Explanation:
An ionic equation will be the one in which all the participating species will be present as ions.
The given reaction will be as follows.
[tex]CrBr_{3} + K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + KBr[/tex]
Balancing this equation by multiplying [tex]CrBr_{3}[/tex] by 2 and [tex]K_{2}CO_{3}[/tex] by 3 on reactant side. Whereas multiply KBr by 6 on product side.
[tex]2CrBr_{3} + 3K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6KBr[/tex]
Hence, the net ionic equation will be as follows.
[tex]2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) + 6K^{+} + 6Br^{-} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6K^{+} + 6Br^{-}[/tex]
As both [tex]K^{+}[/tex] and [tex]Br^{-}[/tex] are spectator ions. Hence, the net ionic equation will be as follows.
[tex]2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) \rightarrow Cr_{2}(CO_{3})_{3}(s)[/tex]
