Oxidation number of plantinum 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 [Pt(NH3)2Cl2]. This indicates that platinum (Pt) is the central metal atom, surrounded by two ammonia (NH3) ligands and two chloride (Cl) ligands.

2. **Assign Oxidation Numbers to Ligands**:
- The oxidation number of ammonia (NH3) is 0 because it is a neutral ligand.
- The oxidation number of each chloride ion (Cl) is -1.

3. **Set Up the Equation**:
Let the oxidation number of platinum be represented as \( x \). The overall charge of the complex is neutral (0). Therefore, we can set up the equation based on the contributions of the oxidation states:
\[ x + 2(-1) + 2(0) = 0 \]
Here, \( 2(-1) \) accounts for the two chloride ions, and \( 2(0) \) accounts for the two ammonia ligands.

4. **Simplify the Equation**:
This simplifies to:
\[ x - 2 = 0 \]

5. **Solve for \( x \)**:
Adding 2 to both sides gives:
\[ x = +2 \]

6. **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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