Question

Match list I with list II and select the correct answer using the codes given below the lists.

List I (Pair of isomers)		List II (Type of isomerism)
p	(I) $[Co(NH_3)_6][Cr(CN)_6]$, (II)$[Cr(NH_3)_6][Co(CN)_6] \}$	1	Ionization
q	(III) $[PtCl_2(NH_3)_4]Br_2$, (IV) $[PtBr_2(NH_3)_4]Cl_2 \}$	2	Hydrate
r	(V) $[Co(SCN)(NH_3)_5]Cl_2$, (VI)$[Co(NCS)(NH_3)_5]Cl_2\}$	3	Coordination
s	(VII) $[Cr(H_2O)_6]Cl_3]$, (VIII) $[CrCl_2(H_2O)_4]Cl.2H_2O\}$	4	Geometrical
		5	Linkage isomerism

• A. $p -4, q- 1, r-5, s-2$
• B. $p -1, q- 3, r-2, s-5$
• C. $p -3, q- 1, r-5, s-2$
• D. $p -1, q- 3, r-5, s-2$

Answer 1

Answer: (C)

Compounds (I) and (II) both contain complex cation and anion, so they show coordination isomerism.

Compounds (III) and (IV) are obtained by the interchange of groups outside the coordination sphere and the ligands within the coordination sphere. Therefore, they given different ions in the solution. Hence, they show ionization isomerism.

Compounds (V) and (VI) are linkage isomers, because the ambident ligands, SCN has two donor atoms, i.e., it has two possibilities of attachment (SCN and NCS) with the central metal. Hence, they show linkage isomerism.

Compounds (VII) and (VIII) are obtained by the replacement of a coordinated group such as $Cl^{-}$ by water of hydration. Hence, they show hydrate isomerism.