In a crystalline solid, anions $B$ are arranged in ccp lattice and cations $A$ occupy $50$ % of the octahedral voids and $50$ % of the tetrahedral voids. What is the formula of the solid?

A B

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A_{3} B_{2}

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A_{2} B_{2}

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A_{2} B_{4}

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Solution

The correct option is B $A_{3} B_{2}$
No of atoms in a CCP unit cell=similar to FCC=4
Therefore no of B atoms present=4
No of A atom=tetrahedral void and octohedral void(4+2)=6
The formula of the compound = $A_{3} B_{2}$

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In a crystalline solid, anions B are arranged in ccp lattice and cations A occupy 50% of the octahedral voids and 50% of the tetrahedral voids. What is the formula of the solid?

The correct option is B A3B2No of atoms in a CCP unit cell=similar to FCC=4Therefore no of B atoms present=4No of A atom=tetrahedral void and octohedral void(4+2)=6The formula of the compound = A3B2

In a crystalline solid, anions $B$ are arranged in ccp lattice and cations $A$ occupy $50$ % of the octahedral voids and $50$ % of the tetrahedral voids. What is the formula of the solid?

The correct option is B $A_{3} B_{2}$
No of atoms in a CCP unit cell=similar to FCC=4
Therefore no of B atoms present=4
No of A atom=tetrahedral void and octohedral void(4+2)=6
The formula of the compound = $A_{3} B_{2}$