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JEE Main Chemistry Coordination Compounds 2026 JEE Main 2026 (02 April Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

Which of the following complexes will show coordination isomerism? A. [Ag(NH_3)_2][Ag(CN)_2] B. [Co(NH_3)_6][Cr(CN)_6] C. [Co(NH_3)_6][Co(CN)_6] D. [Fe(NH_3)_6][Co(CN)_6] E. [Co(NH_3)_6][Fe(CN)_6] Choose the correct answer from the options given below:

Options

  1. A. B, C and D Only
  2. B. B, D and E Only
  3. C. A, C and D Only
  4. D. C, D and E Only

Answer

B. B, D and E Only

Step-by-step solution

Coordination isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. A. [Ag(NH_3)_2][Ag(CN)_2]: Both the complex cation and anion contain the same metal ion (Ag^+). Interchanging one ligand produces a neutral molecule [Ag(NH_3)(CN)], and interchanging both ligands yields the identical compound. Thus, it does not show coordination isomerism. B. [Co(NH_3)_6][Cr(CN)_6]: The complex contains different metal ions (Co^ 3+ and Cr^ 3+ ). Interchange of ligands gives the coordination isomer [Cr(NH_3)_6][Co(CN)_6]. Thus, it shows coordination isomerism. C. [Co(NH_3)_6][Co(CN)_6]: Both the complex cation and anion contain the same metal ion (Co^ 3+ ). According to the strict definition requiring different metal ions, this is generally excluded from being a primary example of coordination isomerism in standard contexts. D. [Fe(NH_3)_6][Co(CN)_6]: The complex contains different metal ions (Fe^ 3+ and Co^ 3+ ). Interchange of ligands gives the coordination isomer [Co(NH_3)_6][Fe(CN)_6]. Thus, it shows coordination isomerism. E. [Co(NH_3)_6][Fe(CN)_6]: This is the coordination isomer of complex D. Since it contains different metal ions, it also shows coordination isomerism by interchanging ligands to form [Fe(NH_3)_6][Co(CN)_6]. Since complexes D and E are coordination isomers of each other, they must both be included together. Therefore, the complexes that show coordination isomerism are B, D, and E. Answer: B, D and E Only

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