Solid State - Result Question 26
26. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with $m$ fraction of octahedral holes occupied by aluminium ions and $n$ fraction of tetrahedral holes occupied by magnesium ions, $m$ and $n$ respectively, are
(a) $\dfrac{1}{2}, \dfrac{1}{8}$
(b) $1, \dfrac{1}{4}$
(c) $\dfrac{1}{2}, \dfrac{1}{2}$
(d) $\dfrac{1}{4}, \dfrac{1}{8}$
(2015 Adv.)
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Answer:
Correct Answer: 26. (a)
Solution:
- Oxide ions are at ccp positions, hence $4 O^{2-}$ ions. Also, there are four octahedral voids and eight tetrahedral voids. Since ’ $m$ ’ fraction of octahedral voids contain $Al^{3+}$ and ’ $n$ ’ fraction of tetrahedral voids contain $Mg^{2+}$ ions, to maintain etectroneutrality $2\left(2 Al^{3+}=+6\right.$ charge $)$ and $\left(Mg^{2+}=+2\right.$ charge $)$, will make unit cell neutral
Hence: $m=\dfrac{2}{4}=\dfrac{1}{2}, n=\dfrac{1}{8}$
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