When an excess of ammonia solution is added to `CuSO_(4)` which solution is formed
(a) `[Cu(NH_(3))_(2)]^(2+)`
(b) `[Cu(NH_(3))_(4)]^(o+)`
(c ) `[Cu(NH_(3))_(2)]^(o+)`
(d) `[Cu(NH_(3))_(4)]^(2+)` .

Explanation

To solve the question of what complex is formed when an excess of ammonia solution is added to `CuSO4`, we can follow these steps:

### Step 1: Identify the Reactants
We start with `CuSO4` (copper(II) sulfate) and ammonia (`NH3`). When ammonia is added to the copper(II) sulfate solution, it acts as a ligand and can form a complex with the copper ions.

### Step 2: Determine the Oxidation State of Copper
In `CuSO4`, copper is in the +2 oxidation state (Cu²⁺). This is important because it will influence the coordination number and the type of complex formed.

### Step 3: Consider the Coordination Number
Copper(II) typically has a coordination number of 4 when it forms complexes with ligands like ammonia. This means that it can bond with four ammonia molecules.

### Step 4: Form the Complex
When excess ammonia is added to the solution, it will coordinate with the copper ion. The resulting complex will be `[Cu(NH3)4]²⁺`. The sulfate ion (SO4²⁻) will remain in the solution as a counter ion.

### Step 5: Write the Final Product
The final product formed when excess ammonia is added to `CuSO4` is `[Cu(NH3)4]²⁺`. This corresponds to option (d) `[Cu(NH3)4]²⁺`.

### Step 6: Naming the Complex
The name of the complex can be derived as follows:
- The ligand is ammonia, which is called "ammine".
- Since there are four ammonia ligands, we use the prefix "tetra".
- The central metal is copper, and since it is in the +2 oxidation state, we denote it as "copper(II)".

Thus, the name of the complex is "tetramine copper(II)".

### Conclusion
The correct answer is (d) `[Cu(NH3)4]²⁺`.

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