Assertion: Gold sol is prepared by Bredig’s arc method.
A. True
B. False
Answer
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Hint: Gold sol is a colloidal solution. Colloidal is a heterogeneous mixture in which a phase remains dispersed in the suspension. The size of colloidal particles varies from $10$ to $1000\,{{\text{A}}^ \circ }$. In gold sol solid particles are dispersed in liquid dispersion phase.
Complete answer:
Gold sol is a colloidal solution of gold. Gold sol is prepared by the dispersion followed by condensation method.
To prepare the gold sol, electrodes of gold are taken into a cell and then an electric arc is supplied. The heat is generated which evaporates the metal. These metal particles are then condensed to form colloidal size gold particles.
Bredig’s arc method is used for the preparation of metals such as gold and copper, so it is true that Gold sol is prepared by Bredig’s arc method.
Therefore, option (A) true, is correct.
Additional information:
Colloidal solutions can be negatively or positively charged. The acidic dyes, metal sol, and metallic sulphide are generally negatively charged sol. The basic dyes and metallic oxide are generally positively charged sol. The haemoglobin is also positively charged. The gold particles that are dispersed phase particles adsorbed negatively charge $({\text{AuO}}_2^ - )$ so, all the gold particles have a negative charge.
Note:
Gold sol is also prepared by reduction method. By the reaction of chloride of gold with reducing agent gold atom forms. The chemical formula of chloride of gold is ${\text{AuC}}{{\text{l}}_{\text{3}}}$ and the reducing agent is tin (II) chloride \[{\text{SnC}}{{\text{l}}_2}\]. By the reaction of ${\text{AuC}}{{\text{l}}_{\text{3}}}$ with reducing agent tin (IV) chloride \[{\text{SnC}}{{\text{l}}_2}\], gold atoms and tin (IV) tetrachloride \[{\text{SnC}}{{\text{l}}_{\text{4}}}\] forms.
The reaction is as follows:
${\text{2}}\,{\text{AuC}}{{\text{l}}_{\text{3}}}\, + \,3\,{\text{SnC}}{{\text{l}}_2}\, \to 2{\text{Au(sol)}}\,\,{\text{ + }}\,3\,{\text{SnC}}{{\text{l}}_{\text{4}}}$
Complete answer:
Gold sol is a colloidal solution of gold. Gold sol is prepared by the dispersion followed by condensation method.
To prepare the gold sol, electrodes of gold are taken into a cell and then an electric arc is supplied. The heat is generated which evaporates the metal. These metal particles are then condensed to form colloidal size gold particles.
Bredig’s arc method is used for the preparation of metals such as gold and copper, so it is true that Gold sol is prepared by Bredig’s arc method.
Therefore, option (A) true, is correct.
Additional information:
Colloidal solutions can be negatively or positively charged. The acidic dyes, metal sol, and metallic sulphide are generally negatively charged sol. The basic dyes and metallic oxide are generally positively charged sol. The haemoglobin is also positively charged. The gold particles that are dispersed phase particles adsorbed negatively charge $({\text{AuO}}_2^ - )$ so, all the gold particles have a negative charge.
Note:
Gold sol is also prepared by reduction method. By the reaction of chloride of gold with reducing agent gold atom forms. The chemical formula of chloride of gold is ${\text{AuC}}{{\text{l}}_{\text{3}}}$ and the reducing agent is tin (II) chloride \[{\text{SnC}}{{\text{l}}_2}\]. By the reaction of ${\text{AuC}}{{\text{l}}_{\text{3}}}$ with reducing agent tin (IV) chloride \[{\text{SnC}}{{\text{l}}_2}\], gold atoms and tin (IV) tetrachloride \[{\text{SnC}}{{\text{l}}_{\text{4}}}\] forms.
The reaction is as follows:
${\text{2}}\,{\text{AuC}}{{\text{l}}_{\text{3}}}\, + \,3\,{\text{SnC}}{{\text{l}}_2}\, \to 2{\text{Au(sol)}}\,\,{\text{ + }}\,3\,{\text{SnC}}{{\text{l}}_{\text{4}}}$
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