Animated Solution for Chemistry - d and f-Block Elements: Which one of the following lanthanoids does not form ${\text{MO}}_2$? [$\text{M}$ is lanthanoid metal]

The question asks us to identify which lanthanoid does not form an oxide of the formula ${\text{MO}}_2$. The first step is to determine the oxidation state of the metal $\text{M}$ in this compound.

Oxygen typically exhibits an oxidation state of $-2$. Since the molecule ${\text{MO}}_2$ is electrically neutral, the sum of the oxidation states must be zero. Let the oxidation state of the metal be $x$.

So, the question boils down to a fundamental property of f-block elements: Which of the given lanthanoids does not exhibit a $+4$ oxidation state?

For all lanthanoids, the most common and thermodynamically stable oxidation state is $+3$. This is the hallmark of the lanthanide series. However, chemistry is rarely without exceptions. Some lanthanoids can exhibit a $+4$ oxidation state, but they only do so if losing that fourth electron leads to a particularly stable electronic configuration.

What counts as stable? An empty f-subshell ($f^0$), a half-filled f-subshell ($f^7$), or a configuration very close to these. The elements that can achieve this and form $+4$ oxides are: Cerium (Ce): Forms ${\text{Ce}}^{4+}$ ($f^0$) Praseodymium (Pr): Forms ${\text{Pr}}^{4+}$ ($f^1$) Neodymium (Nd): Forms ${\text{Nd}}^{4+}$ ($f^2$) Terbium (Tb): Forms ${\text{Tb}}^{4+}$ ($f^7$) Dysprosium (Dy)*: Forms ${\text{Dy}}^{4+}$ ($f^8$)

Looking at our options—Praseodymium (Pr), Dysprosium (Dy), Neodymium (Nd), and Ytterbium (Yb)—we can see that Pr, Dy, and Nd are on our list of $+4$ forming elements. They readily form ${\text{PrO}}_2$, ${\text{DyO}}_2$, and ${\text{NdO}}_2$. This leaves Ytterbium as the odd one out.

Why does Ytterbium refuse to form a $+4$ state? The answer lies in its electronic configuration. Ytterbium has an atomic number of $70$. Its ground state electronic configuration is:

Ytterbium can easily lose its two outermost $6s$ electrons to form the ${\text{Yb}}^{2+}$ ion. This ion is incredibly stable because it possesses a fully filled $4f^{14}$ subshell. It can also lose one more electron to form the common ${\text{Yb}}^{3+}$ state ($4f^{13}$).

However, to form a $+4$ state, Ytterbium would have to lose a fourth electron. This would require breaking into the deeply buried, perfectly symmetrical, and highly stable $4f^{14}$ core. The ionization energy required to remove an electron from a fully filled f-subshell is astronomically high. Therefore, under normal chemical conditions, ${\text{Yb}}^{4+}$ cannot exist, and consequently, Ytterbium will never form ${\text{YbO}}_2$.

A Quick Note on Stability: Even for the lanthanoids that do form $+4$ states (like Ce, Pr, Nd, Tb, Dy), the $+3$ state remains the most stable. Because of this, compounds containing ${\text{M}}^{4+}$ ions have a strong tendency to gain an electron and revert to the $+3$ state. This makes them strong oxidizing agents.