Animated Solution for Chemistry - Coordination Compounds: The pair of metal ions that can given a spin-only magnetic moment of 3.9 BM for the complex $[M(H_{2}O{)}_6]C{l}_2$, is

Let's embark on a journey to unravel the mysteries of coordination compounds. Our starting point is the given complex, $[M(H_{2}O{)}_6]C{l}_2$. Before we dive into the quantum mechanics of electrons, we need to establish the identity of our central metal ion, specifically its oxidation state.

Imagine the complex as a tightly bound family. The chloride ions ($C{l}^{-}$) are the distant relatives, sitting outside the coordination sphere. Since there are two of them, they contribute a total charge of $-2$. To maintain electrical neutrality, the entire coordination sphere $[M(H_{2}O{)}_6]$ must carry a charge of $+2$.

Now, let's look inside the sphere. Water ($H_{2}O$) is a neutral molecule; it brings no charge to the table. Therefore, the entire $+2$ charge must belong solely to the central metal atom, $M$. We have successfully deduced that our metal is in the $+2$ oxidation state ($M^{2+}$).

The problem provides a crucial piece of evidence: the spin-only magnetic moment ($ \mu$) is $3.9\text{ BM}$. This number is not random; it is a direct window into the electronic soul of the metal ion.

The relationship between the magnetic moment and the number of unpaired electrons ($n$) is governed by the elegant formula:

By substituting our known value, we get:

Squaring both sides gives us approximately $15=n(n+2)$. Solving this simple quadratic equation reveals that $n=3$.

This is our master constraint: We are hunting for a pair of metal ions that both possess exactly three unpaired electrons in their $d$-orbitals.

Here is where many students fall into a trap. You cannot simply look at the free metal ion's configuration; you must consider the environment it is in. The ligand dictates the rules of the house.

Our ligand is water ($H_{2}O$). According to the Spectrochemical Series, water is a Weak Field Ligand (WFL). It creates a relatively small crystal field splitting energy (${\Delta }_o$). Because the energy gap between the lower $t_{2g}$ and upper $e_g$ orbitals is small, electrons prefer to jump to the higher energy level rather than pairing up in the lower level. This results in a High-Spin Complex.

Armed with the knowledge that we need $n=3$ in a high-spin octahedral environment, let's interrogate our suspects.

Candidate 1: Vanadium ($V^{2+}$) Vanadium has an atomic number of 23. Losing two electrons gives it a configuration of $[Ar]3d^3$. In our octahedral field, these three electrons will comfortably settle into the three degenerate $t_{2g}$ orbitals.

Configuration: $t_{2g}^3{e}_g^0$ Unpaired electrons: $n=3$. Vanadium is a perfect match!

Candidate 2: Cobalt ($C{o}^{2+}$) Cobalt has an atomic number of 27, leading to a $[Ar]3d^7$ configuration. How do we distribute seven electrons in a high-spin complex?

Following Hund's rule, we first place one electron in each of the five orbitals (three in $t_{2g}$, two in $e_g$). We have two electrons left. These must now pair up in the lowest available energy level, which is the $t_{2g}$ set.

Configuration: $t_{2g}^5{e}_g^2$ Let's count the unpaired electrons: one in $t_{2g}$ and two in $e_g$, giving us a total of $n=3$. Cobalt is also a perfect match!

Both $V^{2+}$ and $C{o}^{2+}$ yield exactly three unpaired electrons under the influence of the weak field water ligands. Consequently, they both exhibit a spin-only magnetic moment of $3.9\text{ BM}$.

This makes option (c) the undisputed correct answer. Always remember, the ligand is the silent director orchestrating the behavior of the electrons. A simple change from water to a strong field ligand like cyanide ($C{N}^{-}$) would completely rewrite the script, forcing pairing and drastically altering the magnetic moment!