Animated Solution for Chemistry - d and f-Block Elements: The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because

Why Actinoids Show More Oxidation States

When we dive into the fascinating world of the f-block elements, we encounter two distinct families: the Lanthanoids and the Actinoids. A classic observation in inorganic chemistry is that actinoids exhibit a much wider variety of oxidation states compared to their lanthanoid cousins. But why does this happen? The answer lies hidden in the spatial geometry of their orbitals.

Lanthanoids vs. Actinoids

Let's start by looking at the electronic configurations. For lanthanoids, the differentiating electrons enter the $4f$ subshell. Their general configuration is $[\text{Xe}]4f^{1-14}5d^{0-1}6s^2$. On the other hand, for actinoids, the electrons fill the $5f$ subshell, giving them a configuration of $[\text{Rn}]5f^{1-14}6d^{0-1}7s^2$.

While both series involve the filling of f-orbitals, the physical characteristics of the $4f$ and $5f$ orbitals are drastically different. This difference in radial distribution is the key to unlocking their chemical behavior.

Buried vs. Extended

Imagine plotting the radial probability distribution—a graph that tells us how likely we are to find an electron at a certain distance $r$ from the nucleus. If we trace the curve for the $4f$ orbitals, we notice something striking: the peak of the curve is very close to the nucleus.

The $4f$ orbitals are deeply "buried" within the atom. Because they are so close to the nucleus, they experience a very strong effective nuclear charge ($Z_{\text{eff}}$). The nucleus holds onto these $4f$ electrons with an iron grip, making it incredibly difficult for them to participate in chemical bonding. This is why lanthanoids predominantly show a $+3$ oxidation state, with only a few exceptions.

Breaking Free

Now, let's shift our focus to the actinoids and trace the curve for the $5f$ orbitals. The principal quantum number has increased from $n=4$ to $n=5$. As a result, the radial probability curve for the $5f$ orbitals is shifted significantly to the right.

The $5f$ orbitals extend much farther from the nucleus compared to the $4f$ orbitals. Because they are further away, the attractive pull from the nucleus is weaker. These electrons are more loosely bound and have a significant presence in the outer "bonding region" of the atom.

Because the $5f$ orbitals extend farther out, their electrons are readily available to interact with other atoms and participate in bond formation. This spatial extension allows actinoids to lose or share multiple electrons, leading to a rich chemistry with a wide array of oxidation states (ranging from $+3$ all the way up to $+7$ in elements like Neptunium and Plutonium).

Furthermore, the energy gap between the $5f$, $6d$, and $7s$ orbitals is remarkably small. This minimal energy difference makes it thermodynamically feasible to excite multiple electrons for bonding.

Therefore, the correct reasoning is that the $5f$ orbitals extend farther from the nucleus than the $4f$ orbitals, making option (d) the perfect answer.