Final Answer:
The complete ionic equation for the reaction between lead perchlorate and sodium sulfide is:
Pb²⁺(aq) + 2ClO₄⁻(aq) + 2Na⁺(aq) + S²⁻(aq) → PbS(s) + 2Na⁺(aq) + 2ClO₄⁻(aq) Option B is answer.
Step-by-step explanation:
Identify the reactants and products:
Reactants: Lead perchlorate (Pb(ClO₄)₂), sodium sulfide (Na₂S)
Products: Lead sulfide (PbS), sodium perchlorate (NaClO₄)
Write the balanced molecular equation:
Pb(ClO₄)₂ + Na₂S → PbS + 2NaClO₄
Separate the electrolytes into their ions:
Pb(ClO₄)₂ → Pb²⁺ + 2ClO₄⁻
Na₂S → 2Na⁺ + S²⁻
Identify spectator ions:
Both sodium and perchlorate ions appear on both sides of the equation, indicating they are not involved in the reaction itself and are considered spectator ions.
Write the complete ionic equation:
Combine the ionic equations for the reactants and exclude spectator ions:
Pb²⁺(aq) + 2ClO₄⁻(aq) + 2Na⁺(aq) + S²⁻(aq) → PbS(s) + 2Na⁺(aq) + 2ClO₄⁻(aq)
Therefore, the complete ionic equation shows the actual participating ions and accurately represents the reaction between lead perchlorate and sodium sulfide.
Option B is answer.