The hybridization and magnetic nature of $$[Mn(CN)_6]^{4-}$$ and $$[Fe(CN)_6]^{3-}$$, respectively are:

We need to determine the hybridization and magnetic nature of $$[Mn(CN)_6]^{4-}$$ and $$[Fe(CN)_6]^{3-}$$.

In $$[Mn(CN)_6]^{4-}$$, manganese is in the +2 oxidation state (since the total charge is $$-4$$ and each $$CN^-$$ contributes $$-1$$, we get $$Mn + 6(-1) = -4$$, so $$Mn = +2$$). The free ion $$Mn^{2+}$$ has the configuration $$[Ar]\,3d^5$$ with five unpaired electrons in the five d-orbitals. Since $$CN^-$$ is a strong field ligand (high in the spectrochemical series), it produces a large crystal field splitting. In the octahedral field, the electrons pair up in the three lower-energy $$t_{2g}$$ orbitals: two electrons go into $$d_{xy}$$, two into $$d_{xz}$$, and one into $$d_{yz}$$, giving the configuration $$t_{2g}^5\,e_g^0$$ with one unpaired electron. The two $$e_g$$ d-orbitals ($$d_{x^2-y^2}$$ and $$d_{z^2}$$) are now empty and participate in bonding. These two inner d-orbitals combine with the 4s and three 4p orbitals to give $$d^2sp^3$$ hybridization. With one unpaired electron, the complex is paramagnetic.

In $$[Fe(CN)_6]^{3-}$$, iron is in the +3 oxidation state ($$Fe + 6(-1) = -3$$, so $$Fe = +3$$). The free ion $$Fe^{3+}$$ also has the configuration $$[Ar]\,3d^5$$. With $$CN^-$$ as a strong field ligand, the same pairing occurs: five electrons fill the three $$t_{2g}$$ orbitals as $$t_{2g}^5\,e_g^0$$, leaving one unpaired electron. The two empty $$e_g$$ orbitals participate in $$d^2sp^3$$ hybridization. With one unpaired electron, this complex is also paramagnetic.

Both $$[Mn(CN)_6]^{4-}$$ and $$[Fe(CN)_6]^{3-}$$ exhibit $$d^2sp^3$$ hybridization and are paramagnetic. The correct answer is Option 2.