The total number of ions produced from the complex \([Cr(NH_3)_6]Cl_3\) in aqueous solution will be _____.

4

The correct answer is option 3. 4.

To determine the total number of ions produced from the complex \([Cr(NH_3)_6]Cl_3\) in aqueous solution, we need to consider the dissociation of the complex into its constituent ions when it is dissolved in water.

The complex ion, \([Cr(NH_3)_6]^{3+}\), will dissociate into its ions in aqueous solution. Each chloride ion (\(Cl^-\)) is already present as an ion in the compound, so it will not dissociate further.

The complex ion \([Cr(NH_3)_6]^{3+}\) contains one chromium ion (\(Cr^{3+}\)) and six ammine ligands (\(NH_3\)) that are not ionized. Therefore, when the complex dissociates, it will produce one \(Cr^{3+}\) ion and six \(NH_3\) ligands that will remain unionized.

So, the total number of ions produced from the complex \([Cr(NH_3)_6]Cl_3\) in aqueous solution is \(1\) \([Cr(NH_3)_6]^{3+}\) ion from the complex plus \(3\) chloride ions (\(Cl^-\)) from the \(Cl_3^-\) part of the compound.

Total number of ions produced = \(4\).

Therefore, the correct answer is option 3: \(4\).