Another of the hard unit cells

Dear Reader,

After my success with the sodium nitrate I will now present tripotassium hexacyanoferrate(III) which is an example of a monoclinic cell. Again we have to start with drawing a box to make our cell. The lengths of the cell are 7.047, 10.400  and 8.384 Å. Two of the angles (alpha and gamma) are 90 degrees while the beta angle is 107.29 degrees.

This will give us a box which is more complex than an orthorhombic box.

We start by adding iron atoms, the fractional coordinates of the iron atoms are 0 0 0 and 0 0.5 0.5 which means we have two iron atoms in the unit cell.

We next have to add the potassiums at the following fraction coordinates 0.000 0.000 0.500, 0.000 0.500 0.000, 0.5025 0.7700 0.6252, 0.5025 0.7300 0.1252, 0.4975 0.2700 0.8748 and 0.4975 0.2300 0.3748.

We should now understand that we will have six potassium atoms to go with our two irons.

Next we add the cyanide carbons

These are at the following fractional coordinates

0,1875	0,9480	0,2102
0,1887	0,945	0,886
0,8795	0,8317	0,9712
0,8795	0,6683	0,4712
0,8113	0,445	0,614
0,8125	0,448	0,2898
0,8125	0,0520	0,7898
0,8113	0,0550	0,1140
0,1205	0,1683	0,0288
0,1205	0,3317	0,5288
0,1887	0,5550	0,3860
0,1875	0,5520	0,7102

We should understand that they are all fully inside the cell so we have 12 cyanide carbons.

Next we do the cyanide nitrogens

0,8004	0,7335	0,9515
0,2974	0,9120	0,8160
0,2962	0,9180	0,3361
0,8004	0,7665	0,4515
0,7026	0,4120	0,6840
0,7038	0,4180	0,1639
0,1996	0,2665	0,0485
0,7026	0,0880	0,1840
0,7038	0,0820	0,6639
0,1996	0,2335	0,5485
0,2974	0,5880	0,3160
0,2962	0,5820	0,8361

Now lets have a look at what we have done, from the side we can see how one of the angles (beta) is a long way from ninety degrees.

Side view of unit cell

Now we will have a look from a different angle.

The front view of the cell

Here are the x,y,z coordinates of the atoms on a conventional 90 degree and 90 degree set of x y and z axis. I have marked on bold the iron atoms which are at the corners of the cell.

46
XYZ file   for : K3Fe(CN)6
K	0.28089	-3.45899	-5.16955
K	-5.00083	0.38661	-3.78117
K	5.38365	0.38134	-3.21339
K	0.10193	4.22694	-1.82501
K	-2.32040	-2.87755	-1.40161
K	2.82526	-0.96562	-0.24795
K	-2.82526	0.96562	0.24795
K	2.32040	2.87755	1.40161
K	-0.10193	-4.22694	1.82501
K	-5.38365	-0.38134	3.21339
K	5.00083	-0.38661	3.78117
K	-0.28089	3.45899	5.16955
Fe	-4.91135	-3.45635	-5.45344
Fe	5.47313	-3.46163	-4.88566
Fe	0.19141	0.38398	-3.49728
Fe	-5.09031	4.22958	-2.10890
Fe	5.29417	4.22430	-1.54112
Fe	-5.29417	-4.22430	1.54112
Fe	5.09031	-4.22958	2.10890
Fe	-0.19141	-0.38398	3.49728
Fe	-5.47313	3.46163	4.88566
Fe	4.91136	3.45635	5.45344
C	4.80935	-2.73005	-3.21574
C	0.69724	-1.37587	-2.85930
C	-1.59727	0.07097	-2.84632
C	0.66992	1.11497	-1.76490
C	-3.38358	3.91480	-1.26682
C	4.72002	2.46500	-0.96219
C	-4.72002	-2.46500	0.96219
C	3.38358	-3.91480	1.26682
C	-0.66992	-1.11497	1.76490
C	1.59728	-0.07097	2.84632
C	-0.69724	1.37587	2.85930
C	-4.80935	2.73005	3.21574
N	0.98970	-2.42716	-2.50303
N	4.41252	-2.27534	-2.24005
N	-2.64379	-0.14067	-2.41469
N	-2.39058	3.70212	-0.72368
N	4.38940	1.41403	-0.63999
N	0.95847	1.56933	-0.75174
N	-4.38940	-1.41403	0.63999
N	2.39058	-3.70212	0.72368
N	-0.95846	-1.56933	0.75174
N	-4.41252	2.27534	2.24005
N	2.64379	0.14067	2.41469
N	-0.98969	2.42716	2.50303

If you want a free program (free to academic users only) then I would suggest ORTEP which I have provided a link for on the right hand side of this blog.

The harder unit cells (NaNO3)

Dear Reader,

A short time ago I chickened (lost my courage) out of the harder unit cells, then it started to rain which stops me doing my project in the garden. I have come back into the house and I am now sitting in front of the PC again.

We will be attempting to deal with a rhombohohdral cell, we will use the suggested example of sodium nitrate from that book which I mentioned recently. This cell has three equal lengths for the sides which are 6.3108 Å long. The angles alpha, beta and gamma are all 47.267 degrees.

N. Elliott, Annales Academiae Scientiarum Fennicae, Series A6: Physica, 1962, 94, 1 to 34 publishes the crystal strcutre. I will use this person’s data to make our cell.

The first part is to draw the box for the cell, this is not so easy as the angles in the cell are not 90 degrees. After you have been able to draw the cell we should start to add the atoms.

The sodiums are easy as we have them at fractional coordinates of 0 0 0 and 0.5 0.5 0.5, those of you who have been paying attention should quickly understand that we have two sodium atoms in each unit cell.

We have nitrogens at two fractional coordinates 0.25 0.25 0.25 and 0.75 0.75 0.75. This means we have two nitrogen atoms in our unit cell.

Finally we have oxygens at six sites, here are the fractional coordinates for these oxygen atoms.

0,0067	0,4933	0,2500
0,2500	0,0067	0,4933
0,4933	0,2500	0,0067
0,9933	0,5067	0,7500
0,7500	0,9933	0,5067
0,5067	0,7500	0,9933

Here is one view of what is inside the cell

A view of a unit cell of sodium nitrate

Sadly one of the sodiums blocks the view of a nitrate anion, so here is a second view.

Second view of the sodium nitrate

Finally here is a table of the cartesian coordinates of the atoms, this will allow some of my readers at home to recreate the unit cell. The sodium at 0 0 0 is the sodium at the centre of the cell. This will help you recreate a cell for yourselves.

17
XYZ	file	for	:
Na	6.90253	2.83847	-3.83540
Na	0.66278	2.74622	-2.89430
Na	2.88070	-1.75338	-2.23276
Na	-3.35906	-1.84563	-1.29166
Na	0.00000	0.00000	0.00000
Na	3.35906	1.84563	1.29166
Na	-2.88070	1.75338	2.23276
Na	-0.66278	-2.74622	2.89430
Na	-6.90253	-2.83847	3.83540
N	3.45126	1.41924	-1.91770
N	-3.45126	-1.41924	1.91770
O	3.33488	0.54360	-2.77519
O	2.91164	2.51399	-2.07865
O	4.10727	1.20012	-0.89926
O	-4.10727	-1.20012	0.89926
O	-2.91164	-2.51399	2.07865
O	-3.33488	-0.54360	2.77519

Non cubic cells

Dear Reader,

When I was at my brother in law’s wedding I took the chance to bring to Sweden a book which I consider to be one of my “secret weapons” in chemistry. It is not the most advanced book in my collection but it is a fun book. It is a 1960s text book which is intended to be a bridge between A-level chemistry and university chemistry.

Now if any of you have a copy of “Chemistry: A Unified Approuch” by Buttle, Daniels and Beckett then I suggest that you turn to page 70 of the third edition. But the rest of you will have to keep on reading my blog.

I want to discuss with you some crystal systems and unit cells which are a little different to the rock salt cubic one which I have already considered.

Now if we make a start with a cube which has been stretched or squeezed in one direction we will have tetragonal. The example which the book gives is mercury(II) cyanide. Now this is a compound which I suspect will get some less mature people angry. The mercury part may outrage them while the cyanide part will also get them in a pickle.

Before we go any further I need to share something with you, sometimes when we learn about chemistry we need to learn about the chemistry of not so nice to handle substances. If I was to try to write a chemistry course which either only included “nice” harmless substances or only contained wild and scary compounds then I would have done the students a great wrong. To understand chemistry you will need to understand both nice innocent compounds and some real nasties. While I do not advocate giving the undergraduates things like plutonium, acrylamide, brucine and large amounts of KCN to do experiments with I still need to teach them about these things as they will need to have an understanding of these things for the good of society.

To fail to teach chemistry students about “horrible” chemicals is as stupid as not teaching law students about “what to do when people break the law”. While law students could be taught about all the laws which relate to nice people who make a point of obeying the law, they need to know about legal reasoning of criminal law, and the details of the laws which deal with bad people.

I have very little sympathy for the idea that I should not teach students about chemicals which they can not handle in their teaching lab. I also have precious little sympathy for people who use (or want to use) the more hazardous chemicals in a reckless manner, I am strongly in favour of responsible chemistry.

Now I am coming down from my soap box, and returning to the chemistry.

O. Reckeweg and A. Simon, Zeitschrift fuer Naturforschung, Teil B. Anorganische Chemie, Organische Chemie, 2002, 57, 895-900 is the most modern determination of mercury(II) cyanide. This compound is in a cube which has been squashed. The cell has two sides which are 9.6922 Å long and one which is 8.9015 Å long. The first two are the x and y axis while the last is the z axis. The z axis is blue while the other two are red and green.

We need to add the mercury atoms at the following locations

0,2500	0,7880	0,8750
0,2500	0,7120	0,3750
0,2120	0,2500	0,1250
0,2880	0,2500	0,6250
0,7500	0,2120	0,1250
0,7500	0,2880	0,6250
0,7880	0,7500	0,8750
0,7120	0,7500	0,3750

Here is what our cell will look like with the metal atoms

The mercury atoms in the cell

Next we should add the cyanide carbons at the following locations

0,0464	0,7950	0,8366
0,0464	0,7050	0,4134
0,4536	0,7950	0,9134
0,4536	0,7050	0,3366
0,7050	0,9536	0,5866
0,7050	0,5464	0,6634
0,7950	0,9536	0,1634
0,7950	0,5464	0,0866
0,9536	0,2050	0,1634
0,9536	0,2950	0,5866
0,5464	0,2050	0,0866
0,5464	0,2950	0,6634
0,2950	0,0464	0,4134
0,2950	0,4536	0,3366
0,2050	0,0464	0,8366
0,2050	0,4536	0,9134

Here is the cell with the carbons and the mercury atoms

The cell with the carbon and mercury atoms in it

Lastly we add the cyanide nitrogens, I know that this might be a lot for you to read but think of how I have to type all the the values !

0,2035	0,9280	0,1785
0,2965	0,9280	0,5715
0,2035	0,5720	0,0715
0,2965	0,5720	0,6785
0,5720	0,7035	0,3215
0,5720	0,7965	0,9285
0,9280	0,7035	0,4285
0,9280	0,7965	0,8215
0,7965	0,0720	0,8215
0,7035	0,0720	0,4285
0,7965	0,4280	0,9285
0,7035	0,4280	0,3215
0,4280	0,2965	0,6785
0,4280	0,2035	0,0715
0,0720	0,2965	0,5715
0,0720	0,2035	0,1785

Now we have all the atoms in the cell, here it is in all its glory.

Here is the cell with all the atoms in it

Now we have the whole of the cell, the interesting question is what is the coordination number of the mercury. Now at first look it might appear that the coordination number of the mercury is four. But it is only two. If we look at the coordination environment of a mercury we will see that we only have two cyanide carbons in the right places.

The coordination environment of a mercury atom

Lead(II) iodide has a series of different hexagonal crystal forms, but the most simple one is described by B. Palosz, W. Steurer and H. Schulz, Journal of Physics: Condensed Matter, 1990, 2, 5285-5295.

After the endurance test of typing the values for the mercury cyanide this one should be quite a nice rest.

The cell is a hexagonal cell, the a and b edges are 4.558 Å long (these are the x and y axis {red and green}) while the z axis is blue and the c edge is 6.986 Å long.

We have a lead atom at each vertex of the cell, so the fractional coordinates of the lead is 0 0 0 (well that was easy).

The first iodine is at 0.3333, 0.6667, 0.2675 while the other is at 0.6667, 0.3333, 0.7325.

Here is what the cell looks like.

Lead(II) iodide cell

Potassium chromate, well I do not know why the book seems to choose toxic examples so much but this is a good example of an orthorhombic cell.

We start with a box which is 7.662 Å by 5.919 Å by 10.391 Å.

Then we add the potassium atoms at the following fractional coordinates

0.1656, 0.2500, 0.0857

0.5100, 0.7500, 0.1998

0.6656, 0.2500, 0.4143

0.0100, 0.7500, 0.3002

0.990, 0.2500, 0.6998

0.3344, 0.7500, 0.5857

0.4900, 0.2500, 0.8002

0.8344, 0.7500, 0.9143

As all of these potassium atoms are totally inside the cell, we have eight potassium atoms in our cell.

Next we need to start to add the chromium atoms these are at

0.7291, 0.2500, 0.0794

0.2291, 0.2500, 0.4206

0.7709, 0.7500, 0.5794

0.2709, 0.7500, 0.9206

As all the chromiums are fully inside the cell we have four chromiums in the unit cell which means that the potassium to chromium ratio is 2:1. This is in agreement with the formula which we are expecting for the solid.

Lastly we add the oxygen atoms to the cell.

We have four groups of four oxygens which are at

0,1982	0,7500	0,0704
0,1971	0,5225	0,8471
0,4846	0,7500	0,9200
0,1971	0,9775	0,8471
0,6971	0,9775	0,6529
0,9846	0,7500	0,5800
0,6982	0,7500	0,4296
0,6971	0,5225	0,6529
0,8018	0,2500	0,9296
0,8029	0,4775	0,1529
0,5154	0,2500	0,0800
0,8029	0,0225	0,1529
0,3029	0,0225	0,3471
0,0154	0,2500	0,4200
0,3018	0,2500	0,5704
0,3029	0,4775	0,3471

I hope by now that the penny will have dropped the cell has a magic point at 0.5, 0.5, 0.5, this is a centre of symmetry. If an atom appears at fractional coordinates q, w, e then another atom will appear at 1-q, 1-w, 1-e.

If we add the oxygens to the cell we will get a tetrahedral arrangement around the chromium atoms, for two of the chromium atoms we have to borrow an oxygen from another unit cell. This finishes off the potassium chromate cell. It also helps explain why chromate is so toxic, the chromate anion is so similar to a sulfate anion that it can pass through cell and nuclear membranes in the same way as a sulfate anion can. The chromate which enters the nucleoplasm can get reduced to chromium(III) which can then form very long lived complexes with the DNA in the cell.

The other crystal systems which have angles which are not right angles are much more of a pain to draw so we will not do them today.

I will put up the pictures of the potassium chromate later.

Cubic unit cells again

Dear Reader,

It is the return of the cubic unit cells ! Now before we get going on how to calculate the radius of a metal atom from the density of the solid we should look at how to make a sodium chloride unit cell. We start by drawing a cube which we are going to decorate with atoms.

Firstly we add the chlorines at the corners of the cube, we have a total of eight chlorines but it is important to note that all eight are shared between eight cells. So in fact we only have one chlorine in the cell.

The cube with the eight chlorines at the corners of the cube.

Next we add the sodium atoms which are half way along the edges of the cube, as each sodium is shared between four cells we have 12 / 4 sodium atoms inside the cell. This works out as three sodium atoms. Here is the cell with the extra atoms added.

We have added the first of the sodium ions to the cell

Next we add the chlorine atoms which are at the centre of the faces of the cube. As these atoms are shared between two cells we now have 6 / 2 = 3 chlorine atoms to add to the cell. Here is a picture of the cell with the new chlorine atoms added.

With the chlorines added which are at the centre of the faces of the cube.

Finally we add the sodium which is right in the centre of the cell, as it is fully within the cell it is not shared with any other cells so it is 1 / 1 = 1 atom in the cell.

The finished unit cell for sodium chloride

Now we should have four chlorines and four sodiums in the unit cell, this gives us a 1:1 ratio which is in agreement with the formula of sodium chloride (NaCl).

Now we should consider the density and radius of atoms in metals.

But back to unit cells, the most simple cubic phase for a metal is the primitive cubic lattice.

The distance along an edge of this cubic cell will be twice the radius of the atom, hence the volume will be given by

V = (2r)³ = 8r³

The cell contains only one atom so the mass will be given by the equation

Mass = M / N_a

Combine the two equations to give us

ρ = M / 8 r³ N_a

Rearrange for r

ρ r³ = M / 8 N_a

r³ = M / ρ 8 N_a

r = (M / ρ 8 N_a) ^⅓

The body centred cubic plutonium has a distance long the edge of a cell which is 3.6375 Å and a density of 16.84

The distance between fractional coordinates 0 0 0 and 1 1 1 will be 4 times the radius (r) of the atoms.

Hence the distance along an edge of the cell will be 4r/3^½

Which means the volume of the cell will be given by

V = (4r/3^½)³ = 64 r³ / (3^1.5)

The cell contains two atoms so the mass of the atoms in the cell are given by

Mass = 2M / N_a

Combine the two equations to give

ρ = 2M 3^1.5 / N_a 64 r³

ρ = M 3^1.5 / N_a 32 r³

Now rearrange for r

ρ r³ = M 3^1.5 / N_a 32

r³ = M 3^1.5 / N_a ρ 32

r = (M 3^1.5 / N_a ρ 32) ^⅓

The delta plutonium is face centred cubic and has a distance along the cell edge of 4.6347 and a density of 16.28

The distance between fractional coordinates 0 0 0 and 1 1 0 will be four times the radius of the plutonium atoms. So the distance along the side of a cell will be

4r/(2^½) = r8^½

So the volume of the cell will be given by

[4r/(2^½)]³ = V = 8^1.5r³

As the cell contains four atoms the density of the solid will be given by

ρ = 4M / N_a8^1.5r³

we can now rearrange it to give us the radius of the atoms

ρ r³ = 4M / N_a8^1.5

r³ = 4M / ρ N_a8^1.5

r = (4M / ρ N_a8^1.5)^⅓

What I suggest that you might want to do is to calculate what the density would be of a primitive cubic plutonium (if it did exist).

Is fusion safe ? and beryllium chemistry

Dear Reader,

I imagine that you have seen the suggestions by fusion experts that nuclear fusion will give us a cheap, safe, clean and green source of energy which will provide power for the world’s needs. I am currently thinking about how green is fusion, right now I have contacted a fusion expert who I know and I am awaiting his views on the matter.

While we are waiting I think it is important to ask the question of what was is the typical cause of a nuclear accident. Is it a issue with management, an “act of god“ or was it a technical failure ?

In the case of the Windscale fire I have seen suggestions that it was human error, poor design of the reactor or mismanagement of the project. I know that before the 1999 Tokaimura that a criticality accident at the JCO site was considered to be a was considered to “be an unrealistic scenario” according to the UN report on the accident.

I have to ask the question, did a failure in a regulatory body (either the external state regulator or the companies own internal regulation) cause the first step to be taken which lead to the accident in 1999.

One model of how accidents occur is the Swiss cheese model, the idea is that a weakness in a system is like a void in a lump of cheese. Due to some event a void can appear in the organisation, this void can grow in size, shrink, vanish or move around. As long as some solid cheese exists which prevents a path existing from one side of the block of cheese to the other then everyone is “safe”.

But when a series of holes align themselves to create a path through the cheese block then an accident occurs and then the airplane crashes, the core melts or some other horrible outcome occurs. In some ways the most important step is for the plant owner or the management is to recognise that a given type of accident is possible.

This first step of admitting that a given accident type is at least a theoretical possibility enables the company to start to take steps to prevent it occurring. For example the understanding that someone could get a body part caught in the moving parts of a machine lead to the idea of the 19th century UK law which requires where possible all moving parts of machines to be fenced off or guarded.

While it is impossible to fence off some moving parts such as the chain of a chain saw or all the parts of a handheld electric drill, this law does improve safety by greatly reducing the number of moving parts which can cause injury to factory workers. In the same way if a fusion reactor is going to be built we need a good understanding of the possible threats which it poses.

One is the beryllium used in the heat transfer fluid in some designs, I was reading recently about fusion reactor safety and I saw that a mixed lithium / beryllium fluoride has been proposed as a tritium breeding layer and as a heat transfer layer. I can tell you that beryllium is a very nasty element, in some ways it is worse than some of the radioactive elements. As a result special care will be needed if beryllium or its compounds are used in fusion reactors.

I have looked at the crystal structure of Li2BeF4 (J.H. Burns and E.K. Gordon, Acta Crystallographica, 1967, 1, 1948-1923), this is an interesting looking 3D network. But before we get stuck into it we should look at some organic salts of “H2BeF4″. L.A.Gerrard and M.T.Weller (Acta Crystallogr.,Sect.C:Cryst.Struct.Commun., 2002, 58, m407) report a nice and simple tetrahedral BeF4 unit which has protonated DABCO as the counterion. Those of you who know VSEPR should have predicted that one OK. Here is a picture of the anion in the solid.

The tetrahedral BeF4 dianion

If we have less fluorides per beryllium centre (to make the Be:F ratio 2:7) then we need to use one of the fluorides as a bridging ligand to give us four electron pairs (eight electrons) around all the metal centres. Then we get the following dinuclear complex. See S. Aleonard and M.-F. Gorius (C.R.Seances Acad.Sci.,Ser.II, 1989, 309, 683)

The BeF7 trianion

If we go a little further and have a Be:F ratio of 1:3 then we will end up with a dinuclear complex which has two bridging flourides. This is shown below. (B. Neumuller and K. Dehnicke, Z.Anorg.Allg.Chem., 2005,631, 2535)

The Be2F6 dianion, note the SiF6 dianion in the right of the picture

And now for something completely different (sorry no monty python for you today) if we mix lithium and beryllium fluorides with an salt of an amine fluoride to give us Li2Be4F14 in the unit cell (L.A. Gerrard and M.T. Weller, Chem.Commun., 2003, 716 ). This network will have a charge of -4 and it will form long strips of metal atoms which are in a 1D coordination polymer. Here is the picture for you of the metal atoms and flouride anions in the unit cell.

The metal and flourine atoms in one unit cell

How here are two strips of metal atoms side by side.

Two strips side by side, note that there are no interconnections between the strips

Now here is four strips viewed from a different angle.

Four strips viewed from a different angle, note that they do not touch each other

Now if we look at Li2BeF4 we will see it is a complex solid, I have looked and all the metal atoms have tetrahedral environments, here are a series of views of the unit cell to show you what the solid looks like. This is going to be hard, it is a 3D coordination polymer. These 3D coordination polymers can turn out to be what I call “atomic fog” but this one is not too bad, I have seen much worse in my time.

Side view of Li2BeF4 cell showing the bonds going in one dirrection

Now after turning by 10 degrees

Now the end view.

One last thing in case any of my readers are thinking of doing beryllium chemistry, my short answer is “do not do it !“. Beryllium is the most toxic non radioactive element, some forms of it are almost as bad gram for gram as Pu-239. In some ways I would like John Hunt (the voice of the UK’s AIDS advert) to dispense advice to you about beryllium chemistry using the voice of doom, but you just have me right now.

I would suggest that if any chemistry students do not want to turn back and do something else then I suggest they talk to your local friendly radiochemist  and learn how to work with gram amounts of plutonium. Then do the beryllium chemistry in the same way using negative pressure boxes and all the other safety precautions which you would use for large scale Pu work.

A herb from india

Dear Reader,

Christopher Maloney (“a brother blogger“) has commented recently on a herb from india which is known to contain a substance called Withaferin A which can be isolated from a plant named “winter cherry” or “ashwagandha“. Now this herb is thought to have some useful medical properties, I have looked and in the Cambridge data base an entry for this substance exists.

P. Bandhoria, V.K. Gupta, P. Kumar, N.K. Satti, P. Dutt and K.A. Suri, Anal. Sci. :X-Ray Struct. Anal. Online, 2006, 22, x89 is a paper which reports the strucutre of the compound.

The compound is a steroid which has an epoxide group and a lactone group in it. One paper suggests that the compound is an antiestrogen (E-R. Hahm, J. Lee, Y. Huang and S.V. Singh, Molecular Carcinogenesis, 2011, 50, pages 614 to 624). I have looked at the strucutre of the drug and here it is as a 2D drawing.

Withaferin A

In real life the molecule has a 3D shape, I looked at it using my trusty copy of ORTEP. I have done a few POVray files for you good people to look at. For those of you at home who have copies of POVray I am tempted just to give the code, but for those of you who are not into POVray art here is the the picture.

Spacefilling model of Withaferin

While a spacefilling model is very useful to be able to hold in your hand and view, it has oftein hard to see the wood for the trees. For that reason we tend to use the ball and stick view of a molecule. Here is the ball and stick view of it.

Withaferin ball and stick view

It is interesting to note that back in 1968, A.T. McPhail and G.A. Sim published (J. Chem. Soc. B, page 962) a strcutre of the para-bromobenzoate / acetate ester of withaferin A. I think that the reason that the para-bromobenzoate was used was that it includes a heavy atom which is more easy to find in X-ray crystallography than a carbon, oxygen or hydrogen atom. The idea of the Patterson method is that you solve the crystal strcutre for the very heavy atom only at first and then during the refinement you can locate the carbons, oxygens and hydrogens using electron density maps. What happens is that the crystallographer looks at the difference in the electron densities obtained from the data with the predictions made from the locations of the atoms which have already been found. In this way new (lighter) atoms can be located which improves the quaility of the result.

Before cheap and powerful computers were avaiable to crystallographers the direct methods (which is more computationally intense) was less popular than it is today. Today crystallographers can use both dirrect and Patterson methods, but I know a lot of crystallographers still like to use the Patterson method when they have a heavy atom in a solid.

Well keep tuned, and if you find any interesting herbs then please tell me. I may be able to look them up to see what interesting chemicals they might contain.

Oxide conduction in Solid Oxide Fuel Cells (SOFC)

Dear Reader,

Recently I was asked about the mechanism by which thorium dioxide is reduced by lithium metal, this made me think more about metal oxides. I also happened to attend a series of talks recently which included some work on fuel cells (SOFC). This made me think more and I have drawn some diagrams to explain things.

Now lets start with the 111 face of zirconium dioxide. Below is a diagram of the oxide with orange for oxygen and blue for zirconium. You can see that the solid has a series of oxygen and zirconium atoms. We have got this by slicing a cube of zirconium dioxide to cut off a corner.

ZrO2 111 face

Now we take a different view of the solid, this is now a 110 face. We get a slightly different arrangement of the atoms on the surface. What we have done is to slice the cube along a diagonal which goes between two opposite edges.

Zirconium dioxide 110 slice

Now to improve the conductivity some of the 4+ zirconium ions have been replaced with 3+ ions such as yttrium ions. These ones are in brown, every time a pair of 3+ ions are used to replace zirconium ions an oxygen is lost from the solid. This is done to keep the solid electrically neutral. Below is a picture of a doped zirconium oxide.

Doped zirconium dioxide with one oxygen now missing, this is a "hole"

By moving the oxygens around we can move the hole through the solid, by this the solid can be made to conduct electricity. Please look at the following pictures.

What you should see is that the hole is moved from one side to the other.

Space dust candy

Dear Reader,

I recently went into a old fashioned type of sweet shop which had a vast range of sweets in jars and some in packets. My wife reminded me about the joys of “space dust” which was the old name of a popping candy which will fizz and pop as you eat it. This made me think about it, a US patent (4,289,794) is the patent for this sweet. A later patent is for food which contains gases in the form of clathrates.

The early patent was for a sugar which contains very small bubbles of carbon dioxide. But the later patent is for foods which contain carbon dioxide which is not in bubbles but spread in a more homogenous manner through the solid. Many crystalline solids contain molecules of solvent in the crystal lattices, but sometimes rather than a “solvent” molecule it is a molecule or atom of a gas.

The classic example of a clathrate is one of a noble gas and water. Recently L. Yang et. al. published a paper (Proceedings of the National Academy of Sciences, 2009, 106(15), pages 6060 to 6064) in which they give details of the crystal structures of xenon and water clathrates. Before we get stuck in with this complex solid it is better to think about normal ice as published by J.D. Bernal and R.H. Fowler, Journal of Chemical Physics, 1933, 1, pages 515 to 548. In this paper by Bernal the arrangement of atoms in normal hexagonal ice is published. The oxygen oxygen distance in normal ice is 2.765 Å. The arrangement of the oxygens is very much like the carbon atoms in diamond. Here is a small part of the crystal lattice of ice.

Arrangement of atoms in ice

In the xenon ice clathrate the oxygen oxygen distances are slightly shorter than those in the ice and the oxygens are arranged in tetrahedrons. Here is a picture of a unit cell of the xenon / water clathrate.  As a result of the fact that the xenon is so heavy I suspect that it not possible to locate the hydrogen atoms on the water molecules.

A picture of a unit cell of the xenon and water clathrate

Now a close up of the central part of the unit cell is shown below. It is clear that the xenon atom is at the centre of a cage of oxygens, as the oxygens are sp3 type atoms none of the orbitals of the oxygen can point towards the major lobes of the oxygen’s orbitals.

The central part of the unit cell of the xenon and water clathrate

I will comment more on clathrates another day.

Prussain blue

Dear Reader,

I have explained how cyanide can bind to metals such as iron to form complexes, these complexes have lone pairs poking out which can bind to other metals. Here is a picture of a unit cell of prussian blue.

A unit cell of a cesium nickel iron cyanide

The carbons are black, the nitrogens are blue, the irons are purple, the nickels are gold coloured and the green atoms are the cesium.

Plastic fantastic

Dear Reader,

In recent times I have shown how a lad can have fun with the unit cells of inorganic solids, but now it is time to move onto something with carbon in it. I choose to look at the solid state structure of a polymer which is a high temperature engineering polymer.

LIMMUP in the crystallographic database is Poly((4,4′-diphenylene)pyromellitimide) which was described by Y. Obata, K. Okuyama, S. Kurihara, Y. Kitano and T. Jinda in Macromolecules, 1995, 28, 1547. This is a solid which is an endless chain of atoms covalently linked to the next. Here is a picture of the unit cell.

Unit cell of the polymer

While here is a picture of five of the polymer chains.

Polymer chains

While looking for examples of the polymer chains I noticed something else, this brings me onto another subject. I hold the view that one of the first steps to maturity is the point at which a person truly accepts that things which they are not interested in, involved in or have experience in can be truly worthwhile and valid. I have to add the warning that there is some work out there which is not worthwhile and is frankly close to worthless, but I do not want to point the finger by naming names well at least not today.

I have spent much of my life working on trying to get molecules to selectively recognise metals; I used to share an office with a man (Zhixue Zhu) who worked for Howard Colquhoun on a project where he was trying to get molecules to recognise short parts of polymer chains. While it might not have been quite the sort of thing that I have done in life, I still hold the view that the work is good work which is worthwhile.

Here what Dr Zhu did was to use a pair of pyrene groups to recognise part of a short chain model of kapton (poly(4,4′-oxydiphenylene-pyromellitimide)); his tweezers recognized the pyromellitimide part of the chain. Here is a picture of the solid which he published in Chemical Communications, 2004 page 2650 together with H.M. Colquhoun, C.J. Cardin and Yu Gan.

Dr Zhu's mini tongs which grip the polymer chain

I suspect that Christine Cardin found this solid interesting as her group have done a lot of work in the past on how things like acridines bond onto DNA through pi-pi effects.