Question

The transition metal (M) complex that can have all isomers (geometric, linkage, and ionization) is

• A. [M(NH₃)4Br₂]SCN
• B. [M(NH₃)₄Cl₂]Br
• C. [M(NH₃)₄(H₂O)₂]Cl₃
• D. [M(NH₃)₄ (H₂O)₂](SCN)₃

Answer 1

The Correct Option is A

To determine which transition metal complex can have all isomers (geometric, linkage, and ionization), we need to analyze the possibilities for each type of isomerism in the given complexes. 

1. Geometric Isomerism: This occurs in coordination complexes with a coordination number of 4 (square planar) or 6 (octahedral). Complexes with different ligands can exhibit cis-trans or facial-meridional isomerism.
2. Linkage Isomerism: This happens when a ligand that can coordinate through different atoms is involved. An example is the thiocyanate ion, SCN^-, which can bind through sulfur or nitrogen.
3. Ionization Isomerism: This is when an ion inside the coordination sphere can interchange with an ion outside the sphere, affecting the formula and properties.

Now, let's evaluate each provided option:

• [M(NH₃)₄Br₂]SCN:
  • Geometric Isomers: This complex can exhibit cis-trans isomerism due to the presence of two different ligands (bromides).
  • Linkage Isomers: The thiocyanate ion can bind through sulfur or nitrogen, allowing for linkage isomerism.
  • Ionization Isomers: The SCN^- ion can interchange with a bromide ligand, generating ionization isomers.
• [M(NH₃)₄Cl₂]Br:
  • Geometric Isomers: Can have cis-trans isomers due to chlorine ligands.
  • Linkage Isomers: No suitable ligand for linkage isomerism.
  • Ionization Isomers: Possible with Br^-.
• [M(NH₃)₄(H₂O)₂]Cl₃:
  • Geometric Isomers: Can have cis-trans isomers due to water ligands.
  • Linkage Isomers: No suitable ligand for linkage isomerism.
  • Ionization Isomers: Possible with chloride ions.
• [M(NH₃)₄ (H₂O)₂](SCN)₃:
  • Geometric Isomers: Can have cis-trans isomers due to water ligands.
  • Linkage Isomers: Possible due to thiocyanate but less likely given coordination.
  • Ionization Isomers: Less favorable as all SCN^- are outside the coordination sphere.

Conclusion: The complex \([M(NH_3)_4Br_2]SCN\) can have all three types of isomers (geometric, linkage, and ionization). Thus, the correct answer is

[M(NH₃)₄Br₂]SCN

.