Animated Solution for Chemistry - Coordination Compounds: The calculated spin only magnetic moments of $[\text{Cr}({\text{NH}}_3{)}_6{]}^{3+}$ and $[{\text{CuF}}_6{]}^{3-}$ in BM, respectively, are (Atomic numbers of Cr and Cu are 24 and 29, respectively)

Welcome to a beautiful exploration of Coordination Chemistry! In this problem, we are tasked with finding the spin-only magnetic moments of two fascinating coordination complexes: $[extCr({\text{NH}}_3{)}_6{]}^{3+}$ and $[ext{CuF}_6{]}^{3-}$.

The master key to unlocking this problem is the spin-only magnetic moment formula:

Here, $n$ represents the number of unpaired electrons in the central metal ion's d-orbitals, and the result is measured in Bohr Magnetons (B.M.). Our mission is simple: determine the oxidation state of the metal, write its electronic configuration, apply Crystal Field Theory to find $n$, and plug it into the formula.

Let's start with our first complex, hexaamminechromium(III), denoted as $[extCr({\text{NH}}_3{)}_6{]}^{3+}$.

Ammonia (${\text{NH}}_3$) is a neutral ligand, contributing zero charge. Therefore, the entire $+3$ charge of the complex sphere rests squarely on the Chromium ion. This means we are dealing with ${\text{Cr}}^{3+}$.

The atomic number of Chromium is 24, giving a neutral atom configuration of $[extAr]3d^{5}4s^1$. When it loses three electrons to become ${\text{Cr}}^{3+}$, it loses the $4s$ electron first, followed by two from the $3d$ subshell, leaving us with a pristine $3d^3$ configuration.

Now, visualize the octahedral crystal field created by the six ammonia ligands. The five degenerate d-orbitals split into two distinct energy levels: a lower energy $t_{2g}$ set (comprising three orbitals) and a higher energy $e_g$ set (comprising two orbitals).

According to Hund's rule of maximum multiplicity, our three electrons will singly occupy the lower energy $t_{2g}$ orbitals before any pairing occurs. This gives us a configuration of $t_{2g}^3{e}_g^0$.

Counting them up, we have exactly 3 unpaired electrons ($n=3$). Substituting this into our formula:

Since we know that $\sqrt{16}=4$, $\sqrt{15}$ must be slightly less than 4. A quick estimation gives us approximately $3.87\text{ B.M.}$

Moving on to the second complex, hexafluoridocuprate(III), denoted as $[ext{CuF}_6{]}^{3-}$.

Fluoride (${\text{F}}^{-}$) is an anionic ligand with a $-1$ charge. With six fluorides, the total ligand charge is $-6$. For the overall complex to have a $-3$ charge, the Copper ion must be in a $+3$ oxidation state. This is a rare and highly oxidizing state for Copper, stabilized here by the highly electronegative fluoride ligands!

Copper normally has an atomic number of 29, with a configuration of $[extAr]3d^{10}4s^1$. Losing three electrons (one from $4s$ and two from $3d$) leaves us with a $3d^8$ configuration.

For a $d^8$ system in an octahedral field, the electron filling is beautifully straightforward. The first six electrons completely fill the lower $t_{2g}$ level. The remaining two electrons must then singly occupy the higher $e_g$ level.

Notice a fascinating fact here: whether the ligand is strong or weak, a $d^8$ configuration in an octahedral field will always result in a $t_{2g}^6{e}_g^2$ arrangement. This leaves us with exactly 2 unpaired electrons ($n=2$).

Let's calculate the magnetic moment for our ${\text{Cu}}^{3+}$ ion:

Knowing that $\sqrt{9}=3$, we can estimate $\sqrt{8}$ to be slightly less than 3, which is approximately $2.84\text{ B.M.}$

Comparing our results ($3.87$ and $2.84$) with the given options, we find a perfect match with Option (A). This problem is a stellar demonstration of how Crystal Field Theory elegantly predicts the magnetic properties of transition metal complexes!