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Question: . If radius of a metal atom A is 5 pm and radius of an electronegative atom B is 20 pm, then in the ...

. If radius of a metal atom A is 5 pm and radius of an electronegative atom B is 20 pm, then in the unit cell:
A.A in octahedral voids, B in FCC unit
B.A in FCC unit, B in tetrahedral void
C.A in BCC unit, B in cubic void
D.A in tetrahedral void, B in FCC unit

Explanation

Solution

Since the metal atom and electronegative atom radius is given, we can say the electronegative atom is anion and other is cation. Radius of these are already given and using the cation-anion radius ratio or radius ratio, we can calculate its coordination which can ultimately tell the type of void present and which will lead to the answer.

Complete step by step answer:
In the given question it is given that metal atom A has a radius of 5 pm and there is another electronegative atom B with 20 pm radius and we have to find the structure of this unit cell i.e. location of A and B in the unit cell.
From the given question, we have predicted one thing that A is a metal atom and B is electronegative atom then, we can say that A will be cationic in nature and B will be anionic in nature (because electronegative atom attracts the pair of electrons towards itself). Now on the basis of cationic and anionic radius, we can find out the radius ratio of the unit cell or also called the Cation-Anion radius ratio.
The ionic radius changes on the basis of coordination number, because if ionic size is more, the atoms will move further and fit into the unit cell. So ionic size can give us an idea about coordination numbers which can further tell the structure.
So, Radius ratio is defined as the ratio of radius of cation to the radius of anion.
If we calculate the radius ratio of the given question, Radius ratio = radius of cationradius of anion=520= 0.25Radius{\text{ }}ratio{\text{ }} = {\text{ }}\dfrac{{radius{\text{ }}of{\text{ }}cation}}{{radius{\text{ }}of{\text{ }}anion}} = \dfrac{5}{{20}} = {\text{ }}0.25
So, the ratio lies in the range of 0.225  0.4140.225{\text{ }}-{\text{ }}0.414 which means coordination number 44 i.e. tetrahedral
We know that tetrahedral void found generally in FCC in which anions are in the FCC structure and cation will occupy tetrahedral voids.

So, the correct answer is D.

Note:
Radius ratio helps to know the coordination number of cation and anion present and on the basis of coordination number, we can predict the structure.
Radius ratio table is as follows which could help in answering the similar questions:

Radius RatioCoordination NumberType of Void
< 0.155 < {\text{ }}0.1552Linear
0.1550.2250.155 - 0.2253Triangular Planar
0.2250.4140.225 - 0.4144Tetrahedral
0.4140.7320.414 - 0.7326Octahedral
0.73210.732 - 18Cubic