Animated Solution for Chemistry - d and f-Block Elements: The radius of ${\text{La}}^{3+}$ (atomic number : $\text{La}=57$) is $1.06\text{ A˚}$. Which one of the following given values will be closest to the radius of ${\text{Lu}}^{3+}$ (atomic number : $\text{Lu}=71$) ?

Imagine you are embarking on a journey across the periodic table, specifically through the fascinating landscape of the f-block elements. As you walk from left to right, from Lanthanum (La) to Lutetium (Lu), you might expect the atoms to grow larger. After all, with each step, you are adding more electrons and more protons. It seems like common sense that a heavier atom with more particles should take up more space.

However, nature often hides its most beautiful secrets behind counterintuitive phenomena. In the case of the lanthanide series, the atoms do not grow larger; they actually shrink! This peculiar behavior is one of the most famous and important concepts in inorganic chemistry, known as Lanthanide Contraction.

To truly understand why this happens, we need to dive deep into the quantum mechanical structure of the atom and explore the invisible forces at play between the nucleus and the electrons.

Lanthanum to Lutetium

Let us first orient ourselves in the periodic table. Lanthanum, with an atomic number of $Z=57$, marks the beginning of the $4f$ transition series. Lutetium, with an atomic number of $Z=71$, sits at the very end of this series.

Between Lanthanum and Lutetium, there are 14 elements. As we move from one element to the next, the atomic number increases by one. This means we are adding exactly one proton to the nucleus and one electron to the electron cloud.

But where do these new electrons go? They do not go into the outermost shell. Instead, they are buried deep within the atom, specifically in the $4f$ subshell. The filling of this inner $4f$ subshell is the root cause of the entire phenomenon.

Shielding Effect and Effective Nuclear Charge

To understand the shrinking of these atoms, we must first understand two critical concepts: the Shielding Effect and the Effective Nuclear Charge ($Z_{\text{eff}}$).

In a multi-electron atom, the positively charged nucleus pulls all the electrons towards itself. However, the electrons in the inner shells act like a shield or a screen. They repel the outermost electrons, effectively blocking a portion of the nuclear pull.

The actual, net attractive force that an outer electron feels from the nucleus is called the Effective Nuclear Charge. It is mathematically expressed as:

Where $Z$ is the actual nuclear charge (number of protons) and $\sigma$ is the shielding constant (the screening effect of inner electrons).

If the inner electrons are good at shielding, the outer electrons feel less pull and the atom expands. If the inner electrons are poor at shielding, the outer electrons feel a stronger pull and the atom contracts.

The Diffused 4f Orbitals

This brings us to the heart of the matter. Not all orbitals shield equally well. The shape of the orbital dictates its shielding power.

The $s$-orbitals are spherical and dense, making them excellent shields. The $p$-orbitals are dumbbell-shaped and are also relatively good shields. The $d$-orbitals are more complex and offer poorer shielding.

But the $f$-orbitals? They are highly diffused, spread out over a large volume of space with complex, multi-lobed shapes. Because they are so spread out, they are terrible at shielding the outer electrons from the pull of the nucleus.

Imagine trying to block the light of a powerful flashlight with a thick, solid piece of cardboard (an $s$-orbital). It works perfectly. Now imagine trying to block that same light with a highly porous, loosely woven net (an $f$-orbital). The light easily passes through.

Lanthanide Contraction

Now, let us put it all together. As we move from Lanthanum to Lutetium, we are adding protons to the nucleus, which increases the nuclear charge $Z$. At the same time, we are adding electrons to the $4f$ subshell.

Because the $4f$ electrons are incredibly poor shields, the shielding constant $\sigma$ does not increase as much as the nuclear charge $Z$ does. As a result, the Effective Nuclear Charge ($Z_{\text{eff}}$) experiences a net increase.

With a stronger effective pull from the nucleus, the outermost electrons are dragged closer to the center. The entire electron cloud is pulled in tighter.

Therefore, as we progress across the lanthanide series, the atomic and ionic radii steadily and continuously decrease. This gradual decrease in size is what we call Lanthanide Contraction.

Beyond the Lanthanides

The story of lanthanide contraction does not end with Lutetium. Its effects ripple through the rest of the periodic table, profoundly impacting the elements that follow it.

Consider the transition metals in the $d$-block. Normally, as you move down a group from the $3d$ series to the $4d$ series, the atomic size increases significantly because a new principal quantum shell is added. You would expect a similar jump in size when moving from the $4d$ series to the $5d$ series.

However, the $5d$ series elements (like Hafnium, Tantalum, Tungsten) come immediately after the lanthanide series. Because the lanthanide contraction has already shrunk the atoms so drastically, the expected increase in size from adding a new shell is almost perfectly canceled out by the contraction.

As a result, the elements of the $4d$ and $5d$ series have nearly identical atomic radii! For example, Zirconium ($4d$) and Hafnium ($5d$) are almost exactly the same size. This makes their chemical properties so incredibly similar that separating them in nature is notoriously difficult. This beautiful symmetry is a direct consequence of the poor shielding of those buried $4f$ electrons.

A Simple Deduction

Armed with this profound understanding of atomic physics, solving the given problem becomes a trivial exercise in logic.

The question provides us with the ionic radius of Lanthanum in its $+3$ oxidation state:

We are asked to find the closest value for the ionic radius of Lutetium in the same $+3$ oxidation state ($r_{{\text{Lu}}^{3+}}$).

Since Lutetium ($Z=71$) comes after Lanthanum ($Z=57$) in the lanthanide series, it has experienced the full brunt of the lanthanide contraction. The poor shielding of its fourteen $4f$ electrons has allowed its nucleus to pull its outer electrons much closer.

Therefore, it is an absolute physical certainty that the radius of ${\text{Lu}}^{3+}$ must be smaller than the radius of ${\text{La}}^{3+}$.

Now, we simply look at the given options: (a) $1.60\text{ A˚}$ (Larger, incorrect) (b) $1.40\text{ A˚}$ (Larger, incorrect) (c) $1.06\text{ A˚}$ (Equal, incorrect) (d) $0.85\text{ A˚}$ (Smaller, correct!)

There is only one option that is mathematically smaller than $1.06\text{ A˚}$. Thus, without needing to memorize any specific data tables, our conceptual understanding leads us directly to the correct answer: $0.85\text{ A˚}$.