Total number of stereoisomers of [Co(acac)2BrCl]- are : — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Determine the configuration and number of unpaired electrons (n) for each complex. Step 2: [Ni(CO)4] is Ni(0) with a 3d^10 configuration (n = 0); [Co(NH3)4(NO2)2]^+ has low-spin Co^3+ with a 3d^6 configuration (n = 0); [Ag(CN)2]^- is Ag^+ with a 4d^10 configuration (n = 0). Step 3: [CuBr4]^2- contains Cu^2+ with a 3d^9 configuration (n = 1). Since it is the only complex with any unpaired electrons, it has the maximum number of unpaired electrons, corresponding to option (d).