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Account for the following (i) Co(II) ...

Account for the following
(i) `Co(II)` is stable in aqueous solution but in the presence of strong ligand and air , it can get oxidized to `Co(III)` .
(ii) `[Ni(CN)_(4)]^(2-)` is square planar and diamagnetic whereas `[NiCl_(4)]^(2-)` is tetrahedral and paramagnetic .

Text Solution

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Co(II) has the configuration `3d^(7)`, i.e. it has three unpaired electrons. Water being a weak ligand, the unpaired electrons do not pair up. In the presence of strong ligands and air, two unpaired electrons in 3d pair up and the third unpaired electron shifts to higher energy sub-shell from where it can be easily lost and hence shows an oxidation state of III.
(ii) `CN^(-)` a strong field ligand forces the electrons to pair up in d-orbitals so it shows dsp hybridization due to which shape is square planar. Further no electron is left unpaired, so it is diamagnetic whileCl^(-) a weak_field ligand-eennot force the electron to pair up showing `sp^(3)` hybridization. So, shape is tetrahedral and due to unpaired electron it is paramagnetic
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