Oxidation number of platinum in cis.platin

Options

A. Zero
B. `+2`
C. `+4`
D. `+6`

Answer

To determine the oxidation number of platinum in cis-platin (cis-diamminedichloroplatinum(II)), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Chemical Formula**:
The chemical formula for cis-platin is given as PtCl2(NH3)2.

2. **Assign Oxidation States to Known Ligands**:
- Chloride ion (Cl) has an oxidation state of -1. Since there are two chloride ions, their total contribution to the oxidation state is:
\[ 2 \times (-1) = -2 \]
- Ammonia (NH3) is a neutral ligand, so its contribution to the oxidation state is 0. Since there are two ammonia molecules, their total contribution is:
\[ 2 \times 0 = 0 \]

3. **Set Up the Equation**:
Let the oxidation number of platinum (Pt) be \( x \). The overall charge of the cis-platin complex is neutral (0). Therefore, we can set up the equation:
\[ x + (-2) + 0 = 0 \]

4. **Solve for the Oxidation Number**:
Rearranging the equation gives:
\[ x - 2 = 0 \\ x = +2 \]

5. **Conclusion**:
The oxidation number of platinum in cis-platin is +2.

### Final Answer:
The oxidation number of platinum in cis-platin is +2.

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