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 . `therefore` No. of O-atoms per unit cell= `4(O^(2-))` .
`therefore` 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 , equation is satisfied only when `m=1/2` and `n=1/8`
(as `12 xx1/2+16xx1/8-8=0`)