TO THE STUDENT
As it was mentioned earlier, problem solving is, probably, the best key to studying chemistry. This work is very often like solving a puzzle; therefore never look at the answer until you succeed (or failed) in your solution. The best way to use this Appendix is to compare your solutions and those given below, or to find explanations for the unsolved problems only after all attempts to solve them have failed.
Appendix 3 presents solutions to all the problems (in-text and additional problems). In many cases not only the answer is given, but also the explanation of the reasoning, even if it has not been questioned. Some problems have more than one solution, but only one is generally given. Thus, your answer may differ from the one presented here. The main thing is that your solution must be correct.
To Chapter 1
1.2. It is similar to methane tetrahedral structure, except for appreciable deviation from the value of 109.5°.
1.3. The CH3 carbon is sp3-hybridized and forms the H-C-H bond angles near 109°. Double-bond carbons are sp2-hybridized and form the H-C-H and H-C-C angles near 120°.
1.4. Both oxygen and nitrogen are sp3-hybridized and have tetrahedral geometry. Three-dimensional structures are similar to those in Figures 1.10, c and 1.11, b, respectively.
1.5. CH3CH2CH2CH2Br, CH3CH2CHBrCH3, (CH3)2CHCH2Br, and (CH3)3CBr.
1.6. (a) CH3NH2; (b) CH3CH=CH2; (c) CH3CH2CH2Cl; (d) (CH3)3CH; (e) CH3CH(OH)CH3. Two isomers (in total) are possible for compounds from (b) to (d), and three isomers - for the compound (e).
1.8. Sodium and oxygen form the ionic bond, CH3CH2O- Na+; other bonds in the molecule are covalent.
1.9. All bonds are covalent, except for CaCl2.