Question

Answer the question on the basis of the passage given below:
Transition metals form a large number of complexes or coordination compounds in which the metal atoms are bound to a number of anions or neutral molecules. The valence bond theory explains the formation, magnetic behaviour, and geometrical shapes, while the crystal field theory explains the effect of different crystal fields on the degeneracy of \( d \)-orbitals and the energies of the central metal atom/ion. This provides for the quantitative estimation of orbital separation energies, magnetic moments, and spectral and stability parameters.
Two complexes of nickel have the same geometry but different magnetic behaviour are:
(A) \( \text{[Ni(CN)}_4]^{2-} \)
(B) \( \text{[Ni(CO)}_4] \)
(C) \( \text{[NiCl}_4]^{2-} \)
(D) \( \text{[Ni(NH}_3)_6]^{2+} \)
Choose the correct answer from the options given below:

• (A) and (B) only
• (B) and (D) only
• (C) and (D) only
• (A) and (D) only

Hint:

Strong field ligands like CN⁻ and CO lead to electron pairing and diamagnetic behavior, while weaker field ligands like Cl⁻ and NH₃ typically result in paramagnetic behavior due to insufficient splitting of the d-orbitals.

Answer 1

The Correct Option is D

- Nickel complexes exhibit different magnetic behaviors due to the nature of the ligands and the crystal field splitting that occurs in the metal's d-orbitals. The geometry of the complex can influence whether it is paramagnetic or diamagnetic.
- (A) \( \text{[Ni(CN)}_4]^{2-} \): The complex of Ni²⁺ with cyanide (CN⁻) is a strong field ligand, causing significant crystal field splitting. This leads to the pairing of electrons in the lower-energy orbitals, resulting in a diamagnetic complex.
- (B) \( \text{[Ni(CO)}_4] \): Carbon monoxide (CO) is another strong field ligand that causes a similar effect to CN⁻ and results in electron pairing, making the complex diamagnetic.
- (C) \( \text{[NiCl}_4]^{2-} \): Chloride ions (Cl⁻) are weaker field ligands, and the crystal field splitting is not sufficient to cause electron pairing.
Thus, [NiCl₄]²⁻ remains paramagnetic.
- (D) \( \text{[Ni(NH}_3)_6]^{2+} \): Ammonia (NH₃) is a medium-strength ligand, which can cause some splitting of the d-orbitals but does not cause full electron pairing. [Ni(NH₃)₆]²⁺ is typically paramagnetic.
Therefore, the correct answer is Option (4), where (A) and (D) are correct. Both cyanide and ammonia cause different magnetic behaviors due to the extent of crystal field splitting.