Problem 57
Question
Compound \(\mathrm{X}\) is an anhydride of sulphuric acid. The number of sigma bonds and the number of pi bonds present in \(\mathrm{X}\) are respectively (a) 3,3 (b) 4,2 (c) 2,4 (d) 4,3
Step-by-Step Solution
Verified Answer
Option (a) 3,3 is correct: \( \text{SO}_3 \) has 3 sigma and 3 pi bonds.
1Step 1: Identify the Anhydride of Sulphuric Acid
Sulphuric acid is \( \text{H}_2\text{SO}_4 \). When we remove water from it, we get its anhydride. The anhydride of sulfuric acid is \( \text{SO}_3 \) (Sulfur trioxide).
2Step 2: Determine the Structure of SO3
\( \text{SO}_3 \) has a trigonal planar structure. It consists of a sulfur atom double bonded to three oxygen atoms. Each double bond is composed of one sigma (σ) bond and one pi (π) bond.
3Step 3: Count the Sigma Bonds in SO3
In \( \text{SO}_3 \), there are three sulfur-oxygen double bonds. Each double bond contains one sigma bond, thus there are three sigma bonds in total.
4Step 4: Count the Pi Bonds in SO3
Each of the three double bonds in \( \text{SO}_3 \) also contains one pi bond. Therefore, there are three pi bonds in total.
5Step 5: Conclusion and Answer Identification
From the analysis, the number of sigma bonds is 3 and the number of pi bonds is 3. Comparing this with the given options, the correct answer is option (a) 3,3.
Key Concepts
Sulfur TrioxideChemical BondingMolecular StructureLewis Structures
Sulfur Trioxide
Sulfur trioxide (\( \text{SO}_3 \)) is a significant compound in inorganic chemistry known for its role as the anhydride form of sulfuric acid. This means it is derived from sulfuric acid (\( \text{H}_2\text{SO}_4 \)) by removing water molecules. Sulfur trioxide is a colorless to white crystalline solid or a gas at higher temperatures, and it's highly reactive.
Ensuring safe handling and correct structural interpretation of \( \text{SO}_3 \) is important due to its reactive nature and corrosive properties.
- It mainly exists as monomer and polymer forms, but the gaseous phase primarily comprises the monomer.
- Upon cooling, it can form trimers, known as sulphur trioxide ice.
Ensuring safe handling and correct structural interpretation of \( \text{SO}_3 \) is important due to its reactive nature and corrosive properties.
Chemical Bonding
Chemical bonding in sulfur trioxide is characterized by its use of covalent bonds. Within \( \text{SO}_3 \), these covalent bonds are crucial for holding together the sulfur and oxygen atoms. The molecule features double bonds, which consist of one sigma (σ) and one pi (π) bond each. Understanding the type and nature of these bonds helps explain many of \( \text{SO}_3 \)'s properties.
- The sigma bond represents the strongest type of covalent bond formed by the head-on overlap of atomic orbitals.
- The pi bond is weaker, formed by the side-to-side overlap of p orbitals above and below the plane of the molecule.
- These bonds are crucial in providing the molecule with both stability and reactivity.
Molecular Structure
The molecular structure of sulfur trioxide is trigonal planar, a common geometry for molecules with three surrounding atoms. The central sulfur atom in \( \text{SO}_3 \) is surrounded symmetrically by oxygen atoms, making the structure planar and giving the molecules angles of 120°.
- In the trigonal planar structure, each oxygen atom is double-bonded to the sulfur atom.
- The equal distribution of electron pairs prevents polarity, making \( \text{SO}_3 \) a nonpolar molecule.
- Despite being nonpolar, \( \text{SO}_3 \) is highly reactive due to the presence of strong double bonds.
Lewis Structures
The Lewis structure is a key tool used to depict the bonding within a molecule like sulfur trioxide. It helps visualize the arrangement of atoms and the distribution of electrons. For \( \text{SO}_3 \), the Lewis structure shows the central sulfur atom double-bonded to three oxygen atoms.
- Each double bond consists of a pair of dots (representing the shared electrons) between sulfur and each oxygen atom.
- Lone pairs are not present on the central sulfur atom, as it makes full use of its available electrons in bonding. However, each oxygen atom completes its octet by forming double bonds with sulfur.
- The structure exemplifies the electron-sharing aspect of covalent bonding, explaining the stable electronic configuration achieved by \( \text{SO}_3 \).
Other exercises in this chapter
Problem 55
Which one of the following is a planar molecule? (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{BCl}_{3}\) (d) \(\mathrm{PCl}_{3}\)
View solution Problem 56
The molecule having largest dipole moment among the following is (a) \(\mathrm{CHI}_{3}\) (b) \(\mathrm{CH}_{4}\) (c) \(\mathrm{CHCl}_{3}\) (d) \(\mathrm{CCl}_{
View solution Problem 58
Which one of the following is a correct set? (a) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{3}\), angular (b) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{2}\), lin
View solution Problem 60
How many sigma and pi bonds are present in toluene? (a) \(3 \pi+15 \sigma\) (b) \(6 \pi+6 \sigma\) (c) \(3 \pi+6 \sigma\) (d) \(3 \pi+8 \sigma\)
View solution