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

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