Question

(a) Write a single Lewis structure for \(\mathrm{SO}_{3}\), and determine the hybridization at the \(\mathrm{S}\) atom. (b) Are there other equivalent Lewis structures for the molecule? (c) Would you expect \(\mathrm{SO}_{3}\) to exhibit delocalized \(\pi\) bonding? Explain.

Short answer

(a) The Lewis structure of SO3 is: O \\ S = O / O The hybridization of the S atom is sp². (b) Yes, there are other equivalent Lewis structures for SO3, representing a resonance hybrid of these structures: O O– –O \\ \• •/ S = O S = O O = S / •/ \• O– –O O (c) Yes, SO3 exhibits delocalized π bonding due to the overlapping of p orbitals and the sharing of π electrons between the sulfur and three oxygen atoms. This stabilizes the molecule and contributes to its overall behavior.

Step-by-step solution

(a) Drawing Lewis structure for SO3 and determining hybridization

To draw the Lewis structure of SO3, we need to follow these steps: 1. Count the total valence electrons: Sulfur (S) has 6 valence electrons, and each oxygen (O) atom has 6 valence electrons. Total valence electrons = 6 + (3x6) = 6 + 18 = 24. 2. Choose the central atom: Sulfur is less electronegative than oxygen, so it will be the central atom. Place the central atom (S) in the center and arrange the remaining oxygen atoms around it. 3. Place the bonds: Sulfur must form bonds with all three oxygen atoms. Since a single bond consists of 2 electrons, we need 6 electrons from our total valence electrons to form the 3 single bonds (S–O). 4. Distribute remaining electrons: We still have 18 valence electrons to distribute (24 - 6). Fill the valence shell of the oxygen atoms with these electrons. It takes the other 18 electrons to complete the valence shells of 3 oxygen atoms, making each oxygen have 6 electrons (2 pairs) in a lone pair. 5. Check formal charges: Formal charges are calculated using the formula: Formal Charge = (valence electrons) - (lone pair electrons + 1/2 bonded electrons). Your Lewis structure for SO3 should look like this: O \\ S = O / O Now, let's determine the hybridization of the S atom. Sulfur has three electron domains around it (3 single bonds with O atoms). It needs three orbitals to hybridize. The three orbitals that hybridize are one 3s and two 3p orbitals, which results in an sp² hybridization. The S atom forms three sigma (σ) bonds with three O atoms.

(b) Are there other equivalent Lewis structures for the molecule?

Yes, there are other equivalent Lewis structures for the SO3 molecule. These resonance structures can be drawn by placing a double bond between sulfur and one of the oxygen atoms, and single bonds with the other two oxygen atoms. We can draw three resonance structures like this, each representing one of the oxygen atoms with a double bond to the sulfur. The true structure of SO3 is a resonance hybrid of these equivalent structures. O O– –O \\ \• •/ S = O S = O O = S / •/ \• O– –O O

(c) Would you expect SO3 to exhibit delocalized π bonding? Explain.

Yes, we would expect SO3 to exhibit delocalized π bonding. This delocalization occurs when the overlapping of p orbitals allows the π electrons to move freely between multiple atoms. In the case of SO3, the π electrons in the double bond (in the different resonance structures) are shared between the sulfur and three oxygen atoms, resulting in delocalized π bonding. This phenomenon stabilizes the molecule and contributes to the molecule's overall behavior.

Key concepts

SO3 Hybridization

Understanding the hybridization of the sulfur trioxide molecule (SO3) gives us insight into its bonding and shape. Hybridization explains how atomic orbitals combine to form new orbitals that can form covalent bonds. In the molecule SO3, the central sulfur (S) atom bonds with three oxygen (O) atoms. To comprehend the SO3 hybridization, we count the regions of electron density around the sulfur atom—each single bond counts as one region. With three single bonds to oxygen atoms, the sulfur central atom has three regions of electron density, which leads to sp² hybridization.

This implies that one 3s orbital and two 3p orbitals from the sulfur atom have merged to form three equivalent sp² hybrid orbitals. These hybridized orbitals overlap with the oxygen atoms' orbitals to form sigma (σ) bonds. The unhybridized 3p orbital remains available for π bond formation, which is crucial to the molecule's resonance. Constructing the SO3's Lewis structure assists in visualizing its arrangement and the central atom's hybridization.

Resonance Structures

When depicting molecules like sulfur trioxide, we encounter the concept of resonance structures. These structures are different ways to draw the same molecule, representing the same distribution of electrons. They illustrate the molecule's electron delocalization, where electrons are not fixed in place but can move between various configurations. For SO3, we can draw multiple Lewis structures that show alternate arrangements of double and single bonds between the sulfur and oxygen atoms.

Understanding Resonance in SO3

The true form of SO3 is not perfectly depicted by any single Lewis structure. Instead, it is a hybrid of all possible resonance structures. We illustrate this by using arrows to switch the position of double and single bonds in the drawings. The equal distribution of electrons in reality means that all S–O bonds in SO3 are identical, and the molecule is more stable than any individual resonance structure suggests. Each oxygen atom in turn forms a double bond with sulfur, leading to the creation of three possible resonance structures. Through these structures, the resonance concept becomes a vital tool in predicting the electronic arrangement and stability in molecules.

Delocalized π Bonding

Delocalized π (pi) bonding is a critical concept when examining the electron structure of molecules like SO3. It involves the sharing of electrons across multiple atoms, not just two. In the context of SO3, this delocalized bonding stems from the p orbitals of the sulfur and oxygen atoms overlapping 'sideways', allowing the π electrons to move freely among them.

Role of π Bonding in Stability

This type of bonding has significant implications for the molecule's stability. In SO3, the delocalization of π electrons across the S–O bonds creates a uniform negative charge cloud above and below the plane of the atoms, lowering the potential energy and thus stabilizing the molecule. The existence of resonance structures in SO3 demonstrates the presence of delocalized π bonding. Instead of being confined to one S=O bond, the π electrons are shared, which results from the molecular orbital formation involving the unhybridized p orbitals. Consequently, delocalized π bonding is critical to understanding why SO3 is more stable and less reactive than it would be if its π electrons were localized.