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(A) To determine the number of ions produced,we look at the dissociation of the complex in water.$A$: $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{

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$[Pt(NH_3)_5Cl]Cl_3$

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(A) To determine the number of ions produced,we look at the dissociation of the complex in water.$A$: $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^{-}$. This yields $1 + 3 = 4$ ions.$B$: $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^{-}$. This yields $1 + 4 = 5$ ions.$C$: $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not ionize.$D$: $[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^{-}$. This yields $1 + 2 = 3$ ions.Therefore,the complex that produces four ions per molecule is $[Pt(NH_3)_5Cl]Cl_3$.

$A$ $Pt$ complex of ammonia and chlorine that produces four ions per molecule in the solution is

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