Question

Consider the following table of three lanthanoid ions X, Y, and Z and their properties.

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. $\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$
• B. $\text{Lu}^{3+}$, $\text{Ce}^{4+}$, and $\text{Eu}^{2+}$
• C. $\text{Eu}^{2+}$, $\text{Ce}^{4+}$, and $\text{Lu}^{3+}$
• D. $\text{Eu}^{2+}$, $\text{Lu}^{3+}$, and $\text{Ce}^{4+}$

Hint:

Lanthanoids have a preferred oxidation state of $+3$. Any $+2$ lanthanoid ion (like $\text{Eu}^{2+}$, $\text{Yb}^{2+}$) will act as a reducing agent, while any $+4$ ion (like $\text{Ce}^{4+}$) will act as an oxidizing agent.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify three specific lanthanoid ions (X, Y, and Z) based on their given magnetic, chemical, and structural properties.

Step 2: Key Formula or Approach:
We must write out the electronic configurations of the neutral lanthanoids and their ionic states to evaluate:
1. Diamagnetism: Occurs when all electrons are paired (e.g., $4\text{f}^0$ or $4\text{f}^{14}$).
2. Reducing behavior: Occurs when an ion can easily lose electrons to achieve a more stable oxidation state (typically $+3$ for lanthanoids).
3. Empty $4\text{f}$ orbitals: Occurs when the configuration is $4\text{f}^0$.

Step 3: Detailed Explanation:

• Let us evaluate each lanthanoid and its ionic electronic configuration:

• Ce ($Z=58$):
itemize

• Neutral configuration: $[\text{Xe}] 4\text{f}^1 5\text{d}^1 6\text{s}^2$

• $\text{Ce}^{4+}$ configuration: $[\text{Xe}] 4\text{f}^0$ (all valence electrons removed)

• This configuration has completely empty $4\text{f}$ orbitals, which matches the property of Z.

Eu ($Z=63$):

• Neutral configuration: $[\text{Xe}] 4\text{f}^7 6\text{s}^2$

• $\text{Eu}^{2+}$ configuration: $[\text{Xe}] 4\text{f}^7$

• Since the $+3$ oxidation state is the most stable state for all lanthanoids, $\text{Eu}^{2+}$ readily oxidizes to $\text{Eu}^{3+}$ by losing one electron. This makes $\text{Eu}^{2+}$ a strong reducing agent in aqueous solution, matching property Y.

Lu ($Z=71$):

• Neutral configuration: $[\text{Xe}] 4\text{f}^{14} 5\text{d}^1 6\text{s}^2$

• $\text{Lu}^{3+}$ configuration: $[\text{Xe}] 4\text{f}^{14}$

• The $4\text{f}$ subshell is completely filled with $14$ electrons, meaning there are no unpaired electrons. Therefore, $\text{Lu}^{3+}$ is diamagnetic, matching property X.

Thus, X is $\text{Lu}^{3+}$, Y is $\text{Eu}^{2+}$, and Z is $\text{Ce}^{4+}$.
itemize

Step 4: Final Answer:
Therefore, the correct assignment of ions is $\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$ (Option A).