In a face-centred cubic (FCC) lattice,atoms o | vedclass

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(B) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Number of atoms of $A$ at corners = $8 - 2 = 6$ atoms.Contribution of each corner

Vedclass · Accepted Answer

$AB_4$

Vedclass · Answer

(B) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Number of atoms of $A$ at corners = $8 - 2 = 6$ atoms.Contribution of each corner atom = $\frac{1}{8}$.Total atoms of $A = 6 	imes \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$.Number of atoms of $B$ at face centres = $6$.Contribution of each face-centred atom = $\frac{1}{2}$.Total atoms of $B = 6 	imes \frac{1}{2} = 3$.Ratio of $A:B = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$.Therefore,the formula of the compound is $AB_4$.

In a face-centred cubic $(FCC)$ lattice,atoms of $A$ form the corner points and atoms of $B$ form the face-centred points. If two atoms of $A$ are missing from the corner points,the formula of the ionic compound is

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