Q32 / 79 · NEST (NISER) 2026 · Chemistry
Consider the following table of three lanthanoid ions X, Y, and Z and their properties.
Lanthanoid Ions Properties X Diamagnetism Y Reducing agent in solution Z Empty 4f orbitals
Atomic numbers of Ce, Eu, and Lu are $58$, $63$, and $71$, respectively. Given these atomic numbers, the lanthanoid ions X, Y, and Z, respectively, are:
| Lanthanoid Ions | Properties |
|---|---|
| X | Diamagnetism |
| Y | Reducing agent in solution |
| Z | Empty 4f orbitals |
Atomic numbers of Ce, Eu, and Lu are $58$, $63$, and $71$, respectively. Given these atomic numbers, the lanthanoid ions X, Y, and Z, respectively, are:
- A$\text{Eu}^{2+}$, $\text{Lu}^{3+}$, and $\text{Ce}^{4+}$
- B$\text{Lu}^{3+}$, $\text{Ce}^{4+}$, and $\text{Eu}^{2+}$
- C$\text{Eu}^{2+}$, $\text{Ce}^{4+}$, and $\text{Lu}^{3+}$
- Correct answerD$\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$
Solution
The electronic configuration of Lu ($Z=71$) is $[\text{Xe}] 4f^{14} 5d^{1} 6s^{2}$.
$\text{Lu}^{3+}$ has the configuration $[\text{Xe}] 4f^{14}$. Since all electrons are paired, it is diamagnetic. Thus, X is $\text{Lu}^{3+}$.
The electronic configuration of Eu ($Z=63$) is $[\text{Xe}] 4f^{7} 6s^{2}$.
$\text{Eu}^{2+}$ has the configuration $[\text{Xe}] 4f^{7}$. It acts as a strong reducing agent in aqueous solution because it readily oxidizes to the more stable $+3$ oxidation state ($\text{Eu}^{3+}$). Thus, Y is $\text{Eu}^{2+}$.
The electronic configuration of Ce ($Z=58$) is $[\text{Xe}] 4f^{1} 5d^{1} 6s^{2}$.
$\text{Ce}^{4+}$ has the configuration $[\text{Xe}] 4f^{0}$. It has empty $4f$ orbitals. Thus, Z is $\text{Ce}^{4+}$.
Therefore, X, Y, and Z are $\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$, respectively.
$\text{Lu}^{3+}$ has the configuration $[\text{Xe}] 4f^{14}$. Since all electrons are paired, it is diamagnetic. Thus, X is $\text{Lu}^{3+}$.
The electronic configuration of Eu ($Z=63$) is $[\text{Xe}] 4f^{7} 6s^{2}$.
$\text{Eu}^{2+}$ has the configuration $[\text{Xe}] 4f^{7}$. It acts as a strong reducing agent in aqueous solution because it readily oxidizes to the more stable $+3$ oxidation state ($\text{Eu}^{3+}$). Thus, Y is $\text{Eu}^{2+}$.
The electronic configuration of Ce ($Z=58$) is $[\text{Xe}] 4f^{1} 5d^{1} 6s^{2}$.
$\text{Ce}^{4+}$ has the configuration $[\text{Xe}] 4f^{0}$. It has empty $4f$ orbitals. Thus, Z is $\text{Ce}^{4+}$.
Therefore, X, Y, and Z are $\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$, respectively.
Hence, the correct answer is option: 'D'.