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 lanthanide element does not show a stable +4 oxidation state among the options given (Eu, Dy, Tb, and Ce), we will analyze the electronic configurations and oxidation states of each element.

### Step-by-Step Solution:

1. **Identify the Elements and Their Atomic Numbers**:
- Europium (Eu) - Atomic Number 63
- Dysprosium (Dy) - Atomic Number 66
- Terbium (Tb) - Atomic Number 65
- Cerium (Ce) - Atomic Number 58

2. **Write the Electronic Configurations**:
- **Europium (Eu)**:
- Electronic configuration: [Xe] 6s² 4f⁷
- **Dysprosium (Dy)**:
- Electronic configuration: [Xe] 6s² 4f⁹
- **Terbium (Tb)**:
- Electronic configuration: [Xe] 6s² 4f⁷ 5d¹
- **Cerium (Ce)**:
- Electronic configuration: [Xe] 6s² 4f¹ 5d¹

3. **Determine Possible Oxidation States**:
- **Europium (Eu)**: Can lose 2 electrons from 6s and 2 from 4f, but the 4f subshell is half-filled (4f⁷), making it stable in the +2 oxidation state. The +4 state is not stable due to the filled nature of the 4f subshell.
- **Dysprosium (Dy)**: Can lose 2 electrons from 6s and can also lose electrons from the 4f subshell, allowing for +4 oxidation state.
- **Terbium (Tb)**: Similar to Dy, it can lose electrons from both the 6s and 4f subshells, allowing for +4 oxidation state.
- **Cerium (Ce)**: Can lose 2 electrons from 6s and 2 from the 4f subshell, allowing for +4 oxidation state.

4. **Conclusion**:
- Among the elements analyzed, **Europium (Eu)** does not show a stable +4 oxidation state due to its electronic configuration and the stability of the half-filled 4f subshell.

### Final Answer:
The lanthanide element that does not show a stable +4 oxidation state is **Europium (Eu)**.

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