Which of the following Lanthanides element do not show stable +4 oxidation state?

Options

A. Eu
B. Dy
C. Tb
D. Ce

Answer

To determine which of the given lanthanide elements does not show a stable +4 oxidation state, we will analyze the electronic configurations of each element and their ability to achieve stable configurations when losing electrons.

### Step-by-Step Solution:

1. **Identify the Elements and Their Configurations:**
- **Europium (Eu)**: Atomic number 63, electronic configuration: [Xe] 4f^7 6s^2
- **Dysprosium (Dy)**: Atomic number 66, electronic configuration: [Xe] 4f^10 6s^2
- **Terbium (Tb)**: Atomic number 65, electronic configuration: [Xe] 4f^9 6s^2
- **Cerium (Ce)**: Atomic number 58, electronic configuration: [Xe] 4f^1 5d^1 6s^2

2. **Determine the +4 Oxidation State Configurations:**
- **Cerium (Ce)**:
- Loses 4 electrons: [Xe] 4f^0 5d^0 6s^0 (Noble gas configuration, stable)
- **Terbium (Tb)**:
- Loses 4 electrons: [Xe] 4f^5 6s^0 (Half-filled configuration, stable)
- **Dysprosium (Dy)**:
- Loses 4 electrons: [Xe] 4f^6 6s^0 (Not half-filled, but stable)
- **Europium (Eu)**:
- Loses 4 electrons: [Xe] 4f^3 6s^0 (Not half-filled, not stable)

3. **Analyze Stability:**
- A stable oxidation state is typically associated with either a fully filled or half-filled subshell.
- **Cerium** shows a stable +4 oxidation state because it achieves a noble gas configuration.
- **Terbium** shows a stable +4 oxidation state due to its half-filled 4f subshell.
- **Dysprosium** can also show a +4 oxidation state, but it is less stable compared to the previous two.
- **Europium**, when in the +4 oxidation state, does not achieve a stable configuration since it does not have a half-filled or fully filled subshell.

4. **Conclusion:**
- The element that does not show a stable +4 oxidation state is **Europium (Eu)**.

### Final Answer:
**Europium (Eu)** does not show a stable +4 oxidation state.