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

Explanation

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

### Step-by-Step Solution:

1. **Identify the components of the complex:**
- The complex consists of chromium (Cr), four water molecules (H₂O), and two chloride ions (Cl).

2. **Assign oxidation states:**
- The oxidation state of water (H₂O) is 0 because it is a neutral molecule.
- The oxidation state of each chloride ion (Cl) is -1.

3. **Set up the equation:**
- Let the oxidation state of chromium be \(x\).
- The total charge of the complex is +1.

4. **Write the equation based on the oxidation states:**
- The equation can be set up as follows:
\[ x + (0 \times 4) + (-1 \times 2) = +1 \]
- Simplifying this gives:
\[ x - 2 = +1 \]

5. **Solve for \(x\):**
- Rearranging the equation:
\[ x = +1 + 2 \]
- Therefore,
\[ x = +3 \]

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

### Summary:
The valency of chromium (Cr) in the complex \([Cr(H_2O)_4Cl_2]^+\) is +3.