How many of the following have a tetrahedral geometry?
\( \ce{[Ni(CO)4]} \), \( \ce{[NiCl4]^2-} \), \( \ce{[Ni(CN)4]^2-} \), \( \ce{[PtCl4]^2-} \), \( \ce{[MnBr4]^2-} \), \( \ce{Fe(CO)5} \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
A four-coordinate entity is either tetrahedral (sp3) or square planar (dsp2), so the number of ligands alone does not decide the shape.
Tetrahedral:
\( \ce{[Ni(CO)4]} \) (nickel in the zero oxidation state; diamagnetic)
\( \ce{[NiCl4]^2-} \) (Ni2+, 3d8, uses one 4s and three 4p orbitals, sp3; two unpaired electrons remain)
\( \ce{[MnBr4]^2-} \) (5.9 BM shows five unpaired electrons, so no 3d orbital is free for dsp2; sp3)
Not tetrahedral:
\( \ce{[Ni(CN)4]^2-} \) — square planar; Ni2+ uses dsp2 hybrid orbitals and the complex is diamagnetic.
\( \ce{[PtCl4]^2-} \) — square planar.
\( \ce{Fe(CO)5} \) — trigonal bipyramidal; it has five CO ligands.
Tetrahedral: 3 entities.
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 below for [NiCl4]2-.” … “Similarly, [Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.”
“In the square planar complexes, the hybridisation involved is dsp2. An example is [Ni(CN)4]2–.”
5.2 (f) Coordination polyhedron – “…[PtCl4]2– is square planar.”
Example 5.7 – “…the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals.”
5.6 Bonding in Metal Carbonyls – “…Tetracarbonylnickel(0) is tetrahedral, pentacarbonyliron(0) is trigonalbipyramidal while hexacarbonyl chromium(0) is octahedral.”