In a fcc lattice, A, B, C and D atoms are arranged at corner, face center, octahedral void and tetrahedral void respectively, then the body diagonal contains :

2 A, C, 2 D

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2 A, 2 B, 2 C

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2 A, 2 B, D

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2 A, 2 D

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Solution

The correct option is A $2 A, C, 2 D$
In fcc, along the body diagonal, $2$ tetrahedral voids and one octahedral void at body center is present. So, a body diagonal contains $2 A, C$ and $2 D$ .

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QUES. In a closed packed arrangement ,A type of atoms are at corners , B type of atoms are at face corners , C type of atoms are at tetrahedral voids and D type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed. Then what will be the formula of compound?

Q. Prove that :

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a) Each body diagonals contains 2 tetrahedral voids
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From the sum of 7a – 2b + 11 and -2a -5, subtract 7a – b + 4.

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In a fcc lattice, A, B, C and D atoms are arranged at corner, face center, octahedral void and tetrahedral void respectively, then the body diagonal contains :

The correct option is A 2A, C, 2DIn fcc, along the body diagonal, 2 tetrahedral voids and one octahedral void at body center is present. So, a body diagonal contains 2A, C and 2D.

The correct option is A $2 A, C, 2 D$
In fcc, along the body diagonal, $2$ tetrahedral voids and one octahedral void at body center is present. So, a body diagonal contains $2 A, C$ and $2 D$ .