Question

You isolate a compound with the formula ${\mathbf{\text{PtCl}}}_{\mathbf{\text{4}}}\mathbf{\cdot }\mathbf{\text{2KCl}}$. From electrical conductance tests of an aqueous solution of the compound, you find that three ions are present, and you also notice that the addition of${\mathbf{\text{AgNO}}}_{\mathbf{\text{3}}}$ does not cause a precipitate. Give the formula for this compound that shows the complex ion present. Explain your findings, and name this compound.

Short answer

The name of the complex anion is hexachloroplatinate (IV) ion. The name of the compound is Potassium hexachloroplatinate (IV).

Step-by-step solution

A concept:

The formula of the hexachloroplatinate complexion is ${\mathbf{\text{K}}}_{\mathbf{\text{2}}}\mathbf{\left[}{\mathbf{\text{PtCl}}}_{\mathbf{\text{6}}}\mathbf{\right]}$.

Give the formula for the given compound that shows the complex ion present:

The formula for this compound that shows the complex ion present is ${\left[{\text{PtCl}}_{\text{6}}\right]}^{2-}$. On addition of${\text{AgNO}}_{\text{3}}$ to aqueous solution of compound i.e.,${\text{PtCl}}_{\text{4}}\cdot \text{2KCl}$ will not yield any chloride ion in solution. Actually, no chloride ion precipitated as$\text{AgCl}$ on addition of ${\text{AgNO}}_{\text{3}}$, because all the chloride ions are in coordination sphere hence no chloride ions are produced in solution.

${\text{PtCl}}_{\text{4}}\cdot \text{2KCl}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\left[{\text{PtCl}}_6\right]}^{2-}\left(aq\right)+2{\text{k}}^{+}\left(aq\right)$

Since the compound contains six chloride ions and two potassium ions therefore platinum must carry a charge of $4+$to produce neutral compound. Thus, platinum has the oxidation state $+4$and you can use platinum (IV) in the name of the compound, the number of ligands are six chloride ions, ${\text{Cl}}^{-}$is designated as chloro and prefix hexa-indicates that six ${\text{Cl}}^{-}$ions are present.

The name of the complex anion is therefore hexachloroplatinate (IV) ion. The name of the compound is Potassium hexachloroplatinate (IV).