In early February I was teaching vibrational motion in water using vibrational mode analysis and group theory. I remember learning about vibrational modes as an undergraduate and I remember distinctly feeling uneasy about it. Why can’t water just vibrate one of the two OH bonds at a time? What is it that forces the linear combination of the two OH stretches into the symmetric and antisymmetric stretches? Even in graduate school, I could do the math and derive the stretches, but it just didn’t ever seem “right.”
Well, I got up early, finished prepping my lecture and discussion points, and hopped on my bike for the ride in. And about halfway in I suddenly had a flash of insight that works pretty well for me as an explanation and answer to my own more than 25 year old question (see the actual moment below). Here comes my logic.
Imagine what the energy would be for a single O-H stretch in water. Lets call it E(1). What would the energy of the other O-H stretch be? It would have to be E(1). The two O-H bonds are equivalent in strength and orientation, and stretching one would be required to have the same energy as stretching the other; they would have degenerate energies. Now, looking at the C_{2v} character table, you notice immediately that all of the representations of symmetry are one-dimensional. In other words, you can not have two degenerate energy levels in water. You could have accidental overlap of energies, but you can not require that two things have the same energy in the C_{2v} point group, as we have just done for the two O-H stretches.
Since each O-H can not be taken independently, we must create linear combinations of the two O-H stretches that remain one-dimensional representations. The linear combinations for two things are trivial to derive: O-H(1) + O-H(2), and O-H(1) - O-H(2), the symmetric and antisymmetric stretches.
Another thing to keep in mind is that any mathematical description of a function must also conform to the symmetry properties of the point group. The O-H bond stretches must be one of the four one-dimensional representations available in C_{2v} (A_{1}, A_{2}, B_{1} or B_{2}).
If we run a single O-H stretch, say O-H1, through the projection operator in C_{2v}, we get the representation shown as line one in table 1 (water in the yz plane). The C_{2} operation (and sigma xz) transforms the O-H1 vector into the O-H2 vector, which returns a character of zero since the vector has moved off its original position. It is not possible to have this representation of symmetry in C_{2v}. When the symmetry operations of the point group map a stretch onto another bond vector, the true representation of that stretch, what we call the vibrational mode, is going to be a linear combination of those two stretches. What about the linear combination of O-H(1) + O-H(2), shown as line two in the table. That linear combination is not changed by any of the symmetry operations of the point group, and is a representation of A_{1} symmetry. Similarly, the antisymmetric stretch, the linear combination of O-H(1) - O-H(2) shown in line three of the table, is a representation of B_{2} symmetry. So, the in-phase and out-of-phase combinations of the 2 OH bond stretches result in a single function. The characters under E are 1. It is not a 2-dimensional representation.
C_{2v} | E | C_{2} | s(xz) | s(yz) |
O-H1 | 1 | 0 | 0 | 1 |
Sym | 1 | 1 | 1 | 1 |
anti | 1 | -1 | -1 | 1 |
The next class period, I was teaching about using CO bond vectors as a basis for a representation in Cr(CO)_{6} in order to determine which were IR active. As I was talking and writing on the board, I suddenly had the fear that I had done something gravely incorrect the previous class period. In the previous class I had said that in water, you can’t just have a single O-H vibration and also the other, because they map onto each other using the symmetry operations of the point group. But when using the 6 CO stretches as a basis in the Cr complex, I was showing them that with 6 M-CO stretches, upon rotation by the C3 axis in the O_{h} point group, all 6 moved to new positions and the character is zero (line 1 in table 2).
O_{h} | E | 8C_{3} | 6C_{2} | 6C_{4} | 3C_{4}^{2} | Etc... |
6-COs | 6 | 0 | 0 | 2 | 2 | ... |
Sum of 6 | 1 | 1 | 1 | 1 | 1 | ... |
As I was talking, in my head I was asking “why do we consider that the two O-H stretches in water give a character of ONE under C_{2 }while the 6 CO stretches give a character of ZERO under C_{3}?”
I didn’t let the class know my conundrum, and kept talking, hoping I would be able to solve my own problem. It wasn’t until I got back to my office that I figured out what was going on. In the water stretch, the linear combination of O-H(1) + O-H(2) is a single function, one-dimensional. It doesn’t move from its position; it is the same function after the C_{2} operation. For the 6 CO stretches, we set that up as a 6-dimensional reducible representation. One of the reduced representations is A_{1g}, which is the linear sum of all 6 CO stretches, and that linear combination is a one-dimensional representation in O_{h} (line 2 in table 2).
Each single stretch of the 6-dimensional representation moved under C_{3}. The difference between it having a ONE or a ZERO as its character during a symmetry operation where the vector moves off its position is whether it is a linear combination (2 O-H stretches considered in-phase as A_{1} in C_{2v} or 6 CO stretches considered simultaneously as A_{1g} in O_{h}) or a multi-dimensional representation (both OH stretches or all 6 CO stretches).
I hope that this blog post helps explain a) the conundrum I was having, and b) the solution I found. And perhaps, it will suggest c) a reason to bike to work. The quiet time I have in my head either walking, running, or biking around helps to get me out of my own head and solve problems. This is another skill I try to instill in my students, but that might have to wait for a future blog post.
Very nice.Yes, completely sound reasoning, and a nice explanation too. Stealing for next year!
Remember when you manipulate characters, the characters are derived from matrices. When C2 operates on OH1, it is transformed into OH2. So you can't have a matrix with just OH1. The basis set must include both OH1 and OH2. You get a character of zero not just because that vector has moved but that it has moved into another basis set vector. To complete your comparison between H2O and Cr(CO)6. The RR with OH1 and OH2 as the basis set is 2 0 0 2, Remember also when a molecule vibrates, it can't cause the molecule to rotate or translate.
The molecular plane of water defines the xz plane, not the yz plane. This changes the RR to 2 0 2 0, which reduces to A1 and B1.
I use Miessler and Tarr, but I'm curious if there is another textbook that uses a different convention for molecules with C2v symmetry.
There are several books that put C2v molecules in the yz plane. I was able to quickly find one: Housecroft and Sharpe. I am very inconsistent with how I orient molecules (not with the z-axis, fortunately) but I always include an axis system and encourage my students to do the same. If their stretches or motions are B1 and mine are B2, I know we switched axis conventions.
Thanks for letting me know I’ve been teaching the wrong convention:( . I’ve found other references as well that are also inline with Housecroft and Sharp (such as Jaffé and Orchin). I’ve been using M&T since the 2nd edition first came out. Anyway, my students also need to define the Cartesian system and I love the projection operator! Nice post.
You aren't using the "wrong" convention. Cotton puts water in the xy plane, and if there is a convention, that is probably it. But there is not universal agreement, which is why one has to assign an axis system.