Which of the following statement is/are correct for complex \([NiCl_4]^{2-}\)

A. \(Ni\) has oxidation state +2

B. \(Cl^-\) is a weak field ligand

C. Compound is paramagnetic

D. \(dsp^2\) hybridization

E. Low spin complex

Choose the correct answer from the options given below:

A, B and C only

The correct answer is option 1. A, B and C only.

Let us break down the analysis of the complex \([NiCl_4]^{2-}\) in detail, considering each statement carefully:

A. Oxidation State of Nickel in \([NiCl_4]^{2-}\)

The statement is correct. In this complex, we know that the chloride ion (\(Cl^-\)) carries a -1 charge, and there are 4 chloride ions. The overall charge of the complex is -2. We can use this information to determine the oxidation state of nickel (\(Ni\)) by setting up the following equation:

\(\text{Oxidation state of Ni} + 4 \times (-1) = -2\)

\(⇒ \text{Oxidation state of Ni} = +2\)

So, in \([NiCl_4]^{2-}\), nickel has an oxidation state of +2.

B. \(Cl^-\) as a Weak Field Ligand

The statement is correct. Chloride ion (\(Cl^-\)) is considered a weak field ligand according to the spectrochemical series. Ligands in this series are ranked by their ability to split the d-orbitals of the central metal ion in a complex. Weak field ligands, like chloride, do not cause significant splitting of the d-orbitals. As a result, the energy gap between the split d-orbitals is relatively small. For weak field ligands, the electrons prefer to remain unpaired and occupy the higher energy orbitals rather than pair up in the lower energy orbitals. This leads to high-spin configurations.

C. Paramagnetism of the Complex 

The statement is correct. To determine whether the complex is paramagnetic or diamagnetic, we need to look at the electron configuration of nickel in its +2 oxidation state and how chloride influences its orbital filling. The electron configuration of \(Ni^{2+}\) (nickel in the +2 state) is \([Ar] 3d^8\). This means there are 8 electrons in the 3d orbitals. Since \(Cl^-\) is a weak field ligand, it causes a small splitting of the d-orbitals. As a result, the electrons prefer to remain unpaired as much as possible. In a tetrahedral arrangement (explained below), the 3d orbitals split into two sets, but the splitting is small, and some d-electrons remain unpaired. This presence of unpaired electrons makes the complex paramagnetic (i.e., attracted to an external magnetic field).

D. \(dsp^2\) Hybridization

The statement is incorrect. The geometry of the complex \([NiCl_4]^{2-}\) is tetrahedral, not square planar. In a tetrahedral complex, the central metal ion (in this case, \(Ni^{2+}\)) is surrounded by four ligands, and the geometry is arranged such that the bond angles between the ligands are close to 109.5°. For a tetrahedral geometry, the hybridization is \(sp^3\), which involves one s orbital and three p orbitals mixing to form four equivalent hybrid orbitals. The hybridization \(dsp^2\) is typically associated with square planar complexes, which is not the case here. Therefore, this statement is false because the complex has \(sp^3\) hybridization, not \(dsp^2\).

E. Low-Spin Complex

The statement is incorrect.  A low-spin complex occurs when the ligands cause a large splitting of the d-orbitals, forcing the electrons to pair up in the lower energy orbitals. In the case of \([NiCl_4]^{2-}\), chloride is a weak field ligand, meaning the splitting of the d-orbitals is small. This small splitting leads to a high-spin complex rather than a low-spin complex. In high-spin complexes, electrons remain unpaired in the higher energy orbitals. Since \([NiCl_4]^{2-}\) is tetrahedral and involves a weak field ligand, it is a high-spin complex.

Thus, the correct answer is 1. A, B, and C only.