The hybridised orbitals used by silver in the complex `[Ag(NH_(3))_(2)]^(+)` are of the type:

Options

A. `sp^(2)`
B. `sp`
C. `sp^(3)`
D. `dsp^(2)`

Answer

To determine the hybridization of silver in the complex \([Ag(NH_3)_2]^+\), we can follow these steps:

### Step 1: Determine the oxidation state of silver in the complex.
- The complex \([Ag(NH_3)_2]^+\) has a +1 charge. Since ammonia (NH₃) is a neutral ligand, the oxidation state of silver (Ag) in this complex is +1.

### Step 2: Write the electron configuration of silver.
- The electron configuration of neutral silver (Ag) is \([Kr] 5s^1 4d^{10}\).
- When silver is in the +1 oxidation state, it loses one electron from the 5s orbital, resulting in the configuration \([Kr] 4d^{10}\).

### Step 3: Identify the available orbitals for hybridization.
- In the +1 oxidation state, silver has the electron configuration \([Kr] 4d^{10}\). The 5s and 5p orbitals are vacant and can be used for hybridization.

### Step 4: Determine the number of ligands and their contribution.
- The complex has two ammonia (NH₃) ligands. Each NH₃ donates a lone pair of electrons to silver. Therefore, silver will coordinate with two lone pairs from the two NH₃ molecules.

### Step 5: Determine the type of hybridization.
- The coordination number of silver in this complex is 2 (due to the two NH₃ ligands).
- To accommodate these two lone pairs, silver will undergo hybridization involving one s orbital and one p orbital, resulting in **sp hybridization**.

### Step 6: Conclude the hybridization type.
- Therefore, the hybridized orbitals used by silver in the complex \([Ag(NH_3)_2]^+\) are of the type **sp**.

### Final Answer:
The hybridised orbitals used by silver in the complex \([Ag(NH_3)_2]^+\) are of the type **sp**.

---