Practice problems for CHE 101 final Approximate breakdown of points: Old Ch. 2: 10 Ch. 3: 15 Ch. 4: 15 Ch. 5: 15 Ch. 6: 15 Ch. 7: 15 Ch. 10: 15 New Ch. 8: 40 Ch. 9: 40 Ch. 11: 20

Final: May 3 at 11:45 am. Rooms posted on UBlearns Check your points entered on posted grade point totals. Keep all assignments.

1) For the following molecules, give the Lewis structures, the molecular geometry, hybridization, and whether the molecule is polar or nonpolar. Where appropriate, indicate resonance structures.

A) XeF4

B) BrF3

C) CO3-2

2) Give Lewis structures for the following. Use formal charges to determine the best ones.

A) NCO-

B) ClO3-

2. The molecular geometry of the hydronium cation (H3O+) is …? A. trigonal B. trigonal pyramidal C. tetrahedral D. trigonal bipyramidal E. bent triatomic

3. Based upon its Lewis structure, how many σ- (“sigma”) and π- (“pi”) bonds are present in the molecule C2H3Cl ? A. 3 σ and 1 π B. 3 σ and 2 π C. 4 σ and 1 π D. 5 σ and 1 π E. 5 σ and 2 π

Also: what types of orbital overlaps occur for formation of bonds?

sp2 hybrid orbitals on carbon with s orbitals of hydrogen or sp3 orbital of chlorine, sp2 hybrid orbitals between carbons form sigma bonds leftover p orbitals on each carbon form a pi bond.

4. Consider the following species – BF4 - , SiH4 , PCl4+ , PCl4- , NF3 , NO2- . In which of these does the central atom violate the “octet rule”? A. PCl4+ , PCl4- , NF3 & NO2- B. BF4 - , SiH4 , PCl4+ C. PCl4+ , PCl4- , NF3 D. PCl4- E. PCl4+ , PCl4- & NO2-

5. Consider the following species SnO2 , O3 , CO32_ , NH3 , ICl4- Which of these have a net molecular dipole moment? A. All of them B. O3 , CO32- & NH3 only C. O3 , NH3 & ICl4 only D. SnO2, O3 & NH3 only E. O3, NH3 only

6. According to Molecular-Orbital Theory, which of the answers below gives the correct bond order for H2 , H2 – and H2 2- (respectively)? A. 1, ½ & 1 B. 1, ½ & 0 C. ½ , 1 and ½ D. 1, 1½ & 2 E. 1, 0 & 0 7. Which of the following gives inconsistent information concerning electron configuration and bond order and magnetic properties? A. N2 : (σ2s)2 (σ*2s)2 (π2p)4 (σ2p)2 bond order = 3, diamagnetic B. O2 : (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)2 bond order = 2, paramagnetic C. O22- : (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)4 bond order = 1, diamagnetic D. B2 : (σ2s)2 (σ*2s)2 (π2p)2 bond order = 1, diamagnetic E. O2+ : (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)1 bond order = 2½, paramagnetic

8. Which is the correct order for boiling points of the noble gases? (a) He < Ne < Ar < Kr < Xe (b) Ar < Ne < He < Xe < Kr (c) Xe < Kr < Ar < Ne < He (d) Xe < Kr < Ne < Ar < He (e) He < Ar < Ne < Xe < Kr

9. Based on a consideration of intermolecular forces, which of the following should have the lowest boiling point? A. Phosphine, PH3 B. Hydrogen sulfide, H2S C. Hydrogen chloride, HCl D. Silane, SiH4 E. Water, H2O

10. The most important intermolecular forces holding molecules of CCl4 together are… A. London dispersion forces B. dipole-dipole interactions C. hydrogen bonding D. covalent bonding E. ionic bonding 11. At 100 ºC and 1 atm, water is in which phase? Phase Diagram for H2O (a) solid (b) liquid (c) gas (d) solid-liquid equilibrium (e) liquid-gas equilibrium

Problem questions from last exam 12. How does the atomic radius of Mg compare with that of Na? (a) Mg is larger because it has more electrons. (b) Mg is larger because its effective nuclear charge is greater. (c) Na is larger because it has fewer electrons. (d) Na is larger because its effective nuclear charge is less. (e) Mg and Na are essentially the same size.

13. A possible set of four quantum numbers (n, ℓ, mℓ, ms) for the highest energy (valence) electron in indium, In, is: (a) 5, 1, -1, 1/2 (b) 4, 2, -3, 1/2 (c) 4, 1, 0, 1/2 (d) 5, 2, -2, 1/2 (e) 4, 3, 1, -1/2

14. The first ionization energy for nitrogen (N) is 1402 kJ/mol while that for oxygen (O) is 1314 kJ/mol. This tells us that: (a) not possible to answer unless the electron affinity is known (b) more energy is released when N accepts an electron than for O (c) more energy is required when N accepts an electron than for O (d) more energy is required to remove an electron from N than for O (e) more energy is released in removing an electron from N than for O

15. (see question 10) The reason for this is: (a) N has an occupied 3d orbital (b) the ionization energy always increases from left to right across a period in the periodic table (c) the ionization energy always decreases from left to right across a period (d) the electron removed from O is a s electron (e) the electron removed from N is from a half-filled p orbital