# A compound has the formula PtCl_4*2KCl...?

## From electrical conductance tests of an aqueous solution of the compound, you ﬁnd that three ions per formula unit are present, and you also notice that addition of $A g N {O}_{3}$ does not cause a precipitate. a) Give the formula for this compound that shows the complex ion present. b) Explain your ﬁndings. c) Name this compound. Any help would be appreciated.

The complex is likely $\text{potassium hexachloroplatinate}$, i.e. ${K}_{2} \left[P t C {l}_{6}\right]$
Because no precipitate is observed with $A {g}^{+}$, there is no free halide ion present, and so all the halides are bound to platinum metal. And thus ${K}_{2} \left[P t C {l}_{6}\right]$, a $\text{platinum IV+}$ species, is likely. In solution it would give 3 equiv of ions: $2 \times {K}^{+} + {\left[P t C {l}_{6}\right]}^{2 -}$, which is consistent with the conductance measurements.