Question

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

Answer 1

f 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

Answer 2

m=1/2 and n=1/8

Explanation:

In CCP, the number of atoms in a unit cell = 4

So, the number of O atoms = 4

We know that in CCP, the number of tetrahedral voids = 2n

Number of octahedral voids = n

Where,

n is the number of atoms.

Number of tetrahedral voids = 8

Number of octahedral voids = 4

The number of octahedral voids is given by m, so the number of $Al^{3+}$ atoms = 4 × m

The number of tetrahedral voids is given by n, so the number of $Mg^{2+}$ atoms = 8 × n

The total negative charge of the cell is -8 because the number of O atoms is 4. Since it is a neutral cell, the negative and positive charges would be the same. Therefore the positive charge = 8

The positive charge is made from 3 aluminium and 2 magnesium ions.

So we have,

3m + 2n = 8

The values that satisfy the equation would be m = 2 and n = 1

n is the fraction occupied by tetrahedral voids whose number is 8, so n = 1/8

m is the fraction occupied by octahedral voids whose number is 4, so m = 2/4 = 1/2

We get the values of m = 1/2 and n = 1/8