Match List-I with List-II : List-I | vedclass

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(A) $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$ exhibit linkage isomerism due to the ambidentate ligand $NO_2^-$. Thus,$a-i$.$(b)$ $[Cr(NH_3)_6][

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$a-i, b-ii, c-iii, d-iv$

Vedclass · Answer

(A) $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$ exhibit linkage isomerism due to the ambidentate ligand $NO_2^-$. Thus,$a-i$.$(b)$ $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$ exhibit coordination isomerism due to the exchange of ligands between cationic and anionic entities. Thus,$b-ii$.$(c)$ $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$ exhibit ionisation isomerism as they produce different ions in solution. Thus,$c-iii$.$(d)$ $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ exhibit solvate (hydrate) isomerism. Thus,$d-iv$.Therefore,the correct match is $a-i, b-ii, c-iii, d-iv$.

List-$I$ (Complexes)	List-$II$ (Types)
$a$. $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$	$i$. Linkage isomerism
$b$. $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$	$ii$. Coordination isomerism
$c$. $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$	$iii$. Ionisation isomerism
$d$. $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$	$iv$. Solvate isomerism

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