Consider the magnetic properties of the following complexes and select the correct statements:
[NiCl4]2−, [Ni(CN)4]2− and [Ni(CO)4].
A. [Ni(CN)4]2− is paramagnetic but [Ni(CO)4] is diamagnetic.
B. Both [Ni(CN)4]2− and [Ni(CO)4] are diamagnetic.
C. Both [Ni(CN)4]2− and [NiCl4]2− are diamagnetic.
D. [NiCl4]2− is paramagnetic.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B and D only
A complex is paramagnetic if it has unpaired electrons and diamagnetic if all electrons are paired. Whether the electrons pair up depends on the oxidation state of the metal and the strength of the ligand. By valence bond theory:
[NiCl4]2−: The charge balance x + 4(−1) = −2 gives Ni(+2). Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8. \( \ce{Cl^{-}} \) is a weak field ligand and does not pair the electrons, so two stay unpaired. The complex uses sp3 orbitals, is tetrahedral and paramagnetic.
[Ni(CN)4]2−: Here too x + 4(−1) = −2, so nickel is \( \ce{Ni^{2+}} \), 3d8. The strong field \( \ce{CN^{-}} \) pairs the two unpaired electrons, freeing one 3d orbital. The complex uses dsp2 orbitals, is square planar and diamagnetic.
[Ni(CO)4]: CO is a neutral ligand, so Ni is in the zero state: Ni (Z = 28) is [Ar]3d84s2. CO pushes the 4s electrons into the 3d orbitals, giving 3d10. The complex uses sp3 orbitals, is tetrahedral and diamagnetic.
A. [Ni(CN)4]2− is paramagnetic but [Ni(CO)4] is diamagnetic. Incorrect. \( \ce{[Ni(CN)4]^{2-}} \) has no unpaired electrons, so it is diamagnetic.
B. Both [Ni(CN)4]2− and [Ni(CO)4] are diamagnetic. Correct. All electrons are paired in both: 3d8 paired in the square planar cyanide complex and 3d10 in the carbonyl.
C. Both [Ni(CN)4]2− and [NiCl4]2− are diamagnetic. Incorrect. \( \ce{[NiCl4]^{2-}} \) has two unpaired electrons, so it is paramagnetic.
D. [NiCl4]2− is paramagnetic. Correct. With two unpaired electrons its spin-only moment is √8 ≈ 2.83 BM.