Solid State Question 181
Question: An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be
[CBSE PMT 2004; AIEEE 2005]
Options:
A) $ AB $
B) $ A_2B $
C) $ AB_3 $
D) $ A_3B $
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Answer:
Correct Answer: C
Solution:
A atoms are at eight corners of the cube. Therefore, the no. of A atoms in the unit cell = $ \frac{8}{8}=1 $ . B atoms are at the face centre of six faces. Therefore, its share in the unit cell = $ \frac{6}{2}=3 $ . The formula is AB3.
BETA
