Let us assume we have an aqueous solution with the ions $\ce{A+}$, $\ce{B+}$, $\ce{C-}$ and $\ce{D-}$. Which ions will form a bond and why? Will the compound formed be $\ce{AC}$ and $\ce{BD}$ or $\ce{AD}$ and $\ce{BC}$?

Is there any general rule to identify the compounds we have at the end?


Your question arises because you believe that ions in an aqueous solution remain bonded with each other. That isn't true though! In fact, ions in an aqueous solution freely roam around throughout the solution.

Note that most ionic compounds have ions bonded to each other in solid state only. Once you dissolve that ionic compound in water, the ions quickly dissociate (the ionic solid dissolves) and become solvated (surrounded by water molecules).

So, to answer your current question: no, there won't be any compound formed, and all the ions would be separate from each other.

Though, another possibility is of precipitation. In that case, the ions involved ($\ce{AD}$ for example) would separate out of the solution and settle down as a solid. Indeed, there may be more than one possible precipitates. So, to decide between any given number of compounds, as to which precipitates first, we'll need to check if their ionic product exceeds their $K_\mathrm{sp}$. The compound for which this happens earlier, due to lower $K_\mathrm{sp}$, would precipitate earlier.

Bonus content: Your question would make more sense if the ions were given in gas phase instead. Now, when the oppositely charged ions would come together, they would stabilize each other via the large amount of electrostatic energy released, and thus forming a stable compound.

Whether $\ce{A+}$ would combine with $\ce{C-}$ or $\ce{D-}$, would depend on their energy of interaction $\left(=\displaystyle -\frac{Kq_1q_2}{r}\right)$. More the energy released for a particular bond, more would be the probability of formation of that bond.

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