In a F.C.C. arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centres. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is

$A_{7} B_{3}$

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$A B_{3}$

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$A_{2} B_{24}$

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$A_{8} B_{21}$

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Solution

The correct option is C
$A_{2} B_{24}$

Effective number of ‘A’ atoms
$= 7 \times \frac{1}{8} = \frac{7}{8}$
Effective number of ‘B’ atoms
$= 6 \times \frac{1}{2} = 3$
$∴$ formula $= A_{\frac{7}{8}} B_{3} = A_{7} B_{24}$

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In a F.C.C. arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centres. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is

The correct option is C A2B24 Effective number of ‘A’ atoms =7×18=78 Effective number of ‘B’ atoms =6×12=3 ∴ formula =A78B3=A7B24

The correct option is C
$A_{2} B_{24}$

Effective number of ‘A’ atoms
$= 7 \times \frac{1}{8} = \frac{7}{8}$
Effective number of ‘B’ atoms
$= 6 \times \frac{1}{2} = 3$
$∴$ formula $= A_{\frac{7}{8}} B_{3} = A_{7} B_{24}$