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
a. `AB`, Frenkel defect
b. `A_(2)B`, Frenkel defect
c. `AB`, Schottky defect
d. `A_(2)B`, Schottky defect

Options

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

Answer

C

Explanation

(i) Since `A^(o+)B^(o+)` has NaCl-type (fcc) structure.
So `B^(o+)` are in fcc arrangement.

Removed atoms from one of the face Number of `B^(o+)` ions=4(corner+face center=1+3=4)
Number of `A^(o+)` ions =4 [in oV formed at body center+edge center =1+3=4]
Number of `B^(o+)` ions removed =4 corner `xx(1)/(8)` ltbr. per corner share+1 face center `xx(1)/(2)` per face
center share `=(4)/(8)+(1)/(2)=1`
Number of `B^(o+)` left `=4-1=3`
Number of `A^(o+)` ions removed=4 edge center `xx(1)/(4)` per edge center share `=(4)/(4)=1`
Number of `A^(o+)` ions removed =4 edge center
`xx(1)/(4)` per edge center share `=(4)/(4)=1`
Number of `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 in Schottky type.