The oxidation of Cr in the complex `[Cr(H_(2)O)_(4)Cl_(2)]^(+)` is

Options

A. `+1`
B. `+3`
C. `+5`
D. `+6`

Explanation

To find the oxidation state of chromium (Cr) in the complex \([Cr(H_2O)_4Cl_2]^+\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the ligands and their charges:**
- The complex contains two types of ligands: water (\(H_2O\)) and chloride (\(Cl^-\)).
- Water is a neutral ligand, meaning it has no charge.
- Chloride has a charge of \(-1\).

2. **Count the number of each ligand:**
- There are 4 water molecules: \(4 \times 0 = 0\) (since water is neutral).
- There are 2 chloride ions: \(2 \times (-1) = -2\).

3. **Set up the equation for the oxidation state:**
- Let the oxidation state of chromium be \(x\).
- The overall charge of the complex is \(+1\).
- Therefore, we can set up the equation:
\[ x + (0) + (-2) = +1 \]

4. **Solve for \(x\):**
- Rearranging the equation gives:
\[ x - 2 = +1 \]
- Adding 2 to both sides results in:
\[ x = +1 + 2 \]
- Thus, we find:
\[ x = +3 \]

5. **Conclusion:**
- The oxidation state of chromium in the complex \([Cr(H_2O)_4Cl_2]^+\) is \(+3\).

### Final Answer:
The oxidation state of Cr in the complex \([Cr(H_2O)_4Cl_2]^+\) is \(+3\).

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