even
senseless; your questions confirm this statement. Very often,
unfortunately, people (scientists) try to extrapolate APPROXIMATE
notions to regions where they are not applicable, i.e., without checking
the limits of validity of the approximate presentation, and this
inevitably raises many questions. In your case the questions sound
reasonable and stimulating.
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Response 7/17
From: gilson (+ at +) indigo12.carb.nist.gov
Dr. Michael K. Gilson
Your questions are good ones which constantly recur. I believe the
following response is valid. One possibly useful reference source is
Stratton's Electromagnetic Theory, McGraw Hill. However, there are
probably other, less obscure, textbooks which would be easier to
study -- just as correct.
Formal answers to your specific questions:
1) Yes.
1a) yes.
1b) not applicable because of 1a.
2) If the C.O.M. were always used for the origin in computing
the dipole moment, it would have formal validity, but its value
in solving any particular problem would depend on the problem.
3) I am not sure how you would define the uncharged species, so
I can't really answer this one.
Further details:
Consider a charge distribution in space, rho(R), where rho is charge
density, and R represents Cartesian coordinates. The electrostatic
potential due to this charge distribution at a place outside of a
sphere containing it (I think this is the right criterion) can be
expressed as the sum of a series of potentials due to multipoles
placed AT THE ORIGIN OF COORDINATES. You may place the origin
wherever you choose, and this will still be true. On the other hand,
there are useful and useless choices. If the lab is on Earth and the
origin on Mars, you will have to include many terms in the the series
before it converges! Or if you are trying to describe the potential
field generated by a molecule, the origin should probably be inside
the molecule, since the multipole expansion is to be an approximation
to the true charge distribution. (More on this below.)
The first moment, the monopole moment, is just the net charge, and its
value is origin-independent. The second term, the dipole moment, is
integral R rho(R)dR over the distribution, and depends upon the choice
of origin if the monopole moment is nonzero. I don't recall the
formula for quadrupole moment, but I am pretty sure it is
origin-dependent if the dipole moment is non-zero.
So: just where to put the origin in a system of non-zero net charge?
It probably depends upon just what one is after. But I suppose that
if one is computing moments for a molecule, one is implicitly trying
to summarize what the molecule "looks like" to a potential-measuring
device outside of it, as noted above. So one wants the multipole
expansion to be centered inside the molecule; i.e., one wants the
origin to be inside the molecule. Then if one approximates the
potential outside the molecule by the first 2 terms of the multipole
expansion (monopole and dipole), one will have an approximation to the
"true" potential produced by the full charge density. The quality of
this approximation will depend upon whether higher order
(e.g. quadrupole, octapole) moments are substantial.
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Response 8/17
From: laidig #*at*# fitz.mchem.washington.edu
Keith E. Laidig
Your question is a good one! Here's probably more information than
you really want:
The moments of the charge distribution are (as you most likely know
already) the coefficients of a Taylor series expansion of the charges
within a system in powers of the position vector from some origin.
This is just a convenient way to summarize the way in which electronic
charge is distributed relative to the nuclear framework. To date, the
charge, dipole, and quadrupole moments have been measured
experimentally.
Only the first non-zero coefficient of the expansion is independent of
the origin from which the moment is determined. This is a consequence
of the expansion series and is pretty straightforward to work through
(See, for example, the review by A. D. Buckingham, in "Intermolecular
Interactions: From Diatomics to Biopolymers", Pullman, B., ed., Wiley,
1978).
In charged species, the zeroth moment (the charge) is non-zero, so
the first and higher moments are origin dependent. So, you can move
the origin about and get different answers.
But this doesn't mean you can't discuss or investigate origin
dependent moments. The quadrupole moment of a system is an example of
a moment which is origin dependent in many cases and is measured and
discussed relative to the center of mass of the system under study.
Charged species also have higher order moments (not all ions are
perfect spheres), which can, in principle, be measured in the lab. So,
you could discuss the dipole moment of a charged species, but always
with reference to a particular origin (otherwise the discussion
is meaningless).
Typically, I choose to reference the multipole moments to the center
of mass of the molecule or a fragment of a molecule (the charge of
which is rarely zero). This provides a readily determined and
physically significant origin from which any discussion of moments
can be made.
Finally, you ask about building the dipole from charges. Basically,
the dipole moment is the result of the charge distribution of the
molecule. So, I don't see how you could add on the molecular charge
at the end. If you plan to build dipole moments from point charges,
the charges should correctly reflect the distribution of the system,
molecular charge and all.
(Just remember that when you fit first order coefficients -dipoles-
from zeroth order coefficients -charges- you won't end up with a
'robust' representation; i.e., the dipole moment is not just a
separation of atomic charges...)
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Response 9/17
From: soperpd { *at * } nylon.es.dupont.com
Paul D. Soper
A charged molecule can also have a dipole moment (as well as higher
multipole moments). All multipole moments are quantum-mechanical
observables and can be calculated from the locations of the nuclei
and the electronic wavefunction (within the Born-Oppenheimer
approximation).
Only the lowest-order non-zero moment is independent of origin,
which is why you see the "inconsistency" in dipole moment for charged
species but not for neutral ones. Rather than the center of mass
as a consistent origin, I'd suggest placing the origin at the point
which minimizes the absolute value of the dipole moment of the
charged species.
Charge is not additive (in the sense of your question) since the
difference in electron density from the "extra" or "missing"
electron(s) is extremely unlikely to be perfectly isotropic.
A discussion of multipole moments is found in some quantum texts
with a spectroscopic bent (e.g., "Molecular Structure and Dynamics"
by W.H. Flygare, Prentice-Hall (1978), pp 303-306.) However, your
best bet would be to look up multipole expansions of continuous
charge distributions in an intermediate or advanced text on
electromagnetic theory. One such is "Intermediate Electromagnetic
Theory" by W.A. Schwarz, Robert E. Krieger Publishing (1973), p 42-46.
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Response 10/17
From: giesen at.at chemsun.chem.umn.edu
David Giesen
Here are a few comments which may (or may not) help clear up a few of
you questions about dipole moments:
Questions 1, 1a, 1b, and 2:
Yes, a charged species can have a dipole moment and even higher
(quadrapole, octapole....) moments. However, according to Quantum
Mechanics (so I was told by my teacher), only the highest, nonzero
'pole' is independent of the chosen origin. This means that for ions,
only the monopole (overall charge) is independant of the origin.
For most neutrals, the dipole is independent of the origin. For
neutrals that have a zero dipole moment by symmetry, the quadrapole
moment is the highest, nonzero 'pole'. (There are probably molecules
out there that have a zero quadrapole moment, making the octapole
moment independent of origin....)
The dipole moment of a charged species can be calculated by any of the
standard electronic structure codes (at least the ones that I have
used). However, I would be rather wary of the results. First of all,
it is an utter necessity to know the origin chosen by the program.
This is not always obvious and is not even the same in all cases!
Secondly, small changes in geometry caused by such things as a
different basis set, or even the use of a different symmetry group can
shift the chose origin be enough to have a dramatic effect on the
dipole moment.
Question 3:
While simple, IHMO this approach would not yield the correct answer.
The dipole moment is directly dependent on the electron distribution
(or the partial charges depending on your method of calculating the
dipole moment) and therefore, ignoring the extra charge (or lack
thereof) will have quite a large effect on the dipole moment.
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Response 11/17
From: mckelvey.,at,.kodak.com
John M. McKelvey
The leading term in the multipole expansion for a system is Q*Z where
Q is the total charge of the system and Z is a reference coordinate....
The reference coordinate can be arbitrary, and will influnce the answer
if Q.ne.0! So, there is no problem with neutral molecules, but there is
one for charged systems!
I know of no simple solutions for the reference point for charged
systems. My personal choice is the center of gravity of the VALENCE
CORE CHARGES. This lets the resulting dipole moment be referenced to
the valence chemistry, in my opinion, which what a lot of chemists
are the most interested in.
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Response 12/17
From: dum -AatT- biosym.com
David U. Martin
By definition the dipole moment of a charge distribution is
dependent upon the origin when there is a net charge. I don't
think the dipole moment for the center of mass coordinate
system is special in any way, but of course the center of
charge coordinate system is special. In particular, the
dipole moment is zero for the center of charge coordinate
system. For any coordinate system, the dipole moment should
be total charge times center of charge.
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Response 13/17
From: lwhung at.at lcbvax.cchem.berkeley.edu
Li-Wei Hung
There are two types of dipole, permanent dipole and induced dipole.
The first does not vary with the external field while the second does.
Your "charged species" should be permanent dipole.
It is a rather simple problem since, once you know the distribution of
your charges, either continuous or discrete, you may calculate the
electric field or potential in terms of 2^n poles. Then you cut it to
the accuracy you want since the 2^n terms depend on r^-(n+1). One of
the assumptions of Electromagnetism is that it obeys "linear
superposition", i.e., the way you said "additive". Check any
standard Electrodynamics textbook, e.g., J.D. Jackson "Electrodynamics"
and you will find everything you want there.
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Response 14/17
From: underhil at.at hp.rmc.ca
Dr. Ross Underhill
1) The simple answer is yes - consider the OH- species.
1a) I don't have a lot of experience with this but I would have
thought that techniques like CHELPG did a satisfactory job.
The real secret is to get an accurate model of the electrostatic
field and work from it - not start from "atomic charges" about
which there has been a lot of discussion on the list.
2) Part of the problem here is that we should probably talk about
average dipole moments, since my experience is that there is
some charge rearrangement resulting form small changes in
conformation. If the "origin" of the ion is changed, with no other
changes in bonds, angles, or torsions, the value for the "dipole"
can also change. Obviously this shouldn't happen.
"If the origin for charged species was defined as always being at
the center of mass, would the resulting value for dipole moment
have any validity?"
This would probably make things worse. Again I would stress
the need to start with the electric field. Actually, it is that
in some cases, especially with anions that a really good ab initio
calculation requires the use of diffuse orbitals (i.e., orbitals
not centred on atomic sites.)
3) All charge moments are independent of all others. It is
impossible to simulate a monopole using a dipole or any
combination of dipoles. Similarly a single dipole can't simulate
a quadrapole or visa versa. The problem with your suggestion above
is that it assumes the "extra" charge will be distributed uniformly
over the species. This is almost never the case.
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Response 15/17
From: gl;at;mdy.univie.ac.at
Gerald Loeffler
1) Yes, because you are representing the charge distribution of
the species by a multipole-expansion, of which the total net charge
is the first term, the dipole moment is the second term, the
quadrupole moment is the third term, ...
1a) The dipole moment of a set of N charges {q_i},
i = 1...N positioned at the spatial coordinates {r_i},
i = 1...N is always:
N
-----
mu = \ q_i * r_i
/
-----
i = 1
(where the r_i and mu are vectors, of course).
2) This is a special case of the rule, that only the first
non-vanishing term of the multipole-expansion is independent of the
center of the expansion. If you have a species with a net charge,
then this charge is the first term of the expansion, and the dipole
moment and all following terms are dependent of the center of the
expansion (and therefore 'unreliable'). If, on the other hand, your
species is has no net charge, then the dipole moment is the first
non-vanishing term of the multipole-expansion, and therefore
independent of the center of this expansion (like the dipole moment
of water, for example).
3) No (see above).
References:
Have a look at: John David Jackson, Classical Electrodynamics,
Second Edition, John Wiley and Sons, 1975.
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Response 16/17
From: hinsenk -x- at -x- ere.umontreal.ca
Hinsen Konrad
1) Of course. Why shouldn't it?
1a) From what? If you are asking about ab-initio calculations,
I can't help you....
2) Of the total charge of a molecule is non-zero, then its dipole
moment depends on the choice of origin. This is a special case
of the general rule that only the lowest non-zero multipole
moment is independent of the choice of origin. This doesn't make the
multipole moments any less valid, however. If you have the dipole
moment for one origin, you can calculate the dipole moment
for any other origin by adding the total charge times the
displacement vector from the old to the new origin. Which origin is
most useful depends on what you want to use the dipole moments for.
3) I am not sure what exactly you are asking about. Charge is indeed an
additive quantity, but that is not necessarily true for multipole
moments. Adding a charge to a molecule will in general modify the
distribution of the previously present charges, which will in turn
change the multipole moments.
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Response 17/17
From: beroza { *at * } scripps.edu
Paul Beroza
I just got back from vacation. If you have not received answers
to your questions let me know.
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Robert W. Zoellner, Ph.D.; Associate Professor of Chemistry
[currently on sabbatical from Northern Arizona University]
ISSECC-CNR; Via Jacopo Nardi, 39; 50132 Firenze; Italy
Office telephone: (39-55) 24.59.90 FAX: (39-55) 24.78.366
Home address: Via di Bellariva, 9; 50136 Firenze; Italy
Home telephone: (39-55) 67.77.98
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