Animated Solution for Chemistry - Metallurgy: The cyanide process of gold extraction involves leaching out gold from its ore with ${\text{CN}}^{-}$ in the presence of $\mathbf{Q}$ in water to form $\mathbf{R}$. Subsequently, $\mathbf{R}$ is treated with $\mathbf{T}$ to obtain $\text{Au}$ and $\mathbf{Z}$. Choose the correct option(s).

Unraveling the Cyanide Process

Gold is a noble metal, famous for its reluctance to react with most chemicals. This chemical inertness is what keeps gold shining for millennia, but it also poses a significant challenge: how do we extract it from its ore when it refuses to dissolve? Enter the Macarthur-Forrest cyanide process, a brilliant application of coordination chemistry and redox reactions that revolutionized gold mining.

Coaxing Gold into Solution

The first major hurdle is getting the solid gold out of the crushed rock and into a liquid solution. This is called leaching. We treat the native gold ore with a dilute aqueous solution of sodium or potassium cyanide (${\text{CN}}^{-}$).

However, cyanide alone isn't enough. Because gold is in a $0$ oxidation state, it needs to lose an electron to form a bond. We need an oxidizing agent. This is where atmospheric oxygen steps in to save the day. The oxygen gas (${\text{O}}_2$) acts as the crucial oxidizing agent, pulling electrons away from the gold atoms.

From this balanced equation, it is crystal clear that the mysterious substance $\mathbf{Q}$ required in the presence of water is Oxygen (${\text{O}}_2$).

A Case of Mistaken Identity

As the gold oxidizes to ${\text{Au}}^{+}$, it immediately gets surrounded by cyanide ions. Cyanide is a strong-field ligand and forms a highly stable, linear coordination complex with gold(I).

This resulting soluble complex, $\mathbf{R}$, is the dicyanoaurate(I) ion, formulated as $[\text{Au}(\text{CN}{)}_2{]}^{-}$.

A common trap for students is confusing this with the gold(III) complex, $[\text{Au}(\text{CN}{)}_4{]}^{-}$. Remember, under these mild leaching conditions, gold only oxidizes to the $+1$ state, giving a coordination number of $2$. Therefore, option (B) is incorrect.

Reclaiming the Gold

Now we have a solution rich in dissolved gold, but we want solid metal! To achieve this, we employ a classic metal displacement reaction, also known as cementation.

We introduce a metal that is significantly more electropositive (more reactive) than gold. Zinc ($\text{Zn}$) is the perfect candidate. Zinc acts as a powerful reducing agent. It eagerly gives up its electrons to the gold ions in the complex, forcing the gold to reduce back to its solid, metallic state (${\text{Au}}^0$).

This identifies our reducing agent $\mathbf{T}$ as Zinc ($\text{Zn}$).

As zinc sacrifices itself to precipitate the gold, it oxidizes to ${\text{Zn}}^{2+}$ and takes gold's place, bonding with the newly freed cyanide ligands. Zinc(II) typically forms a tetrahedral complex with a coordination number of $4$.

Thus, the final soluble complex $\mathbf{Z}$ left in the solution is the tetracyanozincate(II) ion, formulated as $[\text{Zn}(\text{CN}{)}_4{]}^{2-}$.

By carefully tracing the flow of electrons and ligands, we've successfully decoded the entire process, confirming that options A, C, and D are the correct statements.