Which of the following has the highest flocculating power?

The correct answer is Option (1) → \([Fe(CN)_6]^{4-}\)

Coagulation of a sol is caused by the ion carrying a charge opposite to that of the colloidal particles. By the Hardy-Schulze rule, the greater the valence (charge) of this flocculating ion, the greater its power to cause coagulation. The anions given here coagulate a positively charged sol (such as \( \ce{Fe(OH)3} \)), and their flocculating power follows their charge:

    \( \ce{[Fe(CN)6]^{4-}} \) (−4) > \( \ce{PO4^{3-}} \) (−3) > \( \ce{SO4^{2-}} \) (−2) > \( \ce{Cl^{-}} \) (−1)

So [Fe(CN)₆]⁴⁻, with the highest charge, has the highest flocculating power.

Why the other options are incorrect

• Option (2) \(PO_4^{3-}\) — Incorrect. A charge of −3, one less than ferrocyanide, so its flocculating power is lower.

• Option (3) \(SO_4^{2-}\) — Incorrect. A charge of −2; weaker still.

• Option (4) \(Cl^-\) — Incorrect. A charge of only −1; the weakest coagulant of the four.

Note 1: 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 – “… The ion responsible for neutralisation of charge on the particles is called the coagulating ion. A negative ion causes the precipitation of positively charged sol and vice versa. It has been observed that, generally, the greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This is known as Hardy-Schulze rule. In the coagulation of a negative sol, the flocculating power is in the order: Al³⁺ > Ba²⁺ > Na⁺.
Similarly, in the coagulation of a positive sol, the flocculating power is in the order: [Fe(CN)₆]⁴⁻ > PO₄³⁻ > SO₄²⁻ > Cl⁻”

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: