Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The spin-only magnetic moment value of $B_2^{+}$ species is ......... $\times 10^{-2}$ BM. (Nearest integer) [Given, $\sqrt{3}=1.73$]

Imagine you are shrinking down to the subatomic level, staring at a $B_2^{+}$ ion. Our mission today is to uncover its spin-only magnetic moment.

This isn't just a random calculation; it is a beautiful demonstration of how the invisible quantum world dictates the physical properties we can measure in a lab.

To unlock this puzzle, we need our trusty mathematical tool. The spin-only magnetic moment, denoted by $\mu$, is given by the formula:

Here, $n$ represents the number of unpaired electrons, and BM stands for Bohr Magnetons, the standard unit for magnetic moment.

Our entire goal now boils down to finding the exact value of $n$. To do that, we must build the molecular orbital structure of the ion.

Before we can place electrons into orbitals, we need to know how many we have.

A neutral boron atom (atomic number 5) has 5 electrons. Since we have a diatomic molecule, $B_2$, that gives us $5+5=10$ electrons.

But wait! We are dealing with the $B_2^{+}$ cation. The positive charge means it has lost one electron.

Now comes the fun part: Molecular Orbital (MO) Theory.

For lighter elements like boron, carbon, and nitrogen, a phenomenon called s-p mixing occurs. This alters the standard energy order of the molecular orbitals, pushing the ${\sigma }_{2p_z}$ orbital higher in energy than the degenerate ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals.

Let's distribute our 9 electrons. The first 4 electrons fill the inner core orbitals:

This leaves us with $9-4=5$ valence electrons to play with.

Moving up the energy ladder, we fill the 2s bonding and anti-bonding orbitals:

That accounts for 4 of our valence electrons. We have exactly 1 electron left.

Following Hund's rule and the Aufbau principle, this final, lone electron enters the next available energy level, which is the ${\pi }_{2p_x}$ orbital:

Looking at our completed electron configuration, it is clear that there is exactly 1 unpaired electron.

Let's substitute this back into our master equation:

The problem kindly provides the value of $\sqrt{3}$ as $1.73$.

However, the question asks for the answer in a specific format: $x\times 10^{-2}$.

By shifting the decimal point two places to the right, we can rewrite our answer:

Therefore, our final integer value is 173.

This problem is a fantastic reminder of how a single, lonely electron in a quantum orbital can give rise to measurable magnetic properties in the macroscopic world!