JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Coordination Compounds

Paramagnetic compounds have unpaired electrons. $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ is diamagnetic (Ni in zero oxidation state, $d^{10}$). $[{\mathrm{NiCl}}_4{]}^{2-}$ is paramagnetic (Ni${}^{2+}$, $d^8$, tetrahedral, high-spin with 2 unpaired electrons). $[\mathrm{Co}({\mathrm{NH}}_3{)}_4{\mathrm{Cl}}_2]\mathrm{Cl}$ is diamagnetic (Co${}^{3+}$, $d^6$, low-spin). ${\mathrm{Na}}_3[{\mathrm{CoF}}_6]$ is paramagnetic (Co${}^{3+}$, $d^6$, high-spin with F${}^{-}$ weak field ligand, 4 unpaired electrons). ${\mathrm{Na}}_2{\mathrm{O}}_2$ is diamagnetic (O${}_2^{2-}$). ${\mathrm{CsO}}_2$ is paramagnetic (O${}_2^{-}$ superoxide ion has one unpaired electron). Total: 3 paramagnetic compounds.