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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the oxidation states of the other elements in the compound**:
- Sodium (Na) is in Group 1, so its oxidation state is +1.
- Bromine (Br) typically has an oxidation state of -1.
- Chlorine (Cl) also typically has an oxidation state of -1.
- In the nitrite ion (\( \text{NO}_2 \)), the overall charge is -1. The oxidation state of nitrogen in this case is +3 (since each oxygen is -2, giving a total of -4, and to balance it to -1, nitrogen must be +3).
- Ammonia (NH3) has a neutral charge, so the oxidation state of nitrogen is -3, but it does not contribute to the overall charge of the complex.

2. **Set up the equation for the overall charge**:
- The overall charge of the complex ion is 0.
- Let the oxidation state of platinum (Pt) be \( x \).
- We can write the equation based on the oxidation states:
\[ \text{Total charge} = \text{Charge of Na} + \text{Charge of Pt} + \text{Charge of Br} + \text{Charge of Cl} + \text{Charge of NO}_2 + \text{Charge of NH}_3 \]
\[ 0 = (+1) + x + (-1) + (-1) + (-1) + (0) \]

3. **Simplify the equation**:
- Combine the known oxidation states:
\[ 0 = 1 + x - 1 - 1 - 1 + 0 \]
\[ 0 = x - 2 \]

4. **Solve for \( x \)**:
- Rearranging gives:
\[ x = +2 \]

5. **Conclusion**:
- The oxidation state of platinum in \( \text{Na[PtBrCl(NO}_2)(NH_3)] \) is +2.

### Final Answer:
The oxidation state of platinum in \( \text{Na[PtBrCl(NO}_2)(NH_3)] \) is +2.