Q56 / 180 · JIPMER 2015 · Chemistry

Ferrous oxide has a cubic structure. The length of edge of the unit cell is $5 Å$ and the density of the oxide is $4.0 \mathrm{~g} \mathrm{~cm}^{-3}$. Then the number of $\mathrm{Fe}^{2+}$ and $\mathrm{O}^{2-}$ ions present in each unit cell will be

  • A
    four $\mathrm{Fe}^{2+}$ and two $\mathrm{O}^{2-}$
  • Correct answer
    B
    four $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$
  • C
    two Fe ${ }^{2+}$ and four $\mathrm{O}^{2-}$
  • D
    two Fe ${ }^{2+}$ and two O2-

Solution

$\begin{aligned} & \because d=\frac{Z \times M}{N_A \times V} \\ & \therefore Z=\frac{d \times N_A \times V}{M}\end{aligned}$ As $a=5 Å=5 \times 10^{-8} \mathrm{~cm} \therefore V=\left(5 \times 10^{-8}\right)^3 \mathrm{~cm}^3$ Molecular mass, $M=56+16=72 \mathrm{~g}$ $\begin{aligned} Z & =\frac{4 \times 6.023 \times 10^{23} \times\left(5 \times 10^{-8}\right)^3}{72} \\ & =4.16 \approx 4\end{aligned}$ Four molecules of ferrous oxide, FeO per unit cell, it means that $4 \mathrm{Fe}^{2+}$ and $4 \mathrm{O}^{2-}$ ions are present in each unit cell.

Hence, the correct answer is option: 'B'.

More Solid State Questions