The sum of oxidati on states of two silver ions in `[Ag(NH_3)_2[Ag(CN)_2]` complex is

Answer

To determine the sum of oxidation states of the two silver ions in the complex \([Ag(NH_3)_2][Ag(CN)_2]\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Complexes**: The complex can be broken down into two parts: \([Ag(NH_3)_2]^+\) (a cationic complex) and \([Ag(CN)_2]^-\) (an anionic complex).

2. **Determine the Charge of Each Complex**:
- The \([Ag(NH_3)_2]^+\) complex has a +1 charge.
- The \([Ag(CN)_2]^-\) complex has a -1 charge.

3. **Analyze the Ligands**:
- Ammonia (\(NH_3\)) is a neutral ligand, contributing no charge.
- Cyanide (\(CN^-\)) is a negatively charged ligand, contributing -1 charge.

4. **Set Up the Oxidation State Equations**:
- For the first complex \([Ag(NH_3)_2]^+\):
- Let the oxidation state of Ag be \(X\).
- The equation becomes:
\[ X + 0 + 0 = +1 \quad \text{(since \(NH_3\) is neutral)} \]
- Thus, \(X = +1\).

- For the second complex \([Ag(CN)_2]^-\):
- Let the oxidation state of Ag be \(Y\).
- The equation becomes:
\[ Y + (-1) \times 2 = -1 \quad \text{(since there are two \(CN^-\) ligands)} \]
- Simplifying this gives:
\[ Y - 2 = -1 \implies Y = +1. \]

5. **Calculate the Sum of Oxidation States**:
- Now, we have the oxidation states:
- \(X = +1\) (from \([Ag(NH_3)_2]^+\))
- \(Y = +1\) (from \([Ag(CN)_2]^-\))
- Therefore, the sum of the oxidation states of the two silver ions is:
\[ X + Y = +1 + +1 = +2. \]

### Final Answer:
The sum of the oxidation states of the two silver ions in the complex \([Ag(NH_3)_2][Ag(CN)_2]\) is **2**.