Animated Solution for Chemistry - Coordination Compounds: The magnetic moment of an octahedral homoleptic $\text{Mn(II)}$ complex is $5.9\text{ BM}$. The suitable ligand for this complex is

Coordination chemistry is like a microscopic dance between a central metal ion and its surrounding ligands. The nature of this dance—whether the electrons pair up or stay single—is dictated by the strength of the ligands. Let's dive into a classic JEE problem that beautifully connects magnetic properties with Crystal Field Theory (CFT).

Manganese in the Spotlight

Our journey begins with the central metal ion, Manganese. In this octahedral homoleptic complex, Manganese is in the $+2$ oxidation state.

Manganese has an atomic number of $25$. When it loses two electrons to become ${\text{Mn}}^{2+}$, it loses them from the outermost $4s$ orbital first. This leaves us with a valence electronic configuration of $[\text{Ar}]3d^5$.

Imagine those five $3d$ orbitals. According to Hund's Rule, they will each take one electron before any pairing occurs. So, we start with five unpaired electrons.

Spin-Only Magnetic Moment

The problem gives us a crucial piece of evidence: the magnetic moment ($\mu$) of the complex is $5.9\text{ BM}$ (Bohr Magnetons).

We have a powerful tool to decode this—the spin-only magnetic moment formula:

Here, $n$ represents the number of unpaired electrons. Let's substitute our known value into the equation:

If we test $n=5$, we get $\sqrt{5(5+2)}=\sqrt{35}$, which is approximately $5.916$. This perfectly matches our given value!

This is the Eureka moment: The complex has exactly $5$ unpaired electrons.

Strong vs. Weak Field Ligands

Now, let's connect the math to the physical reality. We started with five unpaired electrons in the bare ${\text{Mn}}^{2+}$ ion, and even after the ligands attached themselves to form the complex, we still have five unpaired electrons.

What does this tell us about the ligands?

It means the ligands were not strong enough to force the electrons to pair up against their natural repulsion. In the language of Crystal Field Theory, the crystal field splitting energy (${\Delta }_o$) is less than the pairing energy ($P$). Therefore, the ligand must be a weak field ligand.

Because the inner $3d$ orbitals are fully occupied with single electrons, they are unavailable for bonding. The metal must reach out to its outer shell and use the $4s$, $4p$, and $4d$ orbitals to accommodate the six incoming ligands. This results in an outer orbital complex with $s{p}^3{d}^2$ hybridization.

Our final task is to identify the weak field ligand from the given options: (a) ${\text{CN}}^{-}$ (b) ethylenediamine (c) ${\text{NCS}}^{-}$ (d) $\text{CO}$

According to the spectrochemical series, Carbon Monoxide ($\text{CO}$), Cyanide (${\text{CN}}^{-}$), and ethylenediamine are all notorious strong field ligands. They would have forced the electrons to pair up, drastically reducing the magnetic moment.

Only the thiocyanate ion, ${\text{NCS}}^{-}$, is a weak field ligand. It allows the electrons to remain unpaired, preserving the high magnetic moment of $5.9\text{ BM}$.

And just like that, by following the trail of unpaired electrons, we've solved the mystery!