Meaning of “responds to an external magnetic field”

A species that is paramagnetic responds to an external magnetic field. Paramagnetism arises when there are one or more unpaired electrons.

So, we check each complex for unpaired electrons.

Option A: $[Fe(H_2O)_6]^{3+}$

• Oxidation state of Fe: $x + 6(0) = +3 \Rightarrow x=+3$
• So Fe is $Fe^{3+}$.
• Electronic configuration of Fe: $Fe: [Ar]3d^64s^2$ $Fe^{3+}: [Ar]3d^5$
• $H_2O$ is a weak field ligand, so this is a high-spin octahedral complex.
• For $d^5$ high spin octahedral: $t_{2g}^3 e_g^2$
• This has 5 unpaired electrons.

Therefore, $[Fe(H_2O)_6]^{3+}$ is paramagnetic.

Option B: $[Ni(CN)_4]^{2-}$

• Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Ni is $Ni^{2+}$.
• Electronic configuration: $Ni: [Ar]3d^84s^2$ $Ni^{2+}: [Ar]3d^8$
• $CN^-$ is a strong field ligand.
• For 4-coordinate $d^8$ with strong field, the complex is square planar.
• Square planar $d^8$ complexes have all electrons paired.

Therefore, $[Ni(CN)_4]^{2-}$ is diamagnetic.

Option C: $[Co(CN)_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x=+3$
• So Co is $Co^{3+}$.
• Electronic configuration: $Co: [Ar]3d^74s^2$ $Co^{3+}: [Ar]3d^6$
• $CN^-$ is a strong field ligand, so this is low-spin octahedral.
• For low-spin $d^6$ octahedral: $t_{2g}^6 e_g^0$
• All electrons are paired.

Therefore, $[Co(CN)_6]^{3-}$ is diamagnetic.

Option D: $[Ni(CO)_4]$

• CO is a neutral ligand, so Ni oxidation state is 0.
• Thus Ni is $Ni^0$.
• Electronic configuration: $Ni: [Ar]3d^84s^2$
• In $[Ni(CO)_4]$, CO is a strong field ligand and the complex is tetrahedral with paired electrons effectively giving no unpaired electrons.
• $[Ni(CO)_4]$ is known to be diamagnetic.

Conclusion

Only option A has unpaired electrons and hence responds to an external magnetic field.

$\boxed{A\; [Fe(H_2O)_6]^{3+}}$