Arrange the following actinoid ions in increasing order of unpaired e…

The correct answer is Option (2) → (C), (D), (A), (B)

Actinoids fill the 5f subshell, which has seven orbitals holding up to 14 electrons. To find the unpaired electrons of an M³⁺ ion we need its configuration and then fill the 5f orbitals by Hund’s rule:

• Write the atom’s configuration beyond the radon core (Rn, Z = 86): the number of electrons to place is Z − 86, in the order 5f, 6d, 7s.

• Form M³⁺ by removing three electrons: first the two 7s electrons, then the 6d electron if there is one, otherwise one 5f electron.

• Fill the seven 5f orbitals: one electron in each orbital first (up to 7), and only then start pairing. So fⁿ has n unpaired electrons up to n = 7, and 14 − n after that.

(A) Am³⁺ — Americium, Z = 95; electrons beyond Rn = 95 − 86 = 9.

    Am: [Rn] 5f⁷ 7s² (7 + 2 = 9)
    \( \ce{Am^{3+}} \): remove 2 electrons from 7s and 1 from 5f → [Rn] 5f⁶
    5f⁶: [↑] [↑] [↑] [↑] [↑] [↑] [  ]

Six electrons go singly into six of the seven orbitals and none has to pair, so \( \ce{Am^{3+}} \) has 6 unpaired electrons.

(B) Cm³⁺ — Curium, Z = 96; electrons beyond Rn = 96 − 86 = 10.

    Cm: [Rn] 5f⁷ 6d¹ 7s² (7 + 1 + 2 = 10)
    \( \ce{Cm^{3+}} \): remove 2 electrons from 7s and 1 from 6d → [Rn] 5f⁷
    5f⁷: [↑] [↑] [↑] [↑] [↑] [↑] [↑]

Every one of the seven orbitals holds one electron (a half-filled 5f subshell), so \( \ce{Cm^{3+}} \) has 7 unpaired electrons, the maximum possible for an f ion.

(C) Lr³⁺ — Lawrencium, Z = 103; electrons beyond Rn = 103 − 86 = 17.

    Lr: [Rn] 5f¹⁴ 6d¹ 7s² (14 + 1 + 2 = 17)
    Lr³⁺: remove 2 electrons from 7s and 1 from 6d → [Rn] 5f¹⁴
    5f¹⁴: [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓]

All seven orbitals are completely filled and every electron is paired, so Lr³⁺ has 0 unpaired electrons.

(D) Cf³⁺ — Californium, Z = 98; electrons beyond Rn = 98 − 86 = 12.

    Cf: [Rn] 5f¹⁰ 7s² (10 + 2 = 12)
    Cf³⁺: remove 2 electrons from 7s and 1 from 5f → [Rn] 5f⁹
    5f⁹: [↑↓] [↑↓] [↑] [↑] [↑] [↑] [↑]

The first seven electrons occupy the seven orbitals singly; the 8th and 9th must pair up with two of them. That leaves 7 − 2 = 5 single electrons, so Cf³⁺ has 5 unpaired electrons (equivalently 14 − 9 = 5).

Shortcut: Lr³⁺ (5f¹⁴) has no unpaired electron and \( \ce{Cm^{3+}} \) (half-filled 5f⁷) has the most. Only Option (2) starts with (C) and ends with (B).