Question

Which from following anions has greater power for coagulation of positive sol?

• A. \(\text{Cl}^-\)
• B. \([\text{Fe(CN)}_6]^{4-}\)
• C. \(\text{PO}_4^{2-}\)
• D. \(\text{SO}_4^{2-}\)

Hint:

Hardy-Schulze rule: coagulating power grows with the charge on the oppositely charged ion.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A sol is coagulated by ions carrying a charge opposite to the sol particles. The Hardy-Schulze rule says that the higher the charge on the coagulating ion, the greater its coagulating power.

Step 2: Key Formula or Approach:
A positive sol is coagulated by anions. Compare the charges: \(\text{Cl}^-\) (\(-1\)), \(\text{PO}_4^{2-}\) and \(\text{SO}_4^{2-}\) (as written, \(-2\)) and \([\text{Fe(CN)}_6]^{4-}\) (\(-4\)).

Step 3: Detailed Explanation:
Order of coagulating power: \(\text{Cl}^- < \text{SO}_4^{2-} \approx \text{PO}_4^{2-} < [\text{Fe(CN)}_6]^{4-}\).
The hexacyanoferrate(II) ion has the largest negative charge, \(4-\), so it neutralises the positive charge on the sol particles most effectively and needs the smallest concentration.

Final Answer:
\([\text{Fe(CN)}_6]^{4-}\) has the greatest coagulating power, option (B). \[ \boxed{[\text{Fe(CN)}_6]^{4-}} \]