Animated Solution for Chemistry - d and f-Block Elements: The electronic configuration of bivalent europium and trivalent cerium are (atomic number : $\text{Xe}=54,\text{Ce}=58,\text{Eu}=63$)

When we dive into the f-block of the periodic table, we enter a realm where the standard rules of electron filling often take a fascinating twist. The lanthanide series, spanning from Cerium ($Z=58$) to Lutetium ($Z=71$), is characterized by the progressive filling of the deeply buried $4f$ orbitals.

To master questions involving these elements, you must first anchor yourself to their general electronic configuration. All lanthanides build upon the stable noble gas core of Xenon ($[\text{Xe}]$, $Z=54$). The general template for their valence shell is:

Notice the subtle presence of the $5d$ orbital. Because the energy levels of the $4f$ and $5d$ subshells are incredibly close at the beginning of the series, electrons sometimes "spill over" into the $5d$ orbital before the $4f$ orbital is fully populated. This is the root of many classic exceptions in chemistry.

The Quest for Stability

Let's begin our analysis with Europium (Eu), which has an atomic number of $63$. Our first step is always to write the configuration for the neutral atom.

Subtracting the $54$ electrons of the Xenon core leaves us with $9$ valence electrons to distribute. Following the Aufbau principle, the lower-energy $6s$ orbital fills first, taking $2$ electrons. We are left with $7$ electrons. These remaining electrons drop perfectly into the $4f$ subshell.

Why is there no electron in the $5d$ orbital? The answer lies in the profound stability of half-filled subshells. The $f$-subshell can hold a maximum of $14$ electrons. By placing exactly $7$ electrons in the $4f$ orbital, Europium achieves a perfectly symmetrical, half-filled state ($4f^7$). This configuration maximizes exchange energy and minimizes inter-electronic repulsion, making it highly favorable.

Outside-In

The question asks for the configuration of bivalent Europium, denoted as ${\text{Eu}}^{2+}$. This means the neutral atom has lost two electrons.

Here is where many students make a critical error: they remove electrons in the reverse order of filling. Do not do this. When a transition or inner-transition metal ionizes, electrons are always removed from the outermost principal quantum shell first.

The principal quantum number ($n$) dictates the distance from the nucleus. In Europium, the $6s$ electrons ($n=6$) are physically further from the nucleus than the $4f$ electrons ($n=4$). Therefore, the $6s$ electrons experience less effective nuclear charge and are the first to be stripped away during ionization.

Removing the two $6s$ electrons leaves us with:

A Delicate Energy Balance

Now, let's shift our focus to Cerium (Ce), atomic number $58$. Subtracting the Xenon core leaves us with just $4$ valence electrons.

As always, $2$ electrons fill the $6s$ orbital. We have $2$ electrons left. You might intuitively expect both to go into the $4f$ orbital, giving $4f^2$. However, Cerium is a famous exception. Because the $4f$ and $5d$ energy levels are nearly degenerate (equal in energy) at this specific point in the periodic table, the remaining two electrons split up to minimize repulsion.

One electron enters the $5d$ orbital, and the other enters the $4f$ orbital. Thus, the neutral configuration is:

The problem requires the configuration for trivalent Cerium, or ${\text{Ce}}^{3+}$. We must remove three electrons from the neutral atom.

Applying our "outside-in" rule, we first remove the two electrons from the outermost $6s$ shell ($n=6$). We still need to remove one more electron. We then look to the next outermost shell, which is the $5d$ orbital ($n=5$). Removing the single $5d$ electron leaves us with just one electron remaining in the $4f$ orbital.

By carefully applying the rules of orbital filling, recognizing the stability of half-filled states, and strictly adhering to the outside-in rule for ionization, we arrive at the final answer. The configurations are $[\text{Xe}]4f^7$ and $[\text{Xe}]4f^1$, which corresponds perfectly to option (a).