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).

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.

Chromium oxides

Dear Reader,

I have no idea why my blog came up when someone was searching for chromium oxides, but perhaps I should treat this as a request for me to comment on the chemistry of chromium and related elements.

Before we get going any further it is a good idea to think about the structure of chromium metal, it is a cubic solid with a cell which has sides which are 2.88494 Å long and the chromium atoms are found at the following fractional coordinates

0.0 0.0 0.0

0.5 0.5 0.5

I am sure that you lot can make your own model of the unit cell of the chromium, but I will relent and draw the other solids for you.

The first oxide is chromium(II) oxide, this has a rock salt like lattice which is cubic and the sides of the cell are 4.16 Å long. It is important to make an estimate of the volume change which occurs when the metal is converted into the oxide. I will explain layer why this is important.

The oxygen atoms are at the following fractional coordinates

0.5 0.0 0.0

0.0 0.5 0.0

0.0 0.0 0.5

0.5 0.5 0.5

The chromium atoms are at the following sites

0.0 0.0 0.0

0.5 0.5 0.0

0.5 0.0 0.5

0.0 0.5 0.5

Here is the solid.

Chromium(II) oxide

I looked at the cell of chromium(III) oxide and it is very complex, I will not make an attempt to discuss how to draw it. If you want to draw it then please be my guest. The key thing to note here is that the chromium(III) centres have octahedral coordination environments.

Here is the unit cell.

Chromium(III) oxide

I looked in the inorganic database and I found that chromium(IV) oxide has a tetragonal cell with the following dimensions.

4.419 Å

4.419 Å

2.915 Å

A tetragonal cell is like a cubic cell except it has been squashed, one axis will be a different length to the others. On close examination of the cell it looked like the cell for ruthenium dioxide which is an example of how the same structures appear again and again in inorganic chemistry.

The chromium atoms are at the following fractional coordinates

0.0 0.0 0.0

0.5 0.5 0.5

The oxygen atoms are at the following fractional coordinates

0.3026 0.3026 0.0000

0.6974 0.6974 0.0000

0.1974 0.8026 0.5000

0.8026 0.1974 0.5000

Here is the unit cell

Chromium dioxide AKA chromium(IV) oxide

I could not resist trying out the chrome finish in POVray for the chromium atoms, but it did not turn out well. Here is the picture.

Chromium dioxide with the not so good colour scheme

The next one in our series is chromium trioxide, this has a boxy cell which has three different lengths but thankfully all the angles are ninety degrees.

Chromium trioxide AKA chromium(VI) oxide

Now we will make use of the density measurements to try to make sense of things, The density of chromium metal and the oxides are shown in the following table. Using the relative atomic masses of oxygen (16) and chromium (52) we can make some useful calculations about corrosion.

Species	Density	Percentage Cr (% w/w)	Chromium oxidation state	Electronic state	Pilling-Bedworth ratio
Cr	7.2	100	0	s2 d4	1.00
CrO	6.27	76	II	s0 d4	1.50
Cr2O3	5.22	68	III	s0 d3	2.02
CrO2	4.89	62	VI	s0 d2	2.38
CrO3	2.82	52	VIII	s0 d0	4.91

 

The P-B ratio is the ratio of the volume of the oxide to the metal, it should be clear to you that chromium oxide has a much larger volume than the metal. If chromium metal is heated in air the oxide formed will not be able to stick to the surface the expansion will tend to cause the oxide to spall off, the volume change for the conversion of aluminium to aluminium oxide has a P-B ratio of only 1.28 so when the oxide forms the mechanical stresses on the new oxide coat are less likely to make it spall off. The fact that the P-B ratio is slightly larger than 1 makes the oxide formation seal any cracks in the surface of the metal.

For the formation of calcium oxide the P-B ratio is 0.64. The fact that the P-B ratio is below one causes the oxide to have a smaller volume than the metal, again this imposes mechanical stresses on the system as the oxide forms. So the oxide offers little or no protection to the metal.

Radiation meters and crystals

Dear Reader,

I have noticed that so many people are so willing and so fast to comment on the hotspots which have been found at Fukuashima, a range of comments have been written on one blog have suggested that the used nuclear fuel has relocated up into the stack.

I think that such a upwards relocation is very unlikely, I suspect that Gunderson’s suggestion that during the worst part of the accident that the inner surfaces of the pipe which goes up the stack was coated with cesium and that now the condensation from the humid air from the containments has washed the cesium down the stack back to ground level is reasonable.

One way to tell if the fuel has gone up the pipe or if the cesium salts have come back down the stack is to use gamma spectroscopy. If I was in charge of the Fukuashima site I would want to place a gamma spectrometer close to the bottom of the stack to find out what radioisotopes are inside the pipes.

It is important to understand that radiation detectors are not magic, they work in quite simple ways. One gamma ray detector which is energy dispersive is the sodium iodide detector. Typically a sodium iodide crystal (doped with a little thallium) is placed in a light proof box with a photomultiplier tube attached.

The idea is that when a gamma photon is absorbed by the sodium iodide crystal that its energy is converted to light, the emitted light is then measured by the photomultiplier tube. The greater the energy of the photon then the more photons of light which are generated by the crystal. The photomultiplier tube is then used not only to detect that an event occurred but also how much light was emitted during the event.

The sodium iodide crystal is a cubic crystal which has the same type of arrangement of atoms as found in common salt (sodium chloride). As we have done before we are going to have a go at making a unit cell.

The cell is a cube whose sides are 6.5 Å long

The fractional coordinates of the sodium atoms are

0.0 0.0 0.0

0.0 0.5 0.5

0.5 0.0 0.5

0.5 0.5 0.0

 

The iodine atoms are at the following fractional coordinates

0.0 0.0 0.5

0.0 0.5 0.0

0.5 0.0 0.0

0.5 0.5 0.5

 

Now the clever ones out there should be able to work out that the sodium to iodide ratio is  1:1.

The cell contains 8 sodiums which are shared between eight unit cells

 

8 x 0.125 = 1 sodiums

The cell contains six sodiums which are each shared between two cells

6 x 0.5 = 3 sodiums

Total of four sodiums

The cell contains twelve iodines which are each shared between four unit cells

 

12 x 0.25 = 3 iodines

The cell contains one iodine which is totally inside the box

 

1 x 1 = 1 iodine.

So as the box contains four sodiums and four iodines then the empirical formula is NaI which confirmed what we were expecting.

Back to radiation detectors, what I would want is to have a lead box containing a NaI or better still a germanium detector, the lead box would have a hole facing the hot spot. This device could then be used to collect a gamma spectrum of the hot spot. The germanium (not geranium) detector is a very expensive and large diode which is used as a detector. The idea is that when a gamma photon is absorbed in the germanium it causes free electrons and holes to be formed. The germanium diode is biased with about 2000 or 3000 volts, the charge carriers move through the crystal and charge is passed. The detector measures how much charge is passed during each event, the more energy in the photon the more charge is transported through the crystal. I will tell you more another day about such detectors.

But back to scintillation counters.

Two of the classic solids which have been used in scintillation counters are anthracene and zinc sulphide, these solids can be made to emit light using UV light or X-rays as well as alpha, beta and gamma rays. I have made a short film in which I have exposed the two solids to UV light. The anthracene is on the left and the zinc sulphide on the right. It is important to note that the lifetime of the blue glow for the zinc sulphide is longer than it is for the organic compound.

Enjoy the film, I hope to explain to you soon more about sctintillation counters and other radiation detectors.

Crystals and the home made nuclear reactor

Dear Reader,

I feel that nature has not taken its course yet with the Swedish home made nuclear reactor but it is high time I wrote again on the subject of crystals. So lets do both at once !

I commented on how I thought that it was a bad idea to try to use sulphuric acid to dissolve up the radium which is in solid form. I suspect that the radium in a radioactive source or on the surface of ye olde glow in the dark clock will be as the insoluble radium sulphate.

Much of radium chemistry is shrouded in darkness when compared with other metals, for example only four crystal structures have ever been published which contain radium. One of the key gaps in our knowledge is radium sulphate; we will assume for a moment that radium sulphate is isostructural to barium sulphate. The word isostructural is a big technical sounding word which means that the basic structure is the same, but the exact distances between the atoms in the unit cell might differ.

For example calcium fluoride and uranium dioxide are isostructural, the fractional coordinates of the uranium / oxygen atoms match those for calcium and fluorine atoms. But the size of the cubic unit cells are different. But lets get back to our barium and radium chemistry.

I think that the radium will have a lower solubility in sulphuric acid than it will in tap water. Tap water is normally low in sulphates; this lack of sulphate will become clear in a moment.

For many poorly soluble metal salts a thing called a solubility product is known.

This is often written as Ksp.

For barium sulphate Ksp = [Ba²⁺][SO₄^2-]

[Ba²⁺] and [SO₄^2-] are the concentrations of barium and sulphate in the solution.

Now those of you who paid attention in your GCSE maths lessons should understand that when barium sulphate is placed in pure water and stirred (until it reaches equilibrium) that

[Ba²⁺] = (Ksp)^½

But when the barium sulphate is placed in 0.01 sulphuric acid, then the concentration of the barium will be given by a new equation.

[Ba²⁺] = Ksp / [SO₄^2-]

It should be clear to you that by increasing the sulphate concentration that the equilibrium concentration of the barium will go down. It is very likely that the radium will behave the same way as the barium; Marie Curie isolated the radium from uranium ore together with the barium fraction. As I said yesterday for public safety reasons I will not tell you how she converted the barium / radium fraction into a water soluble form. If you are keen to know, please do not ask me about that chemical step as refusal often offends! If you want to know about other bits of chemistry then feel free to ask.

But now we have thought about solubilities lets look at the solid.

The unit cell of barium sulphate is 8.884 by 5.458 by 7.153 Å and it has atoms with the following fractional coordinates.

Ba	0.1846	0.2500	0.1581
Ba	0.6846	0.2500	0.3419
Ba	0.3154	0.7500	0.6581
Ba	0.8154	0.7500	0.8419
S	0.0630	0.2500	0.6914
S	0.5630	0.2500	0.8086
S	0.4370	0.7500	0.1914
S	0.9370	0.7500	0.3086
O	0.0814	0.0298	0.8190
O	0.1808	0.2500	0.5515
O	0.0814	0.4702	0.8190
O	0.9122	0.2500	0.6062
O	0.4122	0.2500	0.8938
O	0.5814	0.4702	0.6810
O	0.5814	0.0298	0.6810
O	0.6808	0.2500	0.9485
O	0.9186	0.9702	0.1810
O	0.8192	0.7500	0.4485
O	0.9186	0.5298	0.1810
O	0.0878	0.7500	0.3938
O	0.5878	0.75	0.1062
O	0.4186	0.5298	0.319
O	0.4186	0.9702	0.319
O	0.3192	0.75	0.0515

If you build a unit cell with these atoms then I think you need a prize from your teacher! I am not sure how it will apply to those of us who either left school twenty years ago or used a copy of ORTEP or some other computational aid.

For those of you who are not motivated to draw or build a unit cell here is a unit cell for BaSO₄.

A unit cell of barium sulphate, barium is in green, sulphur in orange and oxygen in red

Now the unit cell for strontium sulphate is a 8.377 by 5.350 by 6.873 Å box, all the atoms have the same fractional coordinates except the bariums are now strontiums. I suspect that radium sulphate has the same structure as barium sulphate and that the cell will be slightly bigger than that of barium sulphate. The fluorides of calcium, strontium, barium and radium all have the same fluorite structure, but the unit cells differ in size. Here is a table of the lengths of the sides of the unit cells of the fluorides.

Element	Length of unit cell (Å)
Ca	5.450
Sr	5.800
Ba	6.196
Ra	6.381

Sadly magnesium and beryllium has a different structure so we can not compare it to these other alkaline earth fluorides. Well I suspect that I have given you something to think about for a while.