To determine the spectator ions in the given total ionic equation:
[tex]\[
2 H^{+} + CrO_4^{2-} + Ba^{2+} + 2 OH^{-} \longrightarrow Ba^{2+} + CrO_4^{2-} + 2 H_2O
\][/tex]
we need to identify which ions remain unchanged on both sides of the equation. Spectator ions do not participate in the chemical reaction and appear in the same form on both the reactant and product sides.
1. Identify the reactants and products:
- The reactants are [tex]\(2 H^{+}\)[/tex], [tex]\(CrO_4^{2-}\)[/tex], [tex]\(Ba^{2+}\)[/tex], and [tex]\(2 OH^{-}\)[/tex].
- The products are [tex]\(Ba^{2+}\)[/tex], [tex]\(CrO_4^{2-}\)[/tex], and [tex]\(2 H_2O\)[/tex].
2. Compare ions on both sides of the equation:
- [tex]\( H^{+} \)[/tex] appears on the reactant side but not on the product side. It is not a spectator ion.
- [tex]\(OH^{-}\)[/tex] appears on the reactant side but not on the product side. It is not a spectator ion.
- [tex]\(Ba^{2+}\)[/tex] appears both as a reactant and as a product without any change. It is a spectator ion.
- [tex]\(CrO_4^{2-}\)[/tex] appears both as a reactant and as a product without any change. It is a spectator ion.
From this analysis, the ions that remain unchanged on both sides of the equation are [tex]\(Ba^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex].
Therefore, the spectator ions in this equation are:
[tex]\[
\boxed{CrO_4^{2-} \text{ and } Ba^{2+}}
\][/tex]
The correct option is:
[tex]\( \text{CrO}_4^{2-} \text{ and } \text{Ba}^{2+} \)[/tex], which corresponds to:
[tex]\( CrO_4^{2-} \text{ and } Ba^{2+} \)[/tex] (the 4th option)
