Question

Identify the complex having the highest number of unpaired electrons from following.

• A. \([\text{Co(NH}_3)_6]^{3+}\)
• B. \([\text{CoF}_6]^{3-}\)
• C. \([\text{NiCl}_4]^{2-}\)
• D. \([\text{Ni(CN)}_4]^{2-}\)

Hint:

Compare Co3+ high spin d6 in fluoride with Co3+ low spin and Ni2+ complexes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Unpaired electrons depend on the metal d count and on whether the ligand is strong field (pairs electrons) or weak field.

Step 2: Check (A) and (B)
In \([\text{Co(NH}_3)_6]^{3+}\), Co is +3 (\(d^6\)). \(\text{NH}_3\) is a strong field ligand, so it is low spin with zero unpaired electrons. In \([\text{CoF}_6]^{3-}\), \(\text{F}^-\) is weak field, so it is high spin \(t_{2g}^4 e_g^2\) with 4 unpaired electrons.

Step 3: Check (C) and (D)
In \([\text{NiCl}_4]^{2-}\), Ni is +2 (\(d^8\)) with weak field Cl- and tetrahedral shape, giving 2 unpaired electrons. In \([\text{Ni(CN)}_4]^{2-}\), strong field CN- gives square planar, with 0 unpaired.

Final Answer:
The highest is 4 unpaired electrons in \([\text{CoF}_6]^{3-}\), option (B). \[ \boxed{[\text{CoF}_6]^{3-}} \]