JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Coordination Compounds

Paramagnetism grows with the number of unpaired electrons. In $[\mathrm{Fe}(\mathrm{en})(\mathrm{bpy})(\mathrm{N}{\mathrm{H}}_3{)}_2{]}^{2+}$ the iron is $\mathrm{F}{\mathrm{e}}^{2+}$ ($3d^6$) and the strong-field ligands make it low spin with no unpaired electrons; in $[\mathrm{Pd}(\mathrm{gly}{)}_2]$ the palladium is $\mathrm{P}{\mathrm{d}}^{2+}$ ($4d^8$) and square planar, again with none; and in $[\mathrm{Ti}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$ the titanium is $\mathrm{T}{\mathrm{i}}^{3+}$ ($3d^1$) with a single unpaired electron. In $[\mathrm{Co}(\mathrm{ox}{)}_2(\mathrm{OH}{)}_2{]}^{-}$ the charge balance leaves cobalt in a $d^4$ state, and with ${\Delta }_0>P$ the low-spin $t_{2g}^4$ arrangement carries two unpaired electrons, the most of any option, so this cobalt complex shows the highest paramagnetic behaviour.