The electronic configuration of a transition element is `[Ar]4s^(2)3d^(3)`. The possible oxidation states are

Options

A. `+1, +2 and +3` only
B. `+2 and +3` only
C. `+2, +3, +4 and +5` only
D. `+2 and +5` only

Answer

To determine the possible oxidation states of the transition element with the electronic configuration \([Ar] 4s^2 3d^3\), we can follow these steps:

### Step 1: Identify the Element
The given electronic configuration corresponds to the element Vanadium (V), which has an atomic number of 23.

### Step 2: Understand the Electron Configuration
The electron configuration can be broken down as follows:
- The 4s subshell has 2 electrons.
- The 3d subshell has 3 electrons.

### Step 3: Determine Possible Oxidation States
Transition metals can exhibit multiple oxidation states due to the involvement of both the s and d electrons in bonding.

1. **Removing 4s Electrons**:
- The 4s electrons are generally removed first.
- Removing 1 electron from 4s gives a +1 oxidation state.
- Removing both 4s electrons gives a +2 oxidation state.

2. **Removing 3d Electrons**:
- After the 4s electrons are removed, we can remove electrons from the 3d subshell.
- Since there are 3 electrons in the 3d subshell, we can remove:
- 1 electron from 3d (after removing both 4s) gives a +3 oxidation state.
- 2 electrons from 3d gives a +4 oxidation state.
- 3 electrons from 3d gives a +5 oxidation state.

### Step 4: Compile Possible Oxidation States
From the above analysis, the possible oxidation states for Vanadium are:
- +2 (removing both 4s electrons)
- +3 (removing both 4s and 1 from 3d)
- +4 (removing both 4s and 2 from 3d)
- +5 (removing both 4s and all 3 from 3d)

### Final Answer
Thus, the possible oxidation states of the transition element with the electronic configuration \([Ar] 4s^2 3d^3\) are:
- +2, +3, +4, and +5.

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