Final Exam: Cumulative! W 1-3 pm

1. Final Exam: Cumulative! W 1-3 pm Review Session. M 3 pm Office Hours T 1-3

2. A polar bond is characterized by separation of electrical charge. Polar molecules, therefore, have nonzero dipole moments. For HCl, we can represent the charge separation using d+ and d- to indicate partial charges. Because Cl is more electronegative than H, it has the d- charge, while H has the d+ charge.

3. To determine whether a molecule is polar, we need to determine the electron-dot formula and the molecular geometry. We then use vectors to represent the charge separation. They begin at d+ atoms and go to d- atoms. Vectors have both magnitude and direction. • We then sum the vectors. If the sum of the vectors is zero, the dipole moment is zero. If there is a net vector, the molecule is polar.

4. To illustrate this process, we use arrows with a + on one end of the arrow. We’ll look at CO2 and H2O. CO2 is linear, and H2O is bent. The vectors add to zero (cancel) for CO2. Its dipole moment is zero. For H2O, a net vector points up. Water has a dipole moment.

5. The relationship between molecular geometry and dipole moment is summarized in Table 10.1.

7. Polar molecules experience attractive forces between molecules; in response, they orient themselves in a d+ to d- manner. This has an impact on molecular properties such as boiling point. The attractive forces due to the polarity lead the molecule to have a higher boiling point.

8. We can see this illustrated with two compounds: cis-1,2-dichloroethene trans-1,2-dichloroethene There is no net polarity; this is a nonpolar molecule. The net polarity is down; this is a polar molecule. Boiling point 60°C. Boiling point 48°C.

10. Which of the following molecules would be expected to have a zero dipole moment? a. GeF4 b. SF2 c. XeF2 d. AsF3

11. GeF4: 1(4) + 4(7) = 32 valence electrons. Ge is the central atom. 8 electrons are bonding; 24 are nonbonding. Tetrahedral molecular geometry. GeF4 is nonpolar and has a zero dipole moment.

12. SF2: 1(6) + 2(7) = 20 valence electrons. S is the central atom. 4 electrons are bonding; 16 are nonbonding.Bent molecular geometry. SF2 is polar and has a nonzero dipole moment.

13. XeF2: 1(8) + 2(7) = 22 valence electrons. Xe is the central atom. 4 electrons are bonding; 18 are nonbonding.Linear molecular geometry. XeF2 is nonpolar and has a zero dipole moment.

14. AsF3: 1(5) + 3(7) = 26 valence electrons. As is the central atom. 6 electrons are bonding; 20 are nonbonding.Trigonal pyramidal molecular geometry. AsF3 is polar and has a nonzero dipole moment.

15. Which of the following molecules would be expected to have a zero dipole moment? a. GeF4 tetrahedral molecular geometryzero dipole moment b. SF2 bent molecular geometry nonzero dipole moment c. XeF2 linear molecular geometryzero dipole moment d. AsF3 trigonal pyramidal molecular geometry nonzero dipole moment

16. Valence bond theory is an approximate theory put forth to explain the electron pair or covalent bond by quantum mechanics.

17. A bond forms when • An orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals are said to overlap. • The total number of electrons in both orbitals is no more than two.

18. The greater the orbital overlap, the stronger the bond. • Orbitals (except s orbitals) bond in the direction in which they protrude or point, so as to obtain maximum overlap.

19. To obtain the bonding description about any atom in a molecule: • 1. Write the Lewis electron-dot formula. • 2. Use VSEPR to determine the electron arrangement about the atom. • 3. From the arrangement, deduce the hybrid orbitals. • 4. Assign the valence electrons to the hybrid orbitals one at a time, pairing only when necessary. • 5. Form bonds by overlapping singly occupied hybrid orbitals with singly occupied orbitals of another atom.

20. Let’s look at the methane molecule, CH4. Simply using the atomic orbital diagram, it is difficult to explain its four identical C—H bonds. • The valence bond theory allows us to explain this in two steps: promotion and hybridization.

21. Hybrid orbitals are orbitals used to describe the bonding that is obtained by taking combinations of the atomic orbitals of the isolated atoms. • The number of hybrid orbitals formed always equals the number of atomic orbitals used.

22. Hybrid orbitals are named by using the atomic orbitals that combined: • one s orbital + one p orbital gives two sp orbitals • one s orbital + two p orbitals gives three sp2 orbitals • one s orbital + three p orbitals gives four sp3 orbitals • one s orbital + three p orbitals + one d orbital gives five sp3d orbitals • • one s orbital + three p orbitals + two d orbitals gives six sp3d2 orbitals

23. Hybrid orbitals have definite directional characteristics, as described in Table 10.2.

25. First, the paired 2s electron is promoted to the unfilled orbital. Now each orbital has one electron. Second, these orbitals are hybridized, giving four sp3 hybrid orbitals.

26. Use valence bond theory to describe the bonding about an N atom in N2H4. The Lewis electron-dot structure shows three bonds and one lone pair around each N atom. They have a tetrahedral arrangement. A tetrahedral arrangement has sp3 hybrid orbitals.

27. 1s 1s 2s 2p sp3 The orbital diagram of the ground-state N atom is The sp3 hybridized N atom is Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital.

28. One hybrid orbital is required for each bond (whether a single or a multiple bond) and for each lone pair. • Multiple bonding involves the overlap of one hybrid orbital and one (for a double bond) or two (for a triple bond) nonhybridized p orbitals.

29. To describe a multiple bond, we need to distinguish between two kinds of bonds. • A s bond (sigma) has a cylindrical shape about the bond axis. It is formed either when two s orbitals overlap or with directional orbitals (p or hybrid), when they overlap along their axis. • A p bond (pi) has an electron distribution above and below the bond axis. It is formed by the sideways overlap of two parallel p orbitals. This overlap occurs when two parallel half-filled p orbitals are available after s bonds have formed.

31. Figure A illustrates the s bonds in C2H4. The top of Figure B shows the p orbital on each carbon at 90° to each other, with no overlap. The bottom of Figure B shows parallel p orbitals overlapping to form a p bond.

32. In acetylene, C2H2, each C has two s bonds and two p bonds. The s bonds form using the sp hybrid orbital on C. This is shown in part A. The two p bonds form from the overlap of two sets of parallel p orbitals. This is illustrated in Figure B.

33. cis trans The description of a p bond helps to explain the cis-trans isomers of 1,2-dichloroethene. The overlap of the parallel p orbitals restricts the rotation around the C=C bond. This fixes the geometric positions of Cl: either on the same side (cis) or on different sides (trans) of the C=C bond.

34. One single bond and one triple bond requires two hybrid orbitals and two sets of two parallel p orbitals. That requires sp hybridization.

35. 1s 2s 2p • Describe the bonding about the C atom in formaldehyde, CH2O, using valence bond theory. The electron arrangement is trigonal pyramidal using sp2 hybrid orbitals. The ground-state orbital diagram for C is

36. 1s 1s 2s 2p sp2 2p • After promotion, the orbital diagram is • After hybridization, the orbital diagram is

37. The C—H s bonds are formed from the overlap of two C sp2 hybrid orbitals with the 1s orbital on the H atoms. • The C—O s bond is formed from the overlap of one sp2 hybrid orbital and one O half-filled p orbital. • The C—O p bond is formed from the sideways overlap of the C 2p orbital and an O 2p orbital.