Question

The olivine series of minerals consists of crystal in which $F{e}^{2+}$ and $M{g}^{2+}$ ions may substitute for each-other causing substitutional impurity defect without changing the volume of the unit cell. In olivine series of mineral, oxide ion exist as fcc with $S{i}^{4+}$ occupying $\frac{1}{4}$th of octahedral void and divalent ion occupying $\frac{1}{4}$th of tetrahedral void. The density of forsterite (Magnesium silicate) is $3.21$ g/cc and that of fayalite (Ferrous silicate) is $4.34$ g/cc. If density of olivine is $3.88$ g/cc, then which of the following statement is/are correct?

• A. Forsterite is $M{g}_{2}Si{O}_4$ and Fayalite is $F{e}_{2}Si{O}_4$.
• B. An olivine contains $40.71\%$ Forsterite and $59.29\%$ Fayalite.
• C. Percentage of Forsterite, $M{g}_{2}Si{O}_4$ is $59.29$.
• D. Both $A$ and $B$

Answer 1

Answer: (A), (B)

The correct options are
A Forsterite is $M{g}_{2}Si{O}_4$ and Fayalite is $F{e}_{2}Si{O}_4$.
B An olivine contains $40.71\%$ Forsterite and $59.29\%$ Fayalite.

According to given data :

Number of atoms of $O^{2-}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

Number of atoms of $S{i}^{4+}=\frac{1}{4}\times$ octahedral void $=\frac{1}{4}\times 4=1$

Number of atoms of $M{g}^{2+}=\frac{1}{4}\times$ tetrahedral void $=\frac{1}{4}\times 8=2$

$\because$ Forsterite $=M{g}_{2}Si{O}_4$ and Fayalite $=F{e}_{2}Si{O}_4$

Let the forsterite is $x\%$ and fayalite is $(100-x)\%$, then

$\frac{x\times 3.21+(100-x)\times 4.34}{100}=3.88$ $⟹x=40.71\%$ (Forsterite)

Therefore, percentage of Fayalite is $59.29\%$.