Question

Let $MgTi{O}_3$ exist in a perovskite structure in which $M{g}^{2+}$ ions are present at corner, $Ti$ are present at body centre and $O^{2-}$ are present at face centre in a unit cell. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the density of unit cell if the radius of $M{g}^{2+}$ is 1.0 $\AA$ and the corner ions are touching each other.
[Given atomic mass of $Mg=24,Ti=48$]

• A. New formula is $M{g}_{3}T{i}_4{O}_{10}$
• B. New formula is $M{g}_{3}T{i}_2{O}_5$
• C. Formula mass of new compound = 424 amu
• D. Density = $56{\text{g/cm}}^3$

Answer 1

Answer: (A), (C)

The correct options are
A New formula is $M{g}_{3}T{i}_4{O}_{10}$
C Formula mass of new compound = 424 amu

$\text{Number of}M{g}^{2+}\text{per unit cell}=8\text{(At corners)}\times \frac{1}{8}=1$
$\text{Number of}Ti\text{per unit cell}=1\text{(body centre)}\times \frac{1}{1}=1$
$\text{Number of}O\text{per unit cell}=6\text{(Face centre)}\times \frac{1}{2}=3$
$\text{So, formula}=MgTi{O}_3$
$\text{Atoms are removed along a face diagonal}$
$\text{Number of}M{g}^{2+}=6\text{(At corner)}\times \frac{1}{8}=\frac{6}{8}=\frac{3}{4}$
$\text{Number of}Ti\text{per unit cell}=1\text{(Body centre)}\times \frac{1}{1}=1$
$\text{Number of}O\text{per unit cell}=5\text{(Face centre)}\times \frac{1}{2}=\frac{5}{2}$
$\text{So, formula of compound}=M{g}_{\frac{3}{4}}Ti{O}_{\frac{5}{2}}$
$\text{Or, formula of compound}=M{g}_{3}T{i}_4{O}_{10}$
$\text{Formula mass}=24\times 3+48\times 4+16\times 10=18+48+40=424\text{amu}$
$\text{As corner ion are touching so}=a=2r_{M{g}^{2+}}=2\times 1.0=2\AA$
$d=\frac{\text{mass}}{\text{volume}}=\frac{424\times 1.67\times {10}^{-24}}{(2{)}^3\times {10}^{-24}}{\text{g/cm}}^3=88.51{\text{g/cm}}^3\approx 88.5{\text{g/cm}}^3$