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.

List $I$ (Complex)	List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$	$I$. Optical
$B$. $[Co(en)_3]^{3+}$	$II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$	$III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$	$IV$. Linkage

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

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Which of the following isomerism is not possible for complexes having molecular formulas:
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$(ii) \ CoBr \cdot SO_4 \cdot 5NH_3$
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Match List $I$ with List $II$ :
List $I$ (Complex) List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$ $I$. Optical
$B$. $[Co(en)_3]^{3+}$ $II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$ $III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$ $IV$. Linkage

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