The oxidation number of Cr in `[Cr(NH_(3))_(6)]Cl_(3)` is

Options

A. 8
B. 6
C. 4
D. 3

Explanation

To determine the oxidation number of chromium (Cr) in the complex ion \([Cr(NH_3)_6]Cl_3\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the components of the complex**:
The complex is \([Cr(NH_3)_6]Cl_3\). Here, \([Cr(NH_3)_6]\) is the coordination complex, and \(Cl_3\) indicates that there are three chloride ions outside the coordination sphere.

2. **Determine the charge of the ligands**:
Ammonia (NH₃) is a neutral ligand, which means it does not contribute any charge. Therefore, the total charge contributed by the six ammonia ligands is \(0\).

3. **Set up the equation for the oxidation state**:
Let the oxidation number of chromium be \(X\). The overall charge of the complex ion \([Cr(NH_3)_6]\) must balance the charge of the three chloride ions. Since each chloride ion (Cl⁻) has a charge of \(-1\), the total charge from the three chloride ions is \(-3\).

4. **Write the equation**:
The equation can be written as:
\[ X + 0 = -3 \]
Here, \(X\) is the oxidation state of chromium, and \(0\) is the total charge from the neutral ligands.

5. **Solve for \(X\)**:
Rearranging the equation gives:
\[ X = -3 \]
However, we must remember that the complex itself is neutral, so we need to consider the overall charge of the complex. Since the complex must balance the \(-3\) charge from the three chloride ions, we conclude that:
\[ X = +3 \]

6. **Conclusion**:
Therefore, the oxidation number of chromium in \([Cr(NH_3)_6]Cl_3\) is \(+3\).

### Final Answer:
The oxidation number of Cr in \([Cr(NH_3)_6]Cl_3\) is \(+3\).