A crystalline solid is made of two elements A and B. An ions of B are arranged in ccp lattice and cations of A occupy half of octahedral and half tetrahedral voids . formula of the solid is :

A B

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A_{3} B_{2}

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A_{3} B_{4}

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A_{2} B_{3}

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Solution

The correct option is B $A_{3} B_{2}$
$A$ and $B$
$o c t a = N$
$t e t r a = 2 N$
A occupy $\frac{N}{2} + \frac{2 N}{2}$
$A = \frac{N}{2} + \frac{2 N}{2} = \frac{3 N}{2}$
So, $\frac{A_{3} N}{2} B_{N}$
$\to A_{3} B_{2}$

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A crystalline solid is made of two elements A and B. An ions of B are arranged in ccp lattice and cations of A occupy half of octahedral and half tetrahedral voids . formula of the solid is :

The correct option is B A3B2A and B octa=Ntetra=2NA occupy N2+2N2A=N2+2N2=3N2So,A3N2BN→A3B2

The correct option is B $A_{3} B_{2}$
$A$ and $B$
$o c t a = N$
$t e t r a = 2 N$
A occupy $\frac{N}{2} + \frac{2 N}{2}$
$A = \frac{N}{2} + \frac{2 N}{2} = \frac{3 N}{2}$
So, $\frac{A_{3} N}{2} B_{N}$
$\to A_{3} B_{2}$