$$[Fe(CN)_6]^{4-}$$
$$[Fe(CN)_6]^{3-}$$
$$[Ti(CN)_6]^{3-}$$
$$[Ni(CN)_4]^{2-}$$
$$[Co(CN)_6]^{3-}$$
Among the given complexes, number of paramagnetic complexes is

We need to determine the number of paramagnetic complexes among the given list. A complex is paramagnetic if it has unpaired electrons.

For $$[\text{Fe}(\text{CN})_6]^{4-}$$, Fe is in +2 oxidation state (Fe$$^{2+}$$: 3d$$^6$$) and CN$$^-$$ is a strong field ligand, so all 6 electrons pair up in the t$$_{2g}$$ orbitals: t$$_{2g}^6$$ e$$_g^0$$, giving 0 unpaired electrons and diamagnetic character.

For $$[\text{Fe}(\text{CN})_6]^{3-}$$, Fe is in +3 oxidation state (Fe$$^{3+}$$: 3d$$^5$$) and CN$$^-$$ is a strong field ligand, resulting in t$$_{2g}^5$$ e$$_g^0$$, so it has 1 unpaired electron and is paramagnetic.

For $$[\text{Ti}(\text{CN})_6]^{3-}$$, Ti is in +3 oxidation state (Ti$$^{3+}$$: 3d$$^1$$) because n + 6(−1) = −3 gives n = +3, and CN$$^-$$ being a strong field ligand yields t$$_{2g}^1$$ e$$_g^0$$, so it has 1 unpaired electron and is paramagnetic.

In $$[\text{Ni}(\text{CN})_4]^{2-}$$, Ni is in +2 oxidation state (Ni$$^{2+}$$: 3d$$^8$$) and with CN$$^-$$ the strong field ligand in a square planar complex (dsp$$^2$$ hybridization), all eight d-electrons are paired, giving 0 unpaired electrons and diamagnetic character.

For $$[\text{Co}(\text{CN})_6]^{3-}$$, Co is in +3 oxidation state (Co$$^{3+}$$: 3d$$^6$$) and CN$$^-$$ is a strong field ligand giving t$$_{2g}^6$$ e$$_g^0$$, so there are 0 unpaired electrons and the complex is diamagnetic.

Among these, $$[\text{Fe}(\text{CN})_6]^{3-}$$ and $$[\text{Ti}(\text{CN})_6]^{3-}$$ are paramagnetic, giving a total of 2 paramagnetic complexes. Hence, the answer is 2.