Q131 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Chemistry
Identify the pair of diamagnetic lanthanoid ions by the following?
- Correct answerA$\text{Ce}^{4+}$ and $\text{Yb}^{2+}$
- B$\text{Ce}^{4+}$ and $\text{Eu}^{2+}$
- C$\text{Gd}^{3+}$ and $\text{Eu}^{3+}$
- D$\text{Pm}^{3+}$ and $\text{Sm}^{3+}$
Solution
Diamagnetic species have no unpaired electrons.
The atomic number of Cerium ($\text{Ce}$) is $58$. Its electronic configuration is $[\text{Xe}] 4f^{1} 5d^{1} 6s^{2}$.
For $\text{Ce}^{4+}$, the electronic configuration is $[\text{Xe}] 4f^{0}$. Since there are no unpaired electrons, it is diamagnetic.
The atomic number of Ytterbium ($\text{Yb}$) is $70$. Its electronic configuration is $[\text{Xe}] 4f^{14} 6s^{2}$.
For $\text{Yb}^{2+}$, the electronic configuration is $[\text{Xe}] 4f^{14}$. Since the $4f$ subshell is completely filled, there are no unpaired electrons, making it diamagnetic.
Other ions like $\text{Eu}^{2+}$ ($4f^{7}$), $\text{Gd}^{3+}$ ($4f^{7}$), $\text{Eu}^{3+}$ ($4f^{6}$), $\text{Pm}^{3+}$ ($4f^{4}$), and $\text{Sm}^{3+}$ ($4f^{5}$) have unpaired electrons and are paramagnetic.
Answer: $\text{Ce}^{4+}$ and $\text{Yb}^{2+}$
The atomic number of Cerium ($\text{Ce}$) is $58$. Its electronic configuration is $[\text{Xe}] 4f^{1} 5d^{1} 6s^{2}$.
For $\text{Ce}^{4+}$, the electronic configuration is $[\text{Xe}] 4f^{0}$. Since there are no unpaired electrons, it is diamagnetic.
The atomic number of Ytterbium ($\text{Yb}$) is $70$. Its electronic configuration is $[\text{Xe}] 4f^{14} 6s^{2}$.
For $\text{Yb}^{2+}$, the electronic configuration is $[\text{Xe}] 4f^{14}$. Since the $4f$ subshell is completely filled, there are no unpaired electrons, making it diamagnetic.
Other ions like $\text{Eu}^{2+}$ ($4f^{7}$), $\text{Gd}^{3+}$ ($4f^{7}$), $\text{Eu}^{3+}$ ($4f^{6}$), $\text{Pm}^{3+}$ ($4f^{4}$), and $\text{Sm}^{3+}$ ($4f^{5}$) have unpaired electrons and are paramagnetic.
Answer: $\text{Ce}^{4+}$ and $\text{Yb}^{2+}$
Hence, the correct answer is option: 'A'.