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In a fcc lattice A, B, C and D atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing correspondingly then what is the resulted formula of the compound?

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$A B_{4} C_{4} D_{4}$

$A_{3} B_{3} C_{2} D_{4}$

$A B_{3} C_{4} D_{4}$

$A_{2} B_{3} C_{3} D_{4}$

answer is B.

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Detailed Solution

Without defect formula of the compound would be : ${AB}_{4} C_{4} D_{4}$

Due to defect Missing particles per unit cell are

A B C D
1/4 0 1 1

Hence the formula is $A_{3 / 4} B_{3} C_{3} D_{3} or {AB}_{4} C_{4} D_{4}$

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