Question

Identify a complex having four unpaired electrons in it

• A. \([\text{CoF}_{6}]^{3-}\)
• B. \([\text{Co(NH}_{3})_{6}]^{3+}\)
• C. \([\text{NiCl}_{4}]^{2-}\)
• D. \([\text{Ni(CN)}_{4}]^{2-}\)

Hint:

Find the d electron count of the metal ion in each complex and decide high spin or low spin from the ligand strength.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The number of unpaired electrons depends on the d electron count of the metal ion and on whether the ligand is weak field or strong field.

Step 2: Cobalt complexes
In \([\text{CoF}_6]^{3-}\) cobalt is \(\text{Co}^{3+}\) with \(3d^{6}\). \(\text{F}^{-}\) is a weak field ligand, so the complex is high spin with configuration \(t_{2g}^{4}e_g^{2}\), giving 4 unpaired electrons. In \([\text{Co(NH}_3)_6]^{3+}\), \(\text{NH}_3\) is strong field, so the complex is low spin \(t_{2g}^{6}\) with 0 unpaired electrons.

Step 3: Nickel complexes
In \([\text{NiCl}_4]^{2-}\), \(\text{Ni}^{2+}\) is \(3d^{8}\) in a tetrahedral, \(sp^3\) complex with 2 unpaired electrons. In \([\text{Ni(CN)}_4]^{2-}\), the strong field \(\text{CN}^{-}\) gives a square planar \(dsp^2\) complex with 0 unpaired electrons.

Step 4: Compare the options
Only \([\text{CoF}_6]^{3-}\) has four unpaired electrons.

Final Answer:
The complex with four unpaired electrons is [CoF6]3-. \[ \boxed{\text{(A) }[\text{CoF}_{6}]^{3-}} \]