Animated Solution for Chemistry - d and f-Block Elements: Which one of the following lanthanides exhibits $+2$ oxidation state with diamagnetic nature ? (Given, $Z$ for $\text{Nd}=60$, $\text{Yb}=70$, $\text{La}=57$, $\text{Ce}=58$)

Imagine you are a quantum detective, tasked with finding a very specific suspect among the lanthanides. Our target must satisfy two strict conditions: it must form a stable $+2$ oxidation state, and it must be diamagnetic.

What does it mean to be diamagnetic? In the quantum world, it means perfect harmony. Every single electron must have a partner with an opposite spin. There can be no lone wolves, no unpaired electrons.

To find our suspect, we need to look at their genetic makeup—their electronic configurations. The lanthanides are the elements where the $4f$ subshell is progressively filled. Their general electronic configuration is $[\text{Xe}]4f^{1-14}5d^{0-1}6s^2$.

When these metals ionize to form a $+2$ cation, they don't lose their inner $4f$ electrons first. Instead, they shed their outermost valence electrons, which reside in the $6s$ orbital. So, our $+2$ ion will generally have the configuration $[\text{Xe}]4f^n$.

Let's bring our prime suspect, Ytterbium ($\text{Yb}$), to the interrogation room. With an atomic number of $70$, its neutral ground state configuration is $[\text{Xe}]4f^{14}6s^2$. Notice something beautiful here? The $4f$ subshell is completely filled to its maximum capacity of $14$ electrons.

When Ytterbium forms a $+2$ ion, it gracefully lets go of its two $6s$ electrons. The resulting ${\text{Yb}}^{2+}$ ion has the configuration $[\text{Xe}]4f^{14}$.

Perfect Pairing

Let's visualize this $4f^{14}$ state. The $f$ subshell consists of $7$ distinct orbitals. According to Hund's rule and the Pauli exclusion principle, these $14$ electrons will perfectly fill all $7$ orbitals, with two electrons in each—one spinning up, one spinning down.

Because every single electron is paired, the magnetic fields generated by their spins cancel each other out perfectly. This makes the ${\text{Yb}}^{2+}$ ion diamagnetic.

If we were to investigate the other suspects—Neodymium (${\text{Nd}}^{2+}$), Lanthanum (${\text{La}}^{2+}$), and Cerium (${\text{Ce}}^{2+}$)—we would find that their $4f$ or $5d$ subshells are only partially filled. They harbor unpaired electrons, making them paramagnetic. Thus, Ytterbium is the only one that fits the profile perfectly!