Question

Which of the following pairs is expected to exhibit the same colour?

• A. ₅₈Ce³⁺ and ₆₇Ho³⁺
• B. ₆₀Nd³⁺ and ₆₈Er³⁺
• C. ₆₁Pm³⁺ and ₆₉Tm³⁺
• D. ₆₃Sm³⁺ and ₇₀Yb³⁺

Answer 1

Answer: (B)

B

Answer 2

The color of lanthanide ions ($Ln^{3+}$) is primarily due to f-f electronic transitions. These transitions are largely shielded from the external environment by the 5s and 5p orbitals, resulting in sharp absorption bands and colors that are relatively independent of the ligand.

A key principle for the color of lanthanide ions is that ions with $n$ f-electrons and $(14-n)$ f-electrons often exhibit similar colors. This is because their electronic configurations ($f^n$ and $f^{14-n}$) have a complementary relationship, leading to similar energy level structures and thus similar absorption spectra.

• For Ce³⁺, $n=1$. Complementary is $f^{13}$. Ho³⁺ is $f^{10}$. Not a match.
• For Nd³⁺, $n=3$. Complementary is $f^{11}$. Er³⁺ is $f^{11}$. This is a match.
• For Pm³⁺, $n=4$. Complementary is $f^{10}$. Tm³⁺ is $f^{12}$. Not a match.
• For Sm³⁺, $n=5$. Complementary is $f^9$. Yb³⁺ is $f^{13}$. Not a match.

Thus, Nd³⁺ and Er³⁺ are expected to exhibit similar colors (lilac/reddish-purple and pink/rose, respectively).