A solid `A^(+)B^(-)` has the `B^(-)` ions arranged as below. If the `A^(+)` ions occupy half of the tetrahedral sites in the structure. The formula of solid is :
 

Options

A. AB
B. `AB_(2)`
C. `A_(2)B`
D. `A_(3)B_(4)`

Answer

A

Explanation

F is Fcc lattice
Positions : Face centres `=6xx(1)/(2)=3`, Corners `=8xx(1)/(8)=(1)/(4),Z=4`
So tetrahedral voids `= 8` , Octahedral voids `= 4`
`B^(-)` arranged in Face cneter corner `therefore B_(4)`
`A^(+)` occupies half of tetrahedral , sits `rArr (1)/(2)xx8=4"    "A_(4) therefore A_(4)B_(4)=AB`