Double Replacement Reaction in which everything cancels out

We are currently trying to work out the double replacement solubility reaction for Nickel (III) Nitrate and Rubidium Sulfate.


$$\ce{Ni(NO3)3 + Rb2SO4 -> Ni2(SO4)3 + RbNO3}$$


Balancing gives:


$$\ce{2Ni(NO3)3 + 3Rb2SO4 -> Ni2(SO4)3 + 6RbNO3}$$


We believe all four compounds are salts and therefore split all four into their cations and anions:


$$\ce{2Ni^{3+}(aq) + 6NO3-(aq) + 6Rb+(aq) + 3SO4^{2-}(aq) -> 2Ni^{3\*}(aq) + 3SO4^{2-}(aq) + 6Rb+(aq) + 6NO3-(aq)}$$


At this point, we want to eliminate the matching spectator ions. But they are the same on both sides of the equation and we're left with nothing and no precipitate.


We suspect there is something or things that we are misunderstanding. We don't even know if it's ok for these types of reactions to not have a precipitate and if it is ok, if that is actually the case here with these two compounds.




This problem is impossible, because $\ce{Nickel(III) nitrate}$ and $\ce{Nickel(III) sulfate}$ do not exist. You loose your time looking for their solubilities. The only known compounds where Nickel has an oxidation number $+3$ are Nickel oxide $\ce{Ni\_2O\_3}$ (which is insoluble) and the complex ion $\ce{[NiF\_6]^{3-}}$. Simple $\ce{Ni^{3+}}$ ions do not exist in water. As a consequence, no corresponding nitrate or sulfate exist. And no double replacement may occur with these non-existing substances.


See N. N. Greenwood and A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford, England, 1986, p. 1340


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