(B) In a cubic close-packed $(CCP)$ structure,the number of $O^{2-}$ ions per unit cell is $4$. The number of octahedral voids is equal to the number

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(B) In a cubic close-packed $(CCP)$ structure,the number of $O^{2-}$ ions per unit cell is $4$. The number of octahedral voids is equal to the number of $O^{2-}$ ions,which is $4$. The number of tetrahedral voids is twice the number of $O^{2-}$ ions,which is $8$. Number of $Al^{3+}$ ions = $\frac{1}{2} \times 4 = 2$. Number of $Mn^{2+}$ ions = $\frac{1}{8} \times 8 = 1$. Therefore,the ratio of $Mn : Al : O$ is $1 : 2 : 4$. The chemical formula of the mineral is $MnAl_{2}O_{4}$.

Class 12 Chemistry Solid State Crystal packing

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$A$ mineral consists of a cubic close-packed structure formed by $O^{2-}$ ions,where half the octahedral voids are occupied by $Al^{3+}$ and one-eighth of the tetrahedral voids are occupied by $Mn^{2+}$. The chemical formula of the mineral is

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A
$Mn_{3}Al_{2}O_{6}$
B
$MnAl_{2}O_{4}$
C
$MnAl_{4}O_{7}$
D
$Mn_{2}Al_{2}O_{5}$

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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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$A$ mineral consists of a cubic close-packed structure formed by $O^{2-}$ ions,where half the octahedral voids are occupied by $Al^{3+}$ and one-eighth of the tetrahedral voids are occupied by $Mn^{2+}$. The chemical formula of the mineral is

Similar Questions

In the spinel structure,oxide ions are cubical-closest packed where $1/8$ of tetrahedral voids are occupied by $A^{2+}$ cations and $1/2$ of octahedral voids are occupied by $B^{3+}$ cations. The general formula of the compound having spinel structure is

Calculate the void volume in an $fcc$ unit cell if the total volume of the unit cell is $6.4 \times 10^{-23} \text{ cm}^3$.

If the distance between the corner and the center of a cube is $x$,then the tetrahedral void will be at a distance of .... from the corner of the cube.

$A$ face-centred cubic $(FCC)$ unit cell is made up of two types of atoms $A$ and $B$,in which $A$ occupies the corner positions and $B$ occupies the face centres. If atoms along one of the body diagonals are removed,what is the empirical formula of the remaining solid?

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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