Home
Class 12
CHEMISTRY
In a solid, oxide ions are arranged in c...

In a solid, oxide ions are arranged in ccp. One - sixth of tetrahedral voids are occupied by the cations A and one - third of octahedral voids are occupied by the cations B. What is the formula of the compound ?

Text Solution

Verified by Experts

Let, number of `O^(2-)` ions be 90
Then octahedral voids = 90
Tetrahedral voids = 180
Number of cations `(A)=1//6xx180 = 30`
Number of cations `(B) = 1//3xx90 = 30`
`therefore` Ratio `A : B : O^(2-)=30 : 30 : 90 = 1 : 1 : 3`
The formula of the compound is `ABO_(3)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

In solid oxide are arranged in ccp .One - sixth of tetrabedral voids are occupied by cation A which one third of octahedral voids are occupied by cation B .What is the formula of compound ?

In a compound, oxide ions are arranged in ccp . One-sixth of the tetrahedral voids are occupied by cations. (A) and one-third of OV_(s) are occupieed by cations (B) . (a) What is the formula of the compound of the compound? (b) What are the charges on A and B ?

In a close packed structure of mixed oxides , the lattice is composed of oxide ions , one eighth of tetrahedral voids are occupied by divalent cations while one half of octahedral voids are occupied by trivalent cations . What is the formula of the oxide ?

In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by Zn^(2+) ions and 50% of octahedral holes are occupied by Fe^(3+) ions. The formula of the compound is -

In a solid ,oxide (O^(2-)) ions are arranged in ccp, cations (A^(3+)) occupy one -sixth of tetrahedral void and cations (B^(3+)) occupy one -third of the octahedral voids . What is the formula of the compound?

In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then, R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then, R (radius of octahedral void = 0.414 R). In the spinel structure, oxide ions are cubic close packed whereas 1/8th of tetrahedral voids are occupied by A^(2+) cations and 1/2 of octahedral voids are occupied by B^(3+) cations. The general formula of the compound having spinel structure is

In the spinel structur, oxides ions are cubical-closet packed whereas 1/8th of tetrahedral voids are occupied by A^(2+) cation and 1/2 of octahedral voids are occupied by B^(+) cations. The general formula of the compound having spinel structure is:

In a cubic close packed structure (ccp) of mixed oxides, it is found that lattice has O^(2-) ions and one half of the octahedral voids are occupied by trivalent cations (A^(3+)) and one-eighth of the tetrahedral voids are occupied by divalent cations (B^(2+)) . Derive the formula of the mixed oxide.

In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A^(2+)) while one half of the octahedral voids occupied trivalent ions (B^(3+)) . What is the formula of the oxide ?

In a cubic packed structure of mixed oxides the lattice is made up of oxide ions. One fifth of tetrahedral voids are occupied by divalent (X^(2+)) ions, while one - half of the octahedral voids are occupied by trivalent ions (Y^(3+)) then the formula of the oxide is: