JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Coordination Compounds

In $[\mathrm{FeF}{}_6{]}^{3-}$ the iron(III) centre is $d^5$ and fluoride is a weak-field ligand, so the complex is high-spin with five unpaired electrons and the largest spin-only moment, $\sqrt{35}$ BM. In $[\mathrm{CoF}{}_6{]}^{3-}$ the cobalt(III) centre is $d^6$ and again the weak-field fluoride leaves a high-spin arrangement, now with four unpaired electrons, $\sqrt{24}$ BM. In $[\mathrm{Co}(\mathrm{C}{}_2\mathrm{O}{}_4){}_3{]}^{3-}$ the same $d^6$ cobalt(III) is surrounded by the stronger-field oxalate, which pairs the electrons completely to give a low-spin configuration with no unpaired electrons and zero moment. The spin-only magnetic moments therefore decrease in the order $[\mathrm{FeF}{}_6{]}^{3-}>[\mathrm{CoF}{}_6{]}^{3-}>[\mathrm{Co}(\mathrm{C}{}_2\mathrm{O}{}_4){}_3{]}^{3-}$.