An excess of `Na_(2)S_(2)OO_(3)` react with aqueous `CuSO_(4)` to give

Options

A. CuS2o3
B. Cu2S2O3
C. Na2[Cu(S2O3)2]
D. Na4[Cu6(S2O3)5]

Explanation

To solve the problem of determining what products are formed when an excess of \( \text{Na}_2\text{S}_2\text{O}_3 \) reacts with aqueous \( \text{CuSO}_4 \), we can follow these steps:

### Step 1: Write the initial reaction
When \( \text{CuSO}_4 \) reacts with \( \text{Na}_2\text{S}_2\text{O}_3 \), the first reaction can be represented as:
\[ \text{CuSO}_4 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Cu}_2\text{S}_2\text{O}_3 + \text{Na}_2\text{SO}_4 \]
This reaction indicates that copper(II) sulfate reacts with sodium thiosulfate to form copper(I) thiosulfate and sodium sulfate.

### Step 2: Consider the excess reagent
Since \( \text{Na}_2\text{S}_2\text{O}_3 \) is in excess, it will continue to react with any remaining \( \text{CuSO}_4 \) or the products formed. The reaction can be represented as:
\[ \text{Cu}_2\text{S}_2\text{O}_3 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Cu}_2\text{S}_4\text{O}_6 + \text{Na}_2\text{SO}_4 \]
This indicates that the copper(I) thiosulfate can further react with more thiosulfate ions.

### Step 3: Identify the final product
Continuing the reaction due to the excess of \( \text{Na}_2\text{S}_2\text{O}_3 \), we can form a complex:
\[ \text{Cu}_2\text{S}_4\text{O}_6 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Complex} \, (\text{Cu}_2\text{S}_2\text{O}_3 \cdot \text{Na}_2\text{S}_2\text{O}_3) \]
The final product is a complex that includes copper, sulfur, and oxygen.

### Conclusion
The overall reaction leads to the formation of a complex compound, which can be represented as:
\[ \text{Cu}_2\text{S}_2\text{O}_3 \cdot \text{Na}_2\text{S}_2\text{O}_3 \]
This complex does not react further due to the presence of excess \( \text{Na}_2\text{S}_2\text{O}_3 \).

### Final Answer
The final product formed is a complex of copper(I) thiosulfate and sodium thiosulfate.

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