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 magnesiums ions, m and n respectively, are

Options

A. `1/2 , 1/8`
B. `1,1/4`
C. `1/2, 1/2`
D. `1/4,1/8`

Answer

A

Explanation

In ccp lattice , Z =4 No. of O-atoms per unit cell = `4 ( O^(2-))`
No.of octahedral voids = 4 and No. of tetrahedral voids = 8.
As m fraction of octahedral voids is occupied by `Al^(3+)` ions. Therefore, `Al^(3+)` ions present = 4 m. similarly , ` Mg^(2+)` ions= 8 n. Hence, formula of the mineral is ` Al_(4m) Mg_(8n) O_(4)` . As total charge on the compound is zero, hence.
4 m ( +3) + 8 n ( +2) + 4( -2) =0
or 12 m + 16 n -8 =0
Substituting the given values of m and n m, equation is satisfied only when ` m = 1/2 and n = 1/8`
` ( as 12 xx 1/2 + 16 xx 1/8 -8 =0 )`