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:

E, D, B, A, C

The correct answer is option 3. E, D, B, A, C.

Coagulation (or flocculation) is the process by which colloidal particles come together to form larger aggregates, eventually leading to precipitation. In colloidal systems, the stability of the colloid is maintained by the charge on the particles, which prevents them from coming together.  To destabilize (or coagulate) a colloid, oppositely charged ions can be added. These ions neutralize the charge on the colloidal particles, reducing the repulsion between them and allowing them to aggregate.

The Hardy-Schulze rule is a fundamental principle in colloid chemistry that states:

The coagulating power of an ion is directly related to its charge. The higher the charge (valency) of the ion, the greater its ability to coagulate a colloid.

For example, a trivalent ion (\(Al^{3+}\)) will have a much greater coagulating power than a divalent ion (\(Ca^{2+}\)), which in turn will have more power than a monovalent ion (\(Na^+\)).

Given Ions and Their Charges:

Let us examine the ions in question along with their charges:

(A) \(SO_4^{2-}\): Charge: -2 (Divalent ion)

(B) \(PO_4^{3-}\): Charge: -3 (Trivalent ion)

(C) \(Cl^-\): Charge: -1 (Monovalent ion)

(D) \([Fe(CN)_6]^{4-}\): Charge: -4 (Tetravalent ion)

(E) \([Cu(CN)_6]^{5-}\): Charge: -5 (Pentavalent ion)

Applying the Hardy-Schulze Rule: According to the Hardy-Schulze rule, the coagulating power increases with the charge of the ion. Therefore, the ions should be ranked in decreasing order of their charge:

Pentavalent ion \([Cu(CN)_6]^{5-}\) (Charge = -5): Highest charge, hence the highest coagulating power.

Tetravalent ion \([Fe(CN)_6]^{4-}\) (Charge = -4): Next highest charge, hence the next highest coagulating power.

Trivalent ion \(PO_4^{3-}\) (Charge = -3): Lower than the tetravalent ion, but higher than the divalent ion, so it has intermediate coagulating power.

Divalent ion\(SO_4^{2-}\) (Charge = -2): Lower than the trivalent ion, but higher than the monovalent ion, so it has lower coagulating power

Monovalent ion \(Cl^-\) (Charge = -1): Lowest charge, and thus the lowest coagulating power.

Order of Coagulating Power: Given this analysis, the decreasing order of coagulating power is:

\(\text{(E)} \, [Cu(CN)_6]^{5-} > \text{(D)} \, [Fe(CN)_6]^{4-} > \text{(B)} \, PO_4^{3-} > \text{(A)} \, SO_4^{2-} > \text{(C)} \, Cl^- \)

The correct answer is option 3: E, D, B, A, C.

This option reflects the decreasing order of coagulating power based on the charge of the ions, as dictated by the Hardy-Schulze rule. The pentavalent ion \([Cu(CN)_6]^{5-}\) has the highest coagulating power, followed by the tetravalent ion \([Fe(CN)_6]^{4-}\), then the trivalent ion \(PO_4^{3-}\), the divalent ion \(SO_4^{2-}\), and finally, the monovalent ion \(Cl^-\) with the lowest coagulating power.