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Which one of the following complexes is not expected to exhibit isomerism?
A
$[Ni(NH_3)_4(H_2O)_2]^{2+}$
B
$[Pt(NH_3)_2Cl_2]$
C
$[Ni(NH_3)_2Cl_2]$
D
$[Ni(en)_3]^{2+}$
Detailed Solution
$[Ni(NH_3)_4(H_2O)_2]^{2+}$ is octahedral of the type $[MA_4B_2]$; it shows geometrical (cis–trans) isomerism.
$[Pt(NH_3)_2Cl_2]$ is square planar ($dsp^2$) of the type $[MA_2B_2]$; it shows cis–trans isomerism (cis-platin and trans-platin).
$[Ni(en)_3]^{2+}$ is octahedral with three bidentate ligands; it has no plane of symmetry and shows optical isomerism (d and l forms).
$[Ni(NH_3)_2Cl_2]$: $Ni^{2+}$ with these weak field ligands is $sp^3$ hybridised and tetrahedral. In a tetrahedron all four positions are adjacent to one another, so $[MA_2B_2]$ has only one arrangement; there is no geometrical isomerism, and no optical isomerism either, since it has a plane of symmetry.
Hence $[Ni(NH_3)_2Cl_2]$ is not expected to exhibit isomerism.
$[Pt(NH_3)_2Cl_2]$ is square planar ($dsp^2$) of the type $[MA_2B_2]$; it shows cis–trans isomerism (cis-platin and trans-platin).
$[Ni(en)_3]^{2+}$ is octahedral with three bidentate ligands; it has no plane of symmetry and shows optical isomerism (d and l forms).
$[Ni(NH_3)_2Cl_2]$: $Ni^{2+}$ with these weak field ligands is $sp^3$ hybridised and tetrahedral. In a tetrahedron all four positions are adjacent to one another, so $[MA_2B_2]$ has only one arrangement; there is no geometrical isomerism, and no optical isomerism either, since it has a plane of symmetry.
Hence $[Ni(NH_3)_2Cl_2]$ is not expected to exhibit isomerism.