Question

UseVSEPR to predict bond angles about each atom of carbon, nitrogen, and oxygen in these molecules. (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{N}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}-\mathrm{OH}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) (e) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{O}\) (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\)

Short answer

Question: Using the VSEPR theory, predict the bond angles around each carbon, nitrogen, and oxygen atom in the following molecules: (a) CH3-CH=CH2, (b) CH3-N-CH3, (c) CH3-CH2-C-OH, (d) H2C=C=CH2, (e) H2C=C=O, and (f) CH3-CH=N-OH. Answer: (a) CH3-CH=CH2: 1. C bonded to three H atoms: Approximately 109.5° 2. C with a double bond: 120° 3. C bonded to two H atoms: 120° (b) CH3-N-CH3: 1. N atom: Approximately 107° (c) CH3-CH2-C-OH: 1. C bonded to three H atoms: Approximately 109.5° 2. C bonded to two H atoms: Approximately 109.5° 3. C with a single bond to O: 120° (d) H2C=C=CH2: 1. C bonded to two H atoms: 120° 2. C with two double bonds: 180° (e) H2C=C=O: 1. C bonded to two H atoms: 120° 2. C with a double bond to O: 180° 3. O atom: 180° (f) CH3-CH=N-OH: 1. C bonded to three H atoms: Approximately 109.5° 2. C with a double bond to N: 120° 3. N atom: Approximately 120°

Step-by-step solution

Reviewing VSEPR Theory

VSEPR theory is used to predict the shapes of molecules based on the repulsion between electron domains (which include bonding electron pairs and lone pairs). The bond angle is determined by trying to position these domains as far apart as possible.

Determining Electron Domains

For each molecule, identify the electron domains around each carbon, nitrogen, and oxygen atom. Remember that double bonds are considered as one domain. (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) Around each carbon atom: 1. Carbon bonded to three hydrogen atoms: 3 single bonds. 2. Carbon with a double bond: 1 single bond and 1 double bond. 3. Carbon bonded to two hydrogen atoms: 2 single bonds and 1 double bond. (b) \(\mathrm{CH}_{3}-\mathrm{N}-\mathrm{CH}_{3}\) Around the nitrogen atom: 1. Nitrogen atom: 1 lone pair and 2 single bonds. (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}-\mathrm{OH}\) Around each carbon atom: 1. Carbon bonded to three hydrogen atoms: 3 single bonds. 2. Carbon bonded to two hydrogen atoms: 2 single bonds and 1 single bond to carbon. 3. Carbon with a single bond to oxygen: 1 single bond to hydroxyl group and a double bond to the adjacent carbon. (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) Around each carbon atom: 1. Carbon bonded to two hydrogen atoms: 2 single bonds and 1 double bond. 2. Carbon with two double bonds: 1 double bond to each carbon. (e) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{O}\) Around each carbon and oxygen atom: 1. Carbon bonded to two hydrogen atoms: 2 single bonds and 1 double bond to carbon. 2. Carbon with a double bond to oxygen: 1 double bond to carbon and 1 double bond to oxygen. 3. Oxygen atom: 2 lone pairs and 1 double bond. (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\) Around each carbon and nitrogen atom: 1. Carbon bonded to three hydrogen atoms: 3 single bonds. 2. Carbon with a double bond to nitrogen: 1 single bond and 1 double bond to nitrogen. 3. Nitrogen atom: 1 lone pair, 1 single bond, and 1 double bond.

Determining Bond Angles

Using the electron domain configurations determined in step 1, use VSEPR theory to determine the bond angles around each carbon, nitrogen, and oxygen atom. (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) 1. Tetrahedral geometry (3 single bonds): Approximately \(109.5^{\circ}\) bond angle. 2. Trigonal planar geometry (1 single bond, 1 double bond): \(120^{\circ}\) bond angle. 3. Trigonal planar geometry (2 single bonds, 1 double bond): \(120^{\circ}\) bond angle. (b) \(\mathrm{CH}_{3}-\mathrm{N}-\mathrm{CH}_{3}\) 1. Trigonal pyramidal geometry (1 lone pair, 2 single bonds): Approximately \(107^{\circ}\) bond angle. (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}-\mathrm{OH}\) 1. Tetrahedral geometry (3 single bonds): Approximately \(109.5^{\circ}\) bond angle. 2. Tetrahedral geometry (2 single bonds, 1 single bond to carbon): Approximately \(109.5^{\circ}\) bond angle. 3. Trigonal planar geometry (1 single bond, 1 double bond): \(120^{\circ}\) bond angle. (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) 1. Trigonal planar geometry (2 single bonds, 1 double bond): \(120^{\circ}\) bond angle. 2. Linear geometry (1 double bond to each carbon): \(180^{\circ}\) bond angle. (e) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{O}\) 1. Trigonal planar geometry (2 single bonds, 1 double bond): \(120^{\circ}\) bond angle. 2. Linear geometry (2 double bonds): \(180^{\circ}\) bond angle. 3. Linear geometry (2 lone pairs, 1 double bond): \(180^{\circ}\) bond angle. (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\) 1. Tetrahedral geometry (3 single bonds): Approximately \(109.5^{\circ}\) bond angle. 2. Trigonal planar geometry (1 single bond, 1 double bond): \(120^{\circ}\) bond angle. 3. Trigonal planar geometry (1 lone pair, 1 single bond, 1 double bond): Approximately \(120^{\circ}\) bond angle.

Key concepts

Bond Angles Prediction

Understanding bond angles is crucial in the study of molecular geometry. Valence Shell Electron Pair Repulsion (VSEPR) theory provides a systematic way of predicting bond angles around a central atom in a molecule. According to this theory, electron domains, which include both bonding electron pairs and lone pairs, will repulse each other and try to get as far apart as possible. This repulsion dictates the bond angles in a molecule.

For instance, if a carbon atom is bonded to four other atoms with no lone pairs (tetrahedral electron domain geometry), the bond angles will be approximately 109.5 degrees. However, with fewer bonding domains or with the presence of lone pairs, the angles adjust. For example, in a trigonal planar arrangement, with three bonding domains, the bond angles are around 120 degrees. On the other hand, the presence of lone pairs like in a trigonal pyramidal geometry (as for NH3 with three bonding domains and one lone pair) will slightly reduce the angle to approximately 107 degrees due to the larger repulsion of the lone electron pair.

Electron Domain Geometry

When exploring molecular structures using VSEPR theory, it's crucial to discern the electron domain geometry around the central atom. This relates to the arrangement of both bond pairs and lone pairs of electrons in the valence shell. The electron domain geometry can be tetrahedral, trigonal planar, linear, bent, octahedral, or trigonal bipyramidal, among others.

To determine this geometry, one must count the bonding pairs and lone pairs on the central atom. For instance, four bonding domains and no lone pairs will yield a tetrahedral geometry. Similarly, two bonding domains and no lone pairs lead to a linear arrangement, as seen in carbon dioxide (CO2). The distinction between the types of electron domains — bonding or non-bonding — is critical as it influences the resultant molecular shape and bond angles.

Molecular Shapes

Molecular shape, which describes the three-dimensional arrangement of atoms in a molecule, is predicted using VSEPR theory as well. The shape of a molecule is influenced by the geometry of electron domains around the central atom(s). This shape is directly connected to various properties of the molecule, including reactivity, polarity, and interaction with other molecules.

For example, a molecule with a tetrahedral electron domain geometry might have a tetrahedral molecular shape if all electron domains are bonding pairs. If one of those domains is a lone pair, the molecular shape is called trigonal pyramidal. Again, if there are two lone pairs, the shape becomes bent or angular. It's the balance between the electron domain geometry and the specific distribution of lone pairs and bonding pairs that gives rise to the actual molecular shape observed in real-life compounds.