The flocculating power of which of the following is highest for coagulation of $Al_{2}O_{3}.xH_{2}O$ sol ?

Question

Accepted Answer

The flocculating power of an ion is determined by its ability to neutralize the charge on colloidal particles, leading to coagulation. According to the Schulze-Hardy rule, the effectiveness of an ion in coagulating a sol is directly related to the valency of the ion. The higher the valency, the greater the flocculating power.

1. **Option 3: $[Fe(CN)_{6}]^{4-}$**  
   The ion $[Fe(CN)_{6}]^{4-}$ has a charge of -4. According to the Schulze-Hardy rule, an ion with a higher charge will have a greater ability to neutralize the positive charge on the $Al_{2}O_{3}.xH_{2}O$ sol particles, which are positively charged. Therefore, $[Fe(CN)_{6}]^{4-}$, having the highest negative charge among the given options, will have the highest flocculating power, making it the most effective in coagulating the sol.

2. **Option 1: $Al^{3+}$**  
   The $Al^{3+}$ ion has a charge of +3. While it is effective in coagulating negatively charged sols due to its high positive charge, it is not as effective as $[Fe(CN)_{6}]^{4-}$ for positively charged sols like $Al_{2}O_{3}.xH_{2}O$. The negative charge of $[Fe(CN)_{6}]^{4-}$ is more suitable for neutralizing the positive charge on the sol particles.

3. **Option 2: $SO_{4}^{2-}$**  
   The $SO_{4}^{2-}$ ion has a charge of -2. Although it can neutralize positive charges, its flocculating power is less than that of $[Fe(CN)_{6}]^{4-}$ due to its lower charge magnitude. Therefore, it is less effective in coagulating the $Al_{2}O_{3}.xH_{2}O$ sol compared to $[Fe(CN)_{6}]^{4-}$.

4. **Option 4: $Na^{+}$**  
   The $Na^{+}$ ion has a charge of +1. It is the least effective in coagulating positively charged sols because it has the lowest charge magnitude among the options. It is not suitable for neutralizing the positive charge on the $Al_{2}O_{3}.xH_{2}O$ sol particles.

In conclusion, $[Fe(CN)_{6}]^{4-}$ is the most effective ion for coagulating the $Al_{2}O_{3}.xH_{2}O$ sol due to its high negative charge, which provides the greatest flocculating power according to the Schulze-Hardy rule.