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Question

(A) In a $ccp$ lattice,the number of oxygen atoms per unit cell is $4$.The number of octahedral voids is $4$ and the number of tetrahedral voids is $8

Vedclass · Accepted Answer

$\frac{1}{2}, \frac{1}{8}$

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(A) In a $ccp$ lattice,the number of oxygen atoms per unit cell is $4$.The number of octahedral voids is $4$ and the number of tetrahedral voids is $8$.Let the number of $Al^{3+}$ ions be $4m$ and the number of $Mg^{2+}$ ions be $8n$.For charge neutrality,the total positive charge must equal the total negative charge:$4(-2) + 4m(+3) + 8n(+2) = 0$$-8 + 12m + 16n = 0$$12m + 16n = 8$Dividing by $4$,we get $3m + 4n = 2$.Testing the options,for $m = \frac{1}{2}$ and $n = \frac{1}{8}$:$3(\frac{1}{2}) + 4(\frac{1}{8}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$.Thus,$m = \frac{1}{2}$ and $n = \frac{1}{8}$ satisfies the condition.

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

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