The number of paramagnetic species from the following is
[Ni(CN)$$_4$$]$$^{2-}$$, [Ni(CO)$$_4$$], [NiCl$$_4$$]$$^{2-}$$
[Fe(CN)$$_6$$]$$^{4-}$$, [Cu(NH$$_3$$)$$_4$$]$$^{2+}$$
[Fe(CN)$$_6$$]$$^{3-}$$ and [Fe(H$$_2$$O)$$_6$$]$$^{2+}$$

Seven coordination compounds are given; find the number of paramagnetic species.

[Ni(CN)$$_4$$]$$^{2-}$$: Ni$$^{2+}$$ has configuration [Ar] 3d$$^8$$. CN$$^-$$ is a strong field ligand causing pairing. With 4 ligands, this complex is square planar (dsp$$^2$$ hybridisation). All 8 d-electrons are paired. Diamagnetic.

[Ni(CO)$$_4$$]: Ni is in the 0 oxidation state with configuration [Ar] 3d$$^{10}$$. With 10 d-electrons, all are paired regardless of ligand field. Diamagnetic.

[NiCl$$_4$$]$$^{2-}$$: Ni$$^{2+}$$ (d$$^8$$). Cl$$^-$$ is a weak field ligand, so the complex is tetrahedral (sp$$^3$$ hybridisation). In tetrahedral geometry, d$$^8$$ has 2 unpaired electrons. Paramagnetic.

[Fe(CN)$$_6$$]$$^{4-}$$: Fe$$^{2+}$$ (d$$^6$$). CN$$^-$$ is a strong field ligand causing complete pairing: t$$_{2g}^6$$ e$$_g^0$$. Zero unpaired electrons. Diamagnetic.

[Cu(NH$$_3$$)$$_4$$]$$^{2+}$$: Cu$$^{2+}$$ (d$$^9$$). With 9 d-electrons, there is always 1 unpaired electron regardless of geometry. Paramagnetic.

[Fe(CN)$$_6$$]$$^{3-}$$: Fe$$^{3+}$$ (d$$^5$$). CN$$^-$$ is a strong field ligand: t$$_{2g}^5$$ e$$_g^0$$. One unpaired electron. Paramagnetic.

[Fe(H$$_2$$O)$$_6$$]$$^{2+}$$: Fe$$^{2+}$$ (d$$^6$$). H$$_2$$O is a weak field ligand: t$$_{2g}^4$$ e$$_g^2$$. Four unpaired electrons. Paramagnetic.

The paramagnetic species among these complexes are [NiCl$$_4$$]$$^{2-}$$, [Cu(NH$$_3$$)$$_4$$]$$^{2+}$$, [Fe(CN)$$_6$$]$$^{3-}$$, and [Fe(H$$_2$$O)$$_6$$]$$^{2+}$$, giving a total of 4 paramagnetic complexes.

Answer: 4