Answer
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Hint: Purple of cassius is actually a purple color pigment, which is formed by reaction of gold salts and$Sn\left( II \right)$chloride. It is having purple or even sometimes red color also. This test was first observed by Andreas Cassius who was a German physician.
Step by step solution:
First of all, let’s see preparation of this-
- It is prepared by mixing a very dilute solution of gold chloride with tin chloride (also called stannous chloride) solution. Here, $SnC{{l}_{2}}$is in +2 oxidation state which gets oxidized to $SnC{{l}_{4}}$having +4 oxidation state, which is showing that this is oxidation reaction. And $AuC{{l}_{3}}$ is reduced to Au.
\[2AuC{{l}_{3}}+3SnC{{l}_{2}}\to 2Au\downarrow +3SnC{{l}_{4}}\]
Here we can see that we are able to get gold separately and stannous chloride separately.
- Now the next step in preparation was that, the gold which was precipitated in above part is absorbed by stannic hydroxide , which was formed by reaction of $SnC{{l}_{4}}$ (Tin (1V) Hydroxide) with water. We can see the reaction over here,
\[SnC{{l}_{4}}+4{{H}_{2}}O\to Sn{{\left( OH \right)}_{4}}+4HCl\]
- We are able to see that the$SnC{{l}_{4}}$ formed here, when reacted with water, gives the tin hydroxide and HCl.
-Here, we can say that purple of caussius is formed by the absorption of gold particles by stannic hydroxide
Therefore, we can conclude that purple of caussius is a colloidal solution of gold. Hence, option (B) is correct.
Additional information:
- This is a color generally used in various industrial fields, in the glazing of earthenware, porcelain, and enamels. It is one of the most ancient as well as the most beautiful of colors.
- Purple of Cassius is actually very costly.
- We can find that it is mainly used worldwide to impart red color to glass as well as also used in worldwide laboratories to determine presence of gold as a chemical test.
Note:
- Note that, whenever this is used as a test, at that time the intensity of color correlates with the concentration of gold present.
- One should not be confused with Au and Al elements. Au is gold and Al is aluminium.
Step by step solution:
First of all, let’s see preparation of this-
- It is prepared by mixing a very dilute solution of gold chloride with tin chloride (also called stannous chloride) solution. Here, $SnC{{l}_{2}}$is in +2 oxidation state which gets oxidized to $SnC{{l}_{4}}$having +4 oxidation state, which is showing that this is oxidation reaction. And $AuC{{l}_{3}}$ is reduced to Au.
\[2AuC{{l}_{3}}+3SnC{{l}_{2}}\to 2Au\downarrow +3SnC{{l}_{4}}\]
Here we can see that we are able to get gold separately and stannous chloride separately.
- Now the next step in preparation was that, the gold which was precipitated in above part is absorbed by stannic hydroxide , which was formed by reaction of $SnC{{l}_{4}}$ (Tin (1V) Hydroxide) with water. We can see the reaction over here,
\[SnC{{l}_{4}}+4{{H}_{2}}O\to Sn{{\left( OH \right)}_{4}}+4HCl\]
- We are able to see that the$SnC{{l}_{4}}$ formed here, when reacted with water, gives the tin hydroxide and HCl.
-Here, we can say that purple of caussius is formed by the absorption of gold particles by stannic hydroxide
Therefore, we can conclude that purple of caussius is a colloidal solution of gold. Hence, option (B) is correct.
Additional information:
- This is a color generally used in various industrial fields, in the glazing of earthenware, porcelain, and enamels. It is one of the most ancient as well as the most beautiful of colors.
- Purple of Cassius is actually very costly.
- We can find that it is mainly used worldwide to impart red color to glass as well as also used in worldwide laboratories to determine presence of gold as a chemical test.
Note:
- Note that, whenever this is used as a test, at that time the intensity of color correlates with the concentration of gold present.
- One should not be confused with Au and Al elements. Au is gold and Al is aluminium.
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