Animated Solution for Chemistry - d and f-Block Elements: Which of the following ions has the maximum magnetic moment?

The quest to find the ion with the maximum magnetic moment is essentially a hunt for the most "unpaired" electrons. In the quantum world, every electron acts like a tiny bar magnet due to its spin. When electrons pair up in an orbital, their opposite spins cancel each other out, leaving no net magnetic field. But when they sit alone—unpaired—their magnetic fields add up!

To quantify this magnetic strength, we use the spin-only magnetic moment formula:

Here, $n$ represents the number of unpaired electrons, and the unit is the Bohr Magneton (BM). The math is straightforward: the larger the value of $n$, the greater the magnetic moment $\mu$. Our mission is to write the electronic configuration for each ion and count the unpaired electrons.

Let's start with the first contender, ${\text{Mn}}^{2+}$. Manganese has an atomic number of $Z=25$. Its neutral ground-state configuration is $[\text{Ar}]4s^{2}3d^5$.

When forming a cation, transition metals always lose their outermost $4s$ electrons before touching the $3d$ electrons. So, removing two electrons gives us:

Following Hund's Rule of Maximum Multiplicity, we place one electron in each of the five $3d$ orbitals before any pairing occurs. This gives us exactly $5$ unpaired electrons ($n=5$).

Next up is ${\text{Fe}}^{2+}$. Iron ($Z=26$) starts as $[\text{Ar}]4s^{2}3d^6$. Losing the two $4s$ electrons leaves us with $3d^6$. When we distribute six electrons across five orbitals, one pair must form, leaving us with $4$ unpaired electrons ($n=4$).

What about ${\text{Ti}}^{2+}$? Titanium ($Z=22$) is $[\text{Ar}]4s^{2}3d^2$. Removing the $4s$ electrons leaves $3d^2$. These two electrons will occupy separate orbitals, giving us $2$ unpaired electrons ($n=2$).

Finally, we have ${\text{Cr}}^{2+}$. Chromium ($Z=24$) is a famous exception! Its neutral configuration is $[\text{Ar}]4s^{1}3d^5$ due to the stability of a half-filled $d$-subshell. To form ${\text{Cr}}^{2+}$, we remove the single $4s$ electron and one $3d$ electron, leaving $3d^4$. This results in $4$ unpaired electrons ($n=4$).

Let's summarize our findings: - ${\text{Mn}}^{2+}$ has $n=5$ - ${\text{Fe}}^{2+}$ has $n=4$ - ${\text{Ti}}^{2+}$ has $n=2$ - ${\text{Cr}}^{2+}$ has $n=4$

Manganese takes the crown with the highest number of unpaired electrons! If we plug $n=5$ into our formula, we get:

This is the maximum possible spin-only magnetic moment for any $3d$ transition metal ion. The sheer elegance of Hund's rule and the Aufbau principle leads us directly to the correct answer.