Question

Aliovalent doping of $MgCl_2$ in $NaCl$ leads to the formation of defects. Which one of the following is the correct defect reaction?

• A. $Mg^{\bullet}_{Cl} + Na_{Na} + V^{\prime}_{Cl} = \varnothing$
• B. $Mg^{\bullet}_{Na} + Cl_{Cl} + V^{\prime}_{Na} = \varnothing$
• C. $Mg_{Na} + Cl_{Cl} = \varnothing$
• D. $Mg^{\prime}_{Na} + Cl_{Cl} + V^{\bullet}_{Na} = \varnothing$

Hint:

Aliovalent doping (different valence ion substitution) always creates point defects to preserve charge neutrality. For $MgCl_2$ in NaCl, $Mg^{2+}$ replaces $Na^+$ and generates $Na^+$ vacancies.

Answer 1

The Correct Option is B

Step 1: Understand the doping process.
- Host crystal: NaCl. - Dopant: $MgCl_2$. - In NaCl, the cation is $Na^+$ and the anion is $Cl^-$. - When $MgCl_2$ is added, the $Mg^{2+}$ ions replace $Na^+$ ions on the cation sites.
Step 2: Analyze the charge imbalance.
- Each $Mg^{2+}$ substitutes at a $Na^+$ site. - This creates an extra positive charge (+1 effective charge relative to the replaced ion). - To maintain charge neutrality, a defect must be introduced → typically a $Na^+$ vacancy ($V^{\prime}_{Na}$, negative effective charge).
Step 3: Write the reaction.
\[ MgCl_2 \ \Rightarrow \ Mg^{\bullet}_{Na} + Cl_{Cl} + V^{\prime}_{Na} \] Here: - $Mg^{\bullet}_{Na}$ represents $Mg^{2+}$ ion on a $Na^+$ site with effective +1 charge. - $Cl_{Cl}$ is a chloride ion on a normal lattice site. - $V^{\prime}_{Na}$ is a sodium vacancy with negative effective charge.
Step 4: Match with options.
This matches option (B). Final Answer: \[ \boxed{(B)} \]