Animated Solution for Chemistry - d and f-Block Elements: The lanthanoid that does not show $+4$ oxidation state is

When we dive into the f-block of the periodic table, we encounter the lanthanoids—a fascinating series of elements known for their characteristic $+3$ oxidation state. However, some of these elements like to break the rules and exhibit $+2$ or $+4$ states. The question asks us to identify the lanthanoid among Dysprosium (Dy), Cerium (Ce), Europium (Eu), and Terbium (Tb) that absolutely refuses to show a $+4$ oxidation state. To solve this, we must look at the blueprint of their atoms: their electronic configurations.

Let's put Europium (Eu) under the microscope. Its atomic number is $Z=63$. If we write out its ground state electronic configuration, we get:

Notice something special here? The $4f$ subshell has exactly $7$ electrons. Since an f-subshell can hold a maximum of $14$ electrons, Europium possesses an exactly half-filled $4f$ subshell. In the quantum world, half-filled and fully-filled subshells grant an atom exceptional thermodynamic stability due to maximum exchange energy and symmetrical electron distribution.

When Europium reacts and forms a $+2$ ion, it loses its two outermost electrons from the $6s$ orbital:

This leaves the highly stable $4f^7$ core completely intact. This is why $E{u}^{2+}$ is a well-known and relatively stable ion.

However, the $+3$ state is the hallmark of all lanthanoids. To reach the $+3$ state, Europium must sacrifice one of its precious $4f$ electrons:

Even though it loses that perfect half-filled stability, the overall energetics (like high hydration enthalpy in water or lattice energy in solids) make the $+3$ state the most common and stable state for Europium in nature.

Now, what if we try to push Europium further? What if we want to create $E{u}^{4+}$?

To do this, we would need to rip away a fourth electron, leaving a $4f^5$ configuration. But here is the catch: the nucleus is now holding onto those remaining electrons with a massive effective nuclear charge. The energy required to remove this fourth electron—the 4th Ionization Energy ($I{E}_4$)—is astronomically high, clocking in at around $4140\text{ kJ/mol}$.

There is simply no chemical environment, no oxidizing agent strong enough, to provide this massive amount of energy and stabilize a $E{u}^{4+}$ ion. Therefore, Europium does not show a $+4$ oxidation state.

For context, let's quickly look at the other options: Cerium (Ce): $[\text{Xe}]4f^{1}5d^{1}6s^2$. By losing 4 electrons, it forms $C{e}^{4+}=[\text{Xe}]4f^0$, achieving a highly stable noble gas configuration. Terbium (Tb): $[\text{Xe}]4f^{9}6s^2$. By losing 4 electrons, it forms $T{b}^{4+}=[\text{Xe}]4f^7$, achieving the highly stable half-filled configuration. Dysprosium (Dy):* $[\text{Xe}]4f^{10}6s^2$. It can form $D{y}^{4+}=[\text{Xe}]4f^8$ in certain highly stabilizing solid-state environments (like fluorides).

Thus, Europium stands alone in this list as the element that cannot reach the $+4$ state.