Which of the following statement is/are correct for complex [NiCl_4]^…

The correct answer is Option (1) → A, B and C only

Oxidation state: x + 4(−1) = −2 ⇒ x = +2, so the metal is \( \ce{Ni^{2+}} \).

Electron count: Ni (Z = 28) = [Ar] 3d⁸ 4s². \( \ce{Ni^{2+}} \) loses the two 4s electrons ⇒ 3d⁸.

Filling five 3d orbitals with 8 electrons: \( \ce{Cl^{-}} \) is a weak-field ligand and cannot force pairing, so Hund’s rule applies — one electron in each of the five orbitals (5), then three more pair up (8):
↑↓  ↑↓  ↑↓  ↑  ↑ ⇒ 2 unpaired electrons

No 3d orbital is free, so the four \( \ce{Cl^{-}} \) lone pairs enter one 4s and three 4p orbitals: sp³ hybridisation, tetrahedral, outer-orbital (high spin) complex. μ = √[n(n + 2)] = √8 ≈ 2.83 BM, so it is paramagnetic.

• (A) Ni has oxidation state +2 — True. x + 4(−1) = −2 gives x = +2.

• (B) Cl⁻ is a weak field ligand — True. It lies near the weak end of the spectrochemical series.

• (C) Compound is paramagnetic — True. It has two unpaired electrons.

• (D) dsp² hybridization — False. dsp² (square planar) needs a vacant 3d orbital, which arises only when a strong-field ligand such as \( \ce{CN-} \) pairs the electrons, as in \( \ce{[Ni(CN)4]^2-} \).

• (E) Low spin complex — False. The electrons stay unpaired as far as possible; it is a high spin complex.

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.”

Valence Bond Theory – “… Each Cl⁻ ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons.”

Crystal Field Theory – “… In general, ligands can be arranged in a series in the order of increasing field strength as given below:
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO
Such a series is termed as spectrochemical series.”