Chapter 2: Problem 240
\(s p^{3}\) hybridization is present in: (a) \(\mathrm{SO}_{4}^{2-}\) \(\square\) (b) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (c) \(\mathrm{BO}_{3}^{3-}\) \(\square\) (d) \(\mathrm{NH}_{4}^{+}\)
Short Answer
Expert verified
(a) \(\text{SO}_4^{2-}\) and (d) \(\text{NH}_4^+\).
Step by step solution
01
Understand the Concept of Hybridization
In chemistry, hybridization is a model to explain the arrangement of atomic orbitals during the bonding of atoms in molecules. The term \(sp^3\) hybridization denotes one "s" orbital combining with three "p" orbitals to form four equivalent \(sp^3\) hybrid orbitals. This concept is usually associated with tetrahedral geometry and single bonding (sigma bonds), like in methane (\(CH_4\)).
02
Examine Each Option for \(sp^3\) Hybridization
We have four options. Analyze them based on the tetrahedral geometry requirement and the presence of four equivalent bonds:- **Option (a)** \(\text{SO}_4^{2-}\): This ion has a tetrahedral geometry, suggesting \(sp^3\) hybridization.- **Option (b)** \(\text{C}_2\text{H}_2\): Known as ethyne or acetylene, this molecule exhibits a linear geometry with a triple bond, indicating \(sp\) hybridization, not \(sp^3\).- **Option (c)** \(\text{BO}_3^{3-}\): This is a planar trigonal structure, typically associated with \(sp^2\) hybridization, not \(sp^3\).- **Option (d)** \(\text{NH}_4^+\): Ammonium ion exhibits tetrahedral geometry and is known for \(sp^3\) hybridization.
03
Identify Correct Option(s)
Based on the examination, we find that options (a) \(\text{SO}_4^{2-}\) and (d) \(\text{NH}_4^+\) both exhibit tetrahedral geometry and fulfill the conditions for \(sp^3\) hybridization. Thus, they show \(sp^3\) hybridization.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tetrahedral Geometry
Tetrahedral geometry is a term used in chemistry to describe the spatial arrangement of atoms around a central atom. This occurs when there are four groups of electrons entirely around one central atom, forming structures seen in certain molecules and ions. A perfect tetrahedron has bond angles of 109.5 degrees, providing an even distribution of electron density.
A classic example of a molecule with tetrahedral geometry is methane (\(\text{CH}_4\)). Here, one carbon atom is at the center, bonded to four hydrogen atoms positioned at the corners of a tetrahedron. Each bond is identical and distributed symmetrically around the central atom. Such a structure requires \(sp^3\) hybridization, where one 's' orbital and three 'p' orbitals merge to create four equivalent \(sp^3\) hybrid orbitals.
A classic example of a molecule with tetrahedral geometry is methane (\(\text{CH}_4\)). Here, one carbon atom is at the center, bonded to four hydrogen atoms positioned at the corners of a tetrahedron. Each bond is identical and distributed symmetrically around the central atom. Such a structure requires \(sp^3\) hybridization, where one 's' orbital and three 'p' orbitals merge to create four equivalent \(sp^3\) hybrid orbitals.
- Uniform distribution of electron clouds maintains stability in the tetrahedral shape.
- Commonly observed in molecules and polyatomic ions like ammonium, sulfate, etc.
Ammonium Ion
The ammonium ion (\(\text{NH}_4^+\)) is a positively charged polyatomic ion formed when ammonia (\(\text{NH}_3\)) undergoes protonation. In this process, a lone pair of electrons from the nitrogen atom in ammonia gains a hydrogen ion (\(\text{H}^+\)).
This ion exhibits a tetrahedral geometry due to its \(sp^3\) hybridization. Each of the four hydrogen atoms in ammonium is equidistant from the nitrogen atom, establishing identical \(\text{N-H}\) bonds. Tetrahedral geometry comes into play here, resulting from the mix of one \(s\) orbital and three \(p\) orbitals from the nitrogen, forming four equivalent \(sp^3\) orbitals.
This ion exhibits a tetrahedral geometry due to its \(sp^3\) hybridization. Each of the four hydrogen atoms in ammonium is equidistant from the nitrogen atom, establishing identical \(\text{N-H}\) bonds. Tetrahedral geometry comes into play here, resulting from the mix of one \(s\) orbital and three \(p\) orbitals from the nitrogen, forming four equivalent \(sp^3\) orbitals.
- The resulting structure is symmetric, contributing to the ion's stability.
- Ammonium ions are crucial in biochemical processes and are found in many biological systems.
Sulfate Ion
The sulfate ion (\(\text{SO}_4^{2-}\)) is another example of a polyatomic ion exhibiting tetrahedral geometry and \(sp^3\) hybridization. It consists of a central sulfur atom surrounded by four oxygen atoms, arranged symmetrically at the corners of a tetrahedron.
In the sulfate ion, the sulfur atom uses one 's' and three 'p' orbitals to form the four equivalent \(sp^3\) hybrid orbitals necessary for bonding with oxygen atoms.
In the sulfate ion, the sulfur atom uses one 's' and three 'p' orbitals to form the four equivalent \(sp^3\) hybrid orbitals necessary for bonding with oxygen atoms.
- This symmetric arrangement results in a stable and well-distributed charge across the ion.
- The \(\text{S-O}\) bonds are characterized by resonance, giving sulfate its stability and functional versatility.