Question

In the actinoid series, which oxidation state is the most stable for most actinoids?

• A. \(+2\)
• B. \(+3\)
• C. \(+4\)
• D. \(+6\)

Hint:

Both lanthanoids and actinoids most commonly exhibit the \(+3\) oxidation state due to stability achieved after losing outer electrons.

Answer 1

The Correct Option is B

Concept: Actinoids exhibit variable oxidation states because the energies of \(5f\), \(6d\), and \(7s\) orbitals are very close to one another. However, the \(+3\) oxidation state is the most stable and most commonly observed oxidation state for the majority of actinoids.

Step 1: Understanding the electronic configuration of actinoids.
Actinoids belong to the \(5f\)-series where electrons are progressively filled in the \(5f\)-orbitals. Since the \(5f\), \(6d\), and \(7s\) orbitals have comparable energies, multiple oxidation states become possible.

Step 2: Formation of the \(+3\) oxidation state.
Most actinoids lose: \[ 2 \text{ electrons from }7s \text{ and }1 \text{ electron from }5f/6d \] leading to the formation of: \[ M^{3+} \] This state is particularly stable because removal of three electrons gives a comparatively stable electronic arrangement.

Step 3: Examples of stable \(+3\) oxidation state.
Elements like uranium, neptunium, plutonium, and americium commonly form: \[ U^{3+},\ Np^{3+},\ Pu^{3+},\ Am^{3+} \] Thus, the most stable oxidation state for most actinoids is: \[ \boxed{+3} \]