The total possible coordination isomers for the following compounds respectively are :<br>[Co(en)3][ — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Analyze the ions formed in aqueous solution for each option: pairs (a), (b), and (c) release ions with similar charge magnitudes, resulting in similar conductivities. Step 2: In pair (d), [CoBr(NH3)5]SO4 dissociates into [CoBr(NH3)5]^2+ and SO4^2- (divalent ions, charge factor = 4). [Co(SO4)(NH3)5]Br dissociates into [Co(SO4)(NH3)5]^+ and Br^- (monovalent ions, charge factor = 2). Step 3: Since molar conductivity increases with the charge magnitude of the free ions, these two ionization isomers exhibit significantly different conductivities, matching option (d).