Questions
Match the following columns.
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 |
detailed solution
Correct option is B
(A). A atoms are at the corners and face centers, so number of A atoms =8×18+12×6=1+3=4 corner atoms Face center atoms B atoms are at the edge centers and body center, so number of B atoms per unit cell =12×14+1=3+1=4 Ratio of A and B = 4 : 4 = 1:1 so formula is AB.(B).A atoms are at the corners, so number of A atoms per unit cell =8×18=1 atom. B atoms are at the body diagonals, so number of B atoms per unit cell = 4 x 1 = 4. ∴ Formula of molecule = AB4.(C). When A atoms are at the corners and B atoms are at face centers, the formula is AB3.Similar Questions
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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