What is the charge on the complex `[Cr(C_(2)O_(4))_(2)(H_(2)O)_(2)]` formed by Cr(III) ?

Options

A. `+3`
B. `+1`
C. `+2`
D. `-1`

Answer

To determine the charge on the complex `[Cr(C₂O₄)₂(H₂O)₂]` formed by Cr(III), we can follow these steps:

### Step 1: Identify the oxidation state of chromium
Chromium is in the +3 oxidation state in this complex.

### Step 2: Analyze the ligands
The complex contains two types of ligands:
1. Oxalate ion (C₂O₄²⁻)
2. Water (H₂O)

### Step 3: Determine the charge contributed by the ligands
- The oxalate ion (C₂O₄²⁻) has a charge of -2. Since there are two oxalate ions in the complex, their total contribution to the charge is:
\[ 2 \times (-2) = -4 \]

- The water molecule (H₂O) is a neutral ligand, which means it contributes a charge of 0. Since there are two water molecules, their total contribution is:
\[ 2 \times 0 = 0 \]

### Step 4: Calculate the overall charge of the complex
Now, we can calculate the overall charge of the complex by combining the contributions from chromium and the ligands:
\[ \text{Total charge} = \text{Charge from Cr} + \text{Charge from ligands} \]
Substituting the values we have:
\[ \text{Total charge} = (+3) + (-4) + 0 = 3 - 4 = -1 \]

### Conclusion
The overall charge on the complex `[Cr(C₂O₄)₂(H₂O)₂]` is -1.

### Final Answer
The charge on the complex is -1.