Practice Exam 2 Solutions

Transcription

1 Practice Exam 2 Solutions Questions 1-4 refer to the following substances: (A) CO 2 (B) K (C) N 3 (D) Cl 2 (E) CaO 1. B as the strongest ionic bonding. 2. D as only non-polar bonds. 3. C Consists of polar molecules. 4. D as bonds with equally shared electrons. Questions 5-8 refer to the following compounds: (A) O 2 (B) Kr 2 (C) e 2 (D) I 3 (E) N 3 5. D Is composed of molecules with the smallest bond angles. 6. E Is composed of molecules with one lone pair on the central atom. 7. C Is composed of ions. 8. B Is composed of nonpolar molecules. To tackle this problem it is best to draw the Lewis structures for each. 9. What is the correct way of writing the compound e(no 3 ) 3? (A) Iron (III) Nitride (B) Iron (IV) Nitride (C) Iron (III) Nitrite. (D) Iron (III) Nitrate. (E) Iron (IV) Nitrite. 10. Which of the following compounds exhibits both covalent and ionic bonding? (A) Br 3 (B) 2 O (C) Ca 2 (D) Na 2 CO 3 (E) 2 CO 11. Which of the following bonds would be least polar? (A) C O (B) - (C) N- (D) C- (E) S- 12. Give the correct formula for calcium chloride (A) CaCl (B) CaCl 2 (C) Ca(ClO 4 ) 2 (D) CaClO 4 (E) CaCl Which of the following is the correct formula for the carbonate ion? (A) CO 3 (B) CO 2 (C) CO 3 (D) CO 2 (E) CO What is the electron pair geometry for N 3? (A) trigonal pyramidal (B) trigonal planar (C) tetrahedral (D) linear (E) octahedral Don t confuse electron pair geometry with molecular geometry (which would be trigonal pyramidal) 15. Which of the following explains why the bond length in O 3 is longer than the bond length in O 2? (A) There are more atoms in an O 3 molecule than an O 2 molecule (B) In O 2 there are only single bonds and in O 3 there are only double bonds (C) In O 2 there are only double bonds while in O 3 has delocalized bonds (D) There are multiple resonance structures in O 3 (E) O 3 absorbs UV-light and becomes O 2

2 16. Which of the following best explains why pure water is an electrical insulator (A) Water is not an insulator (B) The valence electrons in water are held between the O and atoms (C) Water is a polar molecule (D) ydrogen only needs 2 electrons to fill its outer shell (E) Oxygen is too electronegative Use the structure below as drawn to answer the following questions. 17. What is the formal charge on oxygen #2? (A) +1 (B) 0 (C) -1 (D) -2 (E) Impossible to determine 18. What is the formal charge on oxygen #1? (A) +1 (B) 0 (C) -1 (D) -2 (E) Impossible to determine 19. Given the structure shown above, why does ozone, O 3, have two bonds of equal length? (A) Ozone is a hybrid of 2 resonance structures (B) All oxygen atoms need 2 bonds to complete their octet (C) Oxygen can never have a formal charge of +1 (D) Oxygen 1 has an expanded octet and there is no d-orbital available (E) Ozone is a linear molecule. 20. Look at the structure below. Lone pairs have been left out of this structure, but you can assume that all formal charges are zero. The structure may or may not be drawn in the correct geometry. What would be the shape (molecular geometry) of the Br 3 molecule? (A) Trigonal planar (B) Trigonal pyramidal (C) Tetrahedral (D) T-shaped (E) See-saw Open Response: 1. Three different isomers for the compound NO are shown below. (A) Draw in all nonzero formal charges on each molecule. (4 pt) Molecule 1: N = -1, O = +1, = 0 Molecule 2: O = 0, N = 0, = 0 Molecule 3: N = -1, = +2, O = -1 (B) Circle the most stable Lewis structure above. Briefly explain your choice. (2 pt) Molecule 2 will be the most stable because it has no formal charges. 2. Consider the molecules ethylene (C 2 4 ) and allene (C 3 4 ) shown below. Each of these molecules is drawn in the correct geometry.

3 (A) Each of the carbon atoms marked with a star (*) is trigonal planar. Identify the hybridization of each of these atoms. (1 pt) If the shape is trigonal planar, it leads to sp 2 hybridization. (B) Identify the hybridization of the central carbon atom in allene. (1 pt) sp (C) Using principles of hybridization and bonding, explain why the four hydrogen atoms in ethylene are all in the same plane, while those in allene are not. (4 pt) In order to make pi bonds, unhybridized p-orbitals must be lined up. In ethylene, both carbon atoms are sp 2 hybridized. The unhybridized p-orbitals are perpendicular to the sp 2 hybrid orbitals. Therefore, when the p-orbitals line up, the remaining orbitals also end up in the same plane, as shown below. In allene, on the other hand, the central carbon atom is involved in two pi bonds, one on each side. In order to do this, it needs to have two unhybridized p-orbitals, which are perpendicular to each other. The pi bond on the left and the pi bond on the right must therefore be perpendicular. This requires the hybrid orbitals that make the sigma bonds to be perpendicular. 3. Consider the nitrate ion, NO 3. (B) Draw a complete Lewis structure for nitrate. Include all non-zero formal charges and any relevant resonance structures. (4 pt)

4 (C) Give the electron pair geometry and molecular geometry of the N atom. (1 pt each) e.p.g: trigonal planar m.g.: trigonal planar 4. The skeleton structures below show how atoms are connected in a molecule, but they ignore multiple bonds, lone pairs, and formal charges. Complete each structure by including all lone pairs (even on terminal atoms), multiple bonds, and nonzero formal charges as appropriate. If more than one resonance structure is possible, draw only the more stable one. Based on your Lewis structure, predict the electron pair geometry and shape of the molecule. Draw a dipole on the Lewis structure where applicable ormula Lewis structure Electron pair geometry Molecular geometry Orbital ybridization 3 O + (3 pt) Tetrahedral Trigonal pyramidal sp 3 Left C: tetrahedral Left C: tetrahedral Left C: sp 3 C 2 5 ON (5 pt) Right C: trigonal planar Right C: trigonal planar Right C: sp 2 N: sp 3 N: tetrahedral N: trigonal pyramidal O XeO 4 (4 pt) Xe Octahedral Square pyramidal sp 3 d 2 5. A) Draw the Lewis Structure for C 2 2. (2 pt) B) ow many sigma bonds does this molecule have? (1 pt) 3_ C) ow many pi bonds? (1 pt) 2

5 6. Consider the substances N 2 4, N 2 2, and N 2. A) Draw the Lewis Structure for each substance. (2 pt) B) Rank the molecules in order of strongest to weakest N-N bond. Explain. (2 pt) N 2, N 2 2, N 2 4 As bond order increases, the strength of the bond increases. Since we know that N 2 has an N-N triple bond, N 2 2 has an N-N double bond, and N 2 4 has only an N-N single bond, we know that they must be in this order. C) Rank the molecules N 2 4 and N 2 2 in order of smallest to largest bond angle. Explain. (2 pt) N 2 4, N 2 2 N 2 4 has a trigonal pyramidal geometry around each N atom. This leads to a bond angle of o (though it is slightly less than this due to the lone pair). Each N atom in N 2 2 has trigonal planar geometry, which leads to a bond angle of 120 o. D) Which of the above molecules is/are polar? (2 pt) N 2 4 would be polar. When we draw the 3-D structure, we know that each N has trigonal pyramidal geometry. Therefore the partial negative on the N atoms will not be surrounded on all sides by partially positive hydrogen atoms. This leads to a dipole.

6 14. Structures similar to the one shown below are important in many biological situations. All bonds and lone pairs are shown correctly. (A) Complete the structure by including any nonzero formal charges. (1 pt) The lower-left N has a formal charge of +1 (B) This molecule has one other significant resonance structure. Draw it. (2 pt) C N N C (C) Is the resonance structure you drew in part (B) less stable, equally stable, or more stable than the original structure shown? Explain. (2 pt) Less stable Equally stable More stable The formal charges are the same (D) Considering your answers above, which of the following correctly puts the C-N bonds in order from shortest to longest? Explain. (3 pt) (i) C 3 -N 2 < C 1 -N 2 < C 3 -N 4 < C 5 -N 4 (ii) C 3 -N 2 = C 3 -N 4 < C 1 -N 2 = C 5 -N 4 C 3 N 4 (iii)c 5 -N 4 < C 3 -N 4 < C 1 -N 2 < C 3 -N 2 C 5 (iv) C 3 -N 2 < C 1 -N 2 = C 3 -N 4 = C 5 -N 4 N 2 C 1 C Due to the fact that the resonance structures above have equal stability, the hybrid between the two will be equal. This means the C 3 -N 2 bond and the C 3 -N 4 bond will have equal length. It also means that C 1 -N 2 and C 5 -N 4 bonds will have equal length. Since the the C 3 - N 2 bond and the C 3 -N 4 bonds exhibit delocalized- bonding, they will have a shorter bond length than the C 1 -N 2 and C 5 -N 4 bonds.