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

Question

Infinity Learn · Accepted Answer

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 $$AB4C5$$.

	Column I (Distribution of particles ,A and B)		Column II (Formula)
A	A - At the corners and face centres B - At the edge centres and body centre	1	AB₃
B	A - At the corners B - One on each body diagonal	2	AB
C	A - At the corners B - At face centres	3	AB₄

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