among the following series of transition metal ions, the one where all ions have some 3d electronic configuration is

Options

A. `Ti^(2+),Cr^(4+), Mn^(5+),V^(3+) `
B. `Ti^(3+), Ni^(2+), Co^+, Zn^(2+)`
C. `Sc^(2+), Ti^(2+), V^(2+), Cr^(2+)`
D. `Mn^(5+), Co^(4+), Ni^(3+),Cu^(2+)`

Answer

To solve the question of which series of transition metal ions has all ions with the same 3d electronic configuration, we will analyze the electronic configurations of each ion step by step.

### Step 1: Identify the Ground State Electronic Configuration
First, we need to find the ground state electronic configuration of each transition metal before they are ionized.

1. **Titanium (Ti)**: Atomic number = 22
- Ground state configuration: \( [Ar] 4s^2 3d^2 \)

2. **Chromium (Cr)**: Atomic number = 24
- Ground state configuration: \( [Ar] 4s^2 3d^5 \)

3. **Manganese (Mn)**: Atomic number = 25
- Ground state configuration: \( [Ar] 4s^2 3d^5 \)

4. **Vanadium (V)**: Atomic number = 23
- Ground state configuration: \( [Ar] 4s^2 3d^3 \)

### Step 2: Determine the Electronic Configuration of Each Ion
Next, we will remove the appropriate number of electrons for each ion based on their oxidation states.

1. **Titanium (Ti²⁺)**:
- Remove 2 electrons from the ground state configuration:
- \( [Ar] 4s^2 3d^2 \) → \( 3d^2 \)

2. **Chromium (Cr⁴⁺)**:
- Remove 4 electrons (2 from 4s and 2 from 3d):
- \( [Ar] 4s^2 3d^5 \) → \( 3d^3 \)

3. **Manganese (Mn⁵⁺)**:
- Remove 5 electrons (2 from 4s and 3 from 3d):
- \( [Ar] 4s^2 3d^5 \) → \( 3d^4 \)

4. **Vanadium (V³⁺)**:
- Remove 3 electrons (2 from 4s and 1 from 3d):
- \( [Ar] 4s^2 3d^3 \) → \( 3d^2 \)

### Step 3: Compare the 3d Configurations
Now, we compare the resulting 3d configurations of each ion:

- **Ti²⁺**: \( 3d^2 \)
- **Cr⁴⁺**: \( 3d^3 \)
- **Mn⁵⁺**: \( 3d^4 \)
- **V³⁺**: \( 3d^2 \)

### Conclusion
From the analysis, we see that only **Ti²⁺** and **V³⁺** have the same 3d configuration of \( 3d^2 \). However, **Cr⁴⁺** and **Mn⁵⁺** have different configurations. Therefore, the series where all ions have the same 3d electronic configuration is **Titanium (Ti²⁺)** and **Vanadium (V³⁺)**.

### Final Answer
The answer is **Titanium (Ti²⁺) and Vanadium (V³⁺)**, as both have the same \( 3d^2 \) electronic configuration.