# What are the crystal field Splitting energy and the spin only moment in Bohr Magneton for the complex K_3[Fe(CN)_6] ?

## How to calculate this?

Nov 5, 2016

Electronic configuration of $\text{_"26} F e \to 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

Electronic configuration of $\text{_"26"Fe^"3+} \to 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5} 4 {s}^{0}$

In ${\left[F e {\left(C N\right)}_{6}\right]}^{\text{3-}}$ ,hexacyanoferrate(III) ion. we have five d-electrons. Here strong cyanide ligand causes greater d-orbital splitting. The octahedral splitting energy,${\Delta}_{o}$ becomes high. During filling up process the first three electrons go into ${t}_{\text{2g}}$ orbitals. Now, however, the splitting energy is much greater so it is less energetically costly for electrons to pair up in the ${t}_{\text{2g}}$ orbitals than to go into the ${e}_{g}$orbitals and LOW SPIN complex is formed.

Here the crystal field Splitting energy

$= - \text{no. of e in " e_gxx0.6Delta_o +"no. of e in " t_"2g} \times 0.4 {\Delta}_{o}$

$= - 0 \times 0.6 {\Delta}_{o} + 5 \times 0.4 {\Delta}_{o} = 2 {\Delta}_{o} \to \text{stabilisation}$

The spin only moment mu_s=sqrt(n(n+2),where n = number of unpaired electron in complex ion system. Here n=1

So ${\mu}_{s} = \sqrt{n \left(n + 2\right)} = \sqrt{1 \times \left(1 + 2\right)} = \sqrt{3} = 1.7 B M$

Nov 5, 2016

The field-splitting energy is $- 2 {\Delta}_{o}$, and the spin-only magnetic moment is $\text{1.732 Bohr magnetons}$, or $\text{BM}$.

In "K"_3["Fe"("CN")_6], the charge of the anion is ${3}^{-}$, while each cyanide ligand is ${1}^{-}$. Thus, we are looking at ${\text{Fe}}^{3 +}$ (atomic number $26$), a $\boldsymbol{{d}^{5}}$ metal, since ${\text{Fe}}^{0}$ originally had an $n {s}^{2} \left(n - 1\right) {d}^{6}$ configuration, and a $+ 3$ oxidation state removes the two $n s$ and one $\left(n - 1\right) d$ electrons.

With six cyanide ligands, which are known to be great $\boldsymbol{\pi}$-acceptor and $\boldsymbol{\sigma}$-donor ligands, there are two effects at work:

• Stabilization of the triply-degenerate $\boldsymbol{{t}_{2 g}}$ $3 d$ molecular orbitals by the back-donation of $\pi$ electrons from the metal simultaneously into the cyanide ligand's $2 {p}_{x}$ and $2 {p}_{y}$ orbitals ($3 {d}_{x z} \to 2 {p}_{x}$, $3 {d}_{y z} \to 2 {p}_{y}$).
• Destabilization of the doubly-degenerate $\boldsymbol{{e}_{g}}$ $3 d$ molecular orbitals by the donation of $\sigma$ electrons from the cyanide ligand's $s p$ orbitals into the metal's $3 {d}_{{z}^{2}}$ (axial) and $3 {d}_{{x}^{2} - {y}^{2}}$ (equatorial) atomic orbitals ($s p \to 3 {d}_{{z}^{2}}$, $s p \to 3 {d}_{{x}^{2} - {y}^{2}}$).

Both of these work to increase the overall ligand field splitting energy, $\Delta$.

Since there are six ligands around iron, we have an octahedral complex, with a splitting energy of $+ \frac{3}{5} {\Delta}_{o}$ and $- \frac{2}{5} {\Delta}_{o}$ from the free-ion energy of the metal (in a spherical field, which affects all orbitals equally). For simplicity, we observe only the $3 d$ orbitals from iron:

The fact that the two previously-mentioned effects are working in coincidence to increase ${\Delta}_{o}$ tells us that ${\text{CN}}^{-}$ is a strong-field ligand.

These make the transition-metal complex a low-spin complex, since as a general rule, it is difficult for the electrons to populate the antibonding ${e}_{g}^{\text{*}}$ orbitals (though simply labeled ${e}_{g}$ in crystal-field theory) when the splitting energy is large.

The ligand-field splitting energy is calculated as each electron in an occupied orbital multiplied by the relevant splitting energy with respect to the free-ion energy. Therefore, we have:

$\textcolor{b l u e}{{E}_{\text{splitting}}} = \left(2 + 2 + 1\right) \times \left(- \frac{2}{5} {\Delta}_{o}\right) = \textcolor{b l u e}{- 2 {\Delta}_{o}}$

The spin-only magnetic moment, ${\mu}_{S}$, is sometimes defined as (Miessler et al., pg. 360):

$\boldsymbol{{\mu}_{S} = g \sqrt{S \left(S + 1\right)}}$,

where $g = 2.00023$ is the gyromagnetic ratio and $S$ is the total spin.

From left to right, we have a total spin of $\left(+ \frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{2}\right) + \frac{1}{2} = \frac{1}{2}$, so we have that:

$\textcolor{b l u e}{{\mu}_{S}} = 2.00023 \sqrt{\frac{1}{2} \left(\frac{1}{2} + 1\right)}$

$=$ $\textcolor{b l u e}{\text{1.732 BM}}$