Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

In \(\mathrm{SO}_{3}^{2-}\) (a) there is sp \(^{3}\) hybridized sulphur atom (b) bonds between \(\mathrm{S}\) and \(\mathrm{O}\) are equivalents (c) \(\mathrm{d} \pi-\mathrm{p} \pi\) bond between \(\mathrm{S}\) and \(\mathrm{O}\) is delocalized (d) all the facts given here are true

Step-by-Step Solution

Answer

Statements (b) and (c) are true; (a) is false.

Step 1: Analyze Hybridization

To determine the hybridization of sulfur in sulfate ion \( \mathrm{SO}_{3}^{2-} \), we first need to count the number of sigma bonds and lone pairs around the sulfur atom. Sulfur forms three sigma bonds with oxygen and has one lone pair, making a total of four regions of electron density. Thus, the hybridization is \( \mathrm{sp}^{3} \), contradicting the statement in (a).

Step 2: Check Bond Equivalency

The structure of \( \mathrm{SO}_{3}^{2-} \) can be represented as a resonance hybrid of three equivalent structures, each having one double bond and two single bonds, which average out to result in equivalent \( \mathrm{S} - \mathrm{O} \) bonds. Hence, statement (b) is true.

Step 3: Examine \( \pi\) Bond Delocalization

In \( \mathrm{SO}_{3}^{2-} \), the \( \mathrm{S} - \mathrm{O} \) bonds involve \( \pi\) bond delocalization due to resonance. The \( d \pi - p \pi\) interactions result in the delocalization of electrons across all the \( \mathrm{S} - \mathrm{O} \) bonds. This means statement (c) is also true.

Step 4: Determine Truth of All Statements

Now that we have concluded that statement (a) is false and statements (b) and (c) are true, we can determine that not all statements are true. Therefore, option (d) is incorrect.

Key Concepts

Understanding Hybridization

Hybridization is a concept that helps us understand the way atomic orbitals combine to form new orbitals. In the case of the sulfur trioxide ion \(\mathrm{SO}_{3}^{2-}\), sulfur is the central atom. To figure out the hybridization of sulfur, count the number of sigma (\(\sigma\)) bonds it forms, plus any lone pairs of electrons. Here, sulfur forms three \(\sigma\) bonds with oxygen atoms, which are robust, single covalent bonds. Additionally, sulfur holds one lone pair of electrons. Thus, the number of regions of electron density around sulfur totals four: three \(\sigma\) bonds plus one lone pair. This gives sulfur \(\mathrm{sp}^{3}\) hybridization. It's essential because it describes the arrangement of orbitals and explains why certain molecular shapes occur, helping make sense of the molecule's geometry.

Exploring Resonance Structures

Resonance structures are multiple ways to depict a molecule’s bonding using Lewis structures. They give us a better understanding of how electrons are distributed in a molecule.For \(\mathrm{SO}_{3}^{2-}\), the concept of resonance structures is vital because it helps us visualize the different possible configurations of bonds within the molecule. In this ion, we can draw three different resonance structures, each depicting one double bond (\(\mathrm{S}=\mathrm{O}\)) with two single bonds (\(\mathrm{S}-\mathrm{O}\)) between sulfur and oxygen atoms. These resonance structures show that the position of the double bond alternates among the different oxygen atoms, providing a simplistic average view of the true state of electron distribution.

Understanding Pi Bond Delocalization

\(\pi\) bond delocalization is an important concept in understanding the behavior of certain molecules, including sulfur compounds. In \(\mathrm{SO}_{3}^{2-}\), \(\pi\) bond delocalization occurs because the electrons that form the \(\pi\) bonds are not confined to just one \(\mathrm{S}=\mathrm{O}\) double bond.Instead, this \(\pi\) bonding extends over all \(\mathrm{S}-\mathrm{O}\) bonds, allowing the electrons to move freely around the molecule. This delocalization comes from the \(d\pi\) - \(p\pi\) interactions between sulfur and oxygen, which promotes a shared electron cloud over the entire molecule. This delocalization is what allows the \(\mathrm{SO}_{3}^{2-}\) ion to exhibit stability and accounts for equivalent bond lengths throughout the ion.

Understanding Equivalent Bonds

Equivalent bonds are bonds that, despite initial appearances, exhibit the same characteristics concerning bond length and strength. In molecules like \(\mathrm{SO}_{3}^{2-}\), although individual resonance structures show different bond types between sulfur and oxygen, the molecule itself has equivalent bonds.This equivalency arises from the nature of the resonance hybrid, which is more representative of the actual molecular structure than any one individual resonance structure. In \(\mathrm{SO}_{3}^{2-}\), average bond lengths are achieved through the constant \(\pi\) bond delocalization, effectively making every \(\mathrm{S}-\mathrm{O}\) bond identical Equivalent bonds thus ensure a uniform distribution of bonding character across the molecule, providing greater stability and an insightful example of resonance's impact on molecular structure.