How many more co-ordination isomers are possible of the compound [Cu(NH3)4][PtCl4] ? — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Determine the number of unpaired electrons (n) in each complex: (I) [Ni(CO)4] is Ni(0) with a 3d^10 configuration (n = 0); (II) [Mn(CN)6]^4- has Mn^2+ (3d^5) in a low-spin state with CN^- (n = 1); (III) [Cr(NH3)6]^3+ has Cr^3+ (3d^3) with NH3 (n = 3). Step 2: For (IV) [CoF6]^3-, Co^3+ is a 3d^6 ion. With the weak-field F^- ligand, it forms a high-spin complex (t_2g^4 eg^2), which has exactly n = 4 unpaired electrons. Step 3: Comparing the number of unpaired electrons (I: 0 < II: 1 < III: 3 < IV: 4) shows that the magnetic moment increases in the order I < II < III < IV, corresponding to option (b).