The highest oxidation state is exhibited by the transition metal with electronic configuration 

Options

A. `(n-1)d^(5)ns^(1)`
B. `(n-1)d^(5)ns^(2)`
C. `(n-1)d^(8)ns^(2)`
D. `(n-1)d^(6)ns^(2)`

Answer

To determine the highest oxidation state exhibited by a transition metal with a given electronic configuration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electronic Configuration**: The question refers to a transition metal with a specific electronic configuration. We need to know the configuration to find the corresponding element.

2. **Determine the Atomic Number**: From the electronic configuration, we can deduce the atomic number of the transition metal. For example, if the configuration is given as \( \text{[Ar]} 3d^5 4s^2 \), we identify that this corresponds to manganese (Mn), which has an atomic number of 25.

3. **Understand the Oxidation States**: Transition metals can exhibit multiple oxidation states. The highest oxidation state is typically achieved when all the valence electrons are removed. For manganese, the valence electrons are in the 4s and 3d orbitals.

4. **Calculate the Highest Oxidation State**: For manganese:
- It has 2 electrons in the 4s subshell and 5 electrons in the 3d subshell.
- When manganese loses all 7 valence electrons (2 from 4s and 5 from 3d), it achieves the oxidation state of +7.

5. **Noble Gas Configuration**: After losing these electrons, manganese achieves a noble gas configuration, which is stable. The electronic configuration after losing 7 electrons is \( \text{[Ar]} \).

6. **Conclusion**: Therefore, the highest oxidation state exhibited by the transition metal with the electronic configuration \( \text{[Ar]} 3d^5 4s^2 \) is +7.

### Final Answer:
The highest oxidation state exhibited is +7.

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