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

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How to calculate this?

How to calculate this?

##### 2 Answers

Electronic configuration of

Electronic configuration of

In **LOW SPIN** complex is formed.

Here the crystal field Splitting energy

The spin only moment **Here n=1**

So

The field-splitting energy is

In **metal**, since

With six **cyanide** ligands, which are known to be great **-acceptor** and **-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,

Since there are six ligands around iron, we have an **octahedral complex**, with a splitting energy of

The fact that the two previously-mentioned effects are working in coincidence to increase **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

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**, *Miessler et al., pg. 360*):

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

where

From left to right, we have a total spin of

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

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