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A compound made of particles A, B, and C...

A compound made of particles `A`, `B`, and `C` forms `ccp` lattice. In the lattice, ions `A` occupy the lattice points and ions `B` and `C` occuphy the alternate `TV_(s)`. If all the ions along one of the body diagonals are removed, then formula of the compound is

A

`A_(3.75)B_(3)C_(3)`

B

`A_(3.75)B_(3)C_(4)`

C

`A_(3)B_(3.75)C_(3)`

D

`A_(3)B_(3)C_(3.75)`

Text Solution

Verified by Experts

The correct Answer is:
A
a. Since the lattice is `c cp (Z_(eff) = 4)`.
`:.` Number of `A` ions `= 4`
(corner `+` face centre `= 1 + 3 = 4`)
Number of `B` ions `=` Number of alternate `TV = 4`
Number of `C` ions `=` Number of alternate `TV = 4`.
Number of `A` ions removed `= 2 xx (1)/(8)` (corner share)
`= (1)/(4)`

One of the body diagonals. Other diagonals are not shown in the figure
Number of `B` ions removed `= 1`
Number of `C` ions removed `= 1`
(Since body diagonal ions are inside the cube so they do not share with other ions).
Number of `A` ions left `= 4 - (1)/(4) = 3.75`
Number of `B` ions left `= 4 - 1 = 3`
Number of `C` ions left `= 4 - 1 = 3`
Thus, formula is: `A_(3.75)B_(3)C_(3)`
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