Identify the reacting species

• $\mathrm{FeCl_3}$ provides $\mathrm{Fe^{3+}}$ ions in solution.
• $\mathrm{K_4[Fe(CN)_6]}$ provides the ferrocyanide ion $\mathrm{[Fe(CN)_6]^{4-}}$.

Nature of Prussian blue

Prussian blue is formed when $\mathrm{Fe^{3+}}$ reacts with ferrocyanide ion $\mathrm{[Fe(CN)_6]^{4-}}$.

The compound formed is ferric ferrocyanide.

Write the neutral formula

Let the formula be: $\mathrm{Fe_x[Fe(CN)_6]_y}$

Since:

• each outer iron is $\mathrm{Fe^{3+}}$
• each ferrocyanide ion has charge $-4$

Charge balance gives: $3x = 4y$

The smallest integers satisfying this are: $x=4, \quad y=3$

Therefore, the formula is: $\mathrm{Fe_4[Fe(CN)_6]_3}$

Match with the options

• A: $\mathrm{Fe[Fe(CN)_6]}$ → not charge balanced for $\mathrm{Fe^{3+}}$ and $\mathrm{[Fe(CN)_6]^{4-}}$
• B: $\mathrm{Fe_4[Fe(CN)_6]_3}$ → correct
• C: $\mathrm{Fe_3[Fe(CN)_6]_2}$ → corresponds to a different charge balance, not Prussian blue here
• D: $\mathrm{K[Fe_2(CN)_6]}$ → incorrect

Conclusion

The prussian blue precipitate is due to formation of: $\boxed{\mathrm{Fe_4[Fe(CN)_6]_3}}$