In a compound ,oxide ions are arranged in cubic close packing arrangement. Cation ‘A’ occupy one-sixth of the tetrahedral voids and ‘B’ occupy one-third of the octahedral voids then the molecular formula of the compound is

In one unit cellnumber of O-2 ions in ccp=4No. of A ions =16×(tetrahedral voids)Number of 'A' ions=16×(8)=43No. of B ions=13×(octahedral voids)Number of ‘B’ ions =13×(4)=43Number of ‘O’ =8x18+6x12=1+3=4Molecular formula of the compound is=A43B43O4 =ABO3