If all the atoms touching one face plane are removed in solid `A^(o+)B^(Θ)` having rock salt type structure, then the formula of the compound left and the defect brought by this removal, respectively, is

Options

A. `AB`, Frenkel defect
B. `A_(2)B`, Frenkel defect
C. `AB`, Schottky defect
D. `A_(2)B`, Schottky defect

Answer

C

Explanation

c. (i) Since `A^(o+)B^(Θ)` has `NaCl`-type `(fc c)` structure.
So `B^(Θ)` are in `(fc c)` arramgement.

Number if `B^(Θ)` ions `= 4`
(Corner `+` face centre `= 1 + 3 = 4)`
Number if `A^(o+)` ions `= 4`
[ in `OV` formed at body centre `+` edge centre `= 1 + 3 = 4]`

Number if `B^(Θ)` ions removed
`= 4` corner `xx (1)/(8)` per corner share
`+ 1` face centre `xx (1)/(2)` per face centre share
`= (4)/(8) + (1)/(2) = 1`

Number if `B^(Θ)` left `= 4 - 1 = 3`

Number if `A^(o+)` ions removed
`= 4` edge centre `xx (1)/(4)` per edge centre share `= (4)/(4) = 1`
Number if `A^(o+)` ions left `= 4 - 1 = 3`
Thus, formula `= A_(3)B_(3) = 3AB.`
ii. As Atoms (or ions) are completely removed.
Hence, the defect in the crystal is Schottky type.