A colloid is a heterogeneous system in which one substance is dispersed as very fine particles in another substance. Their range of diameters is between 1-1000 nm. Based on dispersed phase and dispersion medium, colloids may be of different types like sol, gel, emulsion, aerosol, foam etc. Based on nature of interaction between two phases, sols are divided into Lyophillic and Lyophobic sols. Colloidal solutions exhibit various properties like Tyndall effect, Brownian movement and Electrophoresis. Stability of colloids is due to presence of charge. If somehow charge is removed, the particles will coagulate. Coagulation rate by various solutions containing ions, is decided by Hardy-Schulze rule. Everyday we come across many colloids. The meals we eat, clothes we wear, wooden furniture etc. are largely composed of colloids.

The coagulation of \(TiO_2\) sol can be effectively carried out by the solution containing

\([Fe(CN)_6]^{4-}\)

The correct answer is option 4. \([Fe(CN)_6]^{4-}\).

Coagulation (or flocculation) is the process of destabilizing colloids, leading to the aggregation of dispersed particles so that they form larger clumps and eventually settle out of the solution. This happens when oppositely charged ions neutralize the surface charges of colloidal particles, allowing them to aggregate.

Charge of the Sol: In the case of \(TiO_2\) sol, the dispersed colloidal particles are positively charged. Hence, negatively charged ions (anions) are required to neutralize the positive charge on the particles and bring about coagulation.

Hardy-Schulze Rule:

The Hardy-Schulze rule explains how coagulation occurs based on the charge of the ion. According to this rule

i. The greater the charge on the oppositely charged ion, the more efficient it is at coagulating the colloidal particles.

ii. Multivalent ions are far more effective than monovalent ions.

In the case of positively charged colloids like \(TiO_2\), negatively charged ions (anions) will be responsible for coagulation, and the effectiveness increases with the charge of the anion.

Applying the Rule to the Given Options:

1. \(Cl^-\):

Charge: \(-1\) (monovalent)

Monovalent ions like chloride can cause coagulation but are the least effective because of their single negative charge. They have limited capacity to neutralize the positive charge on the \(TiO_2\) sol.

Effectiveness: Low

2. \(PO_4^{3-}\):

Charge: \(-3\) (trivalent)

Phosphate is a trivalent anion with a charge of \(-3\). According to the Hardy-Schulze rule, trivalent ions are much more effective than monovalent or divalent ions. They can neutralize the charge on colloidal particles more efficiently, causing faster coagulation.

Effectiveness: High

3. \(SO_4^{2-}\):

Charge: \(-2\) (divalent)

Sulfate is a divalent anion with a charge of \(-2\). It is more effective than \(Cl^-\) because it has a higher charge, but less effective than trivalent or tetravalent ions like \(PO_4^{3-}\) or \([Fe(CN)_6]^{4-}\).

Effectiveness: Moderate

4. \([Fe(CN)_6]^{4-}\):

Charge: \(-4\) (tetravalent)

The hexacyanoferrate ion \([Fe(CN)_6]^{4-}\) has a charge of \(-4\), making it a tetravalent ion. According to the Hardy-Schulze rule, the effectiveness of coagulation increases with the charge of the anion. A tetravalent ion will neutralize the positive charge of the \(TiO_2\) particles far more effectively than mono-, di-, or trivalent ions.

Effectiveness: Very High

Relative Coagulating Power Based on Ion Charge:

The order of coagulating power of the given ions (based on the Hardy-Schulze rule) is as follows:

\([Fe(CN)_6]^{4-} > PO_4^{3-} > SO_4^{2-} > Cl^-\)

\([Fe(CN)_6]^{4-}\) is the most effective because it has the highest negative charge (\(-4\)).

\(PO_4^{3-}\) is also highly effective but less so than \([Fe(CN)_6]^{4-}\), as it has a charge of \(-3\).

\(SO_4^{2-}\) has moderate effectiveness due to its charge of \(-2\).

\(Cl^-\) is the least effective, with a charge of only \(-1\).

The coagulation of a positively charged \(TiO_2\) sol will be most effectively carried out by the ion with the highest negative charge, which is \([Fe(CN)_6]^{4-}\).

Therefore, the correct answer is 4. \([Fe(CN)_6]^{4-}\).