A face-centred cubic Lattice Is made up of two types of atoms, atom A and atom B in which A occupies the corner positions and B occupies the face centres. If atoms from one corner is absent per unit cell, the empirical formula of the remaining solid would be:

A_{7} B_{3}

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A_{7} B_{9}

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A_{7} B_{10}

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A_{7} B_{24}

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Solution

The correct option is D $A_{7} B_{24}$
Number of atoms of A from corners of unit cell $= \frac{7}{8}$
Number of atoms of B from face of unit cell $= 3$
Thus
$A : B = \frac{7}{8} : 3$
Or $= 7 : 24$
Therefore the formula is $A_{7} B_{24}$

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A face-centred cubic Lattice Is made up of two types of atoms, atom A and atom B in which A occupies the corner positions and B occupies the face centres. If atoms from one corner is absent per unit cell, the empirical formula of the remaining solid would be:

The correct option is D A7B24Number of atoms of A from corners of unit cell =78Number of atoms of B from face of unit cell =3ThusA:B=78:3Or =7:24Therefore the formula is A7B24

The correct option is D $A_{7} B_{24}$
Number of atoms of A from corners of unit cell $= \frac{7}{8}$
Number of atoms of B from face of unit cell $= 3$
Thus
$A : B = \frac{7}{8} : 3$
Or $= 7 : 24$
Therefore the formula is $A_{7} B_{24}$