A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with up to three electrons in the d orbitals, like Ti3+(d1); V3+(d2); Cr3+(d3); two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund’s rule). Thus, for d4 (Cr2+, Mn3+), d5 (Mn2+, Fe3+), d6 (Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications.
In which of the following do the complex ions [Ni(CN)4]2− and [NiCl4]2− differ?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Geometry
Both ions contain \( \ce{Ni^2+} \) with four ligands; what differs is how the 3d electrons are arranged, and so which orbitals are hybridised.
Oxidation state of Ni: in both ions x + 4(−1) = −2, so x = +2.
Configuration: Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^2+} \): 3d8.
\( \ce{[Ni(CN)4]^2-} \): \( \ce{CN^-} \) is a strong field ligand; the 3d electrons pair up and leave one 3d orbital empty. One 3d, one 4s and two 4p orbitals give dsp2 hybrids: square planar, diamagnetic.
\( \ce{[NiCl4]^2-} \): \( \ce{Cl^-} \) is a weak field ligand; the two unpaired 3d electrons stay unpaired. One 4s and three 4p orbitals give sp3 hybrids: tetrahedral, paramagnetic.
Why the other options are incorrect
Option (1) Oxidation state of nickel — Incorrect. Each ion has four singly charged ligands and a charge of 2− on the ion, so nickel is +2 in both.
Option (2) Coordination number — Incorrect. Each ion has four unidentate ligands bound to nickel, so the coordination number is 4 in both.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. This is illustrated … for [NiCl4]2–. Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d8.”
“In the square planar complexes, the hybridisation involved is dsp2. An example is [Ni(CN)4]2–. Here nickel is in +2 oxidation state and has the electronic configuration 3d8.”