Fe³⁺ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

🧠 Prussian Blue Formula

Prussian blue forms when $\mathrm{Fe^{3+}}$ is added to potassium ferrocyanide $\mathrm{K_4[Fe(CN)_6]}$ (or vice versa with $\mathrm{Fe^{2+}}$ + ferricyanide). The deep-blue precipitate is iron(III) hexacyanoferrate(II): $\mathrm{Fe_4[Fe(CN)_6]_3}$.

Charge balance: outer Fe is +3 (4 cations × +3 = +12); inner $[\mathrm{Fe(CN)_6}]^{4-}$ (3 anions × −4 = −12). Net neutral. ✓

🗺️ Identify the Two Iron Sites

• Outer Fe atoms: Fe(III) (high-spin, d⁵).
• Inner Fe atoms: Fe(II) in $[\mathrm{Fe(CN)_6}]^{4-}$ (low-spin, d⁶ with strong CN⁻).

This mixed-valence structure (Fe(II) + Fe(III) in the same compound) is what gives Prussian blue its intense colour — intervalence charge-transfer between the two Fe sites absorbs strongly in the red region.

⚡ Memorise the Formula

Prussian blue = $\mathrm{Fe_4[Fe(CN)_6]_3}$. Subscripts 4 and 3 — not 2 and 2 (which is option 2). The 4:3 ratio is fixed by charge balance.

⚠️ Don't Confuse With Turnbull's Blue

Turnbull's blue is the older name for the precipitate from $\mathrm{Fe^{2+}}$ + ferricyanide ($[\mathrm{Fe(CN)_6}]^{3-}$). Modern X-ray studies show both Prussian blue and Turnbull's blue are the same compound — $\mathrm{Fe_4[Fe(CN)_6]_3}$. The Fe(II)/Fe(III) sites just relax to the same mixed-valence ground state.

$\boxed{\text{Answer: (4) } \mathrm{Fe_4[Fe(CN)_6]_3}}$