How many ions are furnished by [Pt(NH₃)₆]Cl₄ ?

5 ions

The correct answer is option 1. 5 ions.

To determine how many ions are furnished by \([Pt(NH_3)_6]Cl_4\) in solution, we need to consider how this complex dissociates in water.

The complex \([Pt(NH_3)_6]Cl_4\) consists of the coordination complex \([Pt(NH_3)_6]^{4+}\) and four chloride ions \(Cl^-\).

When \([Pt(NH_3)_6]Cl_4\) dissolves in water, it dissociates as follows:

\([Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-\)

In this dissociation:

The \([Pt(NH_3)_6]^{4+}\) complex ion provides 1 ion. The 4 chloride ions \(Cl^-\) provide 4 ions. So, the total number of ions furnished in solution is:

\(1 \text{ (from } [Pt(NH_3)_6]^{4+} \text{)} + 4 \text{ (from } Cl^- \text{)} = 5 \text{ ions} \)

Therefore, the correct answer is \(5 \text{ ions}\).