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?
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asked Feb 14, 2017 at 8:40
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$\begingroup$ 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. $\endgroup$– Ivan NeretinFeb 14, 2017 at 9:00
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1$\begingroup$ I think you are right about your textbook. Mine clearly says that density is higher $\endgroup$– RaghavFeb 14, 2017 at 9:28
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2 Answers
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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 for $\ce{Nb}$, $\ce{Nb2N}$, $\ce{NbN}$:
$$ \begin{array}{lc} \hline \text{Compound} & \rho/\pu{g cm^-3} \\ \hline \ce{Nb} & 8.57 \\ \ce{Nb2N} & 8.25 \\ \ce{NbN} & 8.47 \\ \hline \end{array} $$
So, you can see that considering different interstitial compounds a general rule cannot be gained.
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2$\begingroup$ 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. $\endgroup$– andselisk ♦Jul 1, 2021 at 13:56
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$\begingroup$ @andselisk The description of density based on crystallographic / diffraction data in $\pu{x Mg/m^{3}}$ with x a real/floating number is pretty typical e.g., for the supplementary information of IUCr's Acta Cryst E. It is somewhat odd, though kg (and not g) is the basis unit in the joint convention of IUPAC and IUPAP (IUPAC's Green Book). For the number, it is the same as for $\pu{g/cm^{3}}$ - though one could say both are off by a factor of 1000 to $\pu{kg/m^3}$ which would use base units only. $\endgroup$ Mar 9 at 7:38
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It depends on what two components of the alloys. As for your question on why adding atoms to the lattice void doesn't increase density, it depends on what the second component of the alloy is. If it is smaller, the decreased distance between electrons and the nucleus of the added component means stronger Coloumbic attractions of negative and positive. This increased attraction means that the atoms are actually closer together than before. On the other hand, if the atoms are bigger, the opposite occurs.
