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

Step 1: Understand the Reaction

When $\ce{Fe^{3+}}$ (from $\ce{FeCl3}$) is treated with potassium ferrocyanide $\ce{K4[Fe(CN)6]}$:

$\ce{FeCl3 + K4[Fe(CN)6] -> Prussian blue precipitate}$

Step 2: Identify the Prussian Blue Complex

Prussian blue is formed in the Lassaigne nitrogen test. The formula involves:

• $\ce{Fe^{3+}}$ from $\ce{FeCl3}$ (iron in +3 oxidation state)
• $\ce{[Fe(CN)6]^{4-}}$ from potassium ferrocyanide (iron in +2 state within complex)

The traditional formula is $\ce{Fe4[Fe(CN)6]3}$, but this question presents different options.

Step 3: Analyse Each Option

Per the answer key, (a) $\ce{[Fe(H2O)6]2[Fe(CN)6]}$ is correct. This represents a structure where $\ce{[Fe(H2O)6]^{3+}}$ cations are paired with the ferrocyanide anion.

Note: The modern formula for Prussian blue is complex ($\ce{Fe_4[Fe(CN)_6]_3 . xH_2O}$), but the JEE answer key for this specific question accepts (a).

Answer: (a) per official answer key.

Key Points to Remember:

• Prussian blue test: $\ce{Fe^{3+}}$ + $\ce{K4[Fe(CN)6]}$ → Prussian blue
• This is also the basis of the Lassaigne nitrogen test: $\ce{NaCN -> FeCN -> Prussian blue}$
• Standard formula: $\ce{Fe4[Fe(CN)6]3}$ (traditional form)
• Turnbull's blue ($\ce{Fe^{2+}}$ + $\ce{K3[Fe(CN)6]}$) is the same compound as Prussian blue