Electronic Configuration of $$_{ 26 }^{ }{ Fe }$$: $$_{ 26 }^{ }{ Fe }\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3{ p }^{ 6 }3{ d }^{ 6 }4{ s }^{ 2 }$$
Electronic Configuration of $$_{ 26 }^{ }{ Fe }{ }^{ 3+ }$$ : $$_{ 26 }^{ }{ Fe }{ }^{ 3+ }\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3p^{ 6 }3{ d }^{ 5 }4{ s }^{ 0 }$$
In $$[fe(CN)_{ 6 }]{ }^{ 3- }$$ ,hexacyanoferrate(iii) ion.We have 5-d electrons.
Here strong cyanide ligand causes greater d-orbital splitting.The octahedral splitting energy,becomes high.
During filling up process,the first 3 electrons go into $${ t }_{ 2g }$$ orbitals.Now however, the splitting energy is much greater so it is lesser energitically 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 electrons in $${ e }_{ g }$$ X0.6$${ \Delta }_{ 0 }$$+no. of electrons in $${ t }_{ 2g }$$ X0.4$${ \Delta }_{ 0 }$$
= -0 X0.6$${ \Delta }_{ 0 }$$+5X0.4$${ \Delta }_{ 0 }$$=2$${ \Delta }_{ 0 }$$
The spin only moment $${ \mu }_{ s }=\sqrt { n(n+2) } $$
So..,
$${ \mu }_{ s }=\sqrt { n(n+2) } $$=$${ \mu }_{ s }=\sqrt { 1X(1+2) } $$=$$\sqrt { 3 } $$=1.7BM