Answer :
(b) The balanced overall ionic equation will be,
![NaI(s)\rightarrow Na^+(aq)+I^-(aq)](https://tex.z-dn.net/?f=NaI%28s%29%5Crightarrow%20Na%5E%2B%28aq%29%2BI%5E-%28aq%29)
(c) The balanced overall ionic equation will be,
![Pb(NO_3)_2(s)\rightarrow Pb^{3+}(aq)+2NO_3^{-}(aq)](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%28s%29%5Crightarrow%20Pb%5E%7B3%2B%7D%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29)
Explanation :
Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
<u>Part B :</u>
When sodium iodide dissolving in water then it dissociates to give sodium ion and iodide ion.
The balanced overall ionic equation will be,
![NaI(s)\rightarrow Na^+(aq)+I^-(aq)](https://tex.z-dn.net/?f=NaI%28s%29%5Crightarrow%20Na%5E%2B%28aq%29%2BI%5E-%28aq%29)
<u>Part C :</u>
When lead(II) nitrate dissolving in water then it dissociates to give lead ion and nitrate ion.
The balanced overall ionic equation will be,
![Pb(NO_3)_2(s)\rightarrow Pb^{3+}(aq)+2NO_3^{-}(aq)](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%28s%29%5Crightarrow%20Pb%5E%7B3%2B%7D%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29)