SATHEE: Solid State - Result Question 23

Solid State - Result Question 23

####23. In a solid $A B$ having the $N a C l$ structure, $A$ atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

(2001, S, 1M)

(a) $A B_{2}$

(b) $A_{2} B$

(d) $A_{3} B_{4}$

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Answer:

Correct Answer: 23. (d)

Solution:

In $N a C l, N a^{+}$ occupies body centre and edge centres while $C l^{-}$ occupies corners and face centres, giving four $N a^{+}$ and four $C l^{-}$ per unit cell. In the present case $A$ represent $C l^{-}$ and $B$ represents $N a^{+}$ . Two face centres lies on one axis.

$\Rightarrow$ Number of $A$ removed $= 2 \times \frac{1}{2} = 1$

Number of $B$ is removed because it is not present on face centres. $\Rightarrow A$ remaining $= 4 - 1 = 3, B$ remaining $= 4$ ,

Formula $= A_{3} B_{4}$

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Solid State - Result Question 23

Answer:

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