Question

Which of the following oxidation state is common for all lanthanoids?

• A. \(+2\)
• B. \(+3\)
• C. \(+4\)
• D. \(+5\)

Hint:

Most common oxidation state of lanthanoids: \[ +3 \] due to stable removal of: \[ 6s^2 \] and one additional electron.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The lanthanoids (elements with atomic numbers 58 to 71) are characterized by the filling of the \( 4f \) subshell.
The electronic configuration of these elements typically follows the pattern \( [Xe] 4f^{1-14} 5d^{0-1} 6s^2 \).
The valence electrons reside in the \( 6s \) and \( 5d \) orbitals, and sometimes the \( 4f \) electrons also participate in bonding.
Step 2: Detailed Explanation:
For all lanthanoids, the most stable and ubiquitous oxidation state is +3.
This state is achieved by losing the two \( 6s \) electrons and one \( 5d \) (or one \( 4f \)) electron.
The ionization energies for removing these first three electrons are relatively low. Once the +3 state is reached, the remaining \( 4f \) electrons are "buried" deep inside the atom and are shielded by the outer \( 5s \) and \( 5p \) electrons. This makes it difficult to remove further electrons under normal conditions.
While some lanthanoids show other oxidation states:
- Cerium (\( Ce \)) can show +4 because \( Ce^{4+} \) has a stable noble gas configuration (\( f^0 \)).
- Europium (\( Eu \)) can show +2 because \( Eu^{2+} \) has a stable half-filled configuration (\( f^7 \)).
- Ytterbium (\( Yb \)) can show +2 because \( Yb^{2+} \) has a stable fully filled configuration (\( f^{14} \)).
However, these +2 and +4 states are exceptions and are often unstable in aqueous solution, eventually reverting to the common +3 state.
The +3 oxidation state is the only one that is shared by {every} member of the lanthanoid series.
Step 3: Final Answer:
The most characteristic and common oxidation state across the entire lanthanoid series is +3.
Thus, the correct option is (B).