The number of unpaired electrons in gaseous species of `Mn^(3+), Cr^(3+)` and `V^(3+)` respectively are………….and most stable species is…………..

Options

A. 4,3 and 2, `V^(3+)`
B. 3,3 and 2, `Cr^(3+)`
C. 4,3 and 2, `Cr^(3+)`
D. 3,3 and 3, `Mn^(3+)`

Explanation

To solve the question regarding the number of unpaired electrons in the gaseous species of \( \text{Mn}^{3+} \), \( \text{Cr}^{3+} \), and \( \text{V}^{3+} \), we will analyze the electronic configurations of each ion step by step.

### Step 1: Determine the electronic configuration of \( \text{Mn}^{3+} \)

1. **Find the atomic number of Manganese (Mn)**: Manganese has an atomic number of 25.
2. **Write the electronic configuration of neutral Mn**:
- The electronic configuration is \( [Ar] 3d^5 4s^2 \).
3. **Determine the configuration for \( \text{Mn}^{3+} \)**:
- To form \( \text{Mn}^{3+} \), we remove 3 electrons: 2 from the \( 4s \) subshell and 1 from the \( 3d \) subshell.
- Thus, the configuration becomes \( [Ar] 3d^4 \).
4. **Count the unpaired electrons**:
- The \( 3d^4 \) configuration has 4 unpaired electrons.

### Step 2: Determine the electronic configuration of \( \text{Cr}^{3+} \)

1. **Find the atomic number of Chromium (Cr)**: Chromium has an atomic number of 24.
2. **Write the electronic configuration of neutral Cr**:
- The electronic configuration is \( [Ar] 3d^5 4s^1 \).
3. **Determine the configuration for \( \text{Cr}^{3+} \)**:
- To form \( \text{Cr}^{3+} \), we remove 3 electrons: 2 from the \( 4s \) subshell and 1 from the \( 3d \) subshell.
- Thus, the configuration becomes \( [Ar] 3d^3 \).
4. **Count the unpaired electrons**:
- The \( 3d^3 \) configuration has 3 unpaired electrons.

### Step 3: Determine the electronic configuration of \( \text{V}^{3+} \)

1. **Find the atomic number of Vanadium (V)**: Vanadium has an atomic number of 23.
2. **Write the electronic configuration of neutral V**:
- The electronic configuration is \( [Ar] 3d^3 4s^2 \).
3. **Determine the configuration for \( \text{V}^{3+} \)**:
- To form \( \text{V}^{3+} \), we remove 3 electrons: 2 from the \( 4s \) subshell and 1 from the \( 3d \) subshell.
- Thus, the configuration becomes \( [Ar] 3d^2 \).
4. **Count the unpaired electrons**:
- The \( 3d^2 \) configuration has 2 unpaired electrons.

### Summary of Unpaired Electrons

- \( \text{Mn}^{3+} \): 4 unpaired electrons
- \( \text{Cr}^{3+} \): 3 unpaired electrons
- \( \text{V}^{3+} \): 2 unpaired electrons

### Step 4: Determine the most stable species

- The most stable species among these is \( \text{Cr}^{3+} \) because it has a half-filled \( t_{2g} \) orbitals configuration, which provides extra stability.

### Final Answer

- The number of unpaired electrons in \( \text{Mn}^{3+} \), \( \text{Cr}^{3+} \), and \( \text{V}^{3+} \) are **4, 3, and 2**, respectively.
- The most stable species is **\( \text{Cr}^{3+} \)**.

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