Animated Solution for Chemistry - d and f-Block Elements: The pair(s) of diamagnetic ions is(are)

Welcome to a fascinating exploration of the f-block elements! In this problem, we are tasked with identifying the pairs of diamagnetic ions from a given list. To solve this, we must first understand what makes an ion diamagnetic or paramagnetic.

Imagine an atom as a tiny magnet. If all of its electrons are paired up in their respective orbitals, their individual magnetic fields perfectly cancel each other out. This state of complete pairing makes the substance diamagnetic. However, if even a single electron is left unpaired, its magnetic field remains uncancelled, rendering the substance paramagnetic. Therefore, our mission is clear: we must determine the exact electronic configurations of these ions and count their unpaired electrons.

The lanthanides, or the 4f series elements, generally follow the electronic configuration pattern of $[\text{Xe}]4f^{1-14}5d^{0-1}6s^2$. The key to finding the correct ionic configuration lies in the order of electron removal during ionization.

When an atom loses electrons to form a cation, the electrons are always removed from the outermost principal quantum shell first. This means we must strip electrons from the $6s$ orbital first, followed by the $5d$ orbital, and finally the $4f$ orbital. Let's apply this golden rule to our suspects.

Option by Option

Let's test the first pair in Option (A): ${\text{La}}^{3+}$ and ${\text{Ce}}^{4+}$. Lanthanum ($\text{La}$) has an atomic number of 57. Its neutral configuration is $[\text{Xe}]5d^{1}6s^2$. Removing three electrons (two from $6s$ and one from $5d$) gives us ${\text{La}}^{3+}$ with a configuration of $[\text{Xe}]4f^0$. Cerium ($\text{Ce}$), with atomic number 58, has a neutral configuration of $[\text{Xe}]4f^{1}5d^{1}6s^2$. Removing four electrons leaves us with ${\text{Ce}}^{4+}$ which is also $[\text{Xe}]4f^0$. Both ions have completely empty valence shells, meaning zero unpaired electrons. Thus, this pair is diamagnetic!

Moving to Option (B): ${\text{Yb}}^{2+}$ and ${\text{Lu}}^{3+}$. Ytterbium ($\text{Yb}$, $Z=70$) is $[\text{Xe}]4f^{14}6s^2$. Removing two $6s$ electrons gives ${\text{Yb}}^{2+}$ as $[\text{Xe}]4f^{14}$. Lutetium ($\text{Lu}$, $Z=71$) is $[\text{Xe}]4f^{14}5d^{1}6s^2$. Removing three electrons leaves ${\text{Lu}}^{3+}$ as $[\text{Xe}]4f^{14}$. Both ions have a fully filled $4f$ subshell, meaning all electrons are perfectly paired. This pair is also diamagnetic.

Now let's check Option (C): ${\text{La}}^{2+}$ and ${\text{Ce}}^{3+}$. For ${\text{La}}^{2+}$, we only remove the two $6s$ electrons, leaving the configuration as $[\text{Xe}]5d^1$. That is one unpaired electron! For ${\text{Ce}}^{3+}$, removing three electrons leaves $[\text{Xe}]4f^1$. Another unpaired electron! Since both possess unpaired electrons, they are paramagnetic.

Finally, Option (D): ${\text{Yb}}^{3+}$ and ${\text{Lu}}^{2+}$. ${\text{Yb}}^{3+}$ requires removing three electrons, leaving it at $[\text{Xe}]4f^{13}$. This means one electron is unpaired in the $f$ subshell. ${\text{Lu}}^{2+}$ means removing only the two $6s$ electrons, leaving $[\text{Xe}]4f^{14}5d^1$. That $5d$ electron is unpaired! Both of these ions are also paramagnetic.

Summarizing our findings, the pairs in options (A) and (B) have zero unpaired electrons, making them completely diamagnetic. Therefore, the correct answers are (A) and (B).

As a bonus thought, if you ever need to calculate the magnetic moment of the paramagnetic ions we found, you can use the spin-only formula:

For our paramagnetic ions with $n=1$ unpaired electron, the magnetic moment would be $\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\text{ BM}$. Keep this formula handy; it is a powerful tool in your chemistry arsenal!