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Draw the Lewis structure for acetamide, \(\mathrm{CH}_{3} \mathrm{CONH}_{2} .\) What are the electron-pair geometry and molecular geometry around the two C atoms? What is the hybridization of each of the \(\mathrm{C}\) atoms? What orbitals overlap to form the \(\sigma\) and \(\pi\) bonds between carbon and oxygen?

Short Answer

Expert verified
Acetamide has tetrahedral geometry/\(sp^3\) for the \(CH_3\) carbon; trigonal planar/\(sp^2\) for the carbonyl carbon. The \(\sigma\)-bond involves \(sp^2/sp^2\) orbitals, and the \(\pi\)-bond involves \(p/p\) orbitals.

Step by step solution

01

Identify Atoms and Electrons

First, determine the number of valence electrons for each atom in acetamide (\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\)). Carbon has 4 valence electrons, hydrogen has 1, oxygen has 6, and nitrogen has 5. Calculate the total valence electrons by adding up each contribution: 2 carbons (2 x 4 = 8), one oxygen (6), one nitrogen (5), and five hydrogens (5 x 1 = 5). This results in 24 valence electrons.
02

Sketch the Basic Structure

Start by arranging the atoms in a reasonable order: \(CH_3\) group, then a carbonyl group (C=O) next, followed by \(NH_2\). Connect them with single bonds initially: \( \text{H}_3\text{C}\text{-C=O-}\text{N}\text{-}\text{H}_2\).
03

Complete Octet on Outer Atoms

Next, complete the octet for the electronegative outer atoms, oxygen and nitrogen, by placing lone pairs: The oxygen will have two lone pairs, and the nitrogen will have one lone pair after forming the \(NH_2\) group.
04

Form Double Bonds where Necessary

Since oxygen is more electronegative, form a double bond between carbon and oxygen to satisfy the octet rule. This is common for carbonyl groups: \(\text{C}-\text{O}\).
05

Count Valence Electrons

Check the usage of valence electrons: \( \text{6 bonds (12 electrons)} + \text{2 lone pairs on O (4 electrons)} + \text{1 lone pair on N (2 electrons)} = 18 electrons\). Be sure all atoms have full valence shells with the remaining six electrons, adjusting bonds or lone pairs if needed.
06

Determine Geometry and Hybridization of C Atoms

Examine the central carbon atoms in the structure. The carbon of \(CH_3\) is bonded to three hydrogens and one other carbon atom (tetrahedral geometry, \( sp^3\) hybridization). The carbon in the carbonyl group is bonded to one oxygen and two other atoms (triangular planar geometry, \( sp^2\) hybridization).
07

Identify Overlapping Orbitals for Bonds

For the \(\sigma\)-bond between carbon and oxygen, the orbitals that overlap are the \(sp^2\)-hybridized orbital of carbon with the \(sp^2\)-hybridized orbital of oxygen. The \(\pi\)-bond is formed from the side-by-side overlap of unhybridized \(p_z\) orbitals of carbon and oxygen.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron-Pair Geometry
Understanding electron-pair geometry is key to predicting the shape and properties of molecules like acetamide. In essence, electron-pair geometry considers all electron pairs surrounding a central atom to determine spatial arrangement. This includes both bonding pairs and lone pairs.

For the two carbon atoms in acetamide:
  • The first carbon (in the (CH_3) group) is surrounded by three hydrogen atoms and one carbon atom from the carbonyl group. This position results in a tetrahedral electron-pair geometry since it accounts for four electron groups.
  • The second carbon (the carbonyl carbon) is bonded to one oxygen and one nitrogen atom. Considering the electron pairs involved, the electron-pair geometry is trigonal planar, consistent with three electron groups.
Grasping this concept helps recognize how electrons influence molecular geometry and create stable molecules.
Molecular Geometry
Molecular geometry focuses on the three-dimensional arrangement of atoms in a molecule, ignoring lone pairs. Its understanding is crucial to predicting how molecules behave and interact.

For acetamide:
  • The (CH_3) carbon maintains a tetrahedral molecular geometry since each of its four groups are bonded to other atoms, similar to the electron-pair geometry.
  • In contrast, the carbonyl carbon exhibits trigonal planar molecular geometry due to its flat layout formed by the oxygen and nitrogen bonds, mirroring its electron-pair setup.
Comprehending molecular geometry is vital for anticipating reactivity and interactions in larger chemical systems.
Hybridization
Hybridization describes how atomic orbitals blend to form new hybrid orbitals, significantly influencing molecular shape.

For acetamide's carbon atoms:
  • The (CH_3) group's carbon undergoes (sp^3) hybridization. Here, one s and three p orbitals merge to form four identical hybrid orbitals, allowing a tetrahedral geometry.
  • The carbonyl carbon experiences (sp^2) hybridization. This involves one s and two p orbitals forming three hybrid orbitals, with its unhybridized p orbital participating in a π bond (double bond) with oxygen.
Understanding hybridization provides insight into bonding types and molecular geometry, assisting with molecular design and prediction.
Sigma and Pi Bonds
In chemistry, sigma and pi bonds are fundamental concepts explaining how atoms bond in molecules, impacting structure and properties.

For acetamide:
  • Sigma (σ) bonds are the primary bonds formed by overlapping hybridized orbitals. For the carbon-oxygen bond in the carbonyl, the (sp^2) hybridized orbitals from both atoms overlap to create a strong, linear σ bond.
  • Pi (π) bonds are secondary bonds resulting from the sideways overlap of unhybridized p orbitals. In acetamide's carbonyl group, the π bond complements the existing σ bond between carbon and oxygen, creating a double bond.
Together, σ and π bonds dictate molecular stability and characteristics, influencing reactions and behavior in diverse chemical environments.

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Most popular questions from this chapter

Sketch the resonance structures for the \(\mathrm{N}_{2} \mathrm{O}\) molecule. Is the hybridization of the N atoms the same or different in each structure? Describe the orbitals involved in bond formation by the central \(\mathrm{N}\) atom.

Consider the following list of small molecules and ions: \(\mathrm{C}_{2}, \mathrm{O}_{2}, \mathrm{CN}^{-}, \mathrm{O}_{2}, \mathrm{CO}, \mathrm{NO}, \mathrm{NO}^{+}, \mathrm{C}_{2}^{2-}, \mathrm{OF}^{-}\) Identify (a) all species that have a bond order of 3 (b) all species that are paramagnetic (c) species that have a fractional bond order

What is the connection between bond order, bond length, and bond energy? Use ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) as examples.

Nitrogen, \(\mathrm{N}_{2}\), can ionize to form \(\mathrm{N}_{2}^{+}\) or add an electron to give \(\mathrm{N}_{2}^{-}\). Using molecular orbital theory, compare these species with regard to (a) their magnetic character, (b) net number of \(\pi\) bonds, (c) bond order, (d) bond length, and (e) bond strength.

Carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) dinitrogen monoxide \(\left(\mathrm{N}_{2} \mathrm{O}\right),\) the azide ion \(\left(\mathrm{N}_{3}^{-}\right),\) and the cyanate ion (OCN \(^{-}\) ) have the same geometry and the same number of valence shell electrons. However, there are significant differences in their electronic structures. (a) What hybridization is assigned to the central atom in each species? Which orbitals overlap to form the bonds between atoms in each structure. (b) Evaluate the resonance structures of these four species. Which most closely describe the bonding in these species? Comment on the differences in bond lengths and bond orders that you expect to see based on the resonance structures.

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