Animated Solution for Chemistry - Coordination Compounds: The value of the 'spin only' magnetic moment for one of the following configurations is $2.84\text{ BM}$. The correct one is

The mystery of the magnetic moment is one of the most fascinating puzzles in coordination chemistry. When we are given a spin-only magnetic moment of $2.84\text{ BM}$, it's like being handed a fingerprint. Our job is to find the exact $d$-orbital configuration that matches this unique print. Let's embark on this quantum detective work!

The journey begins with the spin-only magnetic moment formula, a beautiful equation that connects the macroscopic magnetic properties of a complex to the microscopic world of unpaired electrons. The formula is: where $n$ represents the number of unpaired electrons.

We are given $\mu =2.84\text{ BM}$. By substituting this value into our formula, we get: Squaring both sides gives us approximately $8=n(n+2)$. Solving this simple quadratic equation reveals that $n=2$. This is our golden clue! We are looking for a configuration that houses exactly two unpaired electrons.

Now, let's interrogate our suspects—the given options—using the principles of Crystal Field Theory (CFT). In an octahedral field, the five degenerate $d$-orbitals split into a lower energy $t_{2g}$ set and a higher energy $e_g$ set. The gap between them is the crystal field splitting energy, ${\Delta }_o$.

Suspect 1: $d^5$ in a strong ligand field A strong field means the splitting energy (${\Delta }_o$) is greater than the pairing energy ($P$). The electrons are "afraid" of the high jump to the $e_g$ level, so they prefer to pair up in the lower $t_{2g}$ orbitals. For $d^5$, all five electrons crowd into the $t_{2g}$ level, resulting in a $t_{2g}^5{e}_g^0$ configuration. This leaves only 1 unpaired electron. Not our match!

Suspect 2: $d^3$ in any field For a $d^3$ configuration, the first three electrons comfortably occupy the three separate $t_{2g}$ orbitals singly, following Hund's rule. This gives us 3 unpaired electrons, regardless of whether the field is strong or weak. Still not the one.

Suspect 3: $d^4$ in a weak ligand field In a weak field, the splitting energy (${\Delta }_o$) is small—less than the pairing energy ($P$). The fourth electron finds it easier to jump the small gap to the $e_g$ orbital rather than pair up in the $t_{2g}$ orbital. This leads to a $t_{2g}^3{e}_g^1$ configuration, giving us 4 unpaired electrons. We are getting colder!

Suspect 4: $d^4$ in a strong ligand field Finally, we examine $d^4$ in a strong field. Here, the splitting energy (${\Delta }_o$) is massive. The fourth electron looks at the huge jump to the $e_g$ level and decides it's much more energetically favorable to pair up in the lower $t_{2g}$ level.

This results in a $t_{2g}^4{e}_g^0$ configuration. One orbital in the $t_{2g}$ set has a paired set of electrons, while the other two orbitals have one electron each. This leaves us with exactly 2 unpaired electrons!

Since $n=2$ perfectly yields a magnetic moment of $\sqrt{8}\approx 2.84\text{ BM}$, we have found our culprit. The correct configuration is indeed $d^4$ in a strong ligand field.