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

M n_{3} A l_{2} O_{6}

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M n A l_{2} O_{4}

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M n A l_{4} O_{7}

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M n_{2} A l_{2} O_{5}

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Solution

The correct option is B $M n A l_{2} O_{4}$
As oxide ions are present in ccp so number of $O^{2 -}$ ions=4.
Number of octahedral voids=4
Number of tetrahedral voids=8

As $1 / 8$ of the tetrahedral voids are occupied and $1 / 2$ of the octahedral voids are occupied.

Number of $M n^{2 +}$ ions in tetrahedral voids=1

Number of $A l^{3 +}$ ions in octahedral voids=2

So the formula of the mineral is $M n A l_{2} O_{4}$ .

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Similar questions

O^{2 -}

A l^{3 +}

M n^{2 +}

(A^{2 +})

(B^{3 +})

(A^{2 +})

(B^{3 +})

X^{2 +}

Y^{3 +}

A

B

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