Which of the following complexes is tetrahedral?

The correct answer is Option (1) → [FeCl₄]⁻

A four-coordinate complex is tetrahedral when the metal uses one s and three p orbitals (sp³), and square planar when it uses one inner d, one s and two p orbitals (dsp²), which needs an empty inner d orbital.

• Oxidation state of Fe: in \( \ce{[FeCl4]^-} \), x + 4(−1) = −1, so x = +3.

• Configuration: Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): 3d⁵.

• Filling of the 3d orbitals: \( \ce{Cl^-} \) is a weak field ligand and does not force pairing, so the five 3d electrons stay unpaired, one in each 3d orbital; no 3d orbital is empty for dsp².

• Hybridisation: the four \( \ce{Cl^-} \) electron pairs enter the empty 4s and three 4p orbitals: sp³, tetrahedral.

Why the other options are incorrect

Option (2) [Ni(CN)₄]²⁻ — Incorrect. \( \ce{Ni^2+} \) is 3d⁸; the strong field \( \ce{CN^-} \) ligands pair the 3d electrons and free one 3d orbital, giving dsp², square planar.

Option (3) [PtCl₄]²⁻ — Incorrect. This d⁸ Pt(II) complex is the standard example of a square planar entity.

Option (4) [PdCl₂(NH₃)₂] — Incorrect.

• Pd lies in the same group as Pt, and Pd(II) (4d⁸), like Pt(II), forms square planar four-coordinate complexes.

• [PdCl₂(NH₃)₂] exists as cis and trans isomers, which a tetrahedral complex cannot show.

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.”