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Question

In solid AB, B forms the fcc lattice and A+ occupies all the octahedral voids. If atoms from the edge centers along an axis connecting the opposite edge centers on a same face are removed, what will be the empirical formula of the compound?

A
A8B5
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B
A7B8
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C
A8B5
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D
A3B8
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Solution

The correct option is B A7B8
In AB structure, B ions form fcc and A+ ions are present at edge centers and body centers.
Number of atoms of B per unit cell =4
Number of atoms of A per unit cell =14×12+1=3+1=4
Each atom at the centre of the edge contributes one fourth to the cube.
Now, if atoms from edge centers along an axis joining opposite edges on a same face are removed then,
Effective number of atoms of A removed =14×2=12
Effective number of atoms of A present =412=72
So, A:B72:4=7:8
So, formula of the solid is A7B8

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