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In a solid, oxide ions are arranged in CCP. one-sixth of the tetrahedral voids are occupied by A cations and one-third of the octahedral voids are occupied by B cations. The formula of solid is :-

In a solid, oxide ions are arranged in CCP. one-sixth of the tetrahedral voids are occupied by A cations and one-third of the octahedral voids are occupied by B cations. The formula of solid is :-

Grade:12

3 Answers

Sellavel
11 Points
6 years ago
In CCP,atoms are arranged in corners and Face centred lattice,
and therefore total no of atoms in ccp=4
In ccp no. of tetrahedral voids=2xN,where N is the total No. Of atoms in ccp
Similarly,no. of octahedral Voids=N
Coming to your question,
No. of oxide ions=4 and therefore N=4
 
No. of A atoms=1/6 th tetrahedral voids
No.of A atoms=(1/6)x2x4=4/3
No.of B atoms=(1/3)x4=4/3
The formula is...
A_{\frac{4}{3}}B_{\frac{4}{3}}O_{4}
on simplifying the formula we will get,
A_{4}B_{4}O_{12}
Dhruv mishra
13 Points
4 years ago
Number of oxide ions forming ccp=4
(occupy 8 corners and 6 face centres)
No of tetrahedral voids=8
No of octahedral voids=4
No of cations A occupying tetrahedral voids = 8/6
No of cations B occupying octahedral voids=4/3 
A:B:O—2
8/6:4/3:4
8:8:24
1:1:3
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Number of oxide ions forming ccp=4
(occupy 8 corners and 6 face centres)
No of tetrahedral voids=8
No of octahedral voids=4
No of cations A occupying tetrahedral voids = 8/6
No of cations B occupying octahedral voids=4/3
A:B:O—2
8/6:4/3:4
8:8:24
1:1:3

Thanks and Regards

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