Answer the question on the basis of passage given below:
Transition metals form a large number of complexes or coordination compounds in which the metal atoms are bound to a number of anions or neutral molecules. The valence bond theory explains the formation, magnetic behaviour and geometrical shapes while the crystal field theory explains the effect of different crystal fields on the degeneracy of d-orbital energies of the central metal atom/ion. This provides for the quantitative estimation of orbital separation energies, magnetic moments and spectral and stability parameters.
Which of the following complexes are not correctly matched with hybridization of their central metal ion?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \([MnCl_4]^{2-} : dsp^2\)
dsp2 hybridisation needs one vacant inner (n−1)d orbital. Whether one is free depends on the d-electron count of the metal ion and on whether the ligand is strong enough to pair electrons.
Option (1) [Ni(H2O)6]2+ : sp3d2 — Correctly matched. Ni (Z = 28) [Ar]3d84s2 → \( \ce{Ni^2+} \) 3d8: [↑↓] [↑↓] [↑↓] [↑ ] [↑ ]. Two inner d orbitals cannot be vacated, so the six \( \ce{H2O} \) use 4s, 4p and 4d: outer orbital, octahedral, 2 unpaired.
Option (2) [CoF6]3− : sp3d2 — Correctly matched. Co (Z = 27) [Ar]3d74s2 → \( \ce{Co^3+} \) 3d6. Weak-field \( \ce{F^-} \) causes no pairing: [↑↓] [↑ ] [↑ ] [↑ ] [↑ ], so the outer 4d orbitals are used: octahedral, high spin, 4 unpaired.
Option (3) [Cu(NH3)4]2+ : dsp2 — Correctly matched. Cu (Z = 29) [Ar]3d104s1 → \( \ce{Cu^2+} \) 3d9. The complex is square planar; in valence bond terms the odd electron is moved to a 4p orbital, freeing one 3d orbital for dsp2 bonding, and 1 unpaired electron remains.
Option (4) [MnCl4]2− : dsp2 — Not correctly matched (sp3, tetrahedral). Mn (Z = 25) [Ar]3d54s2 → \( \ce{Mn^2+} \) 3d5: [↑ ] [↑ ] [↑ ] [↑ ] [↑ ]. No 3d orbital is empty and weak-field \( \ce{Cl^-} \) cannot force pairing, so the four \( \ce{Cl^-} \) use the 4s and three 4p orbitals: sp3, tetrahedral, 5 unpaired electrons.
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.”