Coordination Compounds Part 4

Concept:The number of unpaired electrons in a complex depends on the metal's oxidation state, d-electron count, and the ligand field strength which determines geometry (octahedral, tetrahedral, square planar) and spin state.Explanation:For Statement I: Calculate unpaired electrons for each complex.$[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$: $\mathrm{Cu}^{2+}$ is $\mathrm{d}^9$, square planar ($\mathrm{dsp}^2$), one unpaired electron.$[\mathrm{Ni}(\mathrm{en})_3]^{2+}$: $\mathrm{Ni}^{2+}$ is $\mathrm{d}^8$, octahedral ($\mathrm{sp}^3\mathrm{d}^2$), two unpaired electrons (weak field).$[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$: $\mathrm{Ni}^{2+}$ is $\mathrm{d}^8$, octahedral, two unpaired electrons.$[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$: $\mathrm{Mn}^{2+}$ is $\mathrm{d}^5$, octahedral high spin (weak field H$_2$O), five unpaired electrons.Thus $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ has the maximum (5) unpaired electrons. Statement I is true.For Statement II: Check each pair for diamagnetic species (all electrons paired).Pair 1: $[\mathrm{NiCl}_4]^{2-}$ (tetrahedral, $\mathrm{d}^8$, Cl$^-$ weak field, paramagnetic with 2 unpaired) and $[\mathrm{Ni(CO)}_4]$ (tetrahedral, $\mathrm{Ni(0)}$, $\mathrm{d}^{10}$, diamagnetic). Not both diamagnetic.Pair 2: $[\mathrm{NiCl}_4]^{2-}$ (paramagnetic) and $[\mathrm{Ni(CN)}_4]^{2-}$ (square planar, $\mathrm{d}^8$, CN$^-$ strong field, diamagnetic). Not both diamagnetic.Pair 3: $[\mathrm{Ni(CO)}_4]$ (diamagnetic) and $[\mathrm{Ni(CN)}_4]^{2-}$ (diamagnetic). Both are diamagnetic.Only one pair (the third) contains two diamagnetic species. Statement II false (says two).Answer:Statement I is true, Statement II is false. Correct option: D.