In `[Ag(CN)_2]^(-)`, the number of `pi` bonds is

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To determine the number of π (pi) bonds in the complex ion \([Ag(CN)_2]^-\), we can analyze the structure of the cyanide ion (CN⁻) and how it interacts with silver (Ag).

### Step-by-Step Solution:

1. **Identify the Components**:
The complex ion \([Ag(CN)_2]^-\) consists of one silver ion (Ag) and two cyanide ions (CN⁻).

2. **Analyze the Cyanide Ion**:
Each cyanide ion (CN⁻) has a triple bond between the carbon (C) and nitrogen (N). The structure can be represented as:
\[ C \equiv N \]
In a triple bond, there is one sigma (σ) bond and two pi (π) bonds.

3. **Count the Bonds in Each CN⁻**:
- For each CN⁻:
- 1 σ bond
- 2 π bonds
Therefore, each cyanide ion contributes 2 π bonds.

4. **Total π Bonds from Two CN⁻ Ions**:
Since there are two cyanide ions in the complex:
\[ \text{Total π bonds} = 2 \text{ (from first CN⁻)} + 2 \text{ (from second CN⁻)} = 4 \text{ π bonds} \]

5. **Conclusion**:
The total number of π bonds in the complex ion \([Ag(CN)_2]^-\) is 4.

### Final Answer:
The number of π bonds in \([Ag(CN)_2]^-\) is **4**.

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