When `K_(2)CrO_(4)` is added to `CuSO_(4)` solution, there is formation of `CuCrO_(4)` as well as `CuCr_(2)O_(7)`. Formation of `CuCr_(2)O_(7)` is due to

Options

A. Basic nature of `CuSO_(4)` is due to
B. Acidic nature of `CuSO_(4)` solution
C. `CuSo_(4)` oxidizes `CrO_(4)^(2-)` or `Cr_(2)O_(7)^(2-)`
D. It is the tupical property of `CuSO_(4s)`

Explanation

To solve the question regarding the formation of \( CuCr_2O_7 \) when \( K_2CrO_4 \) is added to \( CuSO_4 \) solution, we can follow these steps:

### Step 1: Write the Chemical Reaction
When \( K_2CrO_4 \) is added to \( CuSO_4 \), the following reactions occur:

\[ CuSO_4 + K_2CrO_4 \rightarrow CuCrO_4 + K_2SO_4 \]

Additionally, under certain conditions, \( CuCr_2O_7 \) can also form:

\[ 2CuSO_4 + K_2CrO_4 \rightarrow CuCr_2O_7 + K_2SO_4 \]

### Step 2: Identify Oxidation States
In these compounds, we need to identify the oxidation states of the key elements involved:

- In \( CuSO_4 \), copper (Cu) is in the +2 oxidation state, and sulfur (S) is in the +6 oxidation state.
- In \( K_2CrO_4 \), chromium (Cr) is in the +6 oxidation state.
- In \( CuCrO_4 \), chromium is also in the +6 oxidation state.
- In \( CuCr_2O_7 \), chromium remains in the +6 oxidation state.

### Step 3: Analyze the Behavior of \( CuSO_4 \)
Since sulfur in \( CuSO_4 \) is already in its maximum oxidation state (+6), it cannot be oxidized further. This means that it cannot lose electrons. Instead, it can only accept electrons, behaving as a Lewis acid.

### Step 4: Determine the Formation of \( CuCr_2O_7 \)
The formation of \( CuCr_2O_7 \) occurs because the \( Cu^{2+} \) ions can oxidize the \( CrO_4^{2-} \) ions to form \( Cr_2O_7^{2-} \) ions. The reaction can be viewed as a redox process where \( Cu^{2+} \) acts as an oxidizing agent, allowing for the formation of \( CuCr_2O_7 \).

### Conclusion
Thus, the formation of \( CuCr_2O_7 \) is due to the ability of \( Cu^{2+} \) ions to oxidize \( CrO_4^{2-} \) ions to \( Cr_2O_7^{2-} \) ions, taking advantage of the fact that sulfur cannot be oxidized further.