Answer :
To determine the spectator ions in the reaction [tex]\( Mg (s) + Zn^{2+} + 2NO_3^{-} \rightarrow Mg^{2+} + 2NO_3^{-} + Zn (s) \)[/tex], we need to identify the ions that remain unchanged and do not participate in the chemical change. Let's break down the process step-by-step:
1. Identify the Reactants and Products:
- Reactants: [tex]\( Mg (s) \)[/tex], [tex]\( Zn^{2+} \)[/tex], [tex]\( 2NO_3^{-} \)[/tex]
- Products: [tex]\( Mg^{2+} \)[/tex], [tex]\( 2NO_3^{-} \)[/tex], [tex]\( Zn (s) \)[/tex]
2. Determine Changes in Oxidation State:
- [tex]\( Mg (s) \)[/tex] changes to [tex]\( Mg^{2+} \)[/tex] (Magnesium solid to magnesium ion)
- [tex]\( Zn^{2+} \)[/tex] changes to [tex]\( Zn (s) \)[/tex] (Zinc ion to zinc solid)
- [tex]\( NO_3^{-} \)[/tex] stays [tex]\( NO_3^{-} \)[/tex] (Nitrate ion remains unchanged)
3. Identify Spectator Ions:
- Spectator ions are those ions that appear the same on both reactant and product sides of the equation and do not undergo any change in oxidation state.
- In this reaction, [tex]\( NO_3^{-} \)[/tex] appears unchanged as [tex]\( 2NO_3^{-} \)[/tex] on both sides of the equation.
Therefore, the spectator ions in this reaction are [tex]\( NO_3^{-} \)[/tex].
Based on the options:
A. [tex]\( Zn^{2+} \)[/tex] and [tex]\( Mg^{2+} \)[/tex]: Incorrect, these change during the reaction.
B. [tex]\( Mg (s) \)[/tex] and [tex]\( Zn (s) \)[/tex]: Incorrect, these are not ions and they also change state.
C. [tex]\( NO_3^{-}; Zn^{2+}\)[/tex], and [tex]\( Mg^{2+} \)[/tex]: Incorrect, only [tex]\( NO_3^{-} \)[/tex] is unchanged.
D. [tex]\( NO_3^{-} \)[/tex]: Correct, nitrate ions remain unchanged in the reaction.
Thus, the correct answer is D. [tex]\( NO_3^{-} \)[/tex].
1. Identify the Reactants and Products:
- Reactants: [tex]\( Mg (s) \)[/tex], [tex]\( Zn^{2+} \)[/tex], [tex]\( 2NO_3^{-} \)[/tex]
- Products: [tex]\( Mg^{2+} \)[/tex], [tex]\( 2NO_3^{-} \)[/tex], [tex]\( Zn (s) \)[/tex]
2. Determine Changes in Oxidation State:
- [tex]\( Mg (s) \)[/tex] changes to [tex]\( Mg^{2+} \)[/tex] (Magnesium solid to magnesium ion)
- [tex]\( Zn^{2+} \)[/tex] changes to [tex]\( Zn (s) \)[/tex] (Zinc ion to zinc solid)
- [tex]\( NO_3^{-} \)[/tex] stays [tex]\( NO_3^{-} \)[/tex] (Nitrate ion remains unchanged)
3. Identify Spectator Ions:
- Spectator ions are those ions that appear the same on both reactant and product sides of the equation and do not undergo any change in oxidation state.
- In this reaction, [tex]\( NO_3^{-} \)[/tex] appears unchanged as [tex]\( 2NO_3^{-} \)[/tex] on both sides of the equation.
Therefore, the spectator ions in this reaction are [tex]\( NO_3^{-} \)[/tex].
Based on the options:
A. [tex]\( Zn^{2+} \)[/tex] and [tex]\( Mg^{2+} \)[/tex]: Incorrect, these change during the reaction.
B. [tex]\( Mg (s) \)[/tex] and [tex]\( Zn (s) \)[/tex]: Incorrect, these are not ions and they also change state.
C. [tex]\( NO_3^{-}; Zn^{2+}\)[/tex], and [tex]\( Mg^{2+} \)[/tex]: Incorrect, only [tex]\( NO_3^{-} \)[/tex] is unchanged.
D. [tex]\( NO_3^{-} \)[/tex]: Correct, nitrate ions remain unchanged in the reaction.
Thus, the correct answer is D. [tex]\( NO_3^{-} \)[/tex].
