Question

Pyrite ($Fe{S}_2$), crystallizes in a cubic unit cell with a cell edge of $0.540\text{nm}$. The density of pyrite is $5.02{\text{g/cm}}^3$. How many $F{e}^{2+}$ and $S_2^{2-}$ ions are present in a single unit cell? ($S=32\text{u},Fe=56\text{u}$)

• A. $1F{e}^{2+}\text{and}1S_2^{2-}$
• B. $2F{e}^{2+}\text{and}2S_2^{2-}$
• C. $4F{e}^{2+}\text{and}4S_2^{2-}$
• D. $6F{e}^{2+}\text{and}6S_2^{2-}$

Answer 1

The correct option is C $4F{e}^{2+}\text{and}4S_2^{2-}$

We know that,
$d=\frac{ZM}{a^3{N}_A}$
d - density of the crystal,
Z- Number of unit cell
MM - Molecular mass
a - edge length
$N_A$ - Avagadro number

$5.02=\frac{n\left(120\right)}{6.023\times {10}^{23}(5.4\times {10}^{-8}{)}^3}$
$\Rightarrow n=4$, i.e., $4F{e}^{2+}$ and $4S_2^{2-}$