Question

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?

Answer 1

Electronic configuration of `""_"26"Fe->1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2`

Electronic configuration of `""_"26"Fe^"3+"->1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^0`

In `[Fe(CN)_6]^"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_"2g"` orbitals. Now, however, the splitting energy is much greater so it is less energetically costly for electrons to pair up in the `t_"2g"` orbitals than to go into the `e_g`orbitals and LOW SPIN complex is formed.

Here the crystal field Splitting energy

`=-"no. of e in " e_gxx0.6Delta_o +"no. of e in " t_"2g"xx0.4Delta_o`

`=-0xx0.6Delta_o +5xx0.4Delta_o=2Delta_o->"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(n+2))=sqrt(1xx(1+2))=sqrt3=1.7BM`

Answer 2

The field-splitting energy is `-2Delta_o`, and the spin-only magnetic moment is `"1.732 Bohr magnetons"`, or `"BM"`.

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

With six cyanide ligands, which are known to be great `bb(pi)`-acceptor and `bb(sigma)`-donor ligands, there are two effects at work:

• Stabilization of the triply-degenerate `bb(t_(2g))` `3d` molecular orbitals by the back-donation of `pi` electrons from the metal simultaneously into the cyanide ligand's `2p_x` and `2p_y` orbitals (`3d_(xz) -> 2p_x`, `3d_(yz) -> 2p_y`).
• Destabilization of the doubly-degenerate `bb(e_g)` `3d` molecular orbitals by the donation of `sigma` electrons from the cyanide ligand's `sp` orbitals into the metal's `3d_(z^2)` (axial) and `3d_(x^2 - y^2)` (equatorial) atomic orbitals (`sp -> 3d_(z^2)`, `sp -> 3d_(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 `+3/5Delta_o` and `-2/5Delta_o` from the free-ion energy of the metal (in a spherical field, which affects all orbitals equally). For simplicity, we observe only the `3d` orbitals from iron:

The fact that the two previously-mentioned effects are working in coincidence to increase `Delta_o` tells us that `"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^"*"` 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:

`color(blue)(E_"splitting") = (2 + 2 + 1)xx(-2/5Delta_o) = color(blue)(-2Delta_o)`

The spin-only magnetic moment, `mu_S`, is sometimes defined as (Miessler et al., pg. 360):

`bb(mu_S = gsqrt(S(S+1)))`,

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

From left to right, we have a total spin of `(+1/2 - 1/2) + (1/2 - 1/2) + 1/2 = 1/2`, so we have that:

`color(blue)(mu_S) = 2.00023sqrt(1/2(1/2 + 1))`

`=` `color(blue)("1.732 BM")`