Animated Solution for Chemistry - Coordination Compounds: Considering that ${\Delta }_0>P$, the magnetic moment (in BM) of $[Ru(H_{2}O{)}_6{]}^{2+}$ would be ......... .

Welcome to the fascinating world of Coordination Chemistry, where the invisible arrangement of electrons dictates the physical properties of matter! In this problem, we are tasked with finding the spin-only magnetic moment of the octahedral complex $[Ru(H_{2}O{)}_6{]}^{2+}$.

To embark on this journey, our first step is to identify the oxidation state of the central metal ion, Ruthenium ($Ru$). The ligands surrounding the metal are water ($H_{2}O$) molecules. Since water is a neutral molecule, it contributes zero charge to the complex. Therefore, the entire $+2$ charge of the coordination sphere must belong entirely to the Ruthenium ion.

We are dealing with the $R{u}^{2+}$ ion.

Now, we need to determine the electronic configuration of $R{u}^{2+}$. Ruthenium is a d-block transition metal located in the 4d series, right below Iron, with an atomic number of $Z=44$.

If you follow the standard Aufbau principle blindly, you might guess its ground state configuration to be $[Kr]4d^{6}5s^2$. However, Ruthenium is one of those classic exceptions! Due to complex electron-electron repulsions and nuclear shielding effects in the heavier transition metals, its actual ground state configuration is:

To form the $R{u}^{2+}$ cation, we must strip away two electrons. We always remove electrons from the outermost shell first. So, we take one electron from the $5s$ orbital and one from the $4d$ orbital. This leaves us with:

We have exactly six d-electrons to play with.

According to Crystal Field Theory (CFT), when the six water ligands approach the central $R{u}^{2+}$ ion to form an octahedral geometry, their negative electron clouds repel the electrons in the metal's d-orbitals. Because the d-orbitals have different shapes and orientations, they don't experience this repulsion equally.

The five degenerate d-orbitals split into two distinct energy levels: 1. A lower energy set of three orbitals called $t_{2g}$. 2. A higher energy set of two orbitals called $e_g$.

The energy gap between these two levels is the Crystal Field Splitting Energy, denoted as ${\Delta }_0$.

Now, we face a critical decision: how do we distribute our six electrons? When we reach the fourth electron, it has a choice. It can either pair up with an electron in the lower $t_{2g}$ level (which costs Pairing Energy, $P$), or it can jump the gap to the higher $e_g$ level (which costs ${\Delta }_0$).

The problem explicitly hands us the key: ${\Delta }_0>P$.

This inequality tells us that the energy required to jump the gap is greater than the energy required to pair up. It's like choosing between paying a massive toll to cross a bridge versus paying a small fee to share a room on this side. The electrons will take the path of least resistance and pair up!

Because pairing is energetically favorable, this complex will form a low-spin state. We will completely fill the lower $t_{2g}$ orbitals before even looking at the $e_g$ level.

We have 6 electrons. The $t_{2g}$ level has 3 orbitals, which can hold a maximum of $3\times 2=6$ electrons.

So, all six of our electrons will perfectly pair up in the $t_{2g}$ level. The electronic configuration becomes:

Let's count the number of unpaired electrons ($n$). Since every single electron has found a partner, there are absolutely no unpaired electrons left.

The magnetic properties of a complex are directly tied to its unpaired electrons. Each unpaired electron acts like a tiny bar magnet. We calculate the spin-only magnetic moment ($\mu$) using the formula:

Substituting our value of $n=0$:

The magnetic moment is exactly 0 Bohr Magnetons. Because all the electrons are paired, their individual magnetic fields cancel each other out perfectly, rendering the entire complex diamagnetic.