Question

Which of the following pair(s) of complex ions are diamagnetic and are isoelectronic with $[Fe(CN)_6]^{4-}$?

• A. [Co(en)$_2$ Cl$_2$]$^+$ and $[Co(en)_2$(NH$_3$)$_2$]$^{3+}$
• B. [Ni(CN)$_4$]$^{2-}$ and $[Ni(CO)_4]$
• C. [Fe(CN)$_6$]$^{3-}$ and $[Cu(NH_3)_4]^{2+}$
• D. [V(CO)$_6$]$^-$ and $[Mn(CO)_5]^-$

Answer 1

Answer: (A)

A. $[Co(en)_2 Cl_2]^+$ and $[Co(en)_2(NH_3)_2]^{3+}$

Answer 2

The reference complex is $[Fe(CN)_6]^{4-}$. The oxidation state of Fe is +2, and its electron configuration is $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand, $[Fe(CN)_6]^{4-}$ is a low-spin complex with configuration $t_{2g}^6 e_g^0$, making it diamagnetic. We need to find pairs of complexes that are also diamagnetic and have a metal ion with a $d^6$ electron configuration (isoelectronic with $Fe^{2+}$).

Option A:

• $[Co(en)_2 Cl_2]^+$: Cobalt is in the +3 oxidation state ($Co^{3+}$). Its electron configuration is $[Ar] 3d^6$. Ethylenediamine (en) is a strong field ligand, and $Cl^-$ is a weak field ligand. For $Co^{3+}$ ($d^6$), the complex is generally considered low-spin and thus diamagnetic ($t_{2g}^6 e_g^0$).
• $[Co(en)_2(NH_3)_2]^{3+}$: Cobalt is in the +3 oxidation state ($Co^{3+}$). Its electron configuration is $[Ar] 3d^6$. Both en and $NH_3$ are strong field ligands for $Co^{3+}$. This complex is low-spin and diamagnetic ($t_{2g}^6 e_g^0$). Both complexes in Option A are diamagnetic and isoelectronic with $[Fe(CN)_6]^{4-}$.

Option B:

• $[Ni(CN)_4]^{2-}$: Nickel is in the +2 oxidation state ($Ni^{2+}$). Its electron configuration is $[Ar] 3d^8$. This is not isoelectronic with $Fe^{2+}$ ($d^6$).
• $[Ni(CO)_4]$: Nickel is in the 0 oxidation state ($Ni^0$). Its electron configuration is $[Ar] 3d^8 4s^2$. This is not isoelectronic with $Fe^{2+}$ ($d^6$).

Option C:

• $[Fe(CN)_6]^{3-}$: Iron is in the +3 oxidation state ($Fe^{3+}$). Its electron configuration is $[Ar] 3d^5$. This is not isoelectronic with $Fe^{2+}$ ($d^6$). It is also paramagnetic.
• $[Cu(NH_3)_4]^{2+}$: Copper is in the +2 oxidation state ($Cu^{2+}$). Its electron configuration is $[Ar] 3d^9$. This is not isoelectronic with $Fe^{2+}$ ($d^6$). It is also paramagnetic.

Option D:

• $[V(CO)_6]^-$: Vanadium is in the -1 oxidation state ($V^{-1}$). Its electron configuration is $[Ar] 3d^6$. This is isoelectronic with $Fe^{2+}$ ($d^6$). CO is a strong field ligand, so this complex is low-spin and diamagnetic ($t_{2g}^6 e_g^0$).
• $[Mn(CO)_5]^-$: Manganese is in the -1 oxidation state ($Mn^{-1}$). Its electron configuration is $[Ar] 3d^6 4s^2$. This has 8 valence electrons, not 6, so it is not isoelectronic with $Fe^{2+}$ ($d^6$).

Therefore, only Option A contains a pair of complexes that are diamagnetic and isoelectronic with $[Fe(CN)_6]^{4-}$.