First you need to write out the original equation.
[tex]NiCO_{3(s)}+ HI_{aq} ->NiI_{2(aq)}+CO_{2(g)}+H_2O_{(l)}[/tex]
It needs to be balanced, though, so after balancing it should look like this:
[tex]NiCO_{3(s)}+ 2HI_{aq} ->NiI_{2(aq)}+CO_{2(g)}+H_2O_{(l)}[/tex]
The ionic equation would look like this. (Only aqueous solutions can be split up).
[tex]NiCO_{3(s)}+ 2H^+_{aq}+2I^-_{aq} ->Ni^{2+}_{aq}+2I^-_{(aq)}+CO_{2(g)}+H_2O_{(l)}[/tex]
The net ionic equation only contains the products that are not aqueous. Therefore, the final net ionic equation should be:
[tex]NiCO_{3(s)}+ 2H^+_{aq} ->Ni^{2+}_{aq}+CO_{2(g)}+H_2O_{(l)}[/tex]
So your answer is C.
