In CaF2 crystal, Ca2+ ions are present in fcc arrangement, calculate the number of F- ions in the unit cell.
Dear student
Given that Ca2+ ion are in the fcc arrangement ,
So there are 4 octahedral voids present in the fcc lattice. [8x1/8+ 6x1/2 =4, i,e 8 atoms each with a contribution of 1/8 present in the corners and 6 atom each with a contribution of 1/2 at the edges]
The F- ions occupy the tetrahedral holes which are twice that of octahedral holes = 2 x 4 =8
Thus there are 8 F- present in the CaF2 crystal.
Regards
Given that Ca2+ ion are in the fcc arrangement ,
So there are 4 octahedral voids present in the fcc lattice. [8x1/8+ 6x1/2 =4, i,e 8 atoms each with a contribution of 1/8 present in the corners and 6 atom each with a contribution of 1/2 at the edges]
The F- ions occupy the tetrahedral holes which are twice that of octahedral holes = 2 x 4 =8
Thus there are 8 F- present in the CaF2 crystal.
Regards