Identify the reagent and ion involved

Potassium ferrocyanide is: $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$ Here, the complex ion is ferrocyanide: $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$

The given cation is: $\mathrm{Fe}^{3+}$

Name of the precipitate formed

It is a standard qualitative test that $\mathrm{Fe}^{3+}$ with potassium ferrocyanide gives a Prussian blue precipitate.

Prussian blue has the formula: $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$

Check charge balance

In $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$, the iron is in +2 oxidation state.

So in: $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$

• $4$ outside iron ions are $\mathrm{Fe}^{3+}$, giving total charge: $4 \times (+3) = +12$
• $3$ ferrocyanide ions contribute: $3 \times (-4) = -12$

Net charge: $+12 + (-12) = 0$

So the formula is correct.

Evaluate options

• A: $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]_2[\mathrm{Fe}(\mathrm{CN})_6]$ This does not represent Prussian blue.

• B: $\mathrm{Fe}_2[\mathrm{Fe}(\mathrm{CN})_6]_2$ Charge balance is also not consistent for the Prussian blue composition.

• C: $\mathrm{Fe}_3[\mathrm{Fe}(\mathrm{OH})_2(\mathrm{CN})_4]_2$ This is not the known formula of Prussian blue.

• D: $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ This is the correct formula of Prussian blue.

Final answer

Therefore, the precipitate formed is: $\boxed{\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3}$ So the correct option is D.