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NS rings

Dear Reader,

An event at Chalmers reminded me of a reaction which one of the other PhD students at Loughborough did in the inorganic lab what Tuan Q. Ly did was to heat up a mixture of ammonium chloride and sulfur(I) chloride in a big flask with an air condenser on it. An air condenser is not a think for condensing air it is a condenser cooled by air not water.

What happens is that the ammonium chloride reacts with the sulfur chloride to form SNCl according to the following reaction

NH4Cl + 2S2Cl2 → NSCl + 4HCl + 3S

The NSCl then reacts further to form other things, the vapors of NSCl and S2Clreact to form another product.

2NSCl + S2Cl2 → SCl2  +  N2S3Cl2

The N2S3Cl2 forms a crystals of a solid which appear in the air condensor. Now through the miricle of X-ray crystallography the strucutre of the ions in this solid is known. This solid is an ionic solid which contains chloride anions and N2S3Cl+ cations. Now the cation has the following structure.

N2S3Cl cation

The locations of the double and single bonds in this ion are a little unclear to say the least, based on the bond lengths from the crystallography we can draw the following initial thing.

X2S3Cl cation with double bonds

Now we can add up all the valence electrons which the different atoms give us

Chlorine 7

Three sulfurs (3 x 6 ) = 18

Two nitrogens (2 x 5) = 10

Now to give the chlorine a full octet we need one more electron in the form of the covalent bond to the sulfur. This means we have used up 2 of the valence electrons. We have a total of five more sigma bonds so the sigma bonds will use up 12 valence electrons. Our chlorine will have three lone pairs so we have used up in total 18 electrons.

We started with a total of 35 valence electrons, we now have 17 left.

We subtract one as we have a cation, so we now have 16 to use up.

If we assume that each nitrogen and the sulfur at the top of the ring bears a lone pair pointing out into space away from the centre of the ring then we will have 10 electrons left to use up. Here is the ion with the lone pairs added so far.

S3N2Cl ion with three lone pairs

We next add the lone pair on the bottom right sulfur which points out into space away from the centre of the ring. This will give us a total of 8 electrons left to use. Here is the diagram now.

S3N2Cl ion with four lone pairs

Now we add the second lone pair on the bottom right sulfur to the pi system of the ring and one electron from each of the other atoms in the ring. This gives us two electrons left over. Now this makes the idea of us having a nice aromatic molecule unlikely. Lets try again.

If we have the following bonding where more lone pairs are on the sulfurs that point out into space. To give the following.

S3N2Cl ion lots of lone pairs

Then the lone pairs on the chlorine, nitrogens and sulfurs (10 lone pairs in total) consume 20 electrons. As we have 34 valence electrons after detecting one for the charge, we have a total of 14 left. We have six normal sigma bonds which will use up 12 electrons, the last two will be used either by a long bons (most likely pi) between the nitrogens or by putting two unpaired electrons on the two nitrogens to make the molecule a diradical. Here is the drawing of the diradical version of the cation.

S3N2Cl ion diradical

So it will be hard or impossible to make this molecule aromatic, even while we can draw the single and double bonds to make it look like it is aromatic. I will deal with the N3S3Cl3 molecule later. If you want to see the details of how to make N2S3Cl2 then see W.L. Jolly, K.D. Maguire AND D. Rabixov, Inorganic Chemistry, 1963, 3, 1304-1305. For the crystallography see A. Zalkin, T.E. Hopkins and D.H: Templeton, Inorganic Chemistry, 1966, 5, 1767-1770..


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