# Density of Interstitial compounds

I have read in my textbook (not very reliable) that density of interstitial compounds is lesser than parent compound. But how can this be true?

We add atoms to the lattice voids, so density should increase right?

• Sometimes the atoms we add do not quite fit into the voids, so they push the atoms of the parent compound to make room for themselves. There is no universal answer; the resulting density may be greater, or it may be less. Feb 14 '17 at 9:00
• I think you are right about your textbook. Mine clearly says that density is higher Feb 14 '17 at 9:28

Interstitial compounds are typically obtained when elements such as $$\ce{H},$$ $$\ce{B},$$ $$\ce{C}$$ and $$\ce{N}$$ are located within the interstitial sites of a metallic substructure. Nonetheless, the metallic substructure is not that of the pure metallic element in most cases.

For example let's consider $$\ce{Nb}.$$ We can dissolve some amounts of $$\ce{N}$$ inside the bcc structure of pure $$\ce{Nb}$$. On one hand, this dissolution increases the unit cell volume, thus reducing the density.

On the other hand, if the amount of dissolved $$\ce{N}$$ increases, a hcp structure is first formed $$(\ce{Nb2N}).$$ Further increase of $$\ce{N}$$ content leads to a fcc structure. In both hcp and fcc structure $$\ce{N}$$ occupy interstitial sites of the $$\ce{Nb}$$ substructure and in fact they are both interstitial compounds. But in this case the metallic substructure is not found in pure $$\ce{Nb}.$$

At the end you have the following densities:

$$\begin{array}{lc} \hline \text{Compound} & \rho/\pu{g cm^-3} \\ \hline \ce{Nb} & 8.46 \\ \ce{Nb2N} & 8.25 \\ \ce{NbN} & 8.69 \\ \hline \end{array}$$

So, you can see that considering different interstitial compounds a general rule cannot be gained.

• I took the liberty of correcting units for density (I'm not sure how to interpret original "Mg m-3"). I think you could improve your answer a bit by adding the source the densities have been taken from. Jul 1 at 13:56