Before we get started, let's have a big printout of a major table. This is for the convenience of web users and, at the same time, should infuriate those wasting time, paper, and ink printing these notes!
There is enough information in the answers below to enable you to draw Lewis structures; you are urged to do so!
Problem
7.67: What shape do you expect for molecules that meet the
following descriptions?
| (a) | A central atom with two lone pairs and three bonds to
other atoms.
This would be T-shaped. Refer
to table above.
|
| (b) | A central atom with two lone pairs and two bonds to other
atoms.
This would be bent (e.g.,
water).
|
| (c) | A central atom with two lone pairs and four bonds to
other atoms.
Referring to the table above, you
see that his would be square planar.
|
Problem
7.70: What shape do you expect for each of the following
molecules?
| (a) | H2Se | 2 bonded atoms + 2 lone pairs: Bent |
| (b) | TiCl4 | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (c) | O3 | 2 bonded atoms + 1 lone pair: Bent |
| (d) | GaH3 | 3 bonded atoms + 0 lone pairs: Trigonal planar |
Problem
7.71: What s hape
do you expect for each of the following molecules?
| (a) | XeO4 | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (b) | SO2Cl2 | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (c) | OsO4 | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (d) | SeO2 | 2 bonded atoms + 1 lone pair: Bent |
Problem
7.73: Predict the shape of each of the following ions:
| (a) | NO3- | 3 bonded atoms + 0 lone pairs: Trigonal Planar |
| (b) | NO2+ | 2 bonded atoms + 0 lone pairs: Linear |
| (c) | NO2- | 2 bonded atoms + 1 lone pair: Bent |
Problem
7.74: What shape do you expect for each of the following
anions?
| (a) | PO43- | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (b) | MnO4- | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (c) | SO42- | 4 bonded atoms + 0 lone pairs: Tetrahedral |
| (d) | SO32- | 3 bonded atoms + 1 lone pair: Trigonal pyrimidal |
| (e) | ClO4- | 4 bonded atoms + 0 lone pair: Tetrahedral |
| (f) | SCN- | 2 bonded atoms + 0 lone pair: Linear |
Problem
7.76: What bond angles do you expect for each of the following?
| (a) | The F-S-F bond angle in SF2 | The S is bound to two fluorines and has two lone pairs of electrons. Thus the bond angle would be expected to be about 109o. |
| (b) | The H-N-N angle in N2H2 | N bound two atoms. Each N has a single lone pair. Thus, would expect bond angle of about 120o. |
| (c) | The F-Kr-F angle in KrF4 | Kr bound to four fluorines and contains two extra pairs of electrons. This molecule would be square planar and the bond angle would be 90o. |
| (d) | The Cl-N-O angle in NOCl | N bound to two atoms (single bond to Cl and double bond to O). Has a lone pair. Thus, bent at about 120o. |
Problem
7.77: What bond angles do you expect for each of the following?
| (a) | The Cl-P-Cl angle in PCl6- | Phosphorus is bound to 6 chlorines and, in the Lewis sstructure, you see no lone pairs. Hence, this is octahedral and the bond angle is 90o. |
| (b) | The Cl-I-Cl angle in ICl2- | I is bound to Cl and has 3 lone pairs in addition. Thus, it is a trigonal bipyramidal structure; with just three atoms, they would be on the axis and the bond angle would be linear, namely, 180o. |
| (c) | The O-S-O angle in SO42- | This molecule has found atoms bound
to the central sulfur and no extra lone pairs. Thus it is tetrahedral
and the angle is approximately 109.47122o.
Useful hint: This is cos-1(-1/3) on your calculator. |
| (d) | The O-B-O angle in BO33- | Boron is bound to 3 oxygens and there are no lone pairs. Thus, this is trigonal planar and the bond angle is 120o. |
Here are the charts again for those of you who prefer to scroll down (from above) or don't want to do too much scrolling for the two remaining problems. Again, if you are printing these, well, ... .
Problem
7.78: Acrylonitrile is used as the starting material for
manufacturing acrylic fibers. Predict values for all bond angles
in acrylonitrile.
Let us label the carbons from left to right as a, b, and c. The first two carbons have three objects about them and the third has two. None of the carbons has any spare lone pairs. (Later on, it will seem quite straightforward to view these as sp2, sp2, and sp hybridization, respectively--but we are not that far along yet!)Problem 7.80: Explain why cyclohexane, a substance that contains a six-membered ring of carbon atoms, is not flat but instead has a puckered, nonplanar, shape. Predict the values of the C-C-C bond angles.Anyway, here are the various bond angles. Except for a missing lone pair at the right of the terminal N, teh structure as drawn is the Lewis structure.
Atoms in Bond Approx. Bond Angle H-Ca-H 120o H-Ca-Cb 120o Ca-Cb-Cc 120o Cb-Cc-N 180o Ca-Cb-H 120o H-Cb-Cc 120o Nothing much to this!
All the carbons are surround by four objects. Thus, the ring would be expected to pucker since flatness can occur only if C has three objects around it. The surroundings here are tetrahedral and one would expect the C-C-C typical bond angle to be approximately 109.47122o.