Among the following series of transition metal ions the one where all meal ions have `3d^2` electronic configuration is

Options

A. `Ti^(3+),V^(2+),Cr^(3+),Mn^(4+)`
B. `Ti^(o+),V^(4+),Cr^(6+),Mn^(7+)`
C. `Ti^(4+),V^(3+),Cr^(2+),Mn^(3+)`
D. `Ti^(2+),V^(3+),Cr^(4+),Mn^(5+)`

Explanation

To solve the question of which transition metal ion has a `3d^2` electronic configuration, we will analyze the electronic configurations of the given transition metals and their oxidation states.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers and Electronic Configurations:**
- Titanium (Ti): Atomic number 22, electronic configuration: `1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²`
- Vanadium (V): Atomic number 23, electronic configuration: `1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²`
- Chromium (Cr): Atomic number 24, electronic configuration: `1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹` (note the exceptional configuration)
- Manganese (Mn): Atomic number 25, electronic configuration: `1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²`

2. **Determine the Oxidation States:**
- For each metal, we will find the oxidation states where the remaining electrons correspond to a `3d²` configuration.

3. **Titanium (Ti):**
- In the +2 oxidation state:
- Remove 2 electrons from `4s²`: `3d² 4s⁰` → Configuration: `3d²`
- Thus, Ti²⁺ has a `3d²` configuration.

4. **Vanadium (V):**
- In the +3 oxidation state:
- Remove 3 electrons (2 from `4s²` and 1 from `3d³`): `3d² 4s⁰` → Configuration: `3d²`
- Thus, V³⁺ has a `3d²` configuration.

5. **Chromium (Cr):**
- In the +6 oxidation state:
- Remove 6 electrons (1 from `4s¹` and 5 from `3d⁵`): `3d⁰ 4s⁰` → Configuration: `3d²`
- Thus, Cr⁶⁺ does not have `3d²` configuration.
- In the +4 oxidation state:
- Remove 4 electrons (1 from `4s¹` and 3 from `3d⁵`): `3d² 4s⁰` → Configuration: `3d²`
- Thus, Cr⁴⁺ has a `3d²` configuration.

6. **Manganese (Mn):**
- In the +5 oxidation state:
- Remove 5 electrons (2 from `4s²` and 3 from `3d⁵`): `3d² 4s⁰` → Configuration: `3d²`
- Thus, Mn⁵⁺ has a `3d²` configuration.

7. **Conclusion:**
- The transition metal ions that have `3d²` electronic configuration in their respective oxidation states are:
- Titanium (Ti²⁺)
- Vanadium (V³⁺)
- Chromium (Cr⁴⁺)
- Manganese (Mn⁵⁺)

### Final Answer:
The correct answer is **Titanium (Ti²⁺), Vanadium (V³⁺), Chromium (Cr⁴⁺), and Manganese (Mn⁵⁺)**, where all metal ions have `3d²` electronic configuration.