Q56 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
The Lanthanoid ion which would form coloured compounds is -------------.
Atomic numbers: $\mathrm{Yb}=70, \quad \mathrm{Lu}=71, \mathrm{Pr}=59, \quad \mathrm{La}=57$
- Correct answerA$\operatorname{Pr}^{3+}$
- B$L a^{3+}$
- C$Y b^{2+}$
- D$L u^{3+}$
Solution
Lanthanoid ions exhibit colour due to $f-f$ electronic transitions. Ions with partially filled $f$-orbitals ($f^1$ to $f^{13}$) are coloured, while ions with $f^0$ or $f^{14}$ configurations are colourless.
Electronic configurations of the given ions are:
1. $\mathrm{La}^{3+}$ ($Z=57$): $[\mathrm{Xe}] 4f^0$. Since it has no $f$-electrons, it is colourless.
2. $\mathrm{Lu}^{3+}$ ($Z=71$): $[\mathrm{Xe}] 4f^{14}$. Since the $f$-subshell is completely filled, it is colourless.
3. $\mathrm{Yb}^{2+}$ ($Z=70$): $[\mathrm{Xe}] 4f^{14}$. Since the $f$-subshell is completely filled, it is colourless.
4. $\mathrm{Pr}^{3+}$ ($Z=59$): $[\mathrm{Xe}] 4f^2$. Since it has a partially filled $f$-subshell, it exhibits colour.
Answer: $\operatorname{Pr}^{3+}$
Electronic configurations of the given ions are:
1. $\mathrm{La}^{3+}$ ($Z=57$): $[\mathrm{Xe}] 4f^0$. Since it has no $f$-electrons, it is colourless.
2. $\mathrm{Lu}^{3+}$ ($Z=71$): $[\mathrm{Xe}] 4f^{14}$. Since the $f$-subshell is completely filled, it is colourless.
3. $\mathrm{Yb}^{2+}$ ($Z=70$): $[\mathrm{Xe}] 4f^{14}$. Since the $f$-subshell is completely filled, it is colourless.
4. $\mathrm{Pr}^{3+}$ ($Z=59$): $[\mathrm{Xe}] 4f^2$. Since it has a partially filled $f$-subshell, it exhibits colour.
Answer: $\operatorname{Pr}^{3+}$
Hence, the correct answer is option: 'A'.