Hybridization and shape of \(Cs[FeCl_4]\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(sp^3\), tetrahedral
In \( \ce{Cs[FeCl4]} \) the complex ion is \( \ce{[FeCl4]^{-}} \). Oxidation state of Fe: x + 4(−1) = −1, so x = +3, configuration 3d5.
\( \ce{Cl^{-}} \) is a weak field ligand and does not pair the d electrons, so all five 3d orbitals stay singly occupied. The four \( \ce{Cl^{-}} \) ligands donate their electron pairs into the empty 4s and three 4p orbitals, giving sp3 hybridisation and a tetrahedral ion with five unpaired electrons (paramagnetic), just as in \( \ce{[NiCl4]^{2-}} \).
Why the other options are incorrect
Option (1) \(d^2sp^3\), octahedral — Incorrect. d2sp3 needs six ligands and two empty inner 3d orbitals; \( \ce{[FeCl4]^{-}} \) has four ligands and no vacant 3d orbital.
Option (3) \(dsp^2\), tetrahedral — Incorrect. dsp2 hybridisation gives a square planar, not tetrahedral, shape, and needs an empty 3d orbital that \( \ce{Fe^{3+}} \) (3d5, high spin) does not have.
Option (4) \(sp^3d^2\), octahedral — Incorrect. sp3d2 is for six-coordinate outer orbital complexes such as \( \ce{[FeF6]^{3-}} \); here the coordination number is 4.
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. The hybridisation scheme is as shown in diagram. Each Cl– ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons.”