Question

The correct order of $ [FeF_6]^{3-} $, $ [CoF_6]^{3-} $,$ [Ni(CO)_4] $ and $ [Ni(CN)_4]^{2-} $ complex species based on the number of unpaired electrons present is:

• A. \( [FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CN)_4]^{2-}>[Ni(CO)_4] \)
• B. \( [Ni(CN)_4]^{2-}>[FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CO)_4] \)
• C. \( [CoF_6]^{3-}>[FeF_6]^{3-}>[Ni(CO)_4]>[Ni(CN)_4]^{2-} \)
• D. \( [FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CN)_4]^{2-} = [Ni(CO)_4] \)

Hint:

The number of unpaired electrons in transition metal complexes can be determined based on the electronic configuration of the metal ion and the nature of the ligands.

Answer 1

The Correct Option is D

To determine the correct order of the given complex species based on the number of unpaired electrons, we need to analyze each of them individually. Let's consider each complex:

1. \([FeF_6]^{3-}\):
   • Iron (\(Fe\)) in this complex is in the +3 oxidation state, i.e., \(Fe^{3+}\).
   • The electronic configuration of \(Fe^{3+}\) is \([Ar] \, 3d^5\).
   • \(F^-\) is a weak field ligand, causing high spin, which means all five electrons in the \(3d\) orbital remain unpaired.
   • Therefore, the number of unpaired electrons = 5.
2. \([CoF_6]^{3-}\):
   • Cobalt (\(Co\)) is in the +3 oxidation state in this complex, i.e., \(Co^{3+}\).
   • The electronic configuration of \(Co^{3+}\) is \([Ar] \, 3d^6\).
   • \(F^-\) is a weak field ligand, resulting in a high spin complex with four unpaired electrons.
   • Therefore, the number of unpaired electrons = 4.
3. \([Ni(CO)_4]\):
   • Nickel (\(Ni\)) is in the zero oxidation state, i.e., \(Ni^0\). The electronic configuration is \([Ar] \, 3d^8 \, 4s^2\).
   • The \(CO\) ligand is a strong field ligand and causes \((ni)\) to pair up electrons, resulting in no unpaired electrons.
   • Therefore, the number of unpaired electrons = 0.
4. \([Ni(CN)_4]^{2-}\):
   • Nickel (\(Ni\)) in this case is in the +2 oxidation state, i.e., \(Ni^{2+}\).
   • The electronic configuration of \(Ni^{2+}\) is \([Ar] \, 3d^8\).
   • \(CN^-\) is a strong field ligand leading to the pairing of electrons due to the low spin complex formation, resulting in no unpaired electrons.
   • Therefore, the number of unpaired electrons = 0.

Based on the number of unpaired electrons calculated above, the order is:

\([FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4]\)

Thus, the correct answer is:

\([FeF_6]^{3-}\;>\;[CoF_6]^{3-}\;>\;[Ni(CN)_4]^{2-} = [Ni(CO)_4]\)

Answer 2

- Electronic configuration of each complex: - \( [FeF_6]^{3-} \): \( [Ar] 3d^5 4s^0 \). 
There are 5 unpaired electrons in the \( 3d \)-orbitals. 
- \( [CoF_6]^{3-} \): \( [Ar] 3d^6 4s^0 \). 
There are 4 unpaired electrons in the \( 3d \)-orbitals. 
- \( [Ni(CO)_4] \): \( [Ar] 3d^8 4s^2 \). 
The \( CO \) ligand is a strong field ligand, so the pairing of electrons leads to 0 unpaired electrons. 
- \( [Ni(CN)_4]^{2-} \): \( [Ar] 3d^8 4s^0 \). 
The \( CN \) ligand is a strong field ligand, leading to the pairing of all electrons, so there are 0 unpaired electrons. 
Thus, the order of the unpaired electrons is: \[ [FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CN)_4]^{2-} = [Ni(CO)_4] \]