Solid StateHard
Question
In a compound, oxide ions are arranged in cubic close packing arrangement. Cations A occupy one-sixth of the tetrahedral voids and cations B occupy one-third of the octahedral voids. The formula of the compound is
Options
A.A2BO3
B.AB2O3
C.A2B2O2
D.ABO3
Solution
no. of oxide ions = 4
no. of A particles =
× 8 = 
no. of B particles =
× 4 = 
so formula is A4/3 B4/3O4 or ABO3
no. of A particles =
× 8 = no. of B particles =
× 4 = so formula is A4/3 B4/3O4 or ABO3
Create a free account to view solution
View Solution FreeMore Solid State Questions
Which of the following information(s) is/are incorrect regarding the voids formed in three dimensional HCP of identical ...In an FCC unit cell a cube is formed by joining the centers of all the tetrahedral voids to generate a new cube.Then the...In a face centerd lattice of X and Y, X atoms are present at the corners while Y atoms are at face centers. Then the for...NaCl is doped with 2 × 10–3 mole% SrCl2, the concentration of cation vacancies is (NA = 6 × 1023)...The distance between adjacent oppositely charged ions in rubidium chloride is 328.5 pm, in potassium chloride is 313.9 p...