A cubic solid is made by atoms A forming closed packed arrangement, B occupying one/fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of the compound :-

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

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

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A_{4} B C

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

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Solution

The correct option is B $A_{4} B_{2} C_{2}$
In the given solid,
number of atoms A $= 4$
number of atoms B $= \frac{1}{4} \times 8 = 2$
number of atoms C $= \frac{1}{2} \times 4 = 2$
Therefore, the formula of the compound is $A_{4} B_{2} C_{2}$ or $A_{2} B C$

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A cubic solid is made by atoms A forming closed packed arrangement, B occupying one/fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of the compound :-

The correct option is B A4B2C2In the given solid,number of atoms A =4number of atoms B =14×8=2number of atoms C =12×4=2Therefore, the formula of the compound is A4B2C2 or A2BC

The correct option is B $A_{4} B_{2} C_{2}$
In the given solid,
number of atoms A $= 4$
number of atoms B $= \frac{1}{4} \times 8 = 2$
number of atoms C $= \frac{1}{2} \times 4 = 2$
Therefore, the formula of the compound is $A_{4} B_{2} C_{2}$ or $A_{2} B C$