Question

Nickel combines with a uninegative monodentate ligand (X^-) to form a paramagnetic complex [NiX4]^2-.The hybridization involved and number of unpaired electrons present in the complex and respectively

• A. sp³,two
• B. dsp², zero
• C. dsp²,one
• D. sp³,one

Answer 1

The Correct Option is A

Given complex: [NiX₄]²⁻

Step 1: Oxidation state of Ni
The ligand X⁻ is uninegative and there are 4 of them. The overall charge on the complex is 2−.
Let oxidation state of Ni be +x: 
$x + 4(-1) = -2 \Rightarrow x = +2$

Step 2: Electronic configuration of Ni²⁺
Ni atomic number = 28
Electronic configuration of Ni = [Ar] 4s² 3d⁸
Ni²⁺ = [Ar] 3d⁸

Step 3: Nature of ligand and hybridization
X⁻ is a weak field ligand (like Cl⁻, Br⁻, etc.), so it does not cause pairing of electrons.
Thus, the 3d⁸ configuration remains with two unpaired electrons.
Since the geometry is tetrahedral, the hybridization is sp³.

Answer: sp³ hybridization with two unpaired electrons.

Correct option: (A)

Answer 2

Given: A paramagnetic complex [NiX₄]²⁻ is formed with uninegative monodentate ligands (X⁻).

Step 1: Determine oxidation state of Ni

\[ \text{Let oxidation state of Ni be } x: \quad x + 4(-1) = -2 \Rightarrow x = +2 \]

So, Ni is in the +2 oxidation state → Ni²⁺

Step 2: Electronic configuration of Ni and Ni²⁺

Atomic number of Ni = 28

\[ \text{Ni: } [Ar]\,3d^8\,4s^2 \quad \Rightarrow \quad \text{Ni}^{2+}: [Ar]\,3d^8 \]

Step 3: Nature of ligand (X⁻)

X⁻ is a weak field ligand (like Cl⁻, Br⁻, etc.), so it does not cause pairing of electrons.

Step 4: Determine hybridization and geometry

• With weak field ligands, Ni²⁺ retains its unpaired electrons.
• It uses outer orbital hybridization → sp³ (from 4s and 4p orbitals).
• This leads to a tetrahedral geometry.

Step 5: Unpaired electrons

Ni²⁺ has 8 electrons in 3d → filling 3d orbitals as: ↑↓ ↑↓ ↑ ↑ ↑ (i.e., 2 unpaired electrons)

Final Answer: sp³, two