Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a prussian

Q: Why do we use acidic ferric chloride?

The acidic medium prevents the hydrolysis of ferric ions into ferric hydroxide, which would otherwise form a brown precipitate and interfere with the Prussian blue test.

Q: What is the difference between Prussian blue and Turnbull's blue?

Historically, Prussian blue is formed from Fe3+ and ferrocyanide, while Turnbull's blue is formed from Fe2+ and ferricyanide. However, modern chemistry has shown that both have the exact same chemical composition: $\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$.

Q: How do we balance the charges to get the formula $\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$?

The ferric ion has a +3 charge ($\text{Fe}^{3+}$), and the ferrocyanide complex ion has a -4 charge ($[\text{Fe}(\text{CN})_6]^{4-}$). To balance the charges, we cross-multiply: we need four $\text{Fe}^{3+}$ ions (+12 total) and three $[\text{Fe}(\text{CN})_6]^{4-}$ ions (-12 total).

Q: Is Prussian blue a primary or secondary valency complex?

It involves both! The iron inside the coordination sphere satisfies its secondary valency with cyanide ligands, while the iron outside satisfies the primary valency of the complex anion.

Q: Why does it form a colloidal species instead of a solid precipitate?

The particles of Prussian blue formed are extremely small (nanometer scale). Because of their size and surface charge, they remain suspended in the aqueous medium, forming a colloidal dispersion rather than settling down immediately.

Question

Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a prussian blue coloured colloidal species. It is

Options:

(A) $	ext{Fe}_4[	ext{Fe}(	ext{CN})_6]_3$

(B) $	ext{K}_5	ext{Fe}[	ext{Fe}(	ext{CN})_6]_2$

(C) $	ext{HFe}[	ext{Fe}(	ext{CN})_6]$

(D) $	ext{KFe}[	ext{Fe}(	ext{CN})_6]$

Sigma Percentile · Accepted Answer

### The Chemistry of Colors

Qualitative analysis in chemistry is often a visual treat, and few reactions are as visually striking as the formation of **Prussian Blue**. When we are tasked with identifying the presence of iron in a given sample, this specific reaction serves as a highly reliable confirmatory test. Let's dive into the beautiful coordination chemistry behind this deep blue colloidal species.

### Decoding the Reactants

The problem presents us with two primary reactants. First, we have an acidic solution of **ferric chloride** ($	ext{FeCl}_3$). In an aqueous medium, this dissociates to give us ferric ions, $	ext{Fe}^{3+}$. The acidic medium is crucial here; it prevents the $	ext{Fe}^{3+}$ ions from undergoing hydrolysis to form a brown precipitate of ferric hydroxide, which would mask our desired result.

Our second reactant is **potassium ferrocyanide**, a classic coordination complex with the formula $	ext{K}_4[	ext{Fe}(	ext{CN})_6]$. When dissolved, it yields the ferrocyanide complex ion, $[	ext{Fe}(	ext{CN})_6]^{4-}$. Notice that the iron inside this coordination sphere is in the $+2$ oxidation state.

### The Master Reaction

When these two solutions are mixed, a double displacement reaction occurs. The free $	ext{Fe}^{3+}$ ions from the ferric chloride seek out the negatively charged $[	ext{Fe}(	ext{CN})_6]^{4-}$ complex ions.

To form a neutral compound, nature must balance the charges. We have a $+3$ charge on the ferric ion and a $-4$ charge on the ferrocyanide ion. By simply cross-multiplying the charges, we can determine the stoichiometry of the resulting complex:

$$ 4	ext{Fe}^{3+} + 3[	ext{Fe}(	ext{CN})_6]^{4-} \longrightarrow 	ext{Fe}_4[	ext{Fe}(	ext{CN})_6]_3 $$

### The Final Verdict

The resulting compound, $	ext{Fe}_4[	ext{Fe}(	ext{CN})_6]_3$, is chemically known as **ferric ferrocyanide**. Because the particles formed are extremely fine, they often remain suspended in the solution, creating a mesmerizing, deep blue colloidal dispersion rather than a heavy solid precipitate. This iconic substance is universally known as **Prussian Blue**.

Understanding the oxidation states—ferric ($+3$) outside the sphere and ferro ($+2$) inside the sphere—is the key to mastering this concept. Thus, the correct formula for the Prussian blue species is indeed $	ext{Fe}_4[	ext{Fe}(	ext{CN})_6]_3$.

Animated Solution for Chemistry - Coordination Compounds: Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a prussian blue coloured colloidal species. It is

Select Answer:

Visualized Solution

$Identifying the Reactants$

$The Chemical Reaction$

$The Prussian Blue Complex$

The Sigma Insight: Nomenclature, Isomerism, Importance and Werner's Theory

The Chemistry of Colors

Decoding the Reactants

The Master Reaction

The Final Verdict

Similar Questions

The potassium ferrocyanide solution gives a prussian blue colour, when added to

Which one of the following complexes is violet in colour?

The IUPAC name of the coordination compound $K_{3} [F e (C N)_{6}]$ is

On treatment of $100 mL$ of $0.1 M$ solution of $CoCl_{3} \cdot 6 H_{2} O$ with excess of $AgNO_{3}$ ; $1.2 \times 1 0^{22}$ ions are precipitated. The complex is

The complex that can show optical activity is (ox = oxalate)

On complete reaction of $FeCl_{3}$ with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is ........ . (Round off to the nearest integer).

The oxidation states of Cr, in $[C r (H_{2} O)_{6}] C l_{3}$ , $[C r (C_{6} H_{6})_{2}]$ , and $K_{2} [C r (C N)_{2} (O)_{2} (O_{2}) (N H_{3})]$ respectively are

One mole of the complex compound $C o (N H_{3})_{5} C l_{3}$ , gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $A g N O_{3}$ solution to yield two moles of $A g C l (s)$ . The structure of the complex is

In the coordination compound, $K_{4} [N i (C N)_{4}]$ the oxidation state of nickel is

Complex (A) has a composition of $H_{12} O_{6} Cl_{3} Cr$ . If the complex on treatment with conc. $H_{2} SO_{4}$ loses $13.5%$ of its original mass, the correct molecular formula of (A) is [Given : atomic mass of $Cr = 52 amu$ and $Cl = 35 amu$ ]