In the equatio, ` 4M + 8 CN^(-) + 2 H_2O + O_2 to 4 [M(CN)_2]^(-) + 4 OH ^(-)`, the metal M is

Options

A. Copper
B. Iron
C. Gold
D. Zinc.

Answer

To determine the metal M in the equation:

\[ 4M + 8 \text{CN}^- + 2 \text{H}_2\text{O} + \text{O}_2 \rightarrow 4 [M(\text{CN})_2]^- + 4 \text{OH}^- \]

we can analyze the equation step by step.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants consist of metal M, cyanide ions (CN^-), water (H2O), and oxygen (O2).
- The products are complex ions of the form [M(CN)2]^- and hydroxide ions (OH^-).

2. **Analyze the Reaction**:
- The reaction suggests that metal M forms a complex with cyanide ions. The presence of cyanide indicates that M is likely a transition metal, as these metals commonly form complexes with ligands like CN^-.

3. **Determine the Charge**:
- The product [M(CN)2]^- indicates that the overall charge of the complex ion is -1. Since there are two CN^- ions, each contributing a -1 charge, the metal M must have a +1 oxidation state to balance the charge.

4. **Common Metals with CN^-**:
- Metals that commonly form stable complexes with cyanide include gold (Au), silver (Ag), and platinum (Pt). Given the context of the reaction, we should focus on metals that can form a -1 charge complex with two cyanide ions.

5. **Identify the Metal**:
- The reaction also hints at the extraction of gold from its ores, as cyanide is often used in gold extraction processes. Therefore, it is reasonable to conclude that the metal M is likely to be gold (Au).

### Conclusion:
The metal M in the given equation is **Gold (Au)**.

---