Chemistry 217 Problem Set 1

Transcription

1 hemistry 217 Problem Set 1 Recommended Problems from the Book (1 st ed. in parentheses): 1.1, , , ( , , ) Supplemental Problems: Lewis structures: Weeks, pp ormal harge: Weeks, pp ; Klein, section 1.5. Lone pairs: Klein, section 1.6. Skeletal structures (condensed structures): Klein, section Draw the Lewis structure for each of the following molecules. Show all lone pairs. All molecules are neutral and all formal charges are zero. 2 l 2 l l (b) ( 3 ) 3 (c) ( 3 ) 2 (same as 3 3 ) (d) (f) 3 ( 2 3 ) Each of the following molecules is neutral. Determine the formal charge for each atom in the molecules (All lone pairs are shown). 4-1/2(6)-2=-1 1-1/2(2)-0=0 4-1/2(4)-2=0 (b) 3 P /2(8)=+1 (c) P all atoms have f.c.=0

2 5-1/2(8)-0= /2(2)-6= all atoms have f.c.=0 (d) 3 4-1/2(6)-0=+1 (f) all atoms have f.c.=0 (g) 6-1/2(2)-6=-1 5-1/2(8)-0=+1 4-1/2(6)-2= (h) 5-1/2(4)-4=-1 (i) 5-1/2(8)-0=+1 3. ill in the lone pairs on the following molecules. All non-zero formal charges are shown. (b) (c) no lone pairs 3 Si l (d) 3 3 (d) no lone pairs

3 4. In class, we determined the rules of thumb in determining formal charge for carbon. This information is shown in the first rows of the table below. ill in the remaining blank spaces for,, and. Remember that you cannot violate the octet rule. Atom # of bonds # of lone pairs charge (2 unshared) (4 unshared) (2 unshared) (6 unshared) (2 unshared) (4 unshared) -1 Rationale: We must fulfill the octet rule for each one. Remember:.. = # electrons on free atom ½ (shared) unshared : f.c. = 0 = 1-1 the bonds plus the unshared electrons must total 1. Therefore, there must be one bond, and no unshared pairs. f.c. = +1 = 1 0 the bonds plus the unshared electrons must equal 0. There cannot be any bonds or unshared pairs. f.c. = -1=1-2 the bonds plus the unshared electrons must equal 2. If there are two bonds, or one bond and one unshared electron, the octet rule is violated. Therefore, there must be no bonds and one lone pair. : f.c.=0=6-6 the bonds plus the unshared electrons must total 6. Possibilities: 1 bond, 5 unshared electrons: violates octet 2 bonds, 2 lone pairs: o.k. 3 bonds, 3 unshared electrons: violates octet 4 bonds, one lone pair: violates octet f.c.=+1=6-5 the bonds plus the unshared electrons must total 5. Possibilities: 1 bond, 2 lone pairs: violates octet

4 2 bonds, 3 unshared pairs: violates octet 3 bonds, 1 lone pair: o.k. 4 bonds, 1 unshared pair: vioates octet f.c.=-1=6-7 the bonds plus the unshared electrons must total 7. Possibilities: 1 bond, 3 lone pair: o.k. 2 bonds, 5 unshared: violates octet 3 bonds, two lone pair: violates octet : f.c.=0=5-5 the bonds plus the unshared electrons must total 5. Possibilities: 1 bond, two lone pair: violates octet 2 bonds, 3 unshared: violates octet 3 bonds, one lone pair: o.k. 4 bonds, one unshared: violates octet 5 bonds, no unshared: violates octet f.c.=1=5-4 the bonds plus the unshared electrons must total 4. Possibilities: 1 bond, 3 unshared: violates octet 2 bonds, 2 unshared: violates octet 3 bonds, one unshared: violates octet 4 bonds, no unshared: o.k. f.c.=-1=5-6 the bonds plus the unshared electrons must total 6. Possibilities: 1 bond, 5 unshared: violates octet 2 bonds, 4 unshared: o.k. 3 bonds, 3 unshared: violates octet 4 bonds, two unshared: violates octet 5 bonds, none unshared: violates octet 5. ill in the lone pairs and formal charges on the atoms in the following molecules (all hydrogens are shown). B All atoms have f.c.=0 (b) all atoms have f.c.=0 (c) (d) (f) 3 3

5 (g) 3 3 (h) S (i) Draw valid Lewis structures for the following. or each one, assign all non-zero formal charges and the overall charge on the molecule. (1) Arrange atoms in order: (2) ill in bonds and electrons to fulfill octet. Add in single bonds first: The octet is not fulfilled for the carbon and nitrogen on the right side of the molecule. If we add a double bond between them, the octet will be fulfilled for the carbon. Then, to fulfill the octet for nitrogen, add a lone pair (formal charges shown below): veral charge=+1 If we added lone pairs to the carbon and nitrogen to fulfill their octet, we would get: which has an overall negative charge of -1. This Lewis structure is not as good as the first, because more atoms contain charge. (b) (1) Arrange atoms in order: (2) ill in bonds and electrons to fulfill octet.

6 Add in single bonds first: The octet is not fulfilled for the carbon on the right and both oxygens. If we add a double bond between one of the oxygens and the carbon, it will be fulfilled for the carbon. Then, to fulfill the octet for the oxygens, add in lone pairs (formal charges shown below): or (both valid Lewis structures) overall charge=-1 (c) 2 2 (1) Arrange atoms in order: (2) ill in bonds and electrons to fulfill octet. Add in single bonds first: The octet is not fulfilled for any of the carbons. If we add a double bond between two of them and a lone pair to the third, it will be fullfilled for all three: or overall charge = -1 Another solution to this problem would be to not fulfill the octet for one of the carbons, since a carbon with a positive charge is one of the exceptions to the octet rule. In this case, the answer would be: or overall charge = +1

7 (7) Draw constitutional isomers which fulfill the following criteria: Pentane: 5 12 (three different isomers) (b) 4 10 (c) cyclic: 5 10 (Draw at least four. Two should be cyclic molecules. Two should contain a carbon-carbon double bond. double bonds: (d) 6 10 (Draw at least one isomer of compounds containing the following types of functionality: 1) a compound with two double bonds; 2) a compound with a ring and a double bond; 3) a compound with a triple bond; 4) a compound with two rings

8 A few examples (there are many others) two double bonds: double bond and ring: triple bond: two rings: (Draw as many as you can.) There are many, many constitutional isomers. ere are a few. Just make sure that your compounds have one double bond or ring in them. (8) Give the molecular formula of the following compounds:

9 nicotine vitamin cholesterol (b) (c) 27 46