Question:easy

Actinoids show larger number of oxidation states :

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Actinoids show many oxidation states because the energies of 5f, 6d and 7s orbitals are very close, allowing electrons from all these orbitals to participate in bonding.
Updated On: Jun 29, 2026
  • because they are radioactive in nature
  • because they have large atomic numbers
  • because they have large atomic masses
  • due to comparable energies of 5f, 6d and 7s orbitals
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The Correct Option is D

Solution and Explanation

Step 1: Note the valence orbital set of actinoids.
Actinoids are $5f$-block elements. Their outermost electrons occupy three closely spaced subshells: $5f$, $6d$, and $7s$. For example, uranium has the configuration $[Rn]\,5f^3\,6d^1\,7s^2$.
Step 2: Connect orbital energies to oxidation states.
Because $5f$, $6d$, and $7s$ orbitals have nearly equal energies, electrons from all three subshells can be removed or used in bonding under suitable conditions. This gives actinoids oxidation states ranging from $+2$ up to $+7$ or beyond, a wider range than most transition metals.
Step 3: Contrast with lanthanoids and eliminate other options.
Lanthanoids predominantly show $+3$ because their $4f$ electrons are more shielded and differ significantly in energy from $5d/6s$. Radioactivity, atomic number, and atomic mass do not govern oxidation states. The sole reason for the large number of actinoid oxidation states is the comparable energies of $5f$, $6d$, and $7s$ orbitals.
\[ \boxed{\text{Comparable energies of } 5f,\;6d,\;\text{and }7s \text{ orbitals}} \]
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