Question

The pair of ions with paramagnetic nature and same number of electrons is

• A. $\text{Lu}^{3+}, \text{Yb}^{2+}$
• B. $\text{Eu}^{3+}, \text{Pm}^{2+}$
• C. $\text{Eu}^{2+}, \text{Gd}^{3+}$
• D. $\text{La}^{3+}, \text{Ce}^{4+}$

Hint:

To check for paramagnetism, count the number of unpaired electrons in the ion's electronic configuration. $f^0, f^{14}, d^0, d^{10}$ correspond to diamagnetic ions (no unpaired electrons). For Lanthanides, special stability is conferred by the half-filled ($4f^7$) and full ($4f^{14}$) configurations.

Answer 1

The Correct Option is C

Step 1: Determine the number of electrons and $f$-orbital configuration for each ion.
An ion is paramagnetic if it has unpaired electrons. Both ions in the pair must have the same number of electrons (isoelectronic) and both must have unpaired electrons.
Lanthanides range from $\text{La}(57)$ to $\text{Lu}(71)$ with general configuration $[\text{Xe}]4f^{0-14}5d^{0-1}6s^2$. The $3+$ ions generally lose $6s^2$ and $5d^1$ electrons.

Step 2: Analyze the electron configuration for option (A).
$\text{Lu} (Z=71): [\text{Xe}]4f^{14}5d^16s^2$, so $\text{Lu}^{3+} = [\text{Xe}]4f^{14}$ ($68e^-$, diamagnetic).
$\text{Yb} (Z=70): [\text{Xe}]4f^{14}6s^2$, so $\text{Yb}^{2+} = [\text{Xe}]4f^{14}$ ($68e^-$, diamagnetic).
Isoelectronic (68 $e^-$), but both are diamagnetic. $\Rightarrow$ Incorrect.

Step 3: Analyze the electron configuration for option (B).
$\text{Eu} (Z=63): [\text{Xe}]4f^{7}6s^2$, so $\text{Eu}^{3+} = [\text{Xe}]4f^{6}$ ($60e^-$, 6 unpaired $e^-$, paramagnetic).
$\text{Pm} (Z=61): [\text{Xe}]4f^{4}6s^2$, so $\text{Pm}^{2+} = [\text{Xe}]4f^{5}$ ($59e^-$, 5 unpaired $e^-$, paramagnetic).
Not isoelectronic (60 vs 59 $e^-$). $\Rightarrow$ Incorrect.

Step 4: Analyze the electron configuration for option (C).
$\text{Eu} (Z=63): [\text{Xe}]4f^{7}6s^2$, so $\text{Eu}^{2+} = [\text{Xe}]4f^{7}$ ($61e^-$, 7 unpaired $e^-$, paramagnetic).
$\text{Gd} (Z=64): [\text{Xe}]4f^{7}5d^16s^2$, so $\text{Gd}^{3+} = [\text{Xe}]4f^{7}$ ($61e^-$, 7 unpaired $e^-$, paramagnetic).
Both are isoelectronic ($61e^-$) and paramagnetic ($7$ unpaired $e^-$). $\Rightarrow$ Correct.

Step 5: Analyze the electron configuration for option (D).
$\text{La} (Z=57): [\text{Xe}]5d^16s^2$, so $\text{La}^{3+} = [\text{Xe}]$ ($54e^-$, diamagnetic).
$\text{Ce} (Z=58): [\text{Xe}]4f^15d^16s^2$, so $\text{Ce}^{4+} = [\text{Xe}]$ ($54e^-$, diamagnetic).
Isoelectronic ($54e^-$), but both are diamagnetic. $\Rightarrow$ Incorrect.