A solution of $$FeCl_3$$ when treated with $$K_4[Fe(CN)_6]$$ gives a prussian blue precipitate due to the formation of

We need to identify the compound formed when FeCl$$_3$$ reacts with K$$_4$$[Fe(CN)$$_6$$] to give Prussian blue.

The Prussian blue test is a classic qualitative test for iron ions:

$$Fe^{3+}$$ (from FeCl$$_3$$) + $$[Fe(CN)_6]^{4-}$$ (from K$$_4$$[Fe(CN)$$_6$$]) → Prussian blue precipitate

The reaction is:

$$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$$

The Prussian blue compound is iron(III) hexacyanoferrate(II), with the formula $$Fe_4[Fe(CN)_6]_3$$. This is a dark blue insoluble precipitate. The $$Fe^{3+}$$ ions are the "outer" iron atoms, and the $$Fe^{2+}$$ ions are inside the $$[Fe(CN)_6]^{4-}$$ complex.

The correct answer is Option (4): Fe$$_4$$[Fe(CN)$$_6$$]$$_3$$.