Which one of the following forms with an excess of `CN^(-)` (Cyanide) a complex having coordination number two

Options

A. (a) `Cu^(+)`
B. (b) `Ag^(+)`
C. (c) `Ni^(2+)`
D. (d) `Fe^(2+)`

Explanation

To solve the question of which ion forms a complex with coordination number two when treated with an excess of cyanide ion (CN⁻), we will analyze each option provided: Cu⁺, Ag⁺, Ni²⁺, and Fe²⁺.

### Step-by-Step Solution:

1. **Understanding Coordination Number**:
Coordination number refers to the number of ligand atoms that are bonded to the central metal ion in a complex. In this case, we are looking for a complex that has a coordination number of 2.

2. **Analyzing Each Metal Ion**:
- **Cu⁺ (Copper(I))**: Copper typically forms complexes with higher coordination numbers (like 4 or 6) when bonded to ligands such as CN⁻. Therefore, it is unlikely to form a coordination number of 2 with CN⁻.

- **Ag⁺ (Silver(I))**: Silver ion is known to form a complex with cyanide ions. When treated with an excess of CN⁻, it forms the complex [Ag(CN)₂]⁻. This complex has a coordination number of 2, as it is bonded to two cyanide ligands.

- **Ni²⁺ (Nickel(II))**: Nickel typically forms complexes with coordination numbers of 4 or 6. It is less likely to form a stable complex with a coordination number of 2 with CN⁻.

- **Fe²⁺ (Iron(II))**: Iron(II) can form complexes with various coordination numbers, but it usually prefers higher coordination numbers (like 4 or 6) when interacting with ligands such as CN⁻.

3. **Conclusion**:
After analyzing all the options, we find that only Ag⁺ forms a complex with coordination number 2 when treated with an excess of cyanide ion, resulting in the complex [Ag(CN)₂]⁻.

### Final Answer:
The correct option is **Ag⁺**.