Animated Solution for Chemistry - d and f-Block Elements: The effect of lanthanoid contraction in the lanthanoid series of elements by and large means

Understanding Lanthanoid Contraction

When we dive into the f-block of the periodic table, specifically the lanthanoid series, we encounter a fascinating and highly consequential phenomenon known as Lanthanoid Contraction. But what exactly is it, and why does it happen? Let's break it down step by step.

Imagine you are building an atom from the lanthanoid series, starting from Cerium ($Z=58$) and moving towards Lutetium ($Z=71$). As you move from one element to the next, the atomic number increases by one. This means you are adding one proton to the nucleus and one electron to the electron cloud.

Crucially, these new electrons are not added to the outermost shell. Instead, they are buried deep within the atom, entering the inner $4f$ subshell, while the outermost shell remains $6s^2$.

Poor Shielding of $4f$ Orbitals

In a multi-electron atom, inner electrons act like a shield, protecting the outer electrons from the full attractive force of the positively charged nucleus. This is known as the shielding effect or screening effect. The effectiveness of this shield depends on the shape of the orbital. The order of shielding ability is $s>p>d>f$.

Here lies the core of the problem: the $4f$ orbitals have a highly diffused and complex shape. They are terrible at shielding! Think of it like trying to block the blazing sun with a fishing net instead of a solid umbrella. The outer $6s$ electrons are left highly exposed to the nucleus.

The actual pull felt by an outer electron is called the Effective Nuclear Charge ($Z_{\text{eff}}$), which is mathematically expressed as:

Where $Z$ is the actual nuclear charge (number of protons) and $\sigma$ is the shielding constant. As we move across the lanthanoid series, $Z$ increases steadily. However, because the $4f$ electrons are so poor at shielding, $\sigma$ does not increase enough to compensate for the growing $Z$.

As a result, the effective nuclear charge ($Z_{\text{eff}}$) increases significantly across the series.

Shrinking Radii

What happens when the nucleus pulls harder on the outer electrons? The entire electron cloud gets yanked inward. This strong inward pull draws the outer electron shells closer to the nucleus, causing the atom to shrink.

This steady, progressive decrease in size is exactly what we call Lanthanoid Contraction. Because this fundamental electrostatic pull affects the entire electron cloud, it causes a decrease in both the atomic radii and the ionic radii (such as the common $M^{3+}$ ions) of the lanthanoid elements.

Therefore, the correct answer to our question is that lanthanoid contraction means a decrease in both atomic and ionic radii.