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 magnesium ions, m and n, respectively, are

Step-by-step solution

In ccp, O 2 - ions are 4. Hence total negative charge = -8 Let A l 3 + ions be x, and M g 2 + ions be y. Total positive charge = 3x + 2y ⇒ 3 x + 2 y = 8 This relation is satisfied only by x = 2 and y = 1. Hence number of A l 3 + = 2 and number of M g 2 + = 1 ⇒ n = fraction of octahedral holes occupied by A l 3 + = 2 4 = 1 2 and m = fraction of tetrahedral holes occupied by M g 2 + = 1 8