Discussion
Aliovalent doping of \(MgCl_{2}\) in \(NaCl\) leads to the formation of defects. Which one of the following is the correct defect reaction?
Answer: B
The correct defect reaction is \(Mg_{Na}^{\bullet} + Cl_{Cl} + V_{Na}' = \emptyset\).
In Kroger-Vink notation the subscript is the site and the superscript the charge relative to the perfect lattice (\(\bullet\) positive, \('\) negative). \(Mg^{2+}\) replaces \(Na^{+}\) on a sodium site, so it carries one extra positive charge, \(Mg_{Na}^{\bullet}\). The second chloride ion of \(MgCl_{2}\) occupies a normal chlorine site, \(Cl_{Cl}\), and to keep the lattice neutral and the site ratio right one sodium site is left vacant, a vacancy with effective negative charge, \(V_{Na}'\).
The other options fail on site or charge. Mg on a Cl site is a cation on an anion site. Writing \(Mg_{Na}\) with no charge ignores the valence difference that makes the doping aliovalent. \(Mg_{Na}'\) with \(V_{Na}^{\bullet}\) gives both defects the wrong sign.
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