Arrange the following actinoids in increasing order of the maximum oxidation state they show.
(A) Americium (Am)
(B) Neptunium (Np)
(C) Curium (Cm)
(D) Protactinium (Pa)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (D), (A), (B)
(C) Curium (Cm) — first (lowest). Curium shows only the +3 and +4 states, so its maximum is +4.
(D) Protactinium (Pa) — second. In the first half of the series the maximum state rises from element to element, and Pa reaches +5.
(A) Americium (Am) — third. After neptunium the maximum state falls again; americium shows states up to +6.
(B) Neptunium (Np) — fourth (highest). Np shows the +7 state, the highest among the actinoids.
The order is not that of atomic number (Pa 91, Np 93, Am 95, Cm 96), because the maximum state rises only up to Np and then decreases.
NCERT (2026-27 edition):
4.6.3 Oxidation States – “The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements (Table 4.11).”
Table 4.11: Oxidation States of Actinium and Actinoids (columns for the four elements of the question)
| Pa | Np | Am | Cm |
|---|---|---|---|
| 3 4 5 | 3 4 5 6 7 | 3 4 5 6 | 3 4 |