Explain the magnetic properties of coordination compounds.

(N/A) The magnetic properties of coordination compounds of the first transition series are as follows:
For metal ions with up to three $d$-electrons,such as $Ti^{+3} (d^1)$,$V^{+3} (d^2)$,and $Cr^{+3} (d^3)$,two empty $d$-orbitals are available along with $4s$ and $4p$ orbitals for hybridization to form octahedral geometry. The magnetic behavior of these free ions and their coordination species is identical.
When more than three $d$-electrons are present,the necessary pairing of $3d$-orbitals for octahedral hybridization is not directly available.
For configurations like $d^4$ $(Cr^{+2}, Mn^{+3})$,$d^5$ $(Mn^{+2}, Fe^{+3})$,and $d^6$ $(Fe^{+2}, Co^{+3})$,a pair of empty $d$-orbitals is obtained only by pairing the electrons in the $3d$-orbitals,leaving two,one,and zero unpaired electrons respectively.
Coordination compounds containing $d^6$ ions show maximum spin pairing. However,for species containing $d^4$ and $d^5$ ions:
$[Mn(CN)_6]^{-3}$ has a magnetic moment corresponding to two unpaired electrons.
$[MnCl_6]^{-3}$ has a magnetic moment corresponding to four unpaired electrons.
$[Fe(CN)_6]^{-3}$ has a magnetic moment corresponding to one unpaired electron.
$[FeF_6]^{-3}$ has a magnetic moment corresponding to five unpaired electrons.
$[CoF_6]^{-3}$ is paramagnetic with four unpaired electrons,while $[Co(C_2O_4)_3]^{-3}$ is diamagnetic.
The reason for this discrepancy is that $[Mn(CN)_6]^{-3}$,$[Fe(CN)_6]^{-3}$,and $[Co(C_2O_4)_3]^{-3}$ are inner-orbital complexes involving $d^2sp^3$ hybridization. Among these,$[Mn(CN)_6]^{-3}$ and $[Fe(CN)_6]^{-3}$ are paramagnetic.
$[MnCl_6]^{-3}$,$[FeF_6]^{-3}$,and $[CoF_6]^{-3}$ are outer-orbital complexes involving $sp^3d^2$ hybridization and are paramagnetic,corresponding to $4, 5,$ and $4$ unpaired electrons respectively.