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Which of the following do not show geometrical isomerism? (Assume all ligands are unidentate)
  • a)
    Square planar [MXL3]
  • b)
    Square planar [MX2L2]
  • c)
    Octahedral [MX2L4]
  • d)
    Octahedral [MX3L3]
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Which of the following do not show geometrical isomerism? (Assume all ...
Square planar [MXL3] does not show geometrical isomerism.

Explanation:
Geometrical isomerism occurs when there are different spatial arrangements of ligands around a central metal atom. In order for geometrical isomerism to be possible, there must be at least two different groups or ligands attached to the central metal atom.

a) Square planar [MXL3]:
In this complex, there are three ligands (L) and one monodentate ligand (X) attached to the central metal atom (M). Since all ligands are unidentate, they can only form one bond with the metal atom. Therefore, there is no possibility of having different spatial arrangements of ligands around the central metal atom, and thus no geometrical isomerism is possible.

b) Square planar [MX2L2]:
In this complex, there are two ligands (L) and two monodentate ligands (X) attached to the central metal atom (M). Since all ligands are unidentate, they can only form one bond with the metal atom. However, in this case, there is a possibility of having different spatial arrangements of ligands. The two ligands (L) can be arranged cis (on the same side) or trans (on opposite sides) to each other. This results in two different geometrical isomers.

c) Octahedral [MX2L4]:
In this complex, there are four ligands (L) and two monodentate ligands (X) attached to the central metal atom (M). Since all ligands are unidentate, they can only form one bond with the metal atom. In an octahedral complex, there are several possible spatial arrangements of ligands. The ligands can be arranged cis (on the same side) or trans (on opposite sides) to each other, and they can also be arranged in different positions around the central metal atom. This results in several different geometrical isomers.

d) Octahedral [MX3L3]:
In this complex, there are three ligands (L) and three monodentate ligands (X) attached to the central metal atom (M). Since all ligands are unidentate, they can only form one bond with the metal atom. In an octahedral complex, there are several possible spatial arrangements of ligands. The ligands can be arranged cis (on the same side) or trans (on opposite sides) to each other, and they can also be arranged in different positions around the central metal atom. This results in several different geometrical isomers.

In summary, only the square planar [MXL3] complex does not show geometrical isomerism because all ligands are unidentate and can only form one bond with the central metal atom.
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Community Answer
Which of the following do not show geometrical isomerism? (Assume all ...
In square planar complexes, the type [MXL3] does not have any isomers as there is no pair of ligands that can be arranged adjacent to or opposite each other to form cis or trans forms respectively. All possible combinations result in the same spatial arrangement.
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Which of the following do not show geometrical isomerism? (Assume all ligands are unidentate)a)Square planar [MXL3]b)Square planar [MX2L2]c)Octahedral [MX2L4]d)Octahedral [MX3L3]Correct answer is option 'A'. Can you explain this answer?
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