Question

Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature ? (Given Z for Nd=60, Yb=70, La=57, Ce=58)

• A. Nd
• B. Yb
• C. La
• D. Ce

Hint:

Stability in lanthanide oxidation states usually follows the rule of stable subshells: empty (\( f^0 \)), half-filled (\( f^7 \)), or fully-filled (\( f^{14} \)).

Answer 1

The Correct Option is B

To determine which lanthanide exhibits a +2 oxidation state with diamagnetic nature, we need to consider the electronic configurations and characteristics of the given elements: Neodymium (Nd), Ytterbium (Yb), Lanthanum (La), and Cerium (Ce).

The key to determining diamagnetic nature is to have all electrons paired. Let's examine the elements:

1. Neodymium (Nd): Atomic number 60. The electronic configuration is \text{[Xe]}4f^46s^2. In the +2 oxidation state, the configuration might be \text{[Xe]}4f^4, which still has unpaired electrons. Hence, it is not diamagnetic.
2. Ytterbium (Yb): Atomic number 70. The electronic configuration is \text{[Xe]}4f^{14}6s^2. In the +2 oxidation state, the configuration becomes \text{[Xe]}4f^{14}, where all electrons are paired, leading to a diamagnetic nature.
3. Lanthanum (La): Atomic number 57. The electronic configuration is \text{[Xe]}5d^16s^2. In the +2 oxidation state, the configuration might reduce to \text{[Xe]}5d^0, but it never reaches a state where the f-orbitals are involved, thus not achieving diamagnetic nature from the f-block perspective.
4. Cerium (Ce): Atomic number 58. The electronic configuration is \text{[Xe]}4f^15d^16s^2. In the +2 oxidation state, the configuration might reduce to \text{[Xe]}4f^1, which still has unpaired electrons and is not diamagnetic.

Therefore, among the lanthanides listed, Ytterbium (Yb) is the element that exhibits a +2 oxidation state with diamagnetic nature due to its electronic configuration.

Correct Answer: Yb