Animated Solution for Chemistry - d and f-Block Elements: The spin only magnetic moments (in BM) for free $T{i}^{3+}$, $V^{2+}$, and $S{c}^{3+}$ ions respectively are (Atomic number: Sc = 21, Ti = 22, V = 23)

Welcome to a fascinating journey into the quantum world of transition metals! Today, we are going to unravel the mystery of the spin-only magnetic moment. Imagine you are a quantum detective, and your job is to find out how strongly these tiny ions interact with a magnetic field. The secret lies in their unpaired electrons.

When we talk about the magnetic properties of transition metal ions, we are essentially talking about the spin of their electrons. Every electron acts like a tiny bar magnet. When electrons are paired up in an orbital, their spins cancel each other out. One spins up, the other spins down, and the net magnetic effect is zero.

But when an electron is all alone in an orbital—an unpaired electron—its magnetic field is uncompensated. The more unpaired electrons an ion has, the stronger its magnetic moment. This is the core physical reality we are dealing with.

To quantify this magnetic strength, we use a beautiful and elegant mathematical tool known as 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. This formula is our master key. Our entire mission now boils down to one simple task: finding the exact value of $n$ for each given ion. Let's take them one by one.

Our first candidate is the Titanium ion, $T{i}^{3+}$. To find its unpaired electrons, we must first look at the neutral Titanium atom.

Titanium has an atomic number of 22. Its ground-state electronic configuration is:

Now, the $+3$ charge means the atom has lost three electrons. Where do they go from? They always leave the outermost shell first. So, we strip away the two electrons from the $4s$ orbital, and then we take one more from the $3d$ orbital.

This leaves us with:

Look at that! We have exactly one electron sitting in the $3d$ subshell. Since it's the only one, it must be unpaired. Therefore, $n=1$.

Let's plug this into our master formula:

The square root of 3 is approximately $1.73$. So, the magnetic moment for $T{i}^{3+}$ is $1.73\text{ BM}$.

Next up is Vanadium, $V^{2+}$. Vanadium sits right next to Titanium on the periodic table, with an atomic number of 23.

Its neutral configuration is:

The $+2$ charge tells us to remove two electrons. Again, we take them from the outermost $4s$ orbital.

This gives us the ion's configuration:

We have three electrons in the $3d$ subshell. According to Hund's Rule of Maximum Multiplicity, electrons will fill degenerate orbitals singly before they start pairing up. So, all three of these electrons are unpaired. This means $n=3$.

Let's substitute this into our formula:

We know that the square root of 16 is 4, so the square root of 15 must be just a little bit less than 4. Calculating it gives us approximately $3.87$. Thus, the magnetic moment for $V^{2+}$ is $3.87\text{ BM}$.

Finally, we arrive at Scandium, $S{c}^{3+}$. Scandium is the first element of the 3d transition series, with an atomic number of 21.

Its neutral configuration is:

The $+3$ charge means we need to remove three electrons. We take two from the $4s$ orbital and the single one from the $3d$ orbital.

What are we left with?

The $3d$ subshell is completely empty! There are absolutely no electrons, which means there are zero unpaired electrons. So, $n=0$.

Plugging this into our formula:

With no unpaired electrons, the magnetic moment is exactly $0\text{ BM}$. This ion is perfectly diamagnetic.

We have successfully calculated the magnetic moments for all three ions: - $T{i}^{3+}$ has a moment of $1.73\text{ BM}$. - $V^{2+}$ has a moment of $3.87\text{ BM}$. - $S{c}^{3+}$ has a moment of $0\text{ BM}$.

The question asks for the values in the specific order of $T{i}^{3+}$, $V^{2+}$, and $S{c}^{3+}$.

Matching our results to the given options, we get the sequence: 1.73, 3.87, 0.

This corresponds perfectly to option (b).

Always remember, the key to mastering these problems is writing the electronic configuration carefully and remembering that electrons are always removed from the outermost $ns$ orbital before the $(n-1)d$ orbital. Keep practicing, and you'll be able to solve these in your head!