Question

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

Answer: +2

Please explain.

Answer #1

For nitride ion, it is at the body and corner positions. Each cube has 8 corners. But each corner is shared by 8 cubes, so contribution of each nitride ion becomes 1/8th on corner. So, nitride ions in single cubic unit cell = (8*1/8)+1 = 2.

For metal ions, they are at face positions. Each cubic unit cell has 6 faces and each face is shared by 2 unit cells. So, metal ions in single cubic unit cell = (6*1/2) = 3

Thus, the formula becomes M_{3}N_{2}.

Suppose oxidation state of metal ion is +x.

Oxidation state of nitrogen is -3.

So, 3(+x) + 2(-3) = 0

3x = 6

x = +2

So, oxidation state of M is +2.

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