In a face centred lattice with all the positions occupied by $A$ atoms, the body centred octahedral holes in it is occupied by an atom $B$ of an appropriate size for such a crystal. Predict the formulae of the compound.

A B

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A_{4} B

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A_{4} B_{3}

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A_{4} B_{5}

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Solution

The correct option is B $A B$
$Z_{e f f e c t i v e}$ of $A =$ Face centre position $= \frac{1}{2} \times 6 \times \frac{1}{8} \times 8 = 4$
$Z_{e f f e c t i v e}$ of $B = 1 + \frac{1}{12} \times 4 = 4$ (Effective number of tetrahedral void)
Formula $= A_{4} B_{4} = A B$

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In a face centred lattice with all the positions occupied by A atoms, the body centred octahedral holes in it is occupied by an atom B of an appropriate size for such a crystal. Predict the formulae of the compound.

The correct option is B ABZeffective of A= Face centre position =12×6×18×8=4Zeffective of B=1+112×4=4 (Effective number of tetrahedral void)Formula =A4B4=AB

In a face centred lattice with all the positions occupied by $A$ atoms, the body centred octahedral holes in it is occupied by an atom $B$ of an appropriate size for such a crystal. Predict the formulae of the compound.

The correct option is B $A B$
$Z_{e f f e c t i v e}$ of $A =$ Face centre position $= \frac{1}{2} \times 6 \times \frac{1}{8} \times 8 = 4$
$Z_{e f f e c t i v e}$ of $B = 1 + \frac{1}{12} \times 4 = 4$ (Effective number of tetrahedral void)
Formula $= A_{4} B_{4} = A B$