Consider the eight species [Fe(CN)6]^4-, [Ni(CO)4], [NiCl4]^2-, [Co(NH3)6]^3+, [CoF6]^3-, [MnO4]^-, [Cu(NH3)4]^2+ and [Ni(CN)4]^2-. The number of…

Fix the oxidation state, the $d$-electron count, and the geometry or field strength for each species. $\mathrm{[Fe(CN)_6]^{4-}}$ is $\mathrm{Fe^{2+}}$, $d^6$, with strong-field $\mathrm{CN^-}$ giving low-spin $t_{2g}^6$ with no unpaired electrons, so it is diamagnetic. $\mathrm{[Ni(CO)_4]}$ is $\mathrm{Ni(0)}$, $d^{10}$, tetrahedral, diamagnetic. $\mathrm{[NiCl_4]^{2-}}$ is $d^8$ tetrahedral with weak-field $\mathrm{Cl^-}$, which gives two unpaired electrons, so it is paramagnetic. $\mathrm{[Co(NH_3)_6]^{3+}}$ is $d^6$ low spin and diamagnetic. $\mathrm{[CoF_6]^{3-}}$ is $d^6$ high spin with four unpaired electrons, so it is paramagnetic. $\mathrm{[MnO_4]^-}$ is $\mathrm{Mn(VII)}$, $d^0$, diamagnetic. $\mathrm{[Cu(NH_3)_4]^{2+}}$ is $d^9$ with one unpaired electron, so it is paramagnetic. $\mathrm{[Ni(CN)_4]^{2-}}$ is $d^8$ square planar with strong-field $\mathrm{CN^-}$, which pairs all electrons, so it is diamagnetic. Diamagnetic: Fe, Ni(CO)4, Co(NH3)6, MnO4 and Ni(CN)4, which is 5.