Question:
Which of the following \(f\)-block element exhibits the highest oxidation state?
Which of the following \(f\)-block element exhibits the highest oxidation state?
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The highest commonly observed oxidation state among the early actinides is \(+7\), shown by neptunium in compounds such as \(Np_2O_5\) and related species.
Updated On: Jun 26, 2026
- U
- Np
- Am
- Pa
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The Correct Option is B
Solution and Explanation
Step 1: Recall the oxidation states of actinides.
Actinides exhibit a wide range of oxidation states because the energies of \(5f\), \(6d\), and \(7s\) orbitals are comparable.
Hence, several electrons can participate in bonding.
Step 2: Examine the given elements.
For uranium: \[ \text{Maximum oxidation state}=+6 \] For protactinium: \[ \text{Maximum oxidation state}=+5 \] For americium: \[ \text{Maximum oxidation state}=+6 \] For neptunium: \[ \text{Maximum oxidation state}=+7 \]
Step 3: Compare the oxidation states.
Among the given elements: \[ Np(+7) \gt U(+6) = Am(+6) \gt Pa(+5) \] Thus, neptunium exhibits the highest oxidation state.
Step 4: Final conclusion.
Therefore, the \(f\)-block element exhibiting the highest oxidation state is \[ \boxed{Np} \] Hence, the correct option is \[ \boxed{(2)} \]
Actinides exhibit a wide range of oxidation states because the energies of \(5f\), \(6d\), and \(7s\) orbitals are comparable.
Hence, several electrons can participate in bonding.
Step 2: Examine the given elements.
For uranium: \[ \text{Maximum oxidation state}=+6 \] For protactinium: \[ \text{Maximum oxidation state}=+5 \] For americium: \[ \text{Maximum oxidation state}=+6 \] For neptunium: \[ \text{Maximum oxidation state}=+7 \]
Step 3: Compare the oxidation states.
Among the given elements: \[ Np(+7) \gt U(+6) = Am(+6) \gt Pa(+5) \] Thus, neptunium exhibits the highest oxidation state.
Step 4: Final conclusion.
Therefore, the \(f\)-block element exhibiting the highest oxidation state is \[ \boxed{Np} \] Hence, the correct option is \[ \boxed{(2)} \]
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