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

Question

In \(\mathrm{SO}_{3}{\underline{\phantom{xx}}}^{2-}\) (a) there is sp \(^{3}\) hybridized sulphur atom (b) bonds between \(\mathrm{S}\) and \(\mathrm{O}\) are equivalents (c) 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

All statements are true (Option d).

Step 1: Determine Sulfur Hybridization

To find the hybridization of sulfur (S) in \( \text{SO}_{3}^{2-} \), consider the number of regions of electron density around the sulfur atom. The sulfate ion has 3 double-bonded oxygen atoms surrounding sulfur, with one lone pair on sulfur, making a total of 4 regions. This means the hybridization of sulfur is \( \text{sp}^3 \), corroborating statement (a) as true.

Step 2: Analyze S-O Bond Character

We must check the equivalence between \( \text{S} \) and \( \text{O} \) bonds. In \( \text{SO}_{3}^{2-} \), due to resonance, the double bonds are distributed across all S-O bonds, making them equivalent and double bond in character. Hence, statement (b) is true.

Step 3: Evaluate Delocalization of π Bonds

In \( \text{SO}_{3}^{2-} \), there is delocalization of \( d\pi-p\pi \) bonds. Delocalization occurs because the \( 3d \) orbitals of sulfur can overlap with \( 2p \) orbitals of oxygen, allowing for \( \pi \) bonding that spreads out over the entire ion. Therefore, statement (c) is true.

Step 4: Verify All Statements

Since all the individual statements (a), (b), and (c) have been verified as true, statement (d) must also be true. This makes statement (d) the correct option.

Key Concepts

sp3 Hybridization

When atoms form molecules, their orbitals can mix and form new hybridized orbitals. This is fundamentally what happens in the sulfate ion, \( \text{SO}_{3}^{2-} \), with the sulfur atom. The concept of \(\text{sp}^{3}\) hybridization occurs when one \( s \) and three \( p \) orbitals mix together to form four equivalent \(\text{sp}^{3}\) hybrid orbitals. Each of these hybrid orbitals can hold pairs of electrons and form bonds. In \( \text{SO}_{3}^{2-} \), the sulfur atom is surrounded by four regions of electron density: three oxygen atoms form bonds with sulfur, and there is a lone pair on sulfur. Thus, there are four orbitals available to house electrons, which corresponds to \(\text{sp}^{3}\) hybridization.

• This arrangement allows for the formation of stable bonds with the oxygen atoms.
• The tetrahedral arrangement minimizes the electron-pair repulsion, following VSEPR theory.

By understanding hybridization, we can predict the molecular geometry and bond angles in the sulfate ion.

Resonance in Chemistry

Resonance is a concept in chemistry that describes the delocalization of electrons across multiple structures in a molecule. This is essential for understanding compounds such as \( \text{SO}_{3}^{2-} \), where the distribution of electrons leads to equivalency in all sulfur-oxygen bonds. In resonance, rather than having electrons fixed in one double bond configuration, they are shared across all possible positions.

• In \( \text{SO}_{3}^{2-} \), the electrons originally localized in \( \pi \) (double bonds) are delocalized across all \( \text{S-O} \) bonds.
• This makes each bond between sulfur and oxygen a partial double bond.

Because of resonance, these bonds are equivalent, providing enhanced stability to the sulfate ion.Understanding resonance is critical because it influences the molecular structure, bond lengths, and overall stability of molecules like the sulfate ion.

Delocalization of Pi Bonds

Pi (c) bond delocalization is a crucial concept in chemistry for explaining certain structural and reactive properties of molecules. Delocalization refers to the spreading out of \( \pi \) electrons over several atoms, rather than being localized between just two. For \( \text{SO}_{3}^{2-} \), delocalization happens through \( d\pi - p\pi \) bonding involving the overlap between sulfur's \( 3d \) orbitals and oxygen's \( 2p \) orbitals.

• Such delocalization allows the \( \pi \) electrons to spread throughout the entire ion.
• This results in a network of \( \pi \) bonds that gives the ion additional stability and resonance.

Delocalization not only stabilizes molecules but also plays a role in their reactivity. By understanding how pi bonds delocalize in compounds like \( \text{SO}_{3}^{2-} \), we gain insight into their unique properties and behaviors.