## Willock pp. 1 – 50

These are some very detailed comments, and (hopefully) helpful hints for you as you read  “Moleular Symmetry” by David J. Willock.  For why I’m reading it,  why you should too, and what a group actually is see the first post in this series — https://luysii.wordpress.com/2011/12/11/chemistry-helps-you-understand-group-theory-and-not-vice-versa/

p. 5 — Distinguish in carefully in your mind the difference between a symmetry element and a symmetry operation as you read the book.

A symmetry element  is a geometric structure (point, line, plane) about which a molecule is symmetric.  On p. 23 This definition is clarified — a symmetry element is the set of points which aren’t moved when a symmetry operation takes place.  Points of what? The points of the space in which the molecule is embedded.

A symmetry operation is an action carried out using a symmetry element which leaves the shape of the molecule unchanged (although no atom of the molecule may wind up where it was after the operation).  Ammonia has a 3 fold axis of rotation (e.g. the symmetry element is a line), but 3 distinct symmetry operations about the symmetry element (rotation by 120, 240 and 360 == 0 degrees).

p.  7 — Principal axis — the line of symmetry (axis) with the largest (highest) order of rotation.  Always aligned with the Z axis (vertical).  One of the many important conventions you’ll need to remember is that the highest (largest) symmetry axis defines  the vertical direction.  You’ll need to know what vertical is to appreciate the labelling of the symmetry planes of water on p. 10, and how to distinguish it from horizontal in the case of BF3 (p. 11).

p. 7. — Also very important to keep straight — after a symmetry operation the atoms of a molecule may move, but the axes do not move — why is this important?  — because it’s very easy to get confused about how to do the a second symmetry operation after you’ve done the first. Remembering this will save you later on (see figure 2.4 p. 30).

p. 9 — Another convention — the Greek letter sigma is always used for a plane of symmetry.  If you’re reading this on your own — start yourself a symbol table with the page the symbol is first defined (not the same as when Willock first mentions the term, definitions sometimes come later — this takes some getting used to if you’re used to math books where everything is defined before being used).  Molecular Symmetry should have a symbol table  (newer math books do) but it doesn’t.

p. 12 — The description of when a plane is dihedral and when it is horizontal is ambiguous at best, and confusing at worst.  You’ll need the actual book to follow the following.  In figure 1.15 there are horizontal C2 axes passing through the C-C bond center, but they are not in the mirror planes  shown.  However, there is a third mirror plane which isn’t shown.  When this axis is put in one sees that the dihedral mirror plane splits all 3 C2 rotation axes (which is why it’s called a dihedral plane).   Then another example is given of a dihedral plane which splits the angle with two other planes (called v — for vertical I guess, because the plane contains the vertical (Z) axis).  A horizontal plane is perpendicular to the principal axis.  This could have been made more explicit, so a clear understanding of sigma(v), sigma(h) and sigma(d) emerges.  On p. 23 Willock does make all this explicit.  It was probably better to do this at the outset.

p. 21 “The absorption in an NMR experiment takes a short, but finite time”.  It is truly maddening to find out how long this time is using Google or any of my NMR books.   I’ve speculated about the time it takes for absorption of a wavelength of light in an earlier post.  Here it is again — if any physicist types are reading this, please correct me if I’m wrong, or enlighten me further.

***** The penultimate chapter of Anslyn and Dougherty “Modern Physcal Organic Chemistry” contains an excellent discussion of photochemistry, with lots of physics clearly explained but it leaves one question unanswered which has always puzzled me.  How long does it take for a photon of a given wavelength to be absorbed?
On p. 811 there is an excellent discussion of the way the quantum mechanical operator for kinetic energy (-hBar/2m * del^2) is related to kinetic energy.  The more the wavefunction changes in space, the higher the energy.  It’s like cracking a whip, the faster you move the handle up and down (e.g. the faster the frequency), the more energy you impart to the whip.   Note that the kinetic energy operator applies to particles (like protons, neutrons, electrons) with mass.

Nonetheless, in a meatball sort of way, apply this to the (massless) photon.  Consider light from the classical point of view, as magnetic and electrical fields which fluctuate in time and space.  The fields of course exert force on charged particles, and one can imagine photons exerting forces on the electrons around a nucleus, changing their momentum, hence doing work on them.  Since energy is conserved (even in quantum mechanics), it’s easy to see how the electrons get a higher energy as a result.  The faster the fields fluctuate, the more energy they impart to the electrons.

Now consider a photon going past an atom, and being absorbed by it.  It seems that a full cycle of field fluctuation (e.g. one wavelength) must pass the atom.  So here’s a back of the envelope calculation, which seems to work out.  Figure an atomic diameter around 1 Angstrom (10^-10 meters) for the average atom.  The chapter is about photochemistry, which is absorption of light energetic enough to change electronic energy levels in an atom or a molecule.  All the colored things we see, are colored because changes between their electronic energy levels are absorb photons of visible light — the colors actually result from the photons NOT absorbed.  So choose light of 6000 Angstroms — which has a wavelength of 6 * 10^-7 meters.  It will appear orange to you.

In one second, light moves 3 * 10^8 meters, regardless of how many many wavelengths it contains. If the wavelength were 1 meter it would move past a point in 1/(3 * 10^8) seconds But wavelength of the visible  light  I chose is 6 * 10 ^-7 meters, so the wavelength moves past in 6*10^-7/3 * 10^8 = 2 x 10^-15 seconds, which (I think) is how long it takes visible light to be absorbed.  Have I made a mistake?  Are there any cognoscenti out there to tell me different?

p. 26 — Note that sigma(v’) is in the plane of benzene and water here, while in the case of BF3 (p. 11) the mirror plane in the plane of BF3 is called sigma(h) — this is because the latter is perpendicular to the principal axis of rotation (p. 7), while the former two are parallel to it.   It’s best to have these things clear in your mind when reading further.

p. 26 — Note which of the two operations in C2sigma(v’) is done first — it is always the rightmost — this is standard in the mathematical literature, and probably comes from the way functions within functions are applied — sin(3x) means that you do 3x first and then apply sine to it.  Yet another convention to keep straight and crucial in what follows.

p. 26 — Point group — all the symmetry operations a molecule can undergo. None of the amino acids making up our proteins (except glycine) have more than 1 (the identity operation).

p. 27 — Note the convention for the X, Y and Z axes — Z points in the direction of the principal axis.  Y in the plane of the molecule (assuming there is one) and X perpendicular to Y and Z.  None of this tells you which is positive and negative on each axis — but (without saying so) the X,Y and Z axes are a right handed coordinate system (remember the old right hand rule — use the fingers of your right hand to coil the positive X into the positive Y, and your thumb will point in the positive Z direction.

p. 27 –  I think the bottom diagram of sigma(v’) is incorrect H1 and H2 aren’t changed by it.

p. 27 — The following is also very important to note (and not especially clear),  Distinguish the global coordinate axes (capital X, Y, Z) which aren’t changed by the symmetry operation and the vectors attached to atoms (small x, y, z) which   are (italics) changed by the symmetry operations.  So capital X, Y, Z means global coordinate axis, while small x, y, z means vector.  Easy to confuse the two.

p. 28 — Another important convention — when faced with a table for the multiplication of symmetry operations, do the operation in the row of operations at the top of the table first and the operation in the column of operations on the left side of the table second.

Remember all this advice and you’ll do a lot less looking back.

p. 31 — the Sn operation.  Note that in the example given (staggered ethane), that neither S6 nor C2 in the plane horizontal to it is an actual symmetry element of the molecule.

p. 36 — The term basis is mentioned but not really defined.  It appears to be an arrow attached to an atom which follows the atom through various symmetry operations.  Apparently not the same as vectors attached to the oxygen of water (p. 27) which are labeled x, y and z.

p. 37 — Perhaps the improper rotation on BF3 which requires applications (with a total of 720 degrees) to get things back to where they started, is relative to some rotations in particle space that I recall reading about, that took 720 degrees of rotation to get things back together.

p. 38 — Requiring that an  inverse of a symmetry operation exist is a crucial property of groups, but here it is introduced by fiat without saying why.  For what a group is see the first post in the series.

End of chapter 2 — The fact that the set of symmetry operations of a molecule must be closed, along with the fact that there are only finitely many of them, means that for any symmetry operation (S), applying it over and over eventually gets you back to the original operation.  This means there is some n in the non-negative integers such that for every symmetry operation S, S^n = S (S^n is a Mathematica convention meaning S to the nth power — multiplying S by itself n times — multiplication of symmetry operations just means applying one after the other — rightmost first .  A bit of thought then shows that S^(n-1) = E (the identity).  A bit more thought shows that the powers of S form a group by themselves (they contain 1, and have an inverse).

This means that they are a subgroup of group of all symmetry operations.  Subgroup isn’t in the index, as this is a chemistry book, not a math book.  It’s simple enough concept — a subset of a group of symmetry operations which is also a group (e.g. it contains E, the identity, every element has an inverse and a more obscure characteristic called associativity — which it inherits from the parent group).

This is true for each symmetry operation.  What molecular geometry allows you to do, which isn’t obvious when group theory is studied abstractly, is to see how the powers of each element combine with each other. E.g. For water what is C2*sigma(v)?  The symmetries of water are an example of a very peculiar group called the viergruppe (German — four group).   Nice ! ! ! !

p. 45 — Introduces the character table, and notes that not all the symbols in it have been defined yet.  The term character also isn’t defined at this point.  It’s an interesting teaching technique, and very different from those used in math books.

p. 46 — Pasteur was lucky to find tartaric acid which crystalizes in two form each containing a pure enantiomer.   How often does this happen?

p. 48 — Watch out when reading about point groups. Lots of molecules contain a 2 fold rotation axis (e.g. C2), but the C2 point group contains only that symmetry operation and none other (except the identity) — this is why it must have at least 4 points (because 3 points define a plane, which is a different symmetry element).

p. 48 — The drawing of the C(s) symmetry group is truly terrible and should be improved in a second edition.  Notice that the symmetry plane sigma is labelled sigma(h) even though it appears vertical.

Hopefully this was helpful — I wish I’d written all this down the first time I went through these pages.

• MJ  On January 5, 2012 at 2:09 am

I’ll be interested in hearing your opinion of the text when you’re finished – I had been introduced to the Cotton text on group theory applications to chemistry while in grad school. I haven’t given much thought to group theory since then, although it’s funny how much of it is coming back just from reading your notes.

Regarding NMR – I presume you’ve run a similar calculation to your previous one (example: at ~ 7 Tesla, the Larmor frequency of a proton is ~ 300 MHz, which gives you a 1 meter wavelength, yielding a period of ~ 3.4 ns). If you’re curious about the time period involved in an NMR experiment – about what time period are you inquiring? If it’s the recording of a free induction decay (FID), it can range from milliseconds to seconds, depending on the sample. In many cases, you need to signal average and acquire a number of FIDs, which can be anywhere between a minute to a few days.

Of course, the entire absorption/emission picture for NMR is probably a convenient fiction, as I noted earlier. It works – but there’s that practical vs. theoretical divide I mentioned when one gets down to brass tacks.

• luysii  On January 5, 2012 at 8:54 am

MJ — Primarily interested in the time it takes to absorb a photon. The time period to emit it again is quite different as you point out, and depends on something around to catch it.

I was friendly in Med School with a University of Chicago graduate whose favorite saying was — “That’s how it works in practice, but how does it work in theory?”