Question

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

• A. $4$ and $4$
• B. $2$ and $2$
• C. $1$ and $1$
• D. $3$ and $4$

Answer 1

Answer: (A)

$4$ and $4$

Given that:

The expression for density $\left(d\right)$ is as follows:

$d=\frac{nM}{V{N}_A}⟹n=\frac{dV{N}_A}{M}$

Here, $a$ is the edge length of the unit cell, $M$ is the molecular mass and $V$ is the volume of a unit cell.

Given that:

$a=5\mathring{A}=5\times {10}^{-8}cm,d=3.84gc{m}^{-3}$

putting all the values in the expression of density,

$n=\frac{3.84\times (5\times {10}^{-8}{)}^3\times 6.023\times {10}^{23}}{72}=4$

So, one unit cell contains $4$ $FeO$ molecules or $4F{e}^{2+}$ and $4O^{2-}$ ions.

Hence, the correct option is A.