Animated Solution for Chemistry - d and f-Block Elements: The correct order of atomic radii is

The problem asks us to determine the correct decreasing order of atomic radii for four elements: Nitrogen ($N$), Cerium ($Ce$), Europium ($Eu$), and Holmium ($Ho$). At first glance, comparing a tiny non-metal with heavy f-block metals might seem like comparing apples to oranges. However, this is a classic test of your understanding of periodic trends, specifically the nuances of the lanthanide series.

Let's start by identifying where these elements live on the periodic table. Nitrogen is a p-block element located in the second period. It possesses only two principal electron shells ($n=2$).

On the other hand, Cerium (Atomic No. 58), Europium (Atomic No. 63), and Holmium (Atomic No. 67) are all heavy metals belonging to the lanthanide series in the f-block. They have electrons filling all the way up to the sixth principal shell ($n=6$).

Common sense dictates that an atom with only two shells will be drastically smaller than atoms with six shells. Therefore, Nitrogen is undeniably the smallest atom in this group. Its atomic radius is a mere $65\text{ pm}$. This immediately tells us that Nitrogen must be at the very end of our decreasing order.

Lanthanoid Contraction

Now, let's focus on the three lanthanides: Cerium, Europium, and Holmium. What is the general trend for atomic size across the lanthanide series?

As we move from left to right across the f-block, the atomic number increases. We are adding protons to the nucleus and electrons to the inner $4f$ subshell. However, the $4f$ orbitals have a very diffused shape, which means they are terrible at shielding the outer $6s$ electrons from the increasing positive charge of the nucleus.

This phenomenon is known as Lanthanoid Contraction. Because the effective nuclear charge increases steadily, the outer electrons are pulled closer to the nucleus, causing the atomic radius to shrink as we move across the series.

Following this logic, Cerium ($Z=58$), being earlier in the series, should be larger than Holmium ($Z=67$). Specifically, Cerium has a radius of $183\text{ pm}$, while Holmium has a radius of $176\text{ pm}$.

The Europium Anomaly

If we strictly followed the lanthanoid contraction trend, we would expect Europium ($Z=63$) to have a size somewhere between Cerium and Holmium. But chemistry loves its exceptions, and Europium is a major one!

Let's look at the electronic configuration of Europium:

Notice that $4f^7$ configuration? That is a perfectly half-filled subshell, which grants the atom an extraordinary amount of exchange energy and stability. Europium is very protective of this stable core.

In a solid metal lattice, atoms bond by releasing their valence electrons into a "sea" of delocalized electrons (metallic bonding). Most lanthanides release three electrons (two from $6s$ and one from $4f$ or $5d$) to form strong metallic bonds. Europium, however, refuses to break its stable $4f^7$ core. It only contributes its two $6s$ electrons to the metallic lattice.

Fewer delocalized electrons mean weaker metallic bonding. Because the atoms aren't pulled as tightly together by the electron sea, the metallic lattice expands. This causes Europium's atomic radius to balloon up to $199\text{ pm}$, making it significantly larger than both Cerium and Holmium!

Let's compile all our findings: 1. Europium ($Eu$) is the largest due to its anomalous weak metallic bonding ($199\text{ pm}$). 2. Cerium ($Ce$) is next, being an early lanthanide ($183\text{ pm}$). 3. Holmium ($Ho$) follows, being a later lanthanide affected by lanthanoid contraction ($176\text{ pm}$). 4. Nitrogen ($N$) is the smallest, having only two electron shells ($65\text{ pm}$).

Thus, the correct decreasing order of atomic radii is:

This perfectly matches option (c). Always remember to watch out for those half-filled and fully-filled orbital anomalies!