Question

A compound AB has rock salt type structure. If the formula weight of AB is 6.023 Y amu, and the closest A−B distance is Y1/3 nm (where Y is an arbitrary number), find the density of the lattice.

Answer 1

Hint: AB has rock salt type structure, that means the ratio of atoms of A and B in the lattice is $A:B::1:1$.
Density of the lattice structure is given by the formula:
$\begin{align}
  & d=\dfrac{Z\times M}{{{N}_{0}}\times {{a}^{3}}} \\
 & \Rightarrow d=\dfrac{4\times 6.023Y}{6.023\times {{10}^{23}}\times {{({{Y}^{1/3}})}^{3}}} \\
 & \Rightarrow d=4\times {{10}^{-23}}kg{{m}^{-3}} \\
\end{align}$,
where Z is the number of molecules in the whole unit cell, M is the molecular mass of AB, ${{N}_{0}}$ is Avogadro’s number and a is the length of each edge of the lattice.

Complete step-by-step answer:
Let us first analyse the rock-salt type structure to help solve this question more easily with better understanding of the concepts involved.
The unit cell of NaCl consists of Na+ ions and Cl- ions.
There are four types of sites: unique central position, face site, edge sites and corner site, which are used to determine the number of Na+ ions and Cl- ions in the unit cell of NaCl. When counting the number of ions, a corner site would be shared by 7 other unit cells.
Therefore, 1 corner would be 1/8 of an ion. A similar occurrence happens with the face site and the edge sites. For a face site, it is shared by 1 other unit cell and for an edge site, the ion is shared by 3 other unit cells.
NaCl is a face centred cubic unit cell which has four cations and four anions. This can be shown by counting the number of ions and multiplying them in relation to their position.
\[N{{a}^{+}}={{1}_{center}}+{{12}_{edge}}\times \dfrac{1}{4}={{4}_{t}}_{otal}\] ; \[C{{l}^{-}}={{4}_{face}}\times \dfrac{1}{2}+8{{\text{ }}_{corner}}\times \dfrac{1}{8}={{4}_{total}}\]

Thus, the number of molecules in the whole unit cell of a rock-salt structure (Z) = 4.
Each ion in this lattice has six of the other kinds of ions as its nearest neighbours, and twelve of the same kind of ions as its second nearest neighbours.
There are many ionically bonded molecules that take this structure including all other halides of Na, Li, K and Rb. CsF, AgF, AgCl, BaO, CoO, and SrS are also among many that will form similar structures to NaCl.
Now that we have established the basics, let us apply all this knowledge to the given problem.
For the given values of a, Z and M; let us plug these into the formula for lattice density of rock-salt type structures:
$\begin{align}
  & d=\dfrac{Z\times M}{{{N}_{0}}\times {{a}^{3}}} \\
 & \Rightarrow d=\dfrac{4\times 6.023Y}{6.023\times {{10}^{23}}\times {{({{Y}^{1/3}})}^{3}}} \\
 & \Rightarrow d=4\times {{10}^{-23}}kg{{m}^{-3}} \\
\end{align}$

Therefore, per our analysis, the density of the given lattice is $d=4\times {{10}^{-23}}kg{{m}^{-3}}$

NOTE: Be very careful as far as the particulars of the rock-salt structure and the mathematical solution is concerned, as a mistake in either of them will result in an incorrect answer. Also be very careful of the number of effective atoms present in the NaCl type lattice structure, so as to ensure you don’t make a fairly unnecessary mistake in the calculation part.