The correct 4f sub-shell electronic configuration of \( Yb^{2+} \) ion is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4f^{14}\)
Lanthanoid ions lose their 6s electrons first, and only then 5d or 4f electrons. So begin with the configuration of the neutral atom from its atomic number.
Ytterbium (Z = 70) has the configuration
\( \displaystyle \ce{Yb}: \mathrm{[Xe]\,4f^{14}\,6s^{2}} \)
Forming \( \ce{Yb^2+} \) removes the two 6s electrons:
\( \displaystyle \ce{Yb^2+}: \mathrm{[Xe]\,4f^{14}} \)
So the 4f subshell of \( \ce{Yb^2+} \) is completely filled, 4f14. This filled 4f set is why ytterbium shows a +2 state besides the usual +3.
Why the other options are incorrect
Option (2) \(4f^{13}\) — Incorrect. This is the configuration of \( \ce{Yb^3+} \), which loses one 4f electron besides the two 6s electrons.
Option (3) \(4f^{12}\) — Incorrect. This would need two 4f electrons removed besides the 6s pair, which would be \( \ce{Yb^4+} \), a state ytterbium does not show.
Option (4) \(4f^7\) — Incorrect. This half-filled set belongs to \( \ce{Eu^2+} \) or \( \ce{Gd^3+} \), not to ytterbium.
NCERT (2026-27 edition):
Electronic Configurations – …
