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Among $[Ni(CO)_4]$, $[NiCl_4]^{2-}$, $[Co(NH_3)_4Cl_2]Cl$, $Na_3[Co(C_2O_4)_3]$, $Na_2O_2$, and $CsO_2$, the total number of paramagnetic compounds is:
- 2
- 3
- 4
- 5
Correct answer: 3
Solution
$[Ni(CO)_4]$ is diamagnetic because Ni is in the zero oxidation state and CO is a strong-field ligand. $[NiCl_4]^{2-}$ is paramagnetic, $[Co(NH_3)_4Cl_2]Cl$ is paramagnetic, $Na_3[Co(C_2O_4)_3]$ is diamagnetic, $Na_2O_2$ is diamagnetic, and $CsO_2$ is paramagnetic. Thus, there are 3 paramagnetic compounds.
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