What is the product formed when gold is dissolved in aqua regia?
A) ${H_3}AuC{l_2}{O_2}$
B) $HAuC{l_4}$
C) ${H_2}AuC{l_4}$
D) ${H_2}AuCl{O_2}$
Answer
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Hint: Aqua regia is compound formed by the combination of $HCl$ and $HN{O_3}$ in a molar ratio of 1:3.It is fuming liquid appears to be in yellow-orange and capable of dissolving noble metals like gold and platinum.
Complete answer:
Aqua regia is a mixture of hydrochloric acid ($HCl$) and nitric acid ($HN{O_3}$). It is able to dissolve gold because of the unique action of $HCl$ and $HN{O_3}$. As $HN{O_3}$ is a powerful oxidising agent, it converts gold to its ionic form ($A{u^{3 + }}$). $HCl$makes way for the chlorine anions to react with the gold cations in order to form tetrachloroaurate (III) anions in presence of these $A{u^{3 + }}$ ions in the solution.
Once the above reaction with $HCl$ reaches equilibrium it leads to the formation of chloroaurate anions(${\left[ {AuC{l_4}} \right]^ - }$) and the gold ions are removed from the solution making way for oxidation to occur.
The following represents the reaction between gold and aqua regia:
\[A{u_{(s)}} + 3N{O_3}{^ - _{(aq)}} + 6{H^ + }_{(aq)} \rightleftharpoons A{u^{3 + }}_{(aq)} + 3N{O_2}_{(g)} + 3{H_2}{O_{(s)}}\]
Chloride ions of $HCl$ forms coordination complex with gold
$A{u^{3 + }}_{(aq)} + 4HC{l_{(aq)}} \rightleftharpoons AuC{l_4}$
${H^ + } + AuC{l_4}^ - \rightleftharpoons HAuC{l_4}$
Hence the product is $HAuC{l_4}$ .
$\therefore $Option (B) is the answer.
Additional Information: Aqua regia is widely used in deep cleaning the tubes used in nuclear magnetic resonance or NMR spectroscopy. It is capable of removing all the traces of paramagnetic elements like chromium which can contaminate spectra and ruin research.
Note: By making use of aqua regia it is possible to produce gold having a purity of about 99.999%. The reactivity and strength of aqua regia make it capable of dissolving platinum in the similar manner. Since aqua regia react with its components it loses its effectiveness quickly hence its components are usually only mixed immediately before use.
Complete answer:
Aqua regia is a mixture of hydrochloric acid ($HCl$) and nitric acid ($HN{O_3}$). It is able to dissolve gold because of the unique action of $HCl$ and $HN{O_3}$. As $HN{O_3}$ is a powerful oxidising agent, it converts gold to its ionic form ($A{u^{3 + }}$). $HCl$makes way for the chlorine anions to react with the gold cations in order to form tetrachloroaurate (III) anions in presence of these $A{u^{3 + }}$ ions in the solution.
Once the above reaction with $HCl$ reaches equilibrium it leads to the formation of chloroaurate anions(${\left[ {AuC{l_4}} \right]^ - }$) and the gold ions are removed from the solution making way for oxidation to occur.
The following represents the reaction between gold and aqua regia:
\[A{u_{(s)}} + 3N{O_3}{^ - _{(aq)}} + 6{H^ + }_{(aq)} \rightleftharpoons A{u^{3 + }}_{(aq)} + 3N{O_2}_{(g)} + 3{H_2}{O_{(s)}}\]
Chloride ions of $HCl$ forms coordination complex with gold
$A{u^{3 + }}_{(aq)} + 4HC{l_{(aq)}} \rightleftharpoons AuC{l_4}$
${H^ + } + AuC{l_4}^ - \rightleftharpoons HAuC{l_4}$
Hence the product is $HAuC{l_4}$ .
$\therefore $Option (B) is the answer.
Additional Information: Aqua regia is widely used in deep cleaning the tubes used in nuclear magnetic resonance or NMR spectroscopy. It is capable of removing all the traces of paramagnetic elements like chromium which can contaminate spectra and ruin research.
Note: By making use of aqua regia it is possible to produce gold having a purity of about 99.999%. The reactivity and strength of aqua regia make it capable of dissolving platinum in the similar manner. Since aqua regia react with its components it loses its effectiveness quickly hence its components are usually only mixed immediately before use.
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