Coordination compounds
The coordination compounds are very important in biological systems (like human, plant and animal's body), metallurgy, analytical chemistry, medicine etc. Chlorophyll, Hemoglobin, vitamin B12 etc. are coordination compounds of various metals. They are also called complex compounds and formed by two main entities, metal and ligands. Some ligands are Poly-dentate, like EDTA, which is used in the estimation of hardness of water with Ca2+ and Mg2+ ions. Gold and silver make [Au(CN)2]− and [Ag(CN)2]− type complex ions in aqueous solution during extraction of metals. In photography, developing film form a complex ion [Ag (S2O3)2]3−. A complex compound of platinum (Pt), cis-platin is used in the treatment of cancer. Some metals can be refined by making their complex compounds, like impure Nickel is converted to [Ni(CO)4] complex. All these coordination compounds may be colored or colorless, paramagnetic or diamagnetic, low spin or high spin, and arranged in particular geometry and hybridization. Isomerism is shown by these complexes due to different arrangement of ligands around the central metal atom.
What is geometry, hybridization, and the magnetic nature of complex [Ni(CO)4]?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → sp3, tetrahedral, diamagnetic
In valence bond theory, the geometry and magnetic nature of a complex are found in four steps:
Find the oxidation state of the metal and write its configuration.
Decide whether the ligand is strong field (it pairs up the d electrons) or weak field (it does not).
Find the empty orbitals that take the ligand electron pairs. sp3 gives a tetrahedral shape, dsp2 square planar, and d2sp3 or sp3d2 octahedral.
Unpaired electrons make the complex paramagnetic; if all are paired, it is diamagnetic.
In \( \ce{[Ni(CO)4]} \), CO is a neutral ligand, so nickel is in the 0 oxidation state: Ni (Z = 28) is [Ar]3d84s2. CO is a strong field ligand, so the 4s electrons pair up in the 3d subshell, which becomes 3d10. The empty 4s and three 4p orbitals take the four CO ligands:
\( \displaystyle \text{hybridisation } sp^3\ \Rightarrow\ \text{tetrahedral};\quad \text{no unpaired electrons}\ \Rightarrow\ \text{diamagnetic} \)
Why the other options are incorrect
Option (1) dsp2, square planar, paramagnetic — Incorrect. No nickel complex fits this combination: when \( \ce{Ni^{2+}} \) uses dsp2 hybridisation, all its eight d electrons are paired, so the complex is diamagnetic.
Option (3) dsp2, square planar, diamagnetic — Incorrect. This describes \( \ce{[Ni(CN)4]^{2-}} \). There nickel is \( \ce{Ni^{2+}} \) (3d8), and the strong field \( \ce{CN^{-}} \) pairs the eight electrons into four d orbitals, leaving one 3d orbital free for dsp2 hybridisation.
Option (4) sp3, tetrahedral, paramagnetic — Incorrect. This describes \( \ce{[NiCl4]^{2-}} \). There \( \ce{Cl^{-}} \) is a weak field ligand, so the two unpaired 3d electrons of \( \ce{Ni^{2+}} \) remain and the complex is paramagnetic.
NCERT (2026-27 edition):
Valence Bond Theory – “… In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. This is illustrated below for [NiCl4]2–. Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. … Each Cl– ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons. Similarly, [Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.”