Question

If a solid OQ having NaCl structure is heated so that the ions along one body diagonal are lost and trivalent ions R enter interstitial sites to maintain electrical neutrality so that new empirical formula unit is PxQyRz (x,y,z are smallest possible integers). The value of x+y+z is

Answer 1

Given that :

The solid OQ has NaCl type structure.

The interstitial sites occupied by trivalent ions to maintain neutrality = $R^{+}$

The new empirical formula unit is = $P_x{Q}_y{R}_z$

Step 1:

As we know, NaCl is a face-centred cubic unit cell.

As, OQ = $O^{+}and{Q}^{-}$

For NaCl type structure:

Number of $O^{+}$ ions = 1 body centre + 12 edge $\times \frac{1}{4}=4$

Number of $Q^{-}$ ions = 6 face centre $\times \frac{1}{2}+8corners\times \frac{1}{8}=4$

Step 2:

From the diagram, we conclude that

Number of ions removed = 2 corner atoms + 1 body centre

Therefore, the number of ions become:

Number of $O^{+}$ ions = (1-1) body centre + 12 edge $\times \frac{1}{4}=3$

Number of $Q^{-}$ ions = 6 face centre $\times \frac{1}{2}+(8-2)corners\times \frac{1}{8}=3+\frac{3}{4}=\frac{15}{4}$

Step 3:

The total charge present on the unit cell = $3\times (+1)+\frac{15}{4}\times (-1)=3-\frac{15}{4}=-\frac{3}{4}$

As the trivalent ion occupies the interstitial sites = $\frac{1}{4}\times (-3)$

Thus, $\frac{1}{4}{R}^{+}$ ions will be added.

Hence, the new empirical formula of the compound = $O_3{Q}_{\frac{15}{4}}{R}_{\frac{1}{4}}$

The value of x + y + z = $3+\frac{15}{4}+\frac{1}{4}=\frac{28}{4}=7$