The tetrahedral [CoI_4]^{2-} and square plana | vedclass

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(B) $1$. In $[CoI_4]^{2-}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration). Since $I^-$ is a weak field ligand and the geometry is tetrahedr

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high spin,low spin

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(B) $1$. In $[CoI_4]^{2-}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration). Since $I^-$ is a weak field ligand and the geometry is tetrahedral,the crystal field splitting energy $(\Delta_t)$ is small. Thus,it is a high spin complex.$2$. In $[PdBr_4]^{2-}$,the oxidation state of $Pd$ is $+2$ ($d^8$ configuration). $Pd^{2+}$ is a $4d$ series metal ion. For $4d$ and $5d$ series metal ions,the crystal field splitting energy $(\Delta_o)$ is always large enough to cause pairing of electrons,regardless of the ligand strength. Therefore,square planar complexes of $Pd^{2+}$ are always low spin.

The tetrahedral $[CoI_4]^{2-}$ and square planar $[PdBr_4]^{2-}$ complex ions are respectively:

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