JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Coordination Compounds

In $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$ the iron is $\mathrm{F}{\mathrm{e}}^{3+}$, a $d^5$ ion, and fluoride is a weak-field ligand, so the complex is high spin with five unpaired electrons. In $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$ the cobalt is $\mathrm{C}{\mathrm{o}}^{3+}$, $d^6$, again high spin with weak-field fluoride, giving four unpaired electrons. In $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ the nickel is $\mathrm{N}{\mathrm{i}}^{2+}$, $d^8$, and strong-field cyanide forces a square-planar, fully paired arrangement with no unpaired electrons. In $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ the nickel is in the zero oxidation state, $d^{10}$, which is completely paired with no unpaired electrons. The number of unpaired electrons thus falls in the order $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}>[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}>[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}=[\mathrm{Ni}(\mathrm{CO}{)}_4]$.