Question

Which of the following has maximum coagulation power with ferric hydroxide solution?
A. Cryolite
B. ${K_2}C{r_2}{O_4}$
C. ${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
D. ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$

Answer 1

We know that the Coagulating power depends on the type of colloid. If the Colloid is negative, the ion with highest positive charge will have the highest coagulating power and if the colloid is positive then the ion with highest negative charge will have the highest coagulating power.

Complete step by step answer:
Let us discuss coagulation.
Coagulation is observed in surface chemistry. It is the process in which colloidal particles aggregate with each other and form larger particles and these particles settle down as precipitates. Coagulation process is also referred to as precipitation and flocculation and this process is carried out by the electrolytes which are added to the solution.
Hardy-Schulze rule:
If The valency of oppositely charged ions of the electrolyte being added is greater, faster is the coagulation. If the valency of coagulating ion is greater than greater is the coagulation power.
According to the above stated rule,
Cryolite$\left( {N{a_3}Al{F_6}} \right)$has the maximum coagulating power to coagulate a positive solution of ferric hydroxide because Cryolite has the highly charged opposite ion.
$N{a_3}Al{F_6} \rightleftarrows 3N{a^ + } + A{l^{3 + }} + F_6^{6 + }$
Now we can discuss about ${K_2}C{r_2}{O_4}$ as,
${K_2}C{r_2}{O_4} \rightleftarrows 2{K^ + } + C{r_2}O_4^{2 - }$
Now we can see the ${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ as,
${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftarrows 3{K^ + }{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$
Now we can discuss about ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ as,
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftarrows 4{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$
The coagulation power of the anion falls in the order,
${F_6}^{6 - } > {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} > {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} > C{r_2}{O_4}^{2 - }$

So, the correct answer is Option A.

Note: We need to know that there are two types of colloids, they are positive and negative colloids.
Positive colloids or SOLs:Positive sols have positive charge and all the metal hydroxide are positive sols.
Negative colloids or SOLs:Negative sols have negative charge and all the metal sulfides are negative sols.