The most effective electrolyte for the coagulation \(Fe_2O_3. H_2O / Fe^{3+}\) is
(A) \(AgCl\)
(B) \(FeCl_3\)
(C) \(MgCl_2\)
(D) \(K_4[Fe(CN)_6]\)
(E) \(KCl\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → D only
Hydrated ferric oxide prepared with excess \( \ce{Fe^{3+}} \) adsorbs \( \ce{Fe^{3+}} \) ions, so \( \ce{Fe2O3} \)·xH2O/\( \ce{Fe^{3+}} \) is a positively charged sol. A sol is coagulated by the ion of opposite charge, here the anion of the electrolyte. By the Hardy–Schulze rule, the greater the charge on the coagulating ion, the greater its coagulating power. For a positive sol: \( \ce{[Fe(CN)6]^{4-}} \) > \( \ce{PO4^{3-}} \) > \( \ce{SO4^{2-}} \) > \( \ce{Cl^{-}} \).
(A) AgCl — Incorrect. It is practically insoluble, so it gives almost no ions to the sol.
(B) \( \ce{FeCl3} \) — Incorrect. Its anion is \( \ce{Cl^{-}} \) (charge 1); its \( \ce{Fe^{3+}} \) ion has the same charge as the sol and is the ion that stabilises it.
(C) \( \ce{MgCl2} \) — Incorrect. The coagulating ion is \( \ce{Cl^{-}} \), charge 1.
(D) \( \ce{K4[Fe(CN)6]} \) — Correct. It gives \( \ce{[Fe(CN)6]^{4-}} \), the anion with the highest charge, so it is the most effective.
(E) KCl — Incorrect. The coagulating ion is again \( \ce{Cl^{-}} \), charge 1.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Coagulation or Precipitation – …
