Today's inorganic final makes me very sad because
1) I apparently can't calculate the packing ratio of hcp (something's wrong with my calculation and I know it but can't find the mistake!)
2) I can't do two problems on the exam. Complete blankout. And there aren't that many problems to begin with.
Anyway, I think I've finally figured out how to do the latter part. So here's the solution for the blogosphere to criticize.
(I hope my LaTeX show up properly.)
Question: Find the systematic absences of P2221 and C222.
I'll do C222 just to avoid typing the subscript :P (The other is similar). Furthermore, I mix Hermann-Mauguin and Schonflies symbols. I'm shameless like that.
First thing to notice is the centering: Primitive space groups have no a priori systematic absences. C-centering does: d-spacings with odd h + k Miller indices are not observed. This will pop up formally when I do the final calculation, but it's something worth noticing.
Second thing to notice is that both are orthorhombic space groups. On the exam, I wasn't sure of this, but the trick is that the symmetry cannot be *lower* than orthorhombic - otherwise there'll be a 1 somewhere - and the symmetry cannot be higher, because tetragonal space groups always have a 4 somewhere, trigonal have 3's, hexagonal have 6's, and cubic has a 3 in the middle (space diagonal) term. By these numbers I mean the number itself or some variation thereof (i.e. rotation, rotoinversion, or screw axis). Anyway, of the non-orthorhombic crystal systems, only monoclinic has "monoclinic C" as a legitimate Bravais lattice.
Now I am going to make one rather huge assumption: the space group has standard setting (I don't know the right terminology for this), meaning, "the symmetry operations are located at what it looks like in the crystallography table". I am pretty sure that this *is* an assumption one must make, and this is the assumption I couldn't figure out in the final itself.
Now one must generate the symmetry-equivalent locations. Cheating a bit here (I am really bad at this and this isn't the final anyway), check the lovely Crystallography Tables and one will obtain the following Wyckoff positions:
x, y, z
x, -y, z
-x, y, z
-x, -y, z
x + 1/2, y + 1/2, z
x, 1/2 - y, z
1/2 - x, y + 1/2, z
1/2 - x, 1/2 - y, z
Ok so there's 8 of them and they are paired (applying the criterion that we have c-centering to the top four positions gives the bottom four positions).
Now we can go and look at Miller-index-dependent part of the structure factor. It probably has a name; I call it E(hkl).
F_{hkl} = \displaystyle{\sum_j f_j E(hkl)}where j is the index for all atomic positions, and
E(hkl) = e^{2 \pi i(hx + ky + lz)}Because we have all these symmetry equivalent positions, E(hkl) is a bit more complicated. If we instead index the structure factor over all non-conjugate positions (i.e. positions which are not related by any symmetry operation) and incorporate all the symmetry-related positions into one E(hkl), then we get a much longer expression:
E(hkl) = e^{2 \pi i(hx + ky + lz)} + e^{2 \pi i(hx - ky + lz)} + \ldots + e^{2 \pi i(h(1/2 - x) + k(1/2 - y) + lz)}Do the obvious thing and factor.
E(hkl) = e^{2 \pi i(hx + ky + lz)}\{1 + e^{2 \pi i(-2ky)} + \ldots + e^{2 \pi i[h(1/2 - 2x) + k(1/2 - 2y)]}\}(I hope I got all the parentheses in there). Here's the tricky part. For a systematic absence, E(hkl) = 0. The left factor
e^{2 \pi i(hx + ky + lz)}is nonzero for arbitrary x, y, z. The only way we can get E(hkl) = 0 independently of atomic positions x, y, z is for the right factor to equal 0. If the right factor is actually independent of x, y, z this will be trivial, because Miller indices are always integers and Euler's formula applies:
e^{i \theta} = \cos{\theta} + i\sin{\theta}Sadly, this isn't the case. Let us consider only the right factor. Now what? Factor again. Really. The factor becomes
[1 + e^{\pi i(h + k)}][1 + e^{2\pi i(-2ky)} + e^{2\pi i(-2hx)} + e^{2\pi i(-2hx - 2ky)}]The left sub-factor gives the familiar, well-beloved systematic absence: if h + k is odd, E(hkl) = 0:
0 = 1 + e^{\pi i(h + k)} \Rightarrow (h + k) \equiv 1\,(\mathrm{mod}\,2)The right side is slightly dicier on first glance, but it's actually a sum of the form.
1 + a + b + ab = (a + 1)(b + 1)
So factor again. (Writing out the right factor only:)
[1 + e^{2\pi i(-2hx)}][1 + e^{2\pi i(-2ky)}]Both terms are nonzero in general for arbitrary x and y.
So we are done. The only systematic absence is still the usual condition: d-spacings with Miller index {h,k,l} such that h + k is odd will not be observed. The restrictions imposed by the three C2 axes are position-dependent and therefore do not, in general, give rise to systematic absences.
If the solution's wrong, please let me know. Meanwhile I'll either be studying for my next exam or lamenting about how badly I did in this final. (Knowing me, it's the latter. But I really ought to study*.)
Incidentally (it's probably obvious), I changed some template parameters so that the equations will actually display fully. If it shows up wonky on your computer, let me know.
*For some reason, my sleeping schedule's messed up such that now I sleep during the afternoon/evening and study at night, taking exams in the morning. This isn't good since my next exam is in the afternoon.
Continue...
