QJEEAdv15Ch6e31: JEE Advanced 2015 Chemistry - Class 12 JEEAdv15Ch6e31 Solution: If the unit cell of a mineral has cubic close packed (ccp)…

For ccp, Z = 4 = no. of O-atoms

No. of octahedral voids = 4

No. of tetrahedral voids = 2 $\times$ 4 = 8

No. of Al³⁺ ions = m $\times$ 4

No. of Mg²⁺ ions = n $\times$ 8

Thus, the formula of the mineral is Al_4m Mg_8nO₄

4m(+3) + 8n(+2) + 4( $-$ 2) = 0

12m + 16n $-$ 8 = 0

4(3m + 4n $-$ 2) = 0

3m + 4n = 2

Possible values of m and n are ${1 \over 2}$ and ${1 \over 8}$ respectively.