Arrange the following ions in decreasing order of their coagulating power towards a positively charged colloid:
(A) \(SO_4^{2-}\)
(B) \(PO_4^{3-}\)
(C) \(Cl^-\)
(D) \([Fe(CN)_6]^{4-}\)
(E) \([Cu(CN)_6]^{5-}\)
Choose the correct answers from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → E, D, B, A, C
A positively charged sol is coagulated by the anions of an added electrolyte. By the Hardy-Schulze rule, the greater the valence (charge) of the flocculating ion, the greater its power to cause coagulation.
(E) \( \ce{[Cu(CN)6]^{5-}} \) — charge 5−
(D) \( \ce{[Fe(CN)6]^{4-}} \) — charge 4−
(B) \( \ce{PO4^{3-}} \) — charge 3−
(A) \( \ce{SO4^{2-}} \) — charge 2−
(C) \( \ce{Cl^{-}} \) — charge 1−
Shortcut: Options (2) and (3) start with E, D; only Option (3) puts B (3−) before A (2−).
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 – “… 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.
Similarly, in the coagulation of a positive sol, the flocculating power is in the order: [Fe(CN)6]4– > PO43– > SO42– > Cl–”