logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

In a ccp of mixed oxide, it is found that lattice has $O^{2-}$ ions and one half of the octahedral voids are occupied by trivalent cataions ($X^{3+}$) and one eight of the tetrahedral voids by divalent cataions ($Y^{2+}$). What is the formula of the mixed oxide?

$\begin{array}{1 1} X_2YO_4 \\ X_4YO_4 \\ XY_2O_4 \\ X_2Y_2O_2 \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer: $X_2YO_4$
Number of trivalent catation $= \large\frac{1}{2} $$\times 4 = 2$
Number of divalent catation $= 8 \times \large\frac{1}{8}$$ = 1$
$\Rightarrow $ the formula is $X_2YO_4$
answered Jul 15, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...