The oxidation state of platinum in `Na[PtBrCl(NO_2)(NH_3)]` is

Options

A. `+2`
B. `+4`
C. `+6`
D. 0

Answer

To find the oxidation state of platinum (Pt) in the complex compound \( \text{Na[PtBrCl(NO}_2)(NH_3)] \), we can follow these steps:

### Step 1: Identify the oxidation states of the other components
- Sodium (Na) is an alkali metal and has an oxidation state of +1.
- Bromine (Br) and chlorine (Cl) typically have oxidation states of -1 in complexes unless they are bonded to a more electronegative element.
- The nitro group (NO₂) has an overall charge of -1, as nitrogen typically has an oxidation state of +5 and each oxygen has an oxidation state of -2.
- Ammonia (NH₃) is a neutral ligand, contributing 0 to the overall charge.

### Step 2: Set up the equation
The overall charge of the complex ion must equal the charge of the sodium ion, which is +1. Therefore, we can set up the equation based on the oxidation states:

\[ \text{Oxidation state of Pt} + \text{(oxidation state of Br)} + \text{(oxidation state of Cl)} + \text{(oxidation state of NO}_2) + \text{(oxidation state of NH}_3) = +1 \]

Substituting the known oxidation states:

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

Where \( x \) is the oxidation state of platinum.

### Step 3: Solve for the oxidation state of platinum
Now we simplify the equation:

\[ x - 3 = +1 \]

Adding 3 to both sides gives:

\[ x = +4 \]

### Conclusion
The oxidation state of platinum in \( \text{Na[PtBrCl(NO}_2)(NH_3)] \) is +4.

---