Which one of the following ions has same number of unpared electrons as that of `V^(3+)` ion ?

Options

A. `Cr^(3+)`
B. `Mn^(2+)`
C. `Ni^(2+)`
D. `Fe^(3+)`

Answer

To determine which ion has the same number of unpaired electrons as the \( V^{3+} \) ion, we will follow these steps:

### Step 1: Determine the electronic configuration of Vanadium (V)
- The atomic number of Vanadium (V) is 23.
- The electronic configuration of Vanadium is:
\[ \text{V: } [\text{Ar}] 3d^3 4s^2 \]

### Step 2: Find the electronic configuration of \( V^{3+} \)
- When Vanadium loses 3 electrons to form \( V^{3+} \), the electrons are removed first from the 4s orbital and then from the 3d orbital.
- Therefore, the electronic configuration of \( V^{3+} \) is:
\[ V^{3+}: [\text{Ar}] 3d^2 4s^0 \]
- This means \( V^{3+} \) has 2 unpaired electrons in the 3d orbital.

### Step 3: Analyze the options given
We need to check the number of unpaired electrons in each of the following ions:

1. **Chromium \( Cr^{3+} \)**
- Atomic number of Chromium is 24.
- Electronic configuration:
\[ Cr: [\text{Ar}] 3d^5 4s^1 \]
- For \( Cr^{3+} \), we remove 3 electrons:
\[ Cr^{3+}: [\text{Ar}] 3d^3 4s^0 \]
- Unpaired electrons: 3 (not a match).

2. **Manganese \( Mn^{2+} \)**
- Atomic number of Manganese is 25.
- Electronic configuration:
\[ Mn: [\text{Ar}] 3d^5 4s^2 \]
- For \( Mn^{2+} \), we remove 2 electrons:
\[ Mn^{2+}: [\text{Ar}] 3d^5 4s^0 \]
- Unpaired electrons: 5 (not a match).

3. **Nickel \( Ni^{2+} \)**
- Atomic number of Nickel is 28.
- Electronic configuration:
\[ Ni: [\text{Ar}] 3d^8 4s^2 \]
- For \( Ni^{2+} \), we remove 2 electrons:
\[ Ni^{2+}: [\text{Ar}] 3d^8 4s^0 \]
- Unpaired electrons: 2 (this is a match).

4. **Iron \( Fe^{3+} \)**
- Atomic number of Iron is 26.
- Electronic configuration:
\[ Fe: [\text{Ar}] 3d^6 4s^2 \]
- For \( Fe^{3+} \), we remove 3 electrons:
\[ Fe^{3+}: [\text{Ar}] 3d^5 4s^0 \]
- Unpaired electrons: 5 (not a match).

### Conclusion
The ion that has the same number of unpaired electrons as \( V^{3+} \) is \( Ni^{2+} \), which has 2 unpaired electrons.

### Final Answer
The correct option is **Nickel \( Ni^{2+} \)**.

---