Read the passage carefully and answer the Questions.

According to Valence Bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.

Which among the following are diamagnetic complexes?

$[Ni(CN)_4]^{2-}, [Co(C_2O_4)_3]^{3-}, [Fe(CN)_6]^{4-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$

$[Ni(CN)_4]^{2-}, [Co(C_2O_4)_3]^{3-}, [Fe(CN)_6]^{4-}$

The correct answer is Option (2) → $[Ni(CN)_4]^{2-}, [Co(C_2O_4)_3]^{3-}, [Fe(CN)_6]^{4-}$ **

To determine which complexes are diamagnetic, we look for those where all electrons are paired. This occurs when the metal ion's $d$-electrons are fully paired, typically facilitated by strong-field ligands like $CN^-$ and $C_2O_4^{2-}$.

Analysis of the Complexes

• $[Ni(CN)_4]^{2-}$: Nickel is in the $+2$ oxidation state ($d^8$). $CN^-$ is a very strong-field ligand that causes pairing in a square planar geometry. All 8 electrons pair up in four orbitals, making it diamagnetic.
• $[Co(C_2O_4)_3]^{3-}$: Cobalt is in the $+3$ oxidation state ($d^6$). Oxalate ($C_2O_4^{2-}$) acts as a strong-field ligand with $Co^{3+}$. This causes all 6 electrons to pair up in the lower $t_{2g}$ orbitals ($t_{2g}^6 e_g^0$). It is diamagnetic.
• $[Fe(CN)_6]^{4-}$: Iron is in the $+2$ oxidation state ($d^6$). $CN^-$ is a strong-field ligand, resulting in a low-spin octahedral complex ($t_{2g}^6 e_g^0$) where all electrons are paired. It is diamagnetic.
• $[CoF_6]^{3-}$: Cobalt is $+3$ ($d^6$). However, $F^-$ is a weak-field ligand. It cannot force pairing, resulting in a high-spin complex ($t_{2g}^4 e_g^2$) with 4 unpaired electrons. It is paramagnetic.
• $[Mn(CN)_6]^{3-}$: Manganese is $+3$ ($d^4$). Even though $CN^-$ is a strong-field ligand, a $d^4$ low-spin configuration ($t_{2g}^4 e_g^0$) still leaves 2 unpaired electrons. It is paramagnetic.

Conclusion

The diamagnetic complexes are $[Ni(CN)_4]^{2-}$, $[Co(C_2O_4)_3]^{3-}$, and $[Fe(CN)_6]^{4-}$.