Which of the following transition metal cation has maxi- mum unpaired electrons?

Options

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

Answer

To determine which transition metal cation has the maximum number of unpaired electrons among Magnesium (Mn), Iron (Fe), Cobalt (Co), and Nickel (Ni), we need to follow these steps:

### Step 1: Identify the Atomic Numbers
- **Magnesium (Mn)**: Atomic number = 25
- **Iron (Fe)**: Atomic number = 26
- **Cobalt (Co)**: Atomic number = 27
- **Nickel (Ni)**: Atomic number = 28

### Step 2: Write the Electronic Configurations
- **Mn (neutral)**: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\)
- **Fe (neutral)**: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6\)
- **Co (neutral)**: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\)
- **Ni (neutral)**: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8\)

### Step 3: Determine the Cation Configurations
For each metal cation, we need to remove 2 electrons (as indicated by the 2+ charge):
- **Mn²⁺**: Loses 2 electrons from the 4s orbital, resulting in \(3d^5\)
- **Fe²⁺**: Loses 2 electrons from the 4s orbital, resulting in \(3d^6\)
- **Co²⁺**: Loses 2 electrons from the 4s orbital, resulting in \(3d^7\)
- **Ni²⁺**: Loses 2 electrons from the 4s orbital, resulting in \(3d^8\)

### Step 4: Count Unpaired Electrons
- **Mn²⁺ (3d⁵)**: All 5 electrons are unpaired (1 in each of the 5 d orbitals).
- **Fe²⁺ (3d⁶)**: 4 unpaired electrons (5th electron pairs with one of the previous 5).
- **Co²⁺ (3d⁷)**: 3 unpaired electrons (the 6th and 7th electrons pair up).
- **Ni²⁺ (3d⁸)**: 2 unpaired electrons (the 7th and 8th electrons pair up).

### Step 5: Conclusion
The cation with the maximum number of unpaired electrons is **Mn²⁺**, which has **5 unpaired electrons**.

### Final Answer:
**Mn²⁺ has the maximum number of unpaired electrons (5).**

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