The Eu2+ ion is a strong reducing agent in spite of its ground state electronic configuration (outer

Q: Why do we remove electrons from $6s$ instead of $4f$ when $4f$ is higher in energy during filling?

While the $4f$ orbital is filled after the $6s$ orbital according to the Aufbau principle, the $6s$ orbital is physically further from the nucleus (it has a higher principal quantum number, $n=6$). When an atom ionizes, the outermost electrons are always the first to be removed.

Q: If $4f^7$ is exactly half-filled and very stable, why does $\text{Eu}^{2+}$ want to react at all?

Electronic configuration is only one part of the story. For lanthanides, the $+3$ oxidation state is thermodynamically the most stable state overall, especially in aqueous solutions due to high hydration energy. This thermodynamic drive overpowers the electronic stability of the $+2$ state.

Q: What does it mean to be a "strong reducing agent"?

A reducing agent is a substance that donates electrons to another substance. By giving away an electron, the reducing agent itself gets oxidized. Since $\text{Eu}^{2+}$ desperately wants to become $\text{Eu}^{3+}$, it readily gives away an electron, making it a strong reducing agent.

Q: How do I quickly remember the atomic number of Xenon?

Xenon is the noble gas at the end of the 5th period. A good trick is to remember the noble gas sequence: He (2), Ne (10), Ar (18), Kr (36), Xe (54), Rn (86).

Q: Does this rule apply to all lanthanides?

Yes! All lanthanides predominantly favor the $+3$ oxidation state. Any lanthanide that manages to form a $+2$ ion (like $\text{Eu}^{2+}$ or $\text{Sm}^{2+}$) will act as a reducing agent to reach $+3$, and any that forms a $+4$ ion (like $\text{Ce}^{4+}$) will act as an oxidizing agent to get back down to $+3$.

Question

The $	ext{Eu}^{2+}$ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]

Options:

(A) $4f^7 6s^2$

(B) $4f^6$

(C) $4f^7$

(D) $4f^6 6s^2$

Sigma Percentile · Accepted Answer

The journey into the f-block of the periodic table is like exploring a hidden realm of chemistry. Here, the rules of the game change slightly, and the elements exhibit fascinating behaviors driven by the subtle energies of their inner orbitals.

Today, we are going to unravel the mystery of Europium, a lanthanide that perfectly demonstrates the delicate balance between electronic stability and chemical reactivity.

### The Anatomy of Europium

Imagine you are building an atom of Europium from scratch. With an atomic number of $Z = 63$, we have a lot of electrons to place.

To make things easier, we start with the nearest noble gas, Xenon ($	ext{Xe}$), which neatly packs away $54$ electrons into a highly stable core.

This leaves us with $9$ valence electrons to distribute. According to the Aufbau principle and the specific energy levels of lanthanides, two of these electrons will fill the outermost $6s$ orbital.

The remaining $7$ electrons dive deep into the inner $4f$ subshell.

Why $7$? Because the $f$ subshell can hold a maximum of $14$ electrons. Having exactly $7$ electrons means the $4f$ subshell is **exactly half-filled**.

In the quantum world, half-filled subshells possess a special symmetrical stability. Thus, the ground state electronic configuration of a neutral Europium atom is:

$$ 	ext{Eu} = [	ext{Xe}] 4f^7 6s^2 $$

### The Birth of the Ion

Now, the question asks us about the $	ext{Eu}^{2+}$ ion. To form a cation, an atom must lose electrons.

**Here is where a classic trap lies.**

Even though the $4f$ electrons were the last to be added (energetically), they are not the first to leave. Electrons are always stripped from the outermost shell first—the shell with the highest principal quantum number, $n$.

For Europium, the outermost shell is $n=6$, specifically the $6s$ orbital.

So, to create $	ext{Eu}^{2+}$, we pluck away the two $6s$ electrons.

What remains is the Xenon core and the perfectly half-filled $4f$ subshell:

$$ 	ext{Eu}^{2+} = [	ext{Xe}] 4f^7 $$

### The Paradox of Stability

You might look at the $4f^7$ configuration and think, "Wow, that is incredibly stable! It must be perfectly happy as $	ext{Eu}^{2+}$."

And you would be partially right. The half-filled $4f^7$ state *is* stable, which is why $	ext{Eu}^{2+}$ can exist.

**However, there is a bigger force at play.**

In the world of lanthanides, the $+3$ oxidation state is the undisputed king. It is the thermodynamic "sink" for these elements, especially in aqueous solutions, driven by high hydration enthalpies.

Because the $+3$ state is so overwhelmingly favored, $	ext{Eu}^{2+}$ feels a strong chemical urge to lose one more electron to reach that $+3$ state:

$$ 	ext{Eu}^{2+} ightarrow 	ext{Eu}^{3+} + e^- $$

By losing an electron, $	ext{Eu}^{2+}$ undergoes oxidation. And what do we call a species that easily oxidizes itself to reduce something else? A **strong reducing agent**.

This beautifully explains the premise of the question: $	ext{Eu}^{2+}$ is a strong reducing agent *in spite* of its stable $4f^7$ ground state configuration.

The correct outermost electronic configuration is simply $4f^7$, making option (c) the perfect answer.

Animated Solution for Chemistry - d and f-Block Elements: The $Eu^{2 +}$ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]

Select Answer:

Visualized Solution

\text{Understanding Europium}

\text{Electronic Configuration of Eu}

\text{Formation of } \text{Eu}^{2+}

\text{Electronic Configuration of } \text{Eu}^{2+}

\text{Stability and Reducing Nature}

\text{Final Answer}

The Sigma Insight: Inner Transition Elements

The Anatomy of Europium

The Birth of the Ion

The Paradox of Stability

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