If a cubic unit cell for a metal nitride has nitride ions at all face and corner positions, and metal ions at all edge positions, what is the oxidation state for the metal ion in the nitride?

I think that if something is in a face position, there is half of that something present in the unit cell. So if there are nitride ions at all face positions (6) then there must be 3 total nitride ions. And at the corners, there are eighths of nitride ions. So given 8 corner positions, there must be an additional nitride ion formed by the fragments at the corners.

This gives us a total of 4 nitride ions, or a negative 12 charge that must be balanced by the metal ions at the "edge" positions. But what are edge positions? If I know what it is, then I can visualize how many metal ions there are in the unit cell.

Are the edge positions directly halfway between the corner positions and butted up against the face positions? So is there one edge position per edge of the unit cell? This gives us 12 edge positions, and each fragment is a fourth of an ion, so we have 3 metal ions total. So the oxidation state of the metal ion is +4?


2 Answers 2


Edge positions are those at the centre of an edge of the cube; i.e. in the middle between two closest corner positions. As such, there are 12 edge positions and each belongs to 4 unit cells giving us 3 metal ions.

The way the nitride positions are described is basically a verbose way of saying cubic close-packed or face-centred cubic. Therefore, we can redescribe the cation positions as ‘all octahedral voids except for the one in the centre of the unit cell’. Since there are 4 atoms per unit cell in ccp and 4 octahedral voids, that again gives us a $3:4$ ratio and thus $\ce{M3^{IV}N4}$.


The edge atoms actually occupy the the center of the edge, In the first 2 figures, red atoms show the positions of the cation.enter image description here

The last figure gives a clear idea as to how many cubes actually share one edge. 4 cubes share 1 edge, so effectively, each edge atom contributes 1/4 to the unit cell.

There are 12 edges in total, which means the effective contribution is 12/4=3. Since it is given 3*(oxidation state of metal ion)=12 for neutrality of molecule, we get the O.S. to be +4.


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