Question

If $NaCl$ is doped with ${10^{ - 3}}\;mol$ percent of $SrC{l_2}$, what is the concentration of cation vacancies?

Answer 1

In ionic compounds, charge neutrality has to be maintained and thus introducing a cation with higher charge can create cationic vacancies.

Complete step by step answer:
We know that crystals have ordered structures but there might be some defects in them as well which we can categorize as stoichiometric defects, impurity defects and non-stoichiometric defects.
Stoichiometric defects: Here, the stoichiometry is maintained. For non-ionic crystals, we have simple vacancy and interstitial defects but for ionic ones we have Frenkel and Schottky defects.
Impurity defects: Here, defects are due to the presence of some impurity. In some cases, impurities can be introduced voluntarily to change the properties of the original crystal and it is called doping.
Non-stoichiometric defects: Here, we might have metal in excess or in deficiency which causes disturbed stoichiometry.
Now, let’s have a look at the given crystal of $NaCl$ which has been doped with $SrC{l_2}$. As we can see that initially we had $N{a^ + }$ cations and later we introduced $S{r^{2 + }}$ cations. It means that one mole of $S{r^{2 + }}$ ions would replace two moles of $N{a^ + }$ ions by occupying one mole of cationic positions and leaving one mole cationic vacancies. So, the amount of cationic vacancies is equal to the amount of $S{r^{2 + }}$ ions.
Here, we are given the concentration of $SrC{l_2}$ as $mol\%$. So let’s use that to determine the amount of cationic vacancies. At first we have:
${10^{ - 3}}\;mol\% \;SrC{l_2} = \dfrac{{{{10}^{ - 3}}\;mol\;SrC{l_2}}}{{100{\rm{ }}\;mol}}$
We can assume that the concentration of $SrC{l_2}$ is negligible as compared to that of NaCl which makes the above expression as:
$\dfrac{{{{10}^{ - 3}}\;mol\;SrC{l_2}}}{{100{\rm{ }}mol\;NaCl}}$
We know that the amount of cationic vacancies is equal to the amount of $S{r^{2 + }}$ ions. So, we can write:
$\begin{array}{c} 100{\rm{ }}mol\;NaCl = {10^{ - 3}}\;mol\;{\rm{cationic}}\;{\rm{vacancies}}\\\ 1{\rm{ }}mol\;NaCl = \dfrac{{{{10}^{ - 3}}\;mol\;{\rm{cationic}}\;{\rm{vacancies}}}}{{100}}\\\ \end{array}$
= ${10^{ - 5}}$ mol cationic vacancies
We can convert the amount to number by using the Avogadro constant as follows:
${10^{ - 5}}\;mol\;{\rm{cationic}}\;{\rm{vacancies}} \times \left( {{\rm{6}}{\rm{.023}} \times {\rm{1}}{{\rm{0}}^{23}}\;mo{l^{ - 1}}} \right) = {\rm{6}}{\rm{.023}} \times {\rm{1}}{{\rm{0}}^{18}}\;{\rm{cationic}}\;{\rm{vacancies}}$
Hence, the number of cationic vacancies is ${\rm{6}}{\rm{.023}} \times {\rm{1}}{{\rm{0}}^{18}}$ or in terms of amount it is ${10^{ - 5}}\;mol$ per mole of $NaCl$.

Note: We have to be careful with the units and the charge neutrality in ionic solids. At the end we have to multiply the moles by Avogadro’s number.