Question

In which of the following molecules would you expect to find delocalized molecular orbitals: (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{SO}_{2} ;\) (c) \(\mathrm{H}_{2} \mathrm{CO}\) ? Explain.

Short answer

Delocalized molecular orbitals are present in Ethene (\(\mathrm{C}_{2} \mathrm{H}_{4}\)) and Sulfur Dioxide (\(\mathrm{SO}_{2}\)) but not in Formaldehyde (\(\mathrm{H}_{2} \mathrm{CO}\)).

Step-by-step solution

Understanding Delocalized Orbitals

Delocalized molecular orbitals extend over three or more atoms. They contribute to the stability of molecules and are usually present in molecules that exhibit resonance.

Analyzing Ethene \(\mathrm{C}_{2} \mathrm{H}_{4}\)

In ethene, which is a planar and symmetrical molecule, the six electrons from the carbon atoms are localized in three sigma bonds, two with hydrogen and one with the carbon atom. The remaining electron in each carbon forms a pi bond, which is a delocalized molecular orbital

Analyzing Sulfur Dioxide \(\mathrm{SO}_{2}\)

In sulfur dioxide, sulfur forms one sigma bond with each oxygen atom, and one of the oxygen atoms forms a pi bond with sulfur. However, the molecule has resonance structures, which means that the pi bond electrons are spread over three atoms making them delocalized.

Analyzing Formaldehyde \(\mathrm{H}_{2} \mathrm{CO}\)

In formaldehyde, carbon forms sigma bonds with hydrogen and oxygen, and forms a pi bond with oxygen. The pi bond is localized between carbon and oxygen, and there are no resonating structures. Therefore, there are no delocalized electrons in formaldehyde.

Key concepts

Resonance

Resonance is a fascinating concept in chemistry where molecules can be represented by two or more valid Lewis structures. These structures, known as resonance forms, differ only in the placement of electrons.
Resonance contributes to stability by allowing electrons to be spread over more than one atom, thus delocalizing them. This delocalization reduces a molecule’s potential energy, making it more stable.
In sulfur dioxide (\(\mathrm{SO}_{2}\)), resonance occurs because electrons in the pi bonds are shared between sulfur and both oxygen atoms. The pi bond isn’t fixed between two atoms but extends across all involved atoms, creating delocalized molecular orbitals.

• Resonance structures must have the same arrangement of atoms but differ in electron placement.
• Actual structure is a hybrid of all possible resonance forms.

Recognizing resonance can help you predict molecular stability and chemical reactivity.

Sigma and Pi Bonds

Understanding the differences between sigma and pi bonds is crucial for grasping molecular orbital theory.
Sigma (\(\sigma\)) bonds are the first bonds formed between two atoms. They are characterized by direct overlap of atomic orbitals along the axis connecting two nuclei. These bonds are strong and allow free rotation around the bond axis.

• Formed by head-on overlapping.
• Present in all single bonds.

Pi (\(\pi\)) bonds arise from the sidewise overlap of atomic orbitals. They are generally weaker than sigma bonds and do not allow free rotation.

• Formed by sideways overlapping.
• Present in double and triple bonds, in addition to sigma bonds.

In ethene (\(\mathrm{C}_{2} \mathrm{H}_{4}\)), each carbon has sigma bonds with two hydrogens and the other carbon. The pi bond formed by the sideways overlap of p orbitals between carbons is part of a delocalized molecular orbital. This is key to the molecule's reactivity and properties.

Molecular Stability

Molecular stability is influenced by the extent of delocalization of electrons. Delocalized electrons are less concentrated, lowering the potential energy of a molecule, which enhances stability.
Resonance is a contributor to this effect. When electrons are spread across multiple atoms, as seen in molecules like sulfur dioxide (\(\mathrm{SO}_{2}\)), it results in resonance stabilization. This effect can make a molecule more robust compared to one with localized electrons.

• Delocalization diminishes electron repulsion.
• Enhances chemical and thermal stability.

In formaldehyde (\(\mathrm{H}_{2}\mathrm{CO}\)), the lack of delocalization contributes to a lower stability compared to molecules exhibiting resonance. Recognizing the patterns of stabilization can be crucial in predicting molecular reactions and behaviors.