Animated Solution for Chemistry - d and f-Block Elements: Consider the hydrated ions of ${\text{Ti}}^{2+}$, ${\text{V}}^{2+}$, ${\text{Ti}}^{3+}$ and ${\text{Sc}}^{3+}$. The correct order of their spin-only magnetic moment is

Spin-Only Magnetic Moment

When dealing with transition metal ions, one of the most fundamental properties we analyze is their magnetic behavior. This behavior is primarily dictated by the number of unpaired electrons present in their d-orbitals. The relationship is beautifully captured by the spin-only magnetic moment formula:

Here, $\mu$ represents the magnetic moment measured in Bohr Magnetons (BM), and $n$ is the number of unpaired electrons. The logic is straightforward: the greater the number of unpaired electrons, the higher the magnetic moment. Our task is simply to determine the electronic configuration of each given ion, count the unpaired electrons, and compare them.

Let's break down each ion step-by-step. Remember the golden rule of ionization for transition metals: electrons are always removed from the outermost $4s$ orbital before the $3d$ orbital.

1. Scandium Ion (${\text{Sc}}^{3+}$) Scandium has an atomic number of $Z=21$. Its neutral ground state configuration is $[\text{Ar}]3d^{1}4s^2$. To form the ${\text{Sc}}^{3+}$ ion, we must remove three electrons. We take two from the $4s$ orbital and one from the $3d$ orbital. This leaves us with an empty d-subshell: $[\text{Ar}]3d^0$. Since there are no unpaired electrons ($n=0$), the magnetic moment is:

2. Titanium(III) Ion (${\text{Ti}}^{3+}$) Titanium has an atomic number of $Z=22$, giving it a neutral configuration of $[\text{Ar}]3d^{2}4s^2$. Removing three electrons (two from $4s$, one from $3d$) gives the ${\text{Ti}}^{3+}$ configuration: $[\text{Ar}]3d^1$. Here, we have exactly one unpaired electron ($n=1$). Plugging this into our formula:

3. Titanium(II) Ion (${\text{Ti}}^{2+}$) Starting again from neutral Titanium ($[\text{Ar}]3d^{2}4s^2$), we only need to remove two electrons to form ${\text{Ti}}^{2+}$. We strip away the two $4s$ electrons, leaving the $3d$ electrons untouched. The configuration is $[\text{Ar}]3d^2$. According to Hund's rule, these two electrons will occupy separate orbitals, meaning they are both unpaired ($n=2$).

4. Vanadium(II) Ion (${\text{V}}^{2+}$) Vanadium has an atomic number of $Z=23$, so its neutral configuration is $[\text{Ar}]3d^{3}4s^2$. To form ${\text{V}}^{2+}$, we remove the two $4s$ electrons, resulting in $[\text{Ar}]3d^3$. All three electrons in the d-subshell will be unpaired ($n=3$).

Now that we have the magnetic moments for all four ions, let's line them up: - $\mu ({\text{Sc}}^{3+})=0$ - $\mu ({\text{Ti}}^{3+})=\sqrt{3}$ - $\mu ({\text{Ti}}^{2+})=\sqrt{8}$ - $\mu ({\text{V}}^{2+})=\sqrt{15}$

It is evident that as the number of unpaired electrons increases, the magnetic moment increases. Therefore, the correct increasing order of their spin-only magnetic moments is:

This perfectly matches option (a).