1. Find the oxidation state and $d$-electron count of the metal in each complex

Since $\mathrm{CN^-}$ is a ligand of charge $-1$ and there are 6 ligands, total ligand charge is $-6$.

For each complex $[M(CN)_6]^{3-}$: $x-6=-3 \Rightarrow x=+3$ So in all cases, the metal is in the $+3$ oxidation state.

Now determine the $d$-electron count:

• $\mathrm{Cr}$: $Z=24$, ground state $[Ar]3d^54s^1$ $\mathrm{Cr^{3+}}: 3d^3$
• $\mathrm{Mn}$: $Z=25$, ground state $[Ar]3d^54s^2$ $\mathrm{Mn^{3+}}: 3d^4$
• $\mathrm{Fe}$: $Z=26$, ground state $[Ar]3d^64s^2$ $\mathrm{Fe^{3+}}: 3d^5$
• $\mathrm{Co}$: $Z=27$, ground state $[Ar]3d^74s^2$ $\mathrm{Co^{3+}}: 3d^6$

2. Use the fact that $\mathrm{CN^-}$ is a strong-field ligand

$\mathrm{CN^-}$ causes pairing of electrons in octahedral complexes, so these are low-spin complexes.

Thus, octahedral splitting gives: $t_{2g} < e_g$

3. Write the low-spin configurations and count unpaired electrons

(A) $[Cr(CN)_6]^{3-}$

Metal ion: $\mathrm{Cr^{3+}} = d^3$ $t_{2g}^3e_g^0$ Unpaired electrons $=3$

(B) $[Mn(CN)_6]^{3-}$

Metal ion: $\mathrm{Mn^{3+}} = d^4$ Low spin: $t_{2g}^4e_g^0$ Unpaired electrons $=2$

(C) $[Fe(CN)_6]^{3-}$

Metal ion: $\mathrm{Fe^{3+}} = d^5$ Low spin: $t_{2g}^5e_g^0$ Unpaired electrons $=1$

(D) $[Co(CN)_6]^{3-}$

Metal ion: $\mathrm{Co^{3+}} = d^6$ Low spin: $t_{2g}^6e_g^0$ Unpaired electrons $=0$$

4. Relate unpaired electrons to paramagnetism

Paramagnetic behaviour increases with the number of unpaired electrons. The lowest paramagnetism corresponds to the fewest unpaired electrons.

From above:

• (A): 3 unpaired
• (B): 2 unpaired
• (C): 1 unpaired
• (D): 0 unpaired

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

5. Final answer

The correct option is D.