Oxidation number of platinum in cis.platin

Options

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

Answer

To find the oxidation number of platinum in cis-platin, we can follow these steps:

### Step 1: Identify the formula of cis-platin
Cis-platin has the formula \[ \text{[Pt(NH}_3\text{)}_2\text{Cl}_2] \]. Here, platinum (Pt) is the central metal atom, and it is coordinated by two ammonia (NH₃) ligands and two chloride (Cl⁻) ligands.

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

### Step 3: Set up the equation for the oxidation number
Let the oxidation number of platinum be \( x \). The overall charge of the complex is neutral (0). Therefore, we can set up the equation based on the contributions of the ligands:

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

### Step 4: Simplify the equation
This simplifies to:

\[ x - 2 = 0 \]

### Step 5: Solve for \( x \)
Now, solving for \( x \):

\[ x = 2 \]

### Conclusion
The oxidation number of platinum in cis-platin is +2.