The oxidation state of Cr in `[Cr(NH_(3))_(4)Cl_(2)]^(+)` is

Options

A. 0
B. `+1`
C. `+2`
D. `+3`

Answer

To find the oxidation state of chromium (Cr) in the complex ion \([Cr(NH_3)_4Cl_2]^+\), we can follow these steps:

### Step 1: Identify the components of the complex
The complex consists of:
- 4 ammonia (NH₃) molecules
- 2 chloride (Cl) ions
- The overall charge of the complex is +1.

### Step 2: Assign oxidation states to the ligands
- Ammonia (NH₃) is a neutral ligand, which means it has an oxidation state of 0.
- Each chloride ion (Cl) has an oxidation state of -1.

### Step 3: Set up the equation for the oxidation state of chromium
Let the oxidation state of chromium be \(X\). The contribution to the total charge from the ligands can be calculated as follows:

- Contribution from 4 ammonia molecules: \(4 \times 0 = 0\)
- Contribution from 2 chloride ions: \(2 \times (-1) = -2\)

### Step 4: Write the equation based on the total charge
The total charge of the complex is +1. Therefore, we can set up the equation:

\[ X + 0 + (-2) = +1 \]

### Step 5: Solve for \(X\)
Now, we can simplify the equation:

\[ X - 2 = +1 \]

Adding 2 to both sides gives:

\[ X = +1 + 2 \]
\[ X = +3 \]

### Conclusion
The oxidation state of chromium (Cr) in the complex \([Cr(NH_3)_4Cl_2]^+\) is +3.

### Final Answer
The oxidation state of Cr in \([Cr(NH_3)_4Cl_2]^+\) is **+3**.

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