Q8 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
Which from following anions has greater power for coagulation of positive sol?
- A$\text{Cl}^-$
- Correct answerB$[\text{Fe(CN)}_6]^{4-}$
- C$\text{PO}_4^{2-}$
- D$\text{SO}_4^{2-}$
Solution
According to the Hardy-Schulze rule, the coagulating power of an ion is directly proportional to the magnitude of its charge (valency). For the coagulation of a positive sol, an anion is required.
The given anions and their charge magnitudes are:
$\text{Cl}^-$ : $1$
$\text{SO}_4^{2-}$ : $2$
$\text{PO}_4^{3-}$ (given as $\text{PO}_4^{2-}$ in options) : $3$
$[\text{Fe(CN)}_6]^{4-}$ : $4$
Since $[\text{Fe(CN)}_6]^{4-}$ has the highest negative charge, it possesses the maximum coagulating power among the given options.
The given anions and their charge magnitudes are:
$\text{Cl}^-$ : $1$
$\text{SO}_4^{2-}$ : $2$
$\text{PO}_4^{3-}$ (given as $\text{PO}_4^{2-}$ in options) : $3$
$[\text{Fe(CN)}_6]^{4-}$ : $4$
Since $[\text{Fe(CN)}_6]^{4-}$ has the highest negative charge, it possesses the maximum coagulating power among the given options.
Hence, the correct answer is option: 'B'.