Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour? (At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27)

For an octahedral complex $$[\mathrm{M(CN)_6}]^{3-}$$ the cyanide ion $$\mathrm{CN^-}$$ acts as a very strong field (strong-field, low-spin) ligand. Hence, for every metal ion present, the electrons first pair up in the lower-energy $$t_{2g}$$ set before any electron occupies the higher-energy $$e_g$$ set.

First find the oxidation state and the corresponding $$d$$-electron count for each metal ion.

Each $$\mathrm{CN^-}$$ contributes -1 charge (total -6). The overall charge on the complex is -3, so the metal ion must be in the +3 oxidation state in every option.
Therefore $$\mathrm{M^{3+}}$$ is present in all four complexes.

Electronic configurations of the gaseous atoms (atomic numbers are given):
Cr : $$[Ar]\,3d^{5}4s^{1}$$ → $$\mathrm{Cr^{3+}}$$ : $$3d^{3}$$ (i.e. $$d^{3}$$)
Mn : $$[Ar]\,3d^{5}4s^{2}$$ → $$\mathrm{Mn^{3+}}$$ : $$3d^{4}$$ (i.e. $$d^{4}$$)
Fe : $$[Ar]\,3d^{6}4s^{2}$$ → $$\mathrm{Fe^{3+}}$$ : $$3d^{5}$$ (i.e. $$d^{5}$$)
Co : $$[Ar]\,3d^{7}4s^{2}$$ → $$\mathrm{Co^{3+}}$$ : $$3d^{6}$$ (i.e. $$d^{6}$$)

Now fill these $$d$$ electrons into the octahedral crystal-field levels under strong-field (low-spin) conditions.

Case 1: $$\mathbf{[Cr(CN)_6]^{3-}}$$ ( $$d^{3}$$ )
Distribution: $$t_{2g}^{3}e_g^{0}$$ → three unpaired electrons.

Case 2: $$\mathbf{[Mn(CN)_6]^{3-}}$$ ( $$d^{4}$$ )
Low-spin distribution: $$t_{2g}^{4}e_g^{0}$$ → two unpaired electrons.

Case 3: $$\mathbf{[Fe(CN)_6]^{3-}}$$ ( $$d^{5}$$ )
Low-spin distribution: $$t_{2g}^{5}e_g^{0}$$ → one unpaired electron.

Case 4: $$\mathbf{[Co(CN)_6]^{3-}}$$ ( $$d^{6}$$ )
Low-spin distribution: $$t_{2g}^{6}e_g^{0}$$ → zero unpaired electrons (completely paired, diamagnetic).

The strength of paramagnetism is directly proportional to the number of unpaired electrons. Comparing the four cases:

Number of unpaired electrons: 3 (Cr) > 2 (Mn) > 1 (Fe) > 0 (Co)

Hence the complex with the lowest paramagnetic behaviour is $$[\mathrm{Co(CN)_6}]^{3-}$$.

Option D which is: $$[\text{Co}(\text{CN})_6]^{-3}$$