Question

Which of the following ion has greater coagulating power for negatively charged sol?

• A. $Fe(CN)_{6}^{4-}$
• B. $PO_{4}^{3-}$
• C. $Ba^{2+}$
• D. $Al^{3+}$

Hint:

Chemistry Tip: The coagulating power order for negative sols is typically $Al^{3+}>Mg^{2+} / Ba^{2+}>Na^{+} / K^{+}$. The effect increases exponentially with charge.

Answer 1

The Correct Option is D

Concept: Chemistry (Surface Chemistry) - Coagulation and Hardy-Schulze Rule.

Step 1: Identify the nature of the sol. The question specifies a "negatively charged sol". To cause coagulation (precipitation) of a negative sol, neutralizing positively charged ions (cations) are required.

Step 2: Eliminate the incompatible options. Options (A) $Fe(CN)_{6}^{4-}$ and (B) $PO_{4}^{3-}$ are complex anions (negatively charged). They will not be effective in coagulating a negatively charged sol because like charges repel. We strictly consider the cations: (C) $Ba^{2+}$ and (D) $Al^{3+}$.

Step 3: State the Hardy-Schulze rule. The Hardy-Schulze rule states that the coagulating power of an active flocculating ion is directly proportional to its valency (magnitude of charge). The higher the positive charge on the cation, the greater its power to neutralize and precipitate the negative sol.

Step 4: Compare the charges of the available cations. The Barium ion ($Ba^{2+}$) has a charge of +2. The Aluminum ion ($Al^{3+}$) has a charge of +3.

Step 5: Determine the ion with the greatest coagulating power. Because the $+3$ charge of $Al^{3+}$ is greater than the $+2$ charge of $Ba^{2+}$, $Al^{3+}$ will exhibit a significantly higher coagulating power according to the Hardy-Schulze rule. $$ \therefore \text{The ion with the greater coagulating power is } Al^{3+}. $$