Problem 46
Question
What is the number of sigma and pi bonds present in a molecule of sulphuric acid? (a) \(6 \sigma, 2 \pi\) (b) \(6 \sigma, 0 \pi\) (c) \(2 \sigma, 4 \pi\) (d) \(2 \sigma, 2 \pi\)
Step-by-Step Solution
Verified Answer
The correct answer is (a) \(6 \sigma, 2 \pi\).
1Step 1: Understanding the Structure of Sulphuric Acid
Sulphuric acid has the molecular formula \(H_2SO_4\). Its structure consists of a central sulphur atom bonded to four oxygen atoms, with two of those oxygen atoms also bonded to hydrogen atoms.
2Step 2: Identifying Sigma Bonds
Sigma bonds (\(\sigma\)) are single covalent bonds formed by the direct overlap of atomic orbitals. In sulphuric acid: - Each \(S-O\) bond is a sigma bond. Since there are four \(S-O\) bonds, there are 4 sigma bonds.- Each \(O-H\) bond is also a sigma bond, adding 2 more sigma bonds.Total sigma bonds = 6 (from 4 \(S-O\) and 2 \(O-H\) bonds).
3Step 3: Identifying Pi Bonds
Pi bonds (\(\pi\)) are formed by the sideways overlap of p-orbitals. In sulphuric acid, the sulphur is double-bonded to two oxygen atoms. Each \(S=O\) double bond consists of one sigma bond and one pi bond.Total pi bonds = 2 (from 2 \(S=O\) pi bonds).
4Step 4: Conclusion
Adding the sigma and pi bonds, sulphuric acid has 6 sigma bonds and 2 pi bonds.
Key Concepts
Sigma BondsPi BondsMolecular Structure
Sigma Bonds
Sigma bonds, denoted as \(\sigma\) bonds, are the foundation of molecular structures. They are the strongest type of covalent chemical bond and form the primary connections between atoms. When two atomic orbitals overlap directly, they create a sigma bond. This type of bond allows atoms to rotate freely around the bond axis, leading to increased flexibility within molecules.
- They can form between s-s, s-p, or p-p orbitals.
- Every single bond is a sigma bond.
- Only one sigma bond can exist between any two bonded atoms.
Pi Bonds
Pi bonds, represented as \(\pi\) bonds, contribute to the unique aspects of molecular bonding. These bonds form when the sides of atomic orbitals overlap, which typically involves p-orbitals. Pi bonds are found in double and triple bonds and provide additional strength to molecular connections.
- A double bond consists of one sigma and one pi bond.
- A pi bond restricts the rotation of bonded atoms about the bond axis, creating structural rigidity.
- Pi bonds are generally weaker than sigma bonds due to their sideways overlap.
Molecular Structure
The molecular structure of a compound provides a visual representation of how atoms are arranged in space. This arrangement determines a molecule's physical and chemical properties. In the case of sulphuric acid \((H_2SO_4)\), understanding its molecular structure is essential for recognizing how the bonds and atoms interact.
Sulphuric acid has a central sulfur atom surrounded by four oxygen atoms, with each of the two hydrogens bonded to an oxygen. This structure can be described using a Lewis diagram or structural formula to show the bonding and lone pairs of electrons.
Sulphuric acid has a central sulfur atom surrounded by four oxygen atoms, with each of the two hydrogens bonded to an oxygen. This structure can be described using a Lewis diagram or structural formula to show the bonding and lone pairs of electrons.
- The central position of sulfur allows for coordination with multiple oxygen atoms.
- Two double bonds \((S=O)\) in the structure illustrate the presence of pi bonds.
- Lone pairs on oxygen atoms are crucial for understanding reactivity and polarity.
Other exercises in this chapter
Problem 44
The boiling point of alcohol is high due to (a) hydrogen bonding (b) large size of alcohol (c) presence of \(-\mathrm{OH}\) group (d) high molecular weight
View solution Problem 45
The hybridization of \(\mathrm{I}\) in \(\mathrm{IF}_{3}\) is (a) \(\mathrm{sp}^{3} \mathrm{~d}\) (b) \(\mathrm{sp}^{3}\) (c) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\
View solution Problem 47
The shape of IF \(_{7}\) molecule is (a) pentagonal bipyramidal (b) trigonal pyramidal (c) tetrahedral (d) square planar
View solution Problem 48
The highest dipole moment is of (a) \(\mathrm{CF}_{4}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{CO}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{~F}\)
View solution