Question

Ferrous oxide has a cubic structure with an edge length of the unit cell equal to $5.0~A.$ Assuming the density of ferrous oxide to be $3.84~gc{{m}^{-3}},$ the number of $F{{e}^{2+}}~$ and ${{O}^{2-}}$ ions present in each unit cell will be respectively, (use ${{N}_{A}}=6\times {{10}^{23}}$ )

Answer 1

In the Ferrous oxide structure, both the ions are arranged so that they show a cubic arrangement. The number of $F{{e}^{2+}}$ ions in each unit cell is equal to the number of ${{O}^{2-}}$ ions. Further, one can also say that there is one ${{O}^{2-}}$ ion and one $F{{e}^{2+}}$ ion behind every three vertices.

Complete step by step answer:
First let us understand the cubic structure and then we will move step by step on the answer. So, the cubic structure is one in which atoms are arranged so that they form a cube-like appearance with all sides equal. Ferrous oxide is expected to have a cubic structure. Now, in the question we are given; the expression for density (d) is as follows:
$d=\dfrac{nM}{V{{N}_{A}}}\Rightarrow n=\dfrac{dV{{N}_{A}}}{M}$
Let us move step by step to answer. First we can find volume from the edge. We know that volume of a cubic unit cell is as; Volume of unit cell $={{(side)}^{3}}$
$~a=5A=5\times {{10}^{-8}}cm,d=3.84gc{{m}^{-3}}$
Now, putting all the values in the expression of density,
$n=\dfrac{3.84\times {{\left( 5\times {{10}^{-8}} \right)}^{3}}\times 6.023\times {{10}^{23}}}{72}$
Thus, So, one unit cell contains $4~FeO$ molecules or $4F{{e}^{2+}}$ and $4{{O}^{2-}}$ ions.
Therefore, the correct answer is option A.

Note: Always check that all the values have the same units as in the question the edge length is mentioned in the ${{A}^{o}}$ while other values are in cm. So, we should convert and get one unit first. We know that one mole of a substance always contains ${{N}_{A}}=6\times {{10}^{23}}$ number of molecules or ions or atoms.