Which of the following lanthanoid ions are coloured?

${\mathrm{Lu}}^{+3}$

${\mathrm{Pm}}^{+3}$

${\mathrm{Sm}}^{+3}$

${\mathrm{Eu}}^{+3}$

The correct option is B

${\mathrm{Pm}}^{+3},{\mathrm{Sm}}^{+3},{\mathrm{Eu}}^{+3}$

The explanation for correct option:

(B) ${\mathrm{Pm}}^{+3},{\mathrm{Sm}}^{+3},{\mathrm{Eu}}^{+3}$

• Electronic configuration of ${\mathrm{Pm}}^{+3}$: $\left[\mathrm{Xe}\right]4f^{\mathit{4}}$
• Electronic configuration of ${\mathrm{Sm}}^{+3}$: $\left[\mathrm{Xe}\right]4f^{\mathit{5}}$
• Electronic configuration of ${\mathrm{Eu}}^{+3}$: $\left[\mathrm{Xe}\right]4f^6$
• The reason for some lanthanoids to be colored is $\mathrm{f}-\mathrm{f}$ transition.
• The transition of an electron from a lower-energy f orbital to a higher-energy f orbital is known as a $\mathrm{f}-\mathrm{f}$ transition.
• For the transition to occur, there should be unpaired electrons present in the f-orbital.
• So, ${\mathrm{Pm}}^{+3},{\mathrm{Sm}}^{+3},{\mathrm{Eu}}^{+3}$has unpaired electrons, $\mathrm{f}-\mathrm{f}$ transition is possible, so these three lanthanoids will be colored.

The explanation for incorrect options:

(A) ${\mathrm{Lu}}^{+3},{\mathrm{Pm}}^{+3},{\mathrm{Sm}}^{+3}$

• Electronic configuration of ${\mathrm{Lu}}^{+3}$: $\left[\mathrm{Xe}\right]4f^{14}$
• All the electrons in the f-orbital are paired because it is fully occupied in case of ${\mathrm{Lu}}^{+3}$, so $\mathrm{f}-\mathrm{f}$ transition is not possible.
• ${\mathrm{Lu}}^{+3}$ will be colorless.

(C) ${\mathrm{Lu}}^{+3},{\mathrm{Pm}}^{+3},{\mathrm{Sn}}^{+3}$

• ${\mathrm{Lu}}^{+3}$ will be colorless.

(D) ${\mathrm{Lu}}^{+3},{\mathrm{Sm}}^{+3},{\mathrm{Eu}}^{+3}$

• Again, ${\mathrm{Lu}}^{+3}$ will be colorless.

Therefore, Option (B) is correct, ${\mathrm{Pm}}^{+3},{\mathrm{Sm}}^{+3},{\mathrm{Eu}}^{+3}$ will be colored lanthanoids.