Oxidation state of cobalt in `[Co(NH_(3))_(4)(H_(2)O)Cl]SO_(4)` is

Options

A. 0
B. `+4`
C. `-2`
D. `+3`

Answer

To find the oxidation state of cobalt in the compound \([Co(NH_3)_4(H_2O)Cl]SO_4\), we can follow these steps:

### Step 1: Identify the oxidation states of the other components
- **Ammonia (NH₃)**: The oxidation state of NH₃ is 0 because it is a neutral molecule.
- **Water (H₂O)**: The oxidation state of H₂O is also 0 since it is neutral.
- **Chlorine (Cl)**: In this complex, chlorine typically has an oxidation state of -1.
- **Sulfate (SO₄²⁻)**: The sulfate ion has an oxidation state of -2.

### Step 2: Set up the equation
Let the oxidation state of cobalt (Co) be \(X\). The overall charge of the complex ion \([Co(NH_3)_4(H_2O)Cl]^+\) should equal the charge of the sulfate ion, which is -2. Therefore, we can write the equation as follows:

\[ X + 4(0) + 0 + (-1) = +1 \]

### Step 3: Simplify the equation
Substituting the values we have:

\[ X - 1 = +1 \]

### Step 4: Solve for \(X\)
Now, we can solve for \(X\):

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

### Conclusion
The oxidation state of cobalt in \([Co(NH_3)_4(H_2O)Cl]SO_4\) is +3.

### Final Answer
The oxidation state of cobalt is +3.

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