If the unit cell of a minearl has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahderal holes occupied by aluminium ions and n fraction of tetrahdral holes occupied by magnesium ions, m and n, respectively are

If the unit cell of a minearl has a cubic close packed (ccp) array of

1.	If the unit cell of a minearl has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahderal holes occupied by aluminium ions and n fraction of tetrahdral holes occupied by magnesium ions, m and n, respectively are
Answer» `1/2 , 1/8` `1,1/4` `1/2, 1/2` `1/4,1/8` Solution :In ccp LATTICE , Z =4No. 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 )`