The exhibition of various oxidation states by an element is also related to the outer orbital electornic configuration of its atom. Atom(s) having which of the following outermost electronic confjigurations will exhibit more than one oxidation state in its compounds

Options

A. `3s^(-1)`
B. `3d^(1)4s^(2)`
C. `3d^(2)4s^(2)`
D. `3s^(2)3p^(3)`

Explanation

To determine which outermost electronic configurations will exhibit more than one oxidation state in their compounds, we need to analyze each of the given configurations.

### Step-by-Step Solution:

1. **Understanding Oxidation States**:
- Oxidation states are determined by the number of electrons an atom can lose, gain, or share during chemical reactions. Elements with multiple oxidation states typically have more than one electron in their outermost shell that can participate in bonding.

2. **Analyzing Each Configuration**:
- **Option 1: 3s¹**
- This configuration has only one electron in the outermost shell (3s). It can only lose this one electron, resulting in a +1 oxidation state. Therefore, it cannot exhibit more than one oxidation state.

- **Option 2: 3d¹ 4s²**
- This is a d-block element. It has 2 electrons in the 4s subshell and 1 electron in the 3d subshell.
- It can lose the 2 electrons from the 4s subshell to form a +2 oxidation state and can also lose the 1 electron from the 3d subshell to form a +3 oxidation state. Thus, it can exhibit more than one oxidation state.

- **Option 3: 3d² 4s²**
- This is also a d-block element. It has 2 electrons in the 4s subshell and 2 electrons in the 3d subshell.
- It can lose the 2 electrons from the 4s subshell to form a +2 oxidation state, and it can lose 1 or both electrons from the 3d subshell, allowing for +3 and +4 oxidation states. Therefore, it can exhibit more than one oxidation state.

- **Option 4: 3s² 3p³**
- This configuration corresponds to a p-block element. It has 2 electrons in the 3s subshell and 3 electrons in the 3p subshell.
- It can lose 3 electrons from the 3p subshell to form a +3 oxidation state and can lose 2 electrons from the 3s subshell and 3 from the 3p subshell to form a +5 oxidation state. Thus, it can exhibit more than one oxidation state.

3. **Conclusion**:
- From the analysis, we find that:
- Option 1 (3s¹) can only exhibit +1 oxidation state.
- Options 2 (3d¹ 4s²), 3 (3d² 4s²), and 4 (3s² 3p³) can exhibit more than one oxidation state.
- Therefore, the configurations that can exhibit more than one oxidation state are **Options 2, 3, and 4**.

### Final Answer:
The outermost electronic configurations that will exhibit more than one oxidation state in their compounds are **2, 3, and 4**.

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