Among the following outermost configurations of transition metals, which shows the highest oxidation state

Options

A. `(n-1)d^3ns^2`
B. `(n-1)d^5ns^1`
C. `(n-1)d^8ns^2`
D. `(n-1)d^5ns^2`

Answer

To determine which transition metal shows the highest oxidation state based on their outermost electronic configurations, we can follow these steps:

### Step 1: Understand the Electronic Configuration of Transition Metals
Transition metals have partially filled d-orbitals. The general electronic configuration for transition metals is given by:
\[ (n-1)d^{1-10} ns^{0-2} \]
where \( n \) is the principal quantum number.

### Step 2: Identify the Outermost Configuration
We need to analyze the outermost configurations provided in the question. For example, let's consider the configuration of manganese, which is:
\[ 3d^5 4s^2 \]

### Step 3: Determine the Maximum Oxidation State
The maximum oxidation state of a transition metal can often be determined by the number of unpaired electrons in its d-orbitals. The oxidation state can be calculated as:
\[ \text{Oxidation State} = \text{Number of d-electrons} + \text{Number of s-electrons} \]
For manganese:
- Manganese has 5 unpaired electrons in the 3d subshell and 2 electrons in the 4s subshell.
- Therefore, the maximum oxidation state of manganese can be calculated as:
\[ 5 \text{ (from } 3d\text{)} + 2 \text{ (from } 4s\text{)} = +7 \]

### Step 4: Compare with Other Transition Metals
To confirm that manganese has the highest oxidation state among the given configurations, we would need to compare it with the configurations of other transition metals. However, manganese is known to have one of the highest oxidation states (+7) among transition metals.

### Conclusion
Thus, the transition metal with the highest oxidation state among the given configurations is manganese with an oxidation state of +7.

### Final Answer
The correct option is manganese, with the outermost configuration \( 3d^5 4s^2 \), showing the highest oxidation state of +7.

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