Coordination number of Cr is six A complex with `C_(2)O_(4)^(2-)` en and superoxide `O_(2)`will be in the ration to make complex `[Cr(C_(2)O_(4))_(x')(en)_(y)(O_(2)_(z)]^(Θ)` .

Options

A. `{:(x,,y,,z),(1,,1,,1):}`
B. `{:(x,,y,,z),(1,,1,,2):}`
C. `{:(x,,y,,z),(1,,2,,2):}`
D. `{:(x,,y,,z),(2,,1,,1):}`

Explanation

To solve the problem, we need to determine the values of x, y, and z in the complex \([Cr(C_2O_4)_{x}(en)_{y}(O_2)_{z}]^{\Theta}\) given that the coordination number of Cr is 6 and considering the charges of the ligands involved.

### Step-by-Step Solution:

1. **Identify the Coordination Number and Ligands:**
- The coordination number of Cr is given as 6.
- The ligands involved are:
- \(C_2O_4^{2-}\) (oxalate ion) which has a coordination number of 2.
- \(en\) (ethylenediamine) which is a neutral ligand and has a coordination number of 2.
- \(O_2^{-}\) (superoxide) which has a coordination number of 1.

2. **Set Up the Charge Balance Equation:**
- The overall charge of the complex is given as \(-1\).
- The charge contributions from each component:
- Cr contributes +3 (since it is in +3 oxidation state).
- Each \(C_2O_4^{2-}\) contributes -2, so for x ligands, it contributes \(-2x\).
- Each \(en\) contributes 0, so for y ligands, it contributes \(0y\).
- Each \(O_2^{-}\) contributes -1, so for z ligands, it contributes \(-z\).
- The charge balance equation can be written as:
\[ +3 - 2x - z = -1 \]
- Rearranging gives us:
\[ 2x + z = 4 \quad \text{(Equation 1)} \]

3. **Set Up the Coordination Number Equation:**
- The total coordination number must equal 6:
- From \(C_2O_4^{2-}\), x ligands contribute \(2x\).
- From \(en\), y ligands contribute \(2y\).
- From \(O_2^{-}\), z ligands contribute \(1z\).
- The coordination number equation can be written as:
\[ 2x + 2y + z = 6 \quad \text{(Equation 2)} \]

4. **Substitute Equation 1 into Equation 2:**
- From Equation 1, we can express z in terms of x:
\[ z = 4 - 2x \]
- Substitute this into Equation 2:
\[ 2x + 2y + (4 - 2x) = 6 \]
- Simplifying gives:
\[ 2y + 4 = 6 \]
\[ 2y = 2 \quad \Rightarrow \quad y = 1 \]

5. **Find Values of x and z:**
- Now substitute \(y = 1\) back into Equation 1:
\[ 2x + z = 4 \]
- We can express z in terms of x again:
\[ z = 4 - 2x \]
- Since y must be a positive integer, and both x and z must also be positive integers, we can test integer values for x:
- If \(x = 1\):
\[ z = 4 - 2(1) = 2 \]
- If \(x = 2\):
\[ z = 4 - 2(2) = 0 \quad \text{(not allowed)} \]
- Thus, the only solution is \(x = 1\), \(y = 1\), and \(z = 2\).

### Final Answer:
\[ x = 1, \quad y = 1, \quad z = 2 \]