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
- sp3,two
- dsp2, zero
- dsp2,one
- sp3,one
The Correct Option is A
Approach Solution - 1
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)
Approach Solution -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
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Concepts Used:
d block elements
The elements, in the columns of the periodic table in which d subshells are being occupied are known as d block elements.
These are the elements that have the capability of forming stable cations with incompletely filled d orbitals. Elements like mercury and Zinc are not considered transition metals because they have electronic configurations: (n-1)d10 ns2. These elements have filled d-orbitals in their ground state and, therefore, even in some of their oxidation states.
General Properties Of d-Block Elements
- Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
- Formation of complex compounds- Ligands show a binding behaviour and can form so many stable complexes with the help of transition metals. This property is mainly due to:
- Availability of vacant d orbitals.
- Comparatively small sizes of metals.
- Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
- Melting and boiling points- Melting and boiling points of transition are very high. This is because of the presence of unpaired electrons and partially filled d orbitals. Because of these two things, they form strong bonds and therefore have high melting and boiling points.
- Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
- Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements