Question

Use VSEPR 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) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\) (c) CCC(=O)O (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

Answer: For the following molecules, the bond angles around each C, N, and O atom are: (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\): Each carbon (C) atom has bond angles of approximately 120°. (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\): Each carbon (C) atom has bond angles of approximately 109.5°. (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}(=O)-\mathrm{O}\): Terminal carbon (C) atoms have bond angles of approximately 109.5°, central C and carbonyl oxygen (O) have bond angles of approximately 120°, and the other oxygen (O) has a bond angle of approximately 104.5°. (d) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\): Each carbon (C) atom has bond angles of approximately 120°. (e) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{O}\): Terminal and middle carbon (C) atoms have bond angles of approximately 120°, and the oxygen (O) atom has a bond angle of approximately 104.5°. (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\): Terminal carbon (C) has bond angles of approximately 109.5°, middle C, nitrogen (N) and oxygen (O) have bond angles of approximately 120° (with a deviation for O due to the bent geometry).

Step-by-step solution

For each molecule, find the total number of electron domains (bonds and lone pairs) around each atom of C, N, and O. Based on this, identify the electron domain geometry. (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\): - Both terminal \(\mathrm{C}\) atoms have 3 domains (3 single bonds), so trigonal planar geometry for them. - The central \(\mathrm{C}\) has 3 domains (2 single bonds and 1 double bond), so trigonal planar geometry as well. (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\): - Both \(\mathrm{C}\) atoms have 4 domains (4 single bonds), so tetrahedral geometry for them. (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}(=O)-\mathrm{O}\) (In the SMILES formula, \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}(=O)-\mathrm{O}\) is a standard representation of the given molecule): - Both terminal \(\mathrm{C}\) atoms have 4 domains (4 single bonds), so tetrahedral geometry for them. - The central \(\mathrm{C}\) has 3 domains (2 single bonds and 1 double bond), so trigonal planar geometry for it. - \(\mathrm{O}\) atoms have 3 domains (1 double bond and 1 single bond for carbonyl oxygen and 1 single bond and 2 lone pairs for other oxygen), so trigonal planar for carbonyl oxygen and bent geometry for the other oxygen. (d) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\): - Terminal \(\mathrm{C}\) atoms have 3 domains (2 single bonds and 1 double bond), so trigonal planar geometry for them. - The central \(\mathrm{C}\) has 3 domains (2 six-electron double bonds), so trigonal planar geometry for it. (e) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{O}\): - The terminal \(\mathrm{C}\) has 3 domains (2 single bonds and 1 double bond), so trigonal planar geometry for it. - The middle \(\mathrm{C}\) has 3 domains (3 six-electron double bonds), so trigonal planar geometry for it. - \(\mathrm{O}\) has 3 domains (1 double bond and 2 lone pairs), so bent geometry for it. (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\): - The terminal \(\mathrm{C}\) has 4 domains (4 single bonds), so tetrahedral geometry for it. - The middle \(\mathrm{C}\) has 3 domains (2 single bonds and 1 double bond), so trigonal planar geometry for it. - \(\mathrm{N}\) has 3 domains (1 double bond and 2 single bonds), so trigonal planar geometry for it. - \(\mathrm{O}\) has 3 domains (1 single bond and 2 lone pairs), so bent geometry for it. ##Step 2: Determine the Bond Angles##

Using the electron domain geometries identified in step 1, determine the bond angles around each C, N, and O atom. (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\): - The bond angles around each C are approximately 120° (trigonal planar geometry). (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\): - The bond angles around each C are approximately 109.5° (tetrahedral geometry). (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C}(=O)-\mathrm{O}\): - Bond angles around the terminal C atoms are approximately 109.5° (tetrahedral geometry). - Bond angles around the central C atom and carbonyl O are approximately 120° (trigonal planar geometry). - Bond angle around the other O is approximately 104.5° (bent geometry). (d) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\): - The bond angles around each C are approximately 120° (trigonal planar geometry). (e) \(\mathrm{H}_{2}\mathrm{C}=\mathrm{C}=\mathrm{O}\): - Bond angles around the terminal C and middle C are approximately 120° (trigonal planar geometry). - Bond angle around the O is approximately 104.5° (bent geometry). (f) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{N}-\mathrm{OH}\): - Bond angles around the terminal C are approximately 109.5° (tetrahedral geometry). - Bond angles around the middle C, N and O are approximately 120° (trigonal planar geometry for middle C and N; bent geometry for O with a deviation).