24

Our group uses both methods a lot so here are some examples why you would use X-ray, in addition to NMR, in organic synthesis. Your compound isn't soluble enough: Colleagues produce very large aromatic systems for organic electronics which are basically insoluble in all solvents. With a good NMR machine (600MHz and up) it might still be enough to get ...


20

I agree with @andselisk that this question is quite broad. I will focus on two specific questions asked The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? [...] How do you translate the spots on a detector to electron density plots using Braggs law? Apart from Bragg's law (which tells ...


19

Boron is a covalent solid with high melting point, like diamond (though not quite), and hence its crystals are hard to make. Unlike diamond crystals, they are not nice and probably wouldn't make a great display. The table on http://periodictable.com/Properties/A/MolarVolume.v.log.html seems to corroborate your findings about boron molar volume being the ...


18

When we are talking about the prediction of materials, it is generally about the structure and corresponding free energy, not the crystallization process. That being said there are a huge number of inorganic materials that predicted to be stable yet no one has synthesized them. There are computational databases like Materials Project that contain ...


16

Coming from natural product chemistry, of course the structure elucidation by NMR is the most commonly used method, especially in isolation. NMR requires only a little substance dissolved in whichever deuterated solvent you have on your shelf and is rapidly set up. The only downside is that for very small amounts of sample you will be blocking the strongest ...


15

Arguably, there are more such crystals than those which have been made. It is just that nobody bothered to write them all down, because hey, what's the point. A tiny minority of those theoretical crystals haven't been made for the reasons you envisioned, that is, because the conditions of crystallization are hard to achieve. Here by "conditions" we mean ...


10

Imagine an infantry unit of soldiers marching in file. Each of them may be quite irregular, but together they form a repeating pattern. And that's exactly what happens with protein molecules in a crystal. When we say that a protein is irregular, we mean it on a different level. Indeed, one molecule of our protein is irregular if we are talking about the ...


9

Now, as I understand it, all processes proceed so as to maximize the "randomness" of its constituent particles. (Oversimplified version of the Second Law of Thermodynamics, yes, I know... just don't chew me out in the comments section...) Yes. True. This Law can easily be observed in, and verified by, natural processes. Sure. Still with you. Now the ...


9

IFF I remember correctly... Sometimes you have to "move" the plane, in order to see where the axis crosses. Take example 7, if you look at the example plane, the one actually shown, it is (0 ∞ ∞) not very useful. The ones making the examples have "moved the planes" (not really, but it is a useful fiction) in order to get the plane into the unit cell so ...


8

From http://www.crystallography.net/cod/1000062.html one can see that Sn sits in the middle, without any displacement, and mimics the classic Rutile-type structure: I checked original Wikimedia image, and according to metadata CIF retrieved from The American Mineralogist Crystal Structure Database is no longer available. I also failed to discover this ...


8

From the first picture depicted, I would refrain from stating there are five HCl molecules per unit cell. One thing, while there is some choice in the definition of a unitcell (there are several definitions possible, even under the constrain "it should be the simplest and still complete representation possible"), lower multiplicities occur more often; and ...


8

I've used Reduce before, which is excellent. I'm not sure what you mean by ' does not always seem to be giving the correct result', though? Perhaps you could file a bug report with the PyMOL team if you think there is a bug in h_add.


8

1) Oxygen CAN form 3 or 4 bonds. The most common example is $\ce{H3O+}$ ion This form of bonding is critical to understand basic nature of ammonia solutions, so for exact description of it look that part of the book. 2)Boron also can form more than 3 bonds if at least one its partner provides an electron pair for bonding, not one electron. As a general ...


8

This is a crystallographer’s shorthand for a certain type of space group. You may have heard of the 14 Bravais lattices, e.g. cubic, monoclinic, orthorhombic. These lattices all assume perfect sphere’s at the corners of the elementary cell and potentially also in certain central positions (if the lattice is not a primitive one). In actual X-ray structure ...


8

Ivan basically gave a nice and clear example of what is going in. I’m going to offer a more in-depth explanation. If you were to consider possible crystal structures of just one type of atom, you can boil the possible structures down to a set of similar structures, the Bravais lattices. Only 14 of these exist with different constraints: i.e. the C-centred ...


8

A mole of neutrons in a neutron star would take up about $10^{-20}$ m$^3$. And in a black hole, they would be even smaller.


7

Olex2 (free, available for Windows, MacOS, Linux) can also be used to add hydrogens to any structure with the similar to PyMOL's command hadd. Example: hemoglobin PDB without H-atoms loaded in Olex2: Executing hadd command adds hydrogens to all atoms where it is appropriate (taking hybridization into account), chemdraw assigns proper bonds order: ...


7

The denominator signifies the number of cubes that are needed to completely encompass the whole point. For example, a corner point can be thought of as a center of 8 whole cubes, while a face centre is enclosed by 2 cubes and an edge center by 4. Hence, only 1/8 of a corner atom is in a specific unit cell and so on and so forth. Consequently, the total ...


7

You can think of the body centered cubic lattice as two simple cubic lattice, one with points at coordinates $(ma,na,pa)$ where $m,n,p$ are integers, the other with points at $((m+(1/2))a,(n+(1/2))a,(p+(1/2))a)$. If you work out increasing distances for both omponent cubic lattices you get, in units of $a$: $\sqrt{0^2+0^2+1^2}=1$ $\sqrt{0^2+1^2+1^2}=\sqrt{...


7

With $\sqrt3\over2$ being that close to $1$, BCC packing is better not looked at in terms of coordination spheres. But if you insist... Say you are sitting in the center of a cell. Then: Your first neighbours are at the corners of the same cell. Second neighbours are at the centers of the nearest adjacent cells. Third neighbours: centers of the next ...


6

For your first question, the original paper (J. Chem. Soc., Dalton Trans. 1984, 1349–1356) that described the geometry index $\tau_5$ defined it as an "index of trigonality". For example, they write for a compound with $\tau_5=0.48$ By this criterion, the irregular co-ordination geometry of $\ce{[Cu(bmdhp)(OH2)]2+}$ in the solid state is described as ...


6

Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell. An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell. An atom on the edge is shared by four unit cells and an atom on the corner is shared by eight unit cells What does this have to do with the equations? Well lets ...


6

$\ce{NaH}$ has a rock salt type structure meaning each edge has two $\ce{Na-H}$ bonds. When working out bond lengths in a unit cell, you assume the atoms are point charges (i.e. they take up no volume) and so you do not need to know the radius, only the unit cell parameter which as you have said is $\pu{488 pm}$. Since there are two bonds along each edge, ...


6

The image below gives some example of different rotation inversion operations. For the $\bar 6$ operation rotate by 60 degrees (black dot to open circle in the direction towards 3) then invert through the centre of coordinates to the opposite side. Inversion changes a point at (x, y, z) to (-x, -y, -z) thus, in the diagram point 1 becomes point 2 after the ...


5

What you call a molecular structure is in reality an asymmetric unit. These two terms are not related, and in general are not interchangeable. The same way $\ce{Pb3O4}$ is not a molecular formula, but a formula unit. All pictures you presented are correct in terms that they do represent an asymmetric unit, but actually an IUCr-recommended asymmetric unit (...


5

This question is possibly a little broad- there are entire books dedicated to methods of structure determination and assignment of stereochemistry. The tl;dr answer is that there is no single method by which absolute stereochemisty can be assigned unambiguously, it depends on the nature of the natural product (terpene, polyketide, polypeptide, etc) and the ...


5

If you allow a mole of atoms, then some compounds come to the fore. Like water. Ordinarily, liquid water occupies $6.0\text{ cm}^3/\text{mol atoms}$. Freezing this to ordinary ice (Ice $\text{ I}_h$) increases this volume slightly as water expands upon freezing. But there are high pressure ice phases that are denser and thus give diamond a run for its ...


5

To address you concern about boron, there is a cubic diamond form of boron nitride $\ce{c-BN}$, ICSD #182731 [1], posseses $V_\mathrm{cell} = \pu{7.99 Å3}$, $Z = 2$ and molar volume $$V_\mathrm{m} = \frac{N_\mathrm{A}V_\mathrm{cell}}{Z} = \frac{\pu{6.022e23 mol-1}\cdot\pu{7.99 Å3}}{2} \approx \pu{2.406e-6 m3 mol-1}$$ which is about $30\%$ less than diamond....


4

The reflection with $n=2,\;(h,k,l)=(1,0,0)$ is the same as $n=1,(h,k,l)=(2,0,0)$. Yes, I said "same", not "similar"; they are one thing, there are no "both". For this reason, you may just as well abandon using $n$ altogether. As for why the peak is so small compared to others you've seen, let me respectfully suggest that maybe you haven't seen all that many....


4

"Real" crystals do not grow by stacking unit cells, but by atoms, ions or molecules adsorbing to the surface of the existing crystallite nucleus. The different possible surfaces of the crystal (those that are different, depending on orientation towards the unit cell (symmetry!) and current outer layer) each have some probability to start growing (first ...


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