Question

A compound made of particles `A`, `B`, and `C` forms `ccp` lattice. Ions `A` are at lattice points, `B` occupy `TV_(s) C` occupy `OV_(s)`. If all the ions along one of the edge axis are removed, then formula of the compound is
A. `A_(3.75)B_(8)C_(3.75)`
B. `A_(3.75)B_(4)C_(8)`
C. `A_(4)B_(8),C_(3.75)`
D. `A_(4)B_(3.75)C_(8)`

Answer 1

Since lattice is ccp, `Z_(ff)=4`
`therefore` Number of A ions=4 (corner+face center=1+3=4)
Number of B ions=Number of TVs=8
Number of C ions=Number of Ovs=4
Number of A ions removed `=2xx(1)/(8)`
(corner share) `=(1)/(4)`
Number of C ions removed `=1xx(1)/(4)`
(edge center share `=(1)/(4)`
Number of A ions left `=4-(1)/(4)=3.75`
Number of B ions=8
(Since no B ion has been removed)
Number of C ions left `=4-(1)/(4)=3.75`
Thus, formula is : `A_(3.75)B_(8)C_(3.75)`.