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# Which of the following chemical reactions is correctly shown aqua-regia dissolving in gold?A.$Au + 2\left[ {Cl} \right] \to AuC{l_2}\xrightarrow{{HCl}}{H_2}\left[ {AuC{l_4}} \right]$B.$Ag + 3\left[ {Cl} \right] \to AgC{l_3}\xrightarrow{{HCl}}H\left[ {AgC{l_4}} \right]$C.$Au + 3\left[ {Cl} \right] \to AuC{l_3}\xrightarrow{{HCl}}H\left[ {AuC{l_4}} \right]$D. $Ag + 2\left[ {Cl} \right] \to AgC{l_2}\xrightarrow{{HCl}}{H_2}\left[ {AgC{l_4}} \right]$

Last updated date: 22nd Mar 2023
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Hint: We have to know that water Regia is a yellow-orange (now and then red) seething fluid, so named by chemists since it can break up the respectable metals gold and platinum, however not all metals. Water Regia is basically used to create chloroauric corrosive, the electrolyte in the Wohlwill interaction for refining the greatest ( $99.999\%$ ) gold.

We have to see the water regia breaks down gold, albeit neither constituent corrosive will do so alone. Nitric corrosive is an amazing oxidizer, which will really break up a practically imperceptible measure of gold, framing gold particles ( $A{u^{3 + }}$ ). The hydrochloric corrosive gives a prepared inventory of chloride particles ( $C{l^ - }$ ), which respond with the gold particles to create tetrachloroaurate( $III$ ) anions, additionally in an arrangement. The response with hydrochloric corrosive is a harmony response that favors the arrangement of chloro-urate anions ( $AuC{l^{4 - }}$ ). This outcome, in the expulsion of gold particles from the arrangement, permits further oxidation of gold to occur. The gold disintegrates to become chloro-uric corrosive. Then, the gold might be broken up by the chlorine present in the water regia.
Water regia is a combination of nitric corrosive and hydrochloric corrosive, ideally in a molar proportion of $1:3$. Water regia is a yellow-orange raging fluid. It can disintegrate the honorable metals gold and platinum.
Therefore, Option (C) $Au + 3\left[ {Cl} \right] \to AuC{l_3}\xrightarrow{{HCl}}H\left[ {AuC{l_4}} \right]$ is correct.