Medium MCQ +4 / -1 PYQ · JEE Mains 2020

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :

  1. A AICl<sub>3</sub> &gt; K<sub>3</sub>[Fe(CN)<sub>6</sub>] &gt; K<sub>2</sub>CrO<sub>4</sub> &gt; KBr = KNO<sub>3</sub>
  2. B K<sub>3</sub>[Fe(CN)<sub>6</sub>] &lt; K<sub>2</sub>CrO<sub>4</sub> &lt; KBr = KNO<sub>3</sub> = AlCl<sub>3</sub> Correct answer
  3. C K<sub>3</sub>[Fe(CN)<sub>6</sub>] &lt; K<sub>2</sub>CrO<sub>4</sub> &lt; AlCl<sub>3</sub> &lt; KBr &lt; KNO<sub>3</sub>
  4. D K<sub>3</sub>[Fe(CN)<sub>6</sub>] &gt; AlCl<sub>3</sub> &gt; K<sub>2</sub>CrO<sub>4</sub> &gt; KBr &gt; KNO<sub>3</sub>

Solution

Since, Fe(OH)<sub>3</sub> is positively charged sol, hence, anionic charge will flocculate. <br><br>As per Hardy Schulze rules coagulation power of anion follows the order : <br><br>Fe(CN)<sub>6</sub><sup>3–</sup> $&gt;$ CrO<sub>4</sub><sup>2–</sup> $&gt;$ Cl<sup>–</sup> = Br<sup>–</sup><sup></sup> = NO<sub>3</sub><sub></sub><sup>–</sup> <br><br>Higher the coagulation power lower will be its flocculation value therefore order will be : <br><br>Fe(CN)<sub>6</sub><sup>3–</sup> $&lt;$ CrO<sub>4</sub><sup>2–</sup> $&lt;$ Cl<sup>–</sup> = Br<sup>–</sup><sup></sup> = NO<sub>3</sub><sub></sub><sup>–</sup>

About this question

Subject: Chemistry · Chapter: Surface Chemistry · Topic: Adsorption

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