In the equation:
`M+8CN^(-)+2H_(2)O+O_(2)to4[M(CN)_(2)]^(-)+4OH^(-)` metal M is:

Options

A. `Ag`
B. `Au`
C. `Cu`
D. `Hg`

Answer

To determine the metal M in the given equation, we can follow these steps:

### Step 1: Analyze the Reaction
The equation provided is:
\[ M + 8CN^- + 2H_2O + O_2 \rightarrow 4[M(CN)_2]^- + 4OH^- \]

This reaction involves a metal (M), cyanide ions (CN^-), water (H2O), and oxygen (O2) producing a complex ion and hydroxide ions (OH^-).

### Step 2: Identify the Type of Reaction
This reaction is indicative of a cyanide process, which is commonly used for the extraction of certain metals from their ores. The cyanide process is particularly known for extracting noble metals.

### Step 3: Consider the Properties of Metals
In the context of the cyanide process, we typically consider metals that are less reactive and can form stable complexes with cyanide. The most common metals associated with this process are:
- Gold (Au)
- Silver (Ag)

### Step 4: Confirm the Metal
Given that the reaction produces a soluble complex of the form \([M(CN)_2]^-\), we can conclude that the metal M must be one of the noble metals that can form such complexes.

### Step 5: Conclusion
Based on the analysis, the metal M can be either:
- Silver (Ag)
- Gold (Au)

Thus, the answer is that the metal M is either Ag or Au.

---