Question

.A mixed oxide has ccp arrangement in which the cations 'X' occupy 1/3rd of octahedral voids and the cations 'Y' occupy 1/3rd of tetrahedral voids.The formula of oxide is:

Answer 1

Given that,

The cations 'X' occupy 1/3rd of octahedral voids and the cations 'Y' occupy 1/3rd of tetrahedral voids.

The effective number of oxide ions = 4

We need to calculate the effective number of cations X

Using formula of effective number

$\text{effective number of cations X}=\dfrac{1}{3}\times\text{Number of octahedral voids}$

$\text{effective number of cations X}=\dfrac{1}{3}\times8$

$\text{effective number of cations X}=\dfrac{8}{3}$

We need to calculate the effective number of cations Y

Using formula of effective number

$\text{effective number of cations Y}=\dfrac{1}{3}\times\text{Number of tetrahedral voids}$

$\text{effective number of cations X}=\dfrac{1}{3}\times4$

$\text{effective number of cations X}=\dfrac{4}{3}$

Hence, The formula of the compound will be $A_{2}BO_{3}$

Answer 2

Explanation:

The above answer is wrong. If there are n octahedral voids, there will be 2n tetrahedral voids. So, the correct formula will be XY2O3.