Animated Solution for Chemistry - Coordination Compounds: The correct order of the spin only magnetic moment of metal ions in the following low spin complexes, $[V(CN{)}_6{]}^{4-}$, $[Fe(CN{)}_6{]}^{4-}$, $[Ru(N{H}_3{)}_6{]}^{3+}$, and $[Cr(N{H}_3{)}_6{]}^{2+}$, is

Hello students! Today we are embarking on a fascinating journey into the heart of Coordination Chemistry. Our mission is to determine the correct order of the spin-only magnetic moment for four specific coordination complexes.

The key to unlocking this problem lies in understanding the electronic structure of the central metal ions and how the surrounding ligands influence their d-orbitals. The spin-only magnetic moment, denoted by $\mu$, is directly tied to the number of unpaired electrons ($n$) through the elegant formula:

Since $\mu$ increases as $n$ increases, our primary task is simply to count the unpaired electrons in each complex!

Let's break down our four contenders one by one. We need to find the oxidation state of the central metal to determine its d-electron count.

1. The Vanadium Complex: $[V(CN{)}_6{]}^{4-}$ Let the oxidation state of Vanadium be $x$. Since each cyanide ligand carries a $-1$ charge, we have $x+6(-1)=-4$, which gives $x=+2$. Vanadium (Atomic Number 23) has a neutral configuration of $[Ar]3d^{3}4s^2$. Stripping away two electrons leaves us with $V^{2+}$, a $d^3$ system.

2. The Chromium Complex: $[Cr(N{H}_3{)}_6{]}^{2+}$ Ammonia is a neutral ligand. So, $x+6(0)=+2$, meaning Chromium is in the $+2$ state. Chromium (Atomic Number 24) is an exception with a neutral configuration of $[Ar]3d^{5}4s^1$. Removing two electrons gives $C{r}^{2+}$, a $d^4$ system.

3. The Ruthenium Complex: $[Ru(N{H}_3{)}_6{]}^{3+}$ Again, ammonia is neutral, so Ruthenium is in the $+3$ state. Ruthenium sits right below Iron in the periodic table (Atomic Number 44). Its neutral configuration is $[Kr]4d^{7}5s^1$. Removing three electrons leaves $R{u}^{3+}$, a $d^5$ system.

4. The Iron Complex: $[Fe(CN{)}_6{]}^{4-}$ With cyanide ligands, $x+6(-1)=-4$, giving Iron a $+2$ state. Iron (Atomic Number 26) has a neutral configuration of $[Ar]3d^{6}4s^2$. Removing two electrons yields $F{e}^{2+}$, a $d^6$ system.

Crystal Field vs. Pairing

Now comes the crucial twist! The question explicitly states that these are low spin complexes.

In Crystal Field Theory, the five degenerate d-orbitals split into two sets: the lower energy $t_{2g}$ and the higher energy $e_g$. The energy gap between them is ${\Delta }_o$. A 'low spin' complex forms when this gap is massive—specifically, when ${\Delta }_o>P$ (the pairing energy).

Because jumping the gap is too energetically costly, electrons prefer to pair up in the lower $t_{2g}$ orbitals before even looking at the $e_g$ orbitals. Let's see how this plays out:

$V^{2+}(d^3)$: The first three electrons happily occupy the three $t_{2g}$ orbitals singly. Configuration: $t_{2g}^3{e}_g^0$. $C{r}^{2+}(d^4)$: The fourth electron faces a choice. Because it's low spin, it chooses to pair up in the $t_{2g}$ level rather than jump to $e_g$. Configuration: $t_{2g}^4{e}_g^0$. $R{u}^{3+}(d^5)$: The fifth electron also pairs up in the $t_{2g}$ level. Configuration: $t_{2g}^5{e}_g^0$. $F{e}^{2+}(d^6)$: The sixth electron completes the pairing in the $t_{2g}$ level. Configuration: $t_{2g}^6{e}_g^0$.

Unpaired Electrons

Now, let's count the unpaired electrons ($n$) for each:

$[V(CN{)}_6{]}^{4-}$ ($t_{2g}^3$): $n=3$ $[Cr(N{H}_3{)}_6{]}^{2+}$ ($t_{2g}^4$): $n=2$ $[Ru(N{H}_3{)}_6{]}^{3+}$ ($t_{2g}^5$): $n=1$ $[Fe(CN{)}_6{]}^{4-}$ ($t_{2g}^6$): $n=0$

Since the magnetic moment $\mu$ is directly proportional to the number of unpaired electrons, the order of magnetic moments perfectly mirrors the order of $n$:

This beautiful sequence of logic leads us straight to the correct answer!