Question

What of the following would be the formula of $AB$ solids in which $A - atoms$ are face centred and $B - atoms$ are available on eight edges of sides?
(A) $A{B_2}$
(B) ${A_4}{B_3}$
(C) ${A_2}B$
(D) ${A_3}{B_2}$

Answer 1

We have already learnt that a crystal system is made up of a very large number of unit cells and a very large lattice point occupied by either atoms, ions or molecules. These particles may be present at corners, face-centre and body-centre.

Complete step by step solution: We have already learnt that a crystal system is made up of a very large number of unit cells and a very large lattice point occupied by either atoms, ions or molecules. These particles may be present at corners, face-centre and body-centre. We know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.
In order to calculate the number of points in a unit cell we can use following generalisation:
A point at the corner of a unit cell is shared by eight other unit cells, therefore its contribution to each unit cell is $\dfrac{1}{8}$.
A face is common to two unit cells therefore any point present at the face-centre has a contribution of $\dfrac{1}{2}$ to each unit cell.
A point present on each edge-centre is shared among four unit cells therefore its contribution to each unit cell is $\dfrac{1}{4}$.
Using these conditions we can calculate the formula of the given compound as we are given that $A - atoms$ are face centred and $B - atoms$ are on $8$ edges. So,
Number of atoms from $8$corners of unit cell $= \dfrac{1}{8} \times 8 = 1$
Number of atoms of $A \to face\;centred = 6 \times \frac{1}{2} = 3$
Number of atoms of $B \to eight\;edges = 8 \times \frac{1}{4} = 2$
Therefore we get formula of${A_3}{B_2}$

Hence the correct answer is (D).

Note: Using the same information we can calculate the formula of compounds including body centred atoms and where atoms at corners will be $8$and atoms at centre will be $1$. $NaCl$ has a $AB$ type crystal lattice, $Ca{F_2}$ has $A{B_2}$ type crystal lattice.