A cubic solid is made by atoms A forming close packed arrangement, B occupying one fourth of tetrahedral voids and C occupying half of the octahedral voids. What is the formula of the compound?

A_{4} B_{2} C_{2}

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A_{3} B_{3} C_{2}

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A_{4} B_{3} C_{4}

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None of the above

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Solution

The correct option is A $A_{4} B_{2} C_{2}$
Since A forms close packed arrangement,
Number of atoms A in a unit cell = $4$
B occupies one-fourth of tetrahedral voids,
Number of atoms B in a unit cell = $\frac{1}{4} \times 8 = 2$
C occupies half of octahedral voids,
Number of atoms C in a unit cell = $\frac{1}{2} \times 4 = 2$
Therefore, the formula of the compound is $A_{4} B_{2} C_{2}$

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