Key idea for colour in lanthanoid ions

Lanthanoid ions are generally coloured due to f\!-f electronic transitions.

An ion will be colourless when it has either:

• $4f^0$ configuration, or
• $4f^{14}$ configuration,

because there are no possible f\!-f transitions.

Find the electronic configuration of each $\mathrm{Ln}^{3+}$ ion

Neutral lanthanoids usually lose two $6s$ electrons and one $5d/4f$ electron to form $\mathrm{Ln}^{3+}$.

• $\mathrm{La}^{3+}$

  $\mathrm{La}$: $[\mathrm{Xe}]5d^16s^2$

  So, $\mathrm{La}^{3+} = [\mathrm{Xe}] = 4f^0$ Hence, colourless.

• $\mathrm{Nd}^{3+}$

  $\mathrm{Nd}$: $[\mathrm{Xe}]4f^46s^2$

  So, $\mathrm{Nd}^{3+} = [\mathrm{Xe}]4f^3$ This has partially filled $4f$ orbitals, so coloured.

• $\mathrm{Sm}^{3+}$

  $\mathrm{Sm}$: $[\mathrm{Xe}]4f^66s^2$

  So, $\mathrm{Sm}^{3+} = [\mathrm{Xe}]4f^5$ Partially filled $4f$, so coloured.

• $\mathrm{Eu}^{3+}$

  $\mathrm{Eu}$: $[\mathrm{Xe}]4f^76s^2$

  So, $\mathrm{Eu}^{3+} = [\mathrm{Xe}]4f^6$ Partially filled $4f$, so coloured.

• $\mathrm{Lu}^{3+}$

  $\mathrm{Lu}$: $[\mathrm{Xe}]4f^{14}5d^16s^2$

  So, $\mathrm{Lu}^{3+} = [\mathrm{Xe}]4f^{14}$ Hence, colourless.

Count the colourless ions

Colourless ions are:

• $\mathrm{La}^{3+}$ : $4f^0$
• $\mathrm{Lu}^{3+}$ : $4f^{14}$

Therefore, the number of colourless lanthanoid ions is $2$

Comparison with stored correct answer

Stored correct answer = $2$

My derived answer also = $2$

So, the answer agrees.