Question

The correct order of the number of unpaired electrons in the given complexes is
A. \([Fe(CN)_6]^{3–}\)
B. \([FeF_6]^{3–}\)
C. \([CoF_6]^{3–}\)
D. \([Cr(oxalate)_3]^{3–}\)
E. \([Ni(CO)_4]\)
Choose the correct answer from the options given below:

• A. A < E < D < C < B
• B. E < A < B < D < C
• C. E < A < D < C < B
• D. A < E < C < B < D

Answer 1

The Correct Option is C

 To determine the correct order of unpaired electrons in the given complexes, we need to assess each complex based on its metal oxidation state, ligand strength, and crystal field splitting. Let's analyze each complex one-by-one.

1. \([Fe(CN)_6]^{3–}\): In this complex, iron is in the +3 oxidation state, \(Fe^{3+}\), with an electron configuration of \(3d^5\). Cyanide (CN) is a strong field ligand leading to pairing of electrons. Thus, the config becomes \(t_{2g}^6\) with no unpaired electrons after pairing.
2. \([FeF_6]^{3–}\): Here, iron again is in the +3 oxidation state, \(Fe^{3+}\), with \(3d^5\). Fluoride (F) is a weak field ligand. Thus, no pairing happens, resulting in 5 unpaired electrons as \(t_{2g}^3 e_g^2\).
3. \([CoF_6]^{3–}\): In this complex, cobalt is in the +3 oxidation state, \(Co^{3+}\), with \(3d^6\). Fluoride (F) is a weak field ligand, leading to no pairing, resulting in 4 unpaired electrons as \(t_{2g}^4 e_g^2\).
4. \([Cr(oxalate)_3]^{3–}\): Chromium is in the +3 oxidation state, \(Cr^{3+}\), with \(3d^3\). Oxalate is a moderate ligand causing partial splitting, resulting in 3 unpaired electrons: \(t_{2g}^3\).
5. \([Ni(CO)_4]\): Nickel is in the zero oxidation state, \(Ni^0\), with configuration \(3d^{10} 4s^0\). Carbon monoxide (CO) is a strong field ligand, leading to full electron pairing in the \(3d\) and \(4s\), hence, no unpaired electrons.

We can now list the number of unpaired electrons in each complex:

• \([Fe(CN)_6]^{3–}\): 0 unpaired electrons
• \([FeF_6]^{3–}\): 5 unpaired electrons
• \([CoF_6]^{3–}\): 4 unpaired electrons
• \([Cr(oxalate)_3]^{3–}\): 3 unpaired electrons
• \([Ni(CO)_4]\): 0 unpaired electrons

From least to most number of unpaired electrons, the order is: \([Ni(CO)_4] = 0 < [Fe(CN)_6]^{3–} = 0 < [Cr(oxalate)_3]^{3–} = 3 < [CoF_6]^{3–} = 4 < [FeF_6]^{3–} = 5\).

Therefore, the correct order is: E < A < D < C < B.