(C) In $[Co(CN)_6]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $CN^-$ is a strong field liga

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$[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.

(C) In $[Co(CN)_6]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. This results in all $6$ electrons being paired in the $t_{2g}$ orbitals. Therefore,the complex has no unpaired electrons and is a low-spin complex.

Class 12 Chemistry Coordination Compounds Hybridisation and Geometry

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Which of these statements about $[Co(CN)_6]^{3-}$ is true?

NEET 2015 All NEET

A
$[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a high-spin configuration.
B
$[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
C
$[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
D
$[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a low-spin configuration.

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Which of these statements about $[Co(CN)_6]^{3-}$ is true?

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On the basis of valence bond theory,discuss the geometrical shapes and magnetic properties of the following $4$-coordinated complexes:
$(i)$ $[NiCl_{4}]^{2-}$
$(ii)$ $[Ni(CN)_{4}]^{2-}$
$(iii)$ $[Ni(CO)_{4}]$
$(iv)$ $MnO_{4}^{-}$