Q.

In a unit cell, atoms (A) are present at all corner lattices, (B) are present at alternate faces and all edge centers. Atoms (C) are present at face centers left from (B) and one at each body diagonal at distance of 1/4th of body diagonal from corner. A tetrad axis is passed from the given unit cell and all  the atoms touching the axis are removed. The possible formula of the compound left is

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a

AB3C6 and AB4C5

b

A3B6C7 and A3B6C5

c

A4B5C8 and A4B5C7

d

AB2C and ABC2

answer is A.

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Detailed Solution

Case I If the tetrad axis passes through the face centres where B lies, thenNumber of B atom =4−12×2=3The formula of the compound left is AB3C6.Case II If the tetrad axis passes through the face centre s where C lies, thenNumber of C atoms =6−12×2=5The formula of the compound left is AB4C5.
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