Animated Solution for Chemistry - Coordination Compounds: The pair in which both the species have the same magnetic moment (spin only) is

Have you ever looked at a coordination chemistry problem and felt overwhelmed by the sheer number of complexes you need to analyze? I know the feeling! But here is a secret: most of these problems boil down to a single, elegant concept. In this question, we are on a mission to find a pair of complexes that share the exact same spin-only magnetic moment.

Let's take a deep breath and break this down. What exactly determines the magnetic moment of a complex?

The spin-only magnetic moment, denoted by $\mu$, is given by a beautiful and simple formula: Here, $n$ represents the number of unpaired electrons in the central metal ion. The unit is Bohr Magneton (BM).

Notice something crucial here? The magnetic moment depends entirely on $n$. If two complexes have the same number of unpaired electrons, their magnetic moments will be identical. Our mission just got a lot simpler: we just need to find the pair where $n$ is the same for both!

Let's dive into the options. We need to determine the oxidation state of the metal, its d-electron configuration, and how the ligands split those d-orbitals.

Let's look at option (b), which pairs $[Cr(H_{2}O{)}_6{]}^{2+}$ and $[Fe(H_{2}O{)}_6{]}^{2+}$.

First up is the Chromium complex. The oxidation state of Chromium here is $+2$. A neutral Chromium atom has the configuration $3d^{5}4s^1$. When it loses two electrons to become $C{r}^{2+}$, it becomes a $3d^4$ system.

Now, we look at the ligand. Water ($H_{2}O$) is a classic weak field ligand. It doesn't have the strength to force electrons to pair up against their will. So, the four electrons will occupy the split d-orbitals singly as much as possible. They will fill the lower $t_{2g}$ level with three electrons and push the fourth one up into the $e_g$ level. Counting them up, we have exactly 4 unpaired electrons ($n=4$).

Now, let's examine its partner in option (b): the Iron complex, $[Fe(H_{2}O{)}_6{]}^{2+}$.

Iron is atomic number 26, so neutral Iron is $3d^{6}4s^2$. Losing two electrons gives us $F{e}^{2+}$, which is a $3d^6$ system.

Again, our ligand is water, so we are dealing with a weak field and a high-spin complex. The first five electrons will fill all five d-orbitals singly. Where does the sixth electron go? It has no choice but to pair up in the lowest available energy level, which is one of the $t_{2g}$ orbitals. Let's count the unpaired electrons now. We have two paired electrons in one orbital, leaving two unpaired in the $t_{2g}$ level and two unpaired in the $e_g$ level. That gives us a total of 4 unpaired electrons ($n=4$).

Wow! Both the Chromium complex and the Iron complex have exactly 4 unpaired electrons. Because their $n$ values are identical, their spin-only magnetic moments will be perfectly matched. We have found our answer!

Before we celebrate, I want to point out a classic trap hidden in option (c): $[Fe(N{H}_3{)}_6{]}^{2+}$.

Usually, Ammonia ($N{H}_3$) is a strong field ligand that forces pairing. However, coordination chemistry loves its exceptions! When Ammonia complexes with $F{e}^{2+}$, it actually acts as a weak field ligand. This is a high-yield concept for JEE! Because it acts as a weak field ligand, the $3d^6$ electrons will arrange themselves exactly as they did with water, resulting in 4 unpaired electrons.

Always keep an eye out for these special cases. They are the difference between a good score and a great score. Keep practicing, trust your logic, and you will master coordination chemistry in no time!