Which of the following has the maximum number of unpaired electrons?

Options

A. `Mg^(2+)`
B. `Ti^(3+)`
C. `V^(3+)`
D. `Fe^(2+)`

Explanation

To determine which of the given ions has the maximum number of unpaired electrons, we will analyze the electronic configurations of each ion step by step.

### Step 1: Determine the electronic configuration of Mg²⁺
- Magnesium (Mg) has an atomic number of 12. Its ground state electronic configuration is:
\[ \text{Mg: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \]
- For Mg²⁺, we remove 2 electrons from the outermost shell (3s):
\[ \text{Mg}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^0 \]
- All electrons in the filled orbitals (1s, 2s, 2p) are paired. Thus, Mg²⁺ has **0 unpaired electrons**.

### Step 2: Determine the electronic configuration of Ti³⁺
- Titanium (Ti) has an atomic number of 22. Its ground state electronic configuration is:
\[ \text{Ti: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^2 \, 4s^2 \]
- For Ti³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{Ti}^{3+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^1 \]
- In the 3d subshell, there is **1 unpaired electron**.

### Step 3: Determine the electronic configuration of V³⁺
- Vanadium (V) has an atomic number of 23. Its ground state electronic configuration is:
\[ \text{V: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^3 \, 4s^2 \]
- For V³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{V}^{3+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^2 \]
- In the 3d subshell, there are **2 unpaired electrons**.

### Step 4: Determine the electronic configuration of Fe²⁺
- Iron (Fe) has an atomic number of 26. Its ground state electronic configuration is:
\[ \text{Fe: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \, 4s^2 \]
- For Fe²⁺, we remove 2 electrons from 4s:
\[ \text{Fe}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \]
- In the 3d subshell, there are **4 unpaired electrons**.

### Conclusion
Now we summarize the number of unpaired electrons in each ion:
- Mg²⁺: 0 unpaired electrons
- Ti³⁺: 1 unpaired electron
- V³⁺: 2 unpaired electrons
- Fe²⁺: 4 unpaired electrons

Thus, the ion with the maximum number of unpaired electrons is **Fe²⁺**, which has **4 unpaired electrons**.

### Final Answer
**Fe²⁺ has the maximum number of unpaired electrons (4).**

---