In a crystalline solid, anion B are arranged in CCP lattice and cation...
No of atoms in a CCP unit cell=similar to FCC=4
Therefore no of B atoms present=4
No of A atom=tetrahedral void and octohedral void(4+2)=6
The formula of the compound =A3B2
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In a crystalline solid, anion B are arranged in CCP lattice and cation...
B is 50% of octahedral void = 50% × 6 =3 A is 50% of tetrahedral void = 50% × 4 =2 A3B2 is answer.
In a crystalline solid, anion B are arranged in CCP lattice and cation...
Crystalline solid with CCP lattice
- A crystalline solid is a solid material whose atoms, molecules, or ions are arranged in an ordered pattern extending in all three spatial dimensions.
- CCP (cubic close-packed) lattice is a type of arrangement where the spheres (atoms, molecules, or ions) are packed as close together as possible.
Anion B arranged in CCP lattice
- Anion B is a negatively charged ion.
- In a CCP lattice, the anions are arranged in a cubic array such that each ion is in contact with 12 others.
Cations A occupy 50% of octahedral voids and 50% of tetrahedral voids
- Cations A are positively charged ions.
- Octahedral voids are the spaces between six anions arranged octahedrally.
- Tetrahedral voids are the spaces between four anions arranged tetrahedrally.
- Cations A occupy 50% of the octahedral voids and 50% of the tetrahedral voids, which means that they are present in equal numbers in both types of voids.
Formula of the solid is A3B2
- The formula of the solid can be determined by considering the ratio of cations to anions.
- In this case, there are three cations (A) for every two anions (B), which gives the formula A3B2.
- This formula is consistent with the arrangement of cations in both octahedral and tetrahedral voids.
- Therefore, option B (A3B2) is the correct answer.