A solution of FeCl$$_3$$ when treated with K$$_4$$Fe(CN)$$_6$$ gives a prussian blue precipitate due to the formation of

When $$FeCl_3$$ is treated with $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide), a Prussian blue precipitate is formed. The reaction is:

$$4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$$

The Prussian blue precipitate is ferric ferrocyanide with the formula $$Fe_4[Fe(CN)_6]_3$$. In this compound, $$Fe^{3+}$$ ions (from FeCl$$_3$$) are the external cations, while $$[Fe(CN)_6]^{4-}$$ is the complex anion where Fe is in the +2 oxidation state.

So, the answer is that the Prussian blue precipitate formed is $$Fe_4[Fe(CN)_6]_3$$.