Chemistry 1C Review Problems, Chapters 13, 14 and 16 ...

5. For CH 2 =CH both carbons are 2 sp2 hybridized and all the atoms are in the same plane. For CH 2 =C=CH 2 the two end
carbons are sp 2 and the central carbon is sp hybridized. Not all atoms are in the same plane: the two CH 2 groups are in
perpendicular planes.
6. Molecule A: six σ-bonds, two π-bonds, the hybridization on the left carbon is sp 3 , the other two carbons are sp hybridized;
angle a = 109 o , angle b = 180 o , angle c = 180 o Molecule B: four σ-bonds, one π-bond, the hybridization on the nitrogen
is sp 2 ; angle a = 109 o , angle b = 120 o , angle c = 120 o
7. Br 2 is least polar because it is non-polar.
8. Electronegativity: Rb < O < F Atomic Radius: Ne < Cl < Si < Na Why
9. a) N 2 has a triple bond; one σ-bond and two π-bonds. b) CO has a triple bond; one σ-bond and two π-bonds. c) O 2 has
a double bond; one σ-bond and one π-bond.
10. See problem 62, Chapter 13.
11 The carbons in butane are all sp 3 hybridized. The carbon and hydrogen atoms are not in the same plane. See Fig 22.4, pg.
974. Can you see how all the dipoles in butane cancel and that butane is non-polar Most hydrocarbons are essentially
non-polar.
12. There are three π-bonds and therefore there are six π-electrons. The hybridization on the carbon atoms is sp 2 . All the
carbon and hydrogen atoms are in the same plane because the sp 2 hybridization results in planar geometry. For
resonance to occur the p-orbitals on neighboring carbon atoms must overlap which requires the molecule to be planar.
13. C 2 H 2 has the shortest C–C bond because it is a triple bond whereas C 2 H 4 has a double bond and C 2 H 6 has a single bond.
14. O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 2 Bond order = 2
–
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 3 Bond order = 1.5
2–
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 4 Bond order = 1
2+
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 Bond order = 3 = largest => Largest dissociation energy
15. O (1s) 2 (2s) 2 (2p) 4 Two unpaired electrons Al 3+ [Ne] or (1s) 2 (2s) 2 (2p) 6 No unpaired electrons
C (1s) 2 (2s) 2 (2p) 2 Two unpaired electrons Se 2– [Kr] No unpaired electrons
16. a) Octahedral geometry, sulfur shares 12 electrons. No it is not consistent with the octet rule. b) Sulfur accepts
additional electrons in d-orbitals. Valence shell expansion is not observed for second row elements. Why not
c) d 2 sp 3 hybridization on S.
17. π-bonds are commonly observed for carbon, oxygen and nitrogen. It is less common for sulfur to and phosphorous to
make π-bonds. Sulfur makes π-bonds with oxygen in SO 2 , but S is too big to make strong π-bonds with itself. This is
why unlike O 2 (g), a stable gas at room temperature, S 2 (g) is not observed. Sulfur is a solid at room temperature, S 8 (s).
18. BI 3 has the highest boiling point. These molecules are all covalent non-polar and therefore the larger the molecule, the
more polarizable, the stronger the LDF. 19. sp
20. H 2 has a low molar mass, it has only two electrons => low polarizability => weak LDF => low b.p.
21. a) Xe is the largest in size, greatest LDF=> higher b.p b) NH 3 has a higher b.p. due to H-bonding c) HF has a higher
b.p. due to H-bonding
22. A straight line extrapolation, following the trend of other group members, H 2 Te, H 2 Se and H 2 S, gives a boiling point of
H 2 O of approximately – 123 o C. Without H-bonding water would be a gas at room temperature and pressure.
23. Carbon follows the octet rule, it is not able to expand its valence shell. Sn can expand its octet by putting electrons into
d-orbitals.
24. Carbon has a small enough atomic radius to allow for effective overlap between the p-orbitals. The other group 4
elements are bigger and p-orbitals do not overlap effectively.
25. d 2 sp 3
26. Graphite (covalent network), S 8 (molecular), Na (metallic), H 2 O (molecular), SiO 2 (covalent network) CH 3 NH 2
(molecular), NH 4 Cl (ionic), white phosphorous (P 4 (s) molecular), black phosphorous (covalent network).
27. a) BF trigonal planar, non-polar, LDF b) SO 3 2 bent, polar, dipole-dipole c) C 6 H 6 planar, non-polar, LDF d) Cl 2
linear, non-polar, LDF e) CH 3 OH tetrahedral, polar, H-bonding f) HCl linear, polar, dipole-dipole g)NH 3
trigonal pyramid, polar, H-bonding h) CCl 4 non-polar, LDF 28. Molecular
29. In order to answer this question, determine the hybridization on the carbon atoms. All the carbon atoms in
benzophenone are sp 2 hybridized (trigonal planar geometry) and therefore all the carbon, hydrogen and oxygen atoms
are in the same plane. The planar geormetry results in a maximum surface area of contact between the molecules and
gives rise to relatively large LDF forces. The CO bond is polar so there are some dipole-dipole forces as well but the
LDF forces are dominant in benzophenone. In cyclohexanone, the carbon bound to the oxygen is sp 2 hybridized, the
other 5 carbons are sp 3 hybridized. The molecule is therefor not planar. See Figure 22.6, Pg. 980. LDF forces are
weaker and therefor cyclohexanone has a lower melting point and is a liquid at room temperature.
30. T = 365 K = 92.6 o C. Watch units!!! 1 kJ = 1000 J and you must convert degrees Celcius to degrees Kelvin.
31. a) At point B, solid and liquid CO 2 are in equilibrium. b) The horizontal line drawn is below the triple point and
shows that CO 2 (s) goes directly to a gas at P = 1 atm. c) Solid CO 2 is more dense than liquid CO 2 . You can see this
from the slope of the solid-liquid line. Keeping T constant while increasing P, draw a straight vertical line, liquid CO 2
will go to the more dense phase, solid CO 2 . Water is unusual in this respect, liquid water is more dense than ice. See
Fig. 16, Pg. 780 text. That is why ice floats on water.
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