Question

Match List-I with List-II. 

List-I	Complex
A	[Pt(NH₃)₂Cl₂]
B	[Co(en)₃]Cl₃
C	[Co(NH₃)₅NO₂]Cl₂
D	[Cr(H₂O)₆]Cl₃

 

List-II	Type of isomerism
I	Optical
II	Solvate
III	Geometrical
IV	Linkage

• A. A-III, B-I, C-IV, D-II
• B. A-II, B-I, C-III, D-IV
• C. A-IV, B-III, C-II, D-I
• D. A-I, B-II, C-IV, D-III

Hint:

Square planar complexes show geometrical isomerism, \(en\)-complexes may show optical isomerism, and \(\ce{NO2^-}\) ligand shows linkage isomerism.

Answer 1

The Correct Option is A

Step 1: Identify isomerism in \([Pt(NH_3)_2Cl_2]\).
This complex has two ammonia ligands and two chloride ligands. Such square planar complexes can show cis-trans forms. So: \[ [Pt(NH_3)_2Cl_2] \] shows: \[ \text{Geometrical isomerism}. \] Therefore: \[ A-III \]

Step 2: Identify isomerism in \([Co(en)_3]Cl_3\).
The ligand \(en\) is ethylenediamine. It is a bidentate ligand. Complexes having three bidentate ligands can exist as non-superimposable mirror images. Therefore: \[ [Co(en)_3]Cl_3 \] shows: \[ \text{Optical isomerism}. \] So: \[ B-I \]

Step 3: Identify isomerism in \([Co(NH_3)_5NO_2]Cl_2\).
The ligand \(\ce{NO2^-}\) can coordinate through nitrogen or oxygen. If it coordinates through nitrogen, it is called nitro. If it coordinates through oxygen, it is called nitrito. This gives linkage isomerism. So: \[ C-IV \]

Step 4: Identify isomerism in \([Cr(H_2O)_6]Cl_3\).
Water molecules may be present inside or outside the coordination sphere in hydrated complexes. This gives solvate or hydrate isomerism. So: \[ D-II \]

Step 5: Final matching.
Thus, the correct matching is: \[ A-III,\quad B-I,\quad C-IV,\quad D-II. \]