# Solid state and packing

In NaCl there is a simple cubic unit cell and the coordination number is 6. So i am confused how are both the ions arranged in such a lattice, with such a cordination no as in simple cubic unit cell there are 8 atoms on the edge of a cube with a contribution of 1/8 by each of them to each unit cell. Also how does it manage to have itself satisfied its valency in such a way to have a chemical formula of NaCl.

• There is no such thing as valency. As for the unit cell, it is by no means simple. Imagine a cubic unit cell with the size twice as much as you imagined before, with sodium ions at the corners and at the centers of all faces, and with chlorine atoms at all mid-edges and at the center of the entire cell (or vice versa, that doesn't matter). That would be it. – Ivan Neretin Apr 19 '17 at 20:49
• But in a unit cell an atom always lies on the corners of the cube.so where does this mid-edges come into play! – akash agarwal Apr 19 '17 at 20:51
• You have it all backwards. A unit cell may contain one, or ten, or 100, or 1000 atoms. Whether or not an atom lies at the corners is but a matter of convention. Sometimes it does, sometimes it doesn't. In my case it does, so what's the problem? – Ivan Neretin Apr 19 '17 at 20:54
• Oh yes i get it but how does each atom have a cordination number of 6.Also with regard to valency in NaCl each ,Na+ ion has an electrovalent bond with Cl- ion. So each Na + ion must have only 1 effective bond with all the chlorine ions in its vicinity.I also want to know if a unit cell must effectively show only one molecule or a atom as it says unit cell or there is a different story there. – akash agarwal Apr 19 '17 at 21:04
• Also tell me if a molecule has a simple cubic crystal it can have all type of cubic unit cells like fcc, bcc,etc or not – akash agarwal Apr 19 '17 at 21:07

An unit cell is a periodic structure. There are two types of unit cells: conventional unit cells and primitive unit cells. The conventional unit cell is the one that is used most often. Nevertheless, I suggest that you take a look at what is the difference between the two. There are plenty of resources that explain the difference.

Now with that out of the way. Let's look at the unit cell of NaCl. According to Chem Libre, the unit cell is:

Recall that ionic solids are held together by electrostatic forces. So what matters is that the charges in a unit cell cancel. In other words, we want the net charge to be 0. Let's count the number of sodium atoms and chloride atoms to see whether the charges cancel in this unit cell.

At the top plane, there are 4 chloride atoms that contribute $\frac{1}{8}$ of an atom. There is also 1 chloride atom that contributes $\frac{1}{2}$. On the other hand, there are 4 Na atoms that contribute $\frac{1}{4}$ of an Na atom. Now, each Na atom has a 1+ charge whereas each Cl atom has a 1- charge. Therefore, at the top plane we have that:

$$(+1)(4\,\cdot\,\frac{1\,\,\text{Na atoms}}{4})+(-1)(4\,\cdot\,\frac{1\,\,\text{Cl atoms}}{8})+(-1)(1\,\cdot\,\frac{1\,\,\text{Cl atoms}}{2})=0$$

So we can see that the overall charge for the top layer is 0. The same is true for the plane in the middle. Work through it. It is a good exercise.

Now, lets look at the coordination number. What do we mean by coordination number? The coordination number is simply the number of nearest neighbors. That is all that there is to it. It doesn't matter whether a single sodium ion is in contact with 6 chloride ions. This doesn't imply that the sodium ion will end up with -5 negative charge. This is because even though the sodium ion is in contact with 6 chloride ions, each of these chloride ions is in contact with other sodium ions. As a result, the sodium ion does not really accumulate negative charge.

What determines the coordination number is a bit more complicated. The coordination number is dependent on the size of each atom, the charges, and even the spin. One simple rule that helps you determine the coordination number of binary solids is the radius ration rule. I suggest that you take a look at it: http://minerva.mlib.cnr.it/mod/book/view.php?id=269&chapterid=111