Question

Identify, from the following, the diamagnetic, tetrahedral complex:

• A. [Ni(Cl)₄]²⁻
• B. [Co(C₂O₄)₃]³⁻
• C. [Ni(CN)₄]²⁻
• D. [Ni(CO)₄]

Hint:

Tetrahedral complexes generally form with weak field ligands. However, CO is a strong field ligand that follows sp$^3$ hybridization due to back bonding effects.

Answer 1

The Correct Option is D

Step 1: Identifying the Electronic Configuration 
Nickel (Ni) has an atomic number 28. 
The electronic configuration of Ni is: \[ [Ar] 3d^8 4s^2 \] The oxidation state of Ni in the given complexes needs to be determined. 
Step 2: Evaluating Each Complex 
(A) [Ni(Cl)$_4$]$^{2-}$ 
Cl$^-$ is a weak field ligand.
The tetrahedral geometry follows an sp$^3$ hybridization. 
Paramagnetic due to unpaired electrons. Not the correct answer. 
(B) [Co(C$_2$O$_4$)$_3$]$^{3-}$ 
Co$^{3+}$ has a strong field ligand (oxalate). 
Octahedral structure, not tetrahedral. Incorrect choice. 
(C) [Ni(CN)$_4$]$^{2-}$ 
CN$^-$ is a strong field ligand. 
Follows dsp$^2$ hybridization, giving square planar geometry. 
Diamagnetic, but not tetrahedral. Incorrect choice. 
(D) [Ni(CO)$_4$] 
CO is a strong field ligand.
Causes electron pairing and sp$^3$ hybridization, forming a tetrahedral structure. 
No unpaired electrons = Diamagnetic. Correct answer. 
Final Answer: The correct diamagnetic, tetrahedral complex is [Ni(CO)$_4$].

Answer 2

[Ni(CO)₄] is a tetrahedral complex of nickel in the zero oxidation state.

CO ligands are strong field ligands and cause pairing of the nickel's d-electrons.

Nickel has a d⁸ electronic configuration, but due to strong ligand field from CO, all electrons pair up resulting in no unpaired electrons.

This makes [Ni(CO)₄] diamagnetic.

The tetrahedral shape arises because of four CO ligands arranged around the nickel atom.

Therefore, [Ni(CO)₄] is a diamagnetic, tetrahedral complex.