Certainly! Let's walk through the process of identifying the spectator ions in the given chemical reaction:
[tex]\[ Mg(s) + Zn^{2+} + 2NO_3^{-} \rightarrow Mg^{2+} + 2NO_3^{-} + Zn(s) \][/tex]
1. Write down the initial reactants:
- Magnesium solid [tex]\( Mg(s) \)[/tex]
- Zinc ion [tex]\( Zn^{2+} \)[/tex]
- Two nitrate ions [tex]\( 2NO_3^{-} \)[/tex]
2. Write down the final products:
- Magnesium ion [tex]\( Mg^{2+} \)[/tex]
- Two nitrate ions [tex]\( 2NO_3^{-} \)[/tex]
- Zinc solid [tex]\( Zn(s) \)[/tex]
3. Identify which species change:
- Magnesium [tex]\( Mg \)[/tex] starts as a solid and ends up as an ion [tex]\( Mg^{2+} \)[/tex]. Therefore, it is not a spectator ion.
- Zinc [tex]\( Zn \)[/tex] starts as an ion [tex]\( Zn^{2+} \)[/tex] and ends up as a solid. Therefore, it is not a spectator ion.
4. Compare the remaining species:
- The nitrate ions [tex]\( NO_3^{-} \)[/tex] remain unchanged throughout the reaction; they appear as [tex]\( 2NO_3^{-} \)[/tex] on both the reactant and product sides.
Species that remain unchanged during the course of the reaction are called spectator ions. They do not participate in the actual chemical reaction and can be ignored when writing the net ionic equation.
So, the spectator ions are:
[tex]\[ NO_3^{-} \][/tex]
Therefore, the correct answer is:
B. [tex]\( NO_3^{-} \)[/tex]
