Animated Solution for Chemistry - Coordination Compounds: The spin only magnetic moment value for the complex $[Co(CN{)}_6{]}^{4-}$ is ...... BM. [Atomic number of Co = 27]

Welcome, future engineers and doctors! Today, we are going to dive into a fascinating problem from Coordination Chemistry. This question tests your understanding of crystal field theory, hybridization, and magnetic properties of coordination complexes. Let's unravel the mystery of the hexacyanidocobaltate(II) ion!

Our journey begins with the complex ion $[Co(CN{)}_6{]}^{4-}$. The first step in analyzing any coordination complex is to determine the oxidation state of the central metal atom.

Let the oxidation state of Cobalt ($Co$) be $x$. We know that the cyanide ion ($C{N}^{-}$) is a uninegative ligand. Since there are six cyanide ligands, their total charge is $-6$. The overall charge on the complex is $-4$.

Setting up the equation: $x+6(-1)=-4$ $x=+2$

So, Cobalt is present in the $+2$ oxidation state, giving us the $C{o}^{2+}$ ion.

Next, we need to look at the electronic configuration. The atomic number of Cobalt is 27. Its ground state electronic configuration is $[Ar]3d^{7}4s^2$. When it loses two electrons to form $C{o}^{2+}$, the configuration becomes $[Ar]3d^7$.

Now, here is where the magic happens! The cyanide ion ($C{N}^{-}$) is a strong field ligand. According to Crystal Field Theory, strong field ligands cause a large splitting of the d-orbitals, forcing the electrons to pair up against Hund's rule.

Out of the 7 electrons in the 3d subshell, 6 will pair up completely, leaving exactly one unpaired electron.

But wait, there's a catch! The complex has a coordination number of 6, which means it needs to form an octahedral geometry. Because $C{N}^{-}$ is a strong field ligand, it prefers to form an inner orbital complex using $d^{2}s{p}^3$ hybridization. This requires two empty 3d orbitals.

To make room, the single unpaired electron in the 3d orbital is excited and transferred to the higher energy 4d orbital. Even after this transference, the number of unpaired electrons ($n$) remains exactly 1.

Finally, we calculate the spin-only magnetic moment ($\mu$) using the formula: $\mu =\sqrt{n(n+2)}\text{ BM}$

Substituting $n=1$: $\mu =\sqrt{1(1+2)}=\sqrt{3}\text{ BM}$

The value of $\sqrt{3}$ is approximately 1.73 BM.

The question asks for the nearest integer. Since 1.73 is closer to 2 than to 1, we round it off to 2.

And there you have it! By carefully analyzing the oxidation state, the strength of the ligand, and the required hybridization, we successfully navigated through this beautiful problem. Keep visualizing the electrons, and you'll master coordination chemistry in no time!