Solve the following

Both the compounds i.e., $\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ fall under the category of addition compounds with only one major difference i.e., the former is an example of a double salt, while the latter is a coordination compound.

A double salt is an addition compound that is stable in the solid state but that which breaks up into its constituent ions in the dissolved state. These compounds exhibit individual properties of their constituents. For e.g.

$\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ breaks into $\mathrm{Fe}^{2+}, \mathrm{NH}^{4+}$, and $\mathrm{SO}_{4}{ }^{2-}$ ions. Hence, it gives a positive test for $\mathrm{Fe}^{2+}$ ions.

A coordination compound is an addition compound which retains its identity in the solid as well as in the dissolved state. However, the individual properties of the constituents are lost. This happens because $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ does not show the test for $\mathrm{Cu}^{2+}$. The ions present in the solution of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ are $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ and $\mathrm{SO}_{4}{ }^{2-}$.