Question

give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of iron(ii) chloride and lead(ii) nitrate are mixed.\
\\(\ce{pb^{2+}(aq) + 2 cl^-(aq) \
ightarrow pbcl2 (s)}\\)\
\\(\ce{no reaction occurs.}\\)\
\\(\ce{fe^{2+}(aq) + 2 no3^-(aq) \
ightarrow fe(no3)2 (s)}\\)\
\\(\ce{fe^{2+}(aq) + 2 cl^-(aq) + pb^{2+}(aq) + 2 no3^-(aq) \
ightarrow pbcl2 (s) + fe^{2+}(aq) + 2 no3^-(aq)}\\)

Explanation:

Step1: Identify Reactants and Products

The reactants are aqueous iron(II) chloride ($\ce{FeCl2(aq)}$) and aqueous lead(II) nitrate ($\ce{Pb(NO3)2(aq)}$). When dissolved, $\ce{FeCl2}$ dissociates into $\ce{Fe^{2+}(aq)}$ and $\ce{2Cl^-(aq)}$, and $\ce{Pb(NO3)2}$ dissociates into $\ce{Pb^{2+}(aq)}$ and $\ce{2NO3^-(aq)}$.

Step2: Determine Precipitate

We check the solubility of possible products. $\ce{Fe(NO3)2}$ is soluble (nitrates are soluble), and $\ce{PbCl2}$ is insoluble (chlorides of $\ce{Pb^{2+}}$ are insoluble). So the precipitate is $\ce{PbCl2(s)}$.

Step3: Write Net Ionic Equation

Spectator ions (ions that don't form the precipitate) are $\ce{Fe^{2+}(aq)}$ and $\ce{NO3^-(aq)}$ (they remain in solution). The net ionic equation includes only the ions that form the precipitate: $\ce{Pb^{2+}(aq) + 2Cl^-(aq) -> PbCl2(s)}$.

Answer:

$\ce{Pb^{2+}(aq) + 2Cl^-(aq) -> PbCl2(s)}$ (corresponding to the first option: $\ce{Pb^{2+}(aq) + 2 Cl^-(aq) -> PbCl2(s)}$)