Choose the correct pair regarding `IE_(3)`

Options

A. `Mn gt Cr`
B. `Mn gt Fe`
C. `Zn gt Cu`
D. All of these

Answer

To solve the question regarding the third ionization energy (IE₃) of the given pairs, we will analyze each pair step by step.

### Step 1: Understand Ionization Energy
**Definition**: Ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. The third ionization energy (IE₃) refers to the energy required to remove the third electron from a cation.

### Step 2: Analyze the First Pair (Manganese vs. Chromium)
1. **Manganese (Mn)**:
- Atomic number: 25
- Electron configuration: [Ar] 4s² 3d⁵
- For Mn²⁺: Configuration becomes [Ar] 3d⁵ (removal from 4s).
- This configuration is stable (half-filled d-orbitals).

2. **Chromium (Cr)**:
- Atomic number: 24
- Electron configuration: [Ar] 4s¹ 3d⁵
- For Cr²⁺: Configuration becomes [Ar] 3d⁴.
- This configuration is less stable than Mn²⁺.

**Conclusion**: The removal of the third electron from Mn²⁺ is more difficult than from Cr²⁺ due to the stability of Mn²⁺. Thus, **IE₃ (Mn) > IE₃ (Cr)**.

### Step 3: Analyze the Second Pair (Manganese vs. Iron)
1. **Iron (Fe)**:
- Atomic number: 26
- Electron configuration: [Ar] 4s² 3d⁶
- For Fe²⁺: Configuration becomes [Ar] 3d⁶.
- This configuration is not particularly stable.

**Conclusion**: Since Mn²⁺ is more stable than Fe²⁺, it is harder to remove the third electron from Mn²⁺ compared to Fe²⁺. Thus, **IE₃ (Mn) > IE₃ (Fe)**.

### Step 4: Analyze the Third Pair (Zinc vs. Copper)
1. **Zinc (Zn)**:
- Atomic number: 30
- Electron configuration: [Ar] 4s² 3d¹⁰
- For Zn²⁺: Configuration becomes [Ar] 3d¹⁰ (removal from 4s).
- This configuration is stable (fully filled d-orbitals).

2. **Copper (Cu)**:
- Atomic number: 29
- Electron configuration: [Ar] 4s¹ 3d¹⁰
- For Cu²⁺: Configuration becomes [Ar] 3d⁹.
- This configuration is less stable.

**Conclusion**: The removal of the third electron from Zn²⁺ is more difficult than from Cu²⁺ due to the stability of Zn²⁺. Thus, **IE₃ (Zn) > IE₃ (Cu)**.

### Final Conclusion
- All three pairs are correct regarding their third ionization energies:
- **IE₃ (Mn) > IE₃ (Cr)**
- **IE₃ (Mn) > IE₃ (Fe)**
- **IE₃ (Zn) > IE₃ (Cu)**

Therefore, the answer to the question is that all pairs are correct.