Step 1: Recall the Hardy-Schulze rule for coagulation.
The Hardy-Schulze rule states that the coagulating power of an ion is directly proportional to the magnitude of its charge. Higher the charge on the ion (opposite to the sol charge), greater its ability to neutralise the charge on colloidal particles and cause coagulation.
Step 2: Identify the type of sol being coagulated.
The question says we are coagulating a positive sol. For positive sols, negatively charged ions (anions) cause coagulation. Higher the negative charge on the anion, greater the coagulating power.
Step 3: Determine the charges of the given ions.
Ion I: $[Fe(CN)_6]^{4-}$ carries a charge of $-4$. Ion II: $Cl^-$ carries a charge of $-1$. Ion III: $SO_4^{2-}$ carries a charge of $-2$.
Step 4: Apply the Hardy-Schulze rule.
Since coagulating power increases with increasing charge magnitude for anions coagulating a positive sol: \[ [Fe(CN)_6]^{4-} > SO_4^{2-} > Cl^- \] That is, I $>$ III $>$ II.
Step 5: Match with the options.
The order I $>$ III $>$ II matches option 4.
Step 6: Confirm the reasoning.
This is a straightforward application of the Hardy-Schulze rule. The ion with the highest charge ($[Fe(CN)_6]^{4-}$, charge $= -4$) is the most effective coagulant, followed by sulphate ($-2$), then chloride ($-1$).
\[ \boxed{\text{I} > \text{III} > \text{II}} \]