Which is not the correct Statement? (At. nos. Ce = 58, Lu = 71, La = 57, Yb = 70)

The colour (or absence of colour) of lanthanide ions depends on the presence of electrons in the 4f subshell. Only when at least one vacancy exists in the 4f set can an $$f \rightarrow f$$ electronic transition lie in the visible region and impart colour to the ion.

Let us analyse the given statements one by one.

Statement A: “Colour of $$\text{Yb}^{3+}$$ ion is pink.”
  • Atomic number of Yb = 70. Neutral Yb: $$[Xe]\,4f^{14}\,6s^{2}$$ (the 5d level is empty).
  • $$\text{Yb}^{3+}$$ loses two 6s electrons and one 4f electron → $$[Xe]\,4f^{13}$$.
  • An $$f^{13}$$ configuration has just one vacancy in the 4f subshell. The corresponding $$f \rightarrow f$$ transition lies in the near-infra-red region, so almost no light is absorbed in the visible region; the solution therefore appears colourless (very faint yellow at most, never pink).
Hence the statement is incorrect.

Statement B: “$$\text{La}^{3+}$$ is diamagnetic.”
  • Atomic number of La = 57. Neutral La: $$[Xe]\,5d^{1}\,6s^{2}$$.
  • $$\text{La}^{3+}$$ loses 2(6s) + 1(5d) electrons → $$[Xe]$$, i.e. $$4f^{0}$$.
  • With all shells either completely filled or empty, no unpaired electron remains, so it is diamagnetic.
The statement is correct.

Statement C: “$$\text{Ce}^{4+}$$ has $$f^{0}$$ configuration.”
  • Atomic number of Ce = 58. Neutral Ce: $$[Xe]\,4f^{1}\,5d^{1}\,6s^{2}$$.
  • $$\text{Ce}^{4+}$$ loses 2(6s) + 1(5d) + 1(4f) electrons → $$[Xe]$$ → $$4f^{0}$$.
The statement is correct.

Statement D: “$$\text{Lu}^{3+}$$ has $$f^{14}$$ configuration.”
  • Atomic number of Lu = 71. Neutral Lu: $$[Xe]\,4f^{14}\,5d^{1}\,6s^{2}$$.
  • $$\text{Lu}^{3+}$$ loses 2(6s) + 1(5d) electrons → $$[Xe]\,4f^{14}$$.
The statement is correct.

Only Statement A is wrong.

Option A which is: Colour of $$\text{Yb}^{3+}$$ ion is pink