Question:medium

The ′f′ orbitals are half and completely filled, respectively in lanthanide ions
[Given: Atomic no. Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66]

Updated On: Apr 19, 2026
  • \(Eu^{2+}\) and \(Tm^{2+}\)
  • \(Sm^{2+}\) and \(Tm^{3+}\)
  • \(Tb^{4+}\) and \(Yb^{2+}\)
  • \(Dy^{3+}\) and \(Yb^{3+}\)
Show Solution

The Correct Option is C

Solution and Explanation

 To determine which lanthanide ions have half-filled and completely filled f orbitals, we must consider the electronic configurations of the elements and their ions.

  1. The lanthanides have the general electron configuration \([Xe] 4f^n 6s^2\).
  2. Half-filled Orbitals: A half-filled f subshell has 7 electrons (since the f subshell can hold up to 14 electrons).
  3. Completely filled Orbitals: A completely filled f subshell has 14 electrons.

Let's analyze the provided atomic numbers:

  1. Europium (\(Eu\)): Atomic number 63, configuration is \([Xe] 4f^7 6s^2\).
    \([Eu^{2+}]:\) Removing 2 electrons gives \(4f^7\), a half-filled configuration.\)
  2. Thulium (\(Tm\)): Atomic number 69, configuration is \([Xe] 4f^{13} 6s^2\).
    \([Tm^{2+}]:\) Removing 2 electrons gives \(4f^{13}\), not full nor half.\)
  3. Ytterbium (\(Yb\)): Atomic number 70, configuration is \([Xe] 4f^{14} 6s^2\).
    \([Yb^{2+}]:\) Removing 2 electrons gives \(4f^{14}\), a completely filled configuration.\)
  4. Terbium (\(Tb\)): Atomic number 65, configuration is \([Xe] 4f^9 6s^2\).
    \([Tb^{4+}]:\) Removing 4 electrons results in \(4f^7\), a half-filled configuration.\)

Let's evaluate the answer choices:

  • \(Eu^{2+}\) and \(Tm^{2+}\): Eu has a half-filled, but Tm does not have either half or full-filled.
  • \(Sm^{2+}\) and \(Tm^{3+}\): Sm also doesn't result in a half or fully filled f-orbital, and neither does Tm in its 3+ state.
  • \(Tb^{4+}\) and \(Yb^{2+}\): Tb results in a half-filled, and Yb results in a completely filled configuration, fulfilling the requirement.
  • \(Dy^{3+}\) and \(Yb^{3+}\): Dy does not have half or fully filled f-orbital either.

Thus, the correct answer is \(Tb^{4+}\) and \(Yb^{2+}\) because they satisfy the conditions of half-filled and completely filled f-orbitals, respectively.

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