Question

A compound made of particles `A` and `B`. A forms `fc c` packing and `B` occupies all the `OV_(s)`. If all the particles along the plane as shown in the figure below are removeed, then the simplest formula of the compound is

A. `A_(5)B_(7)`
B. `A_(7)B_(5)`
C. `AB`
D. `AB_(3.75)`

Answer 1

Correct Answer - C
c. Since the lattice is `fc c (Z_(eff) = 4)`, therefore, numbered of `A` ions `= 4 (1` corner `+ 3` face centre)
`(OV_(s)` are formed at body centre andedge centre). ltbr. Lons are removed from the plane `(PQRS)` as shown in the figure. From the figure, it is clear, `4` corner ions and two edge centre ions on `PQ` and `RS` are alos removed.
Also two face centre ions lying on `PR` and `QS` along with one ions on present in the body centre are removed as shown in the figure below.
b

`:.` Number of `A` ions removed
`= 4 xx (1)/(8)` (corner share) `+ 2 xx (1)/(2)` (face centre share)
`= (1)/(2) + 1 = 1(1)/(2)`
Number of `B` ions removed
`2 xx (1)/(4)` (edge centre share) `+ (1)/(1)` (body centre share)
`= (1)/(2) + 1 = 1(1)/(2)`
Number of `A` ions left `= 4 -1(1)/(2) = 2.5`

Number of `B` ions left `= 4 -1(1)/(2) = 2.5`
Thus, formula `= A_(2.5) B_(2.5) = 2.5 AB`
Simplest formula `= AB`