Explain magnetic property of coordination compound. - Vidyadip

Question

Explain magnetic property of coordination compound.

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Answer

$\rightarrow$ Metal ions with upto three electrons in the d orbitals, like $Ti ^{3+}\left( d ^1\right) ; V ^{3+}\left( d ^2\right) ; Cr ^{3+}\left( d ^3\right) ;$ tuo vacant $d$ orbitals are available for octahed $d^3$ hybridization with $4 s$ and $4 p$ orbitals.
$\rightarrow$ The magnetic behaviour of these free jons and their coordination entities is similar.
$\rightarrow$ When more than three $3d$ electrons are present, the required pair of $3d$ orbitals for octahedral hybridization is not directly available (as a consequence of Hund's rule).
$\rightarrow$ Thus, for $d ^4\left( Cr ^{2+}, Mn ^{3+}\right), d ^5\left( Mn ^{2+}, Fe ^{3+}\right), d ^6$
$\left( Fe ^{2+}, Co ^{3+}\right)$ cases, a vacant pair of $d$ orbitals results only by pairing of $3 d$ electrons which leaves two, one and zero unpaired electrons, respectively.
$\rightarrow$ The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing $d^6$ ions. However, with species containing $d^4$ and $d^5$ ions there are complications.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-}$ has magnetic moment of two unpaired electrons while $\left[ MnCl _6\right]^{3-}$ has a paramagnetic moment of four unpaired electrons.
$\rightarrow$ $\left[ Fe ( CN )_6\right]^{3-}$ has magnetic moment of a single unpaired electron while $\left[ FeF _6\right]^{3-}$ has a paramagnetic moment of five unpaired electrons.
$\rightarrow$ $\left[ CoF _6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ is diamagnetic.
$\rightarrow$ This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-},\left[ Fe ( CN )_6\right]^{3-}$ and $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ are inner orbital complexes involving $d ^2 sp ^3$ hybridisation, the former two complexes are paramagnetic and the latter diamagnetic.
$\rightarrow$ On the other hand, $\left[ MnCl _6\right]^{3-},\left[ FeF _6\right]^{3-}$ and $\left[ CoF _6\right]^{3-}$ are outer orbital complexes involving $sp ^3 d^2$ hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.

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