Carbon donor ligands are strong ligands and u | vedclass

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Question

(C) The complex among the given complexes which shows high spin is $[Fe(C_2O_4)_3]^{3-}$.This is because the oxalate ion,$(C_2O_4)^{2-}$,is a weak fie

Vedclass · Accepted Answer

$[Fe(C_2O_4)_3]^{3-}$

Vedclass · Answer

(C) The complex among the given complexes which shows high spin is $[Fe(C_2O_4)_3]^{3-}$.This is because the oxalate ion,$(C_2O_4)^{2-}$,is a weak field ligand according to the spectrochemical series.In $[Fe(C_2O_4)_3]^{3-}$,the oxidation state of $Fe$ is $+3$ ($d^5$ configuration).Since $(C_2O_4)^{2-}$ is a weak field ligand,it does not cause pairing of electrons,resulting in a high spin complex.In contrast,$CN^-$ is a strong field ligand that causes pairing of electrons,leading to low spin complexes in $[Fe(CN)_6]^{3-}$ and $[Fe(CN)_6]^{4-}$.

Carbon donor ligands are strong ligands and usually form low spin complexes. Among the complexes given below,select the high spin complex.

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$[Fe(CN)_{6}]^{4-}$ and $[Fe(H_{2}O)_{6}]^{2+}$ are of different colours in dilute solutions. Why?

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The number of unpaired $d$-electrons in $[Co(H_2O)_6]^{3+}$ is $\qquad$

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