Consider the following complexes.<br>(i) [FeIF(CN)(H2O)(en)]<br>(ii) [MoCl2F2(gly)]^2-<br>Then, calc — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Evaluate the options: (I) K3[Fe(CN)6] contains Fe^3+ (3d^5) with strong-field CN^-, which forms a low-spin t_2g^5 eg^0 complex with 1 unpaired electron (paramagnetic). (II) [Ni(CO)4] contains Ni(0) (3d^10) and is diamagnetic. Step 2: (III) [Cr(NH3)6]^3+ contains Cr^3+ (3d^3) with 3 unpaired electrons. For d^1, d^2, and d^3 ions, there is no distinction between high and low spin configurations because the electrons always fill the three t_2g orbitals singly first. (IV) [Mn(CN)6]^4- contains Mn^2+ (3d^5) with strong-field CN^-, forming a low-spin t_2g^5 eg^0 complex with 1 unpaired electron (paramagnetic). Step 3: Therefore, only complexes I and IV are both paramagnetic and classified as low-spin complexes, corresponding to option (c).