CHEM 115 EXAM #3 Practice Exam Fall 2013

Transcription

1 Name CHEM 115 EXAM #3 Practice Exam Fall 2013 Circle the correct answer. (numbers 1-9, 3 points each) 1. Choose the correct statement about the compound SO 2. a. the S O bonds are ionic in character b. the molecule cannot exist since the bonding does not obey the octet rule (typo corrected) c. the S atom has a lone pair of electrons d. the S O bonds are different lengths since one is a single and the other a double bond e. none of the above 2. An expanded octet may occur: a. in the first and second period only b. in groups 1A, 2A, and 3A c. in periods 3 and greater d. in all groups except 1A e. none of the above 3. Which chloride should exhibit the most covalent (least ionic) type of bond? a. NaCl b. KCl c. BaCl 2 d. BeCl 2 e. CaCl 2 4. Which of the following bonds should be the most polar? a. N Cl b. N F c. F F d. H I e. N Br 5. Which of the following species has a central atom with sp 2 hybridization? a. CO 2 b. NO 3 - c. CH 4 d. PCl 5 e. none of the above 6. Which of the following bonds should be the shortest and strongest? a. C=C b. C C c. O=O d. C O e. C _ C 7. Which of the following sets contains the greatest number of bent molecules? a. CO 2, HCN, O 2 b. H 2 S, HCN, CO 2 c. H 2 O, CO, H 2 S d. H 2 S, CO, CO 2 8. Which of the following is a tetrahedral molecule (or ion)? a. CO 2 b. NO 3 - c. CH 4 d. PCl 5 e. none of the above 9. Which of the following is a molecule (or ion) that contains at least one π bond? a. CO 2 b. NO 3 - c. CH 4 d. PCl 5 e. none of the above [note: nitrate also has π bonding, but it is delocalized... it has less than a full π bond on any one bond axis]

2 2 Complete the following as directed (point values as indicated). (16 points) 10. For the following two species complete the tasks specified below in a - d. i. CH3COOH ii. C 2 H 2 a. Draw the best Lewis structures. b. Specify the hybridization of both the C atoms and the O atom specified in (i) and the C atoms in (ii)? c. Show the energy level diagram for both the C atoms and the O atom in (i) and the C atoms in (ii), fill the diagrams with the correct number of valence electrons, and identify the types of electrons present (i.e. σ, π, lone pair) d. What are the bond angles around both the C atoms and the O atom in (i) and each of the C atoms in (ii)? C in CH 3 sp 3 C in CO sp 2 O between CO and H sp 3 C in CH 3 sp 3 ( )( )( )( ) 4 σ bonds bold are electrons from that C C in CO p ( ) sp 2 ( )( )( ) σ σ σ bold are electrons from that C O between CO and H sp 3 ( )( )( )( ) 2 lone pairs & 2 σ bonds bold or are O electrons C in CH 3 about C in CO about 120 O [the COH angle] a bit < C in CH sp C in CH π π p ( ) ( ) sp ( )( ) σ σ bold are electrons from that C C in CH 180 (10 points) 11. (a) Draw Lewis structures for CN -, CO, N 2, and NO +. Label all atoms with their formal charges. ALL OF THESE HAVE TEN (10) VALENCE ELECTRONS [:C N:] - :C O: :N N: [:N O:] (b) Select the species from part (a) that you would expect to have the shortest, strongest bond. In order to receive credit you must EXPLAIN the reasoning for your selection. The triple bond between the C and O should be the shortest and strongest. In addition to being a triple bond, the -1 and +1 formal charges indicate that a strong partial ionic character will exist in this bond (adding to the strength of the bond).

3 3 (15 points) 12. Fill in the MO energy diagrams for the species listed below using valence electrons ONLY. The answer the questions that appear under the MO energy diagram of each species. Oxygen Ο 2 12 e - Peroxide Ο e - σ 2p σ 2p π 2p π 2p π 2p π 2p π2p π2p π2p π2p σ2p σ2p σ 2s σ 2s σ2s σ2s Show the calculation of the bond order for oxygen. BO = ½ (bonding antibonding) BO = ½ (8-4) = 2 Is oxygen diamagnetic or paramagnetic? Explain your answer. Molecular oxygen, O 2, is paramagnetic because it possesses unpaired electrons. Draw the Lewis structure of oxygen. Show the calculation of the bond order for the peroxide ion. BO = ½ (bonding antibonding) BO = ½ (8-6) = 1 Is the peroxide ion diamagnetic or paramagnetic? Explain your answer. The peroxide ion, O 2 2-, is diamagnetic because it possesses all paired electrons. Draw the Lewis structure for the peroxide ion Does the bond order for oxygen from the Lewis structure match the bond order from the MO diagram? Yes it does! Does the bond order for the peroxide ion from the Lewis structure match the bond order from the MO diagram? Yes it does! Which should have the longer, weaker bond; oxygen or the peroxide ion? Explain your choice! The peroxide ion is expected to have the longer weaker bond. This is due to two factors. First, the peroxide ion has a bond order of ONE vs. the bond order of TWO found in molecular oxygen. Also, the peroxide ion has a formal charge of minus one (-1) on each oxygen which will result in a repulsive force lengthening the bond.

4 4 Answer the following. Provide as much information as required to answer the question clearly. 13. Balance the following reactions AND draw Lewis symbol structures for part (e): (a) 2 C 6 H 14 (l) + 19 O 2 (g) 12 CO 2 (g) + 14 H 2 O (l) (b) 5 O 2 (g) + 4 NH 3 (g) 4 NO (g) + 6 H 2 O (l) (c) 3 NO 2 (g) + H 2 O (l) 2 HNO 3 (aq) + NO (g) (d) Fe 2 O 3 (s) + 2 Al (s) 2 Fe (l) + Al 2 O 3 (s) Don t forget to draw the Lewis symbols for each of the compounds in this next reaction! (e) CaC 2 (s) + 2 H 2 O (l) C 2 H 2 (g) + Ca(OH) 2 (s) Ca Write balanced chemical equations for each of the following word descriptions of a chemical process. (a) Solid magnesium metal reacts with carbon dioxide to produce magnesium oxide and carbon monoxide. Mg (s) + CO 2 (g) MgO (s) + CO (g) (b) Solid magnesium oxide reacts with water to form aqueous magnesium hydroxide. MgO (s) + H 2 O (l) Mg(OH) 2 (aq) (c) Liquid bromine reacts (quite violently) with aluminum metal to form aluminum bromide. 2 Al (s) + 3 Br 2 (l) 2 AlBr 3 (s) (d) Butane gas combusts completely with oxygen gas to form the expected products. 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O ( g) (e) Aqueous hydrochloric acid reacts with solid zinc to form zinc chloride and hydrogen. Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g)

5 5 BONUS (for up to 10 points) Place the correct atomic symbols into the correct blocks for elements 1 through 30. Label the following correctly on the table as well: (a) noble gases, (b) alkali metals, (c) halogens, (d) alkaline earth metals, (e) transition metals, (f) lanthanides, (g) actinides, (h) the coinage metals, and (i) the chalcogens. (fixed another typo) coinage metals are Cu, Ag, Au (group 11) chalcogens ( ore formers ) the oxygen family (group 16) principally O, S, Se See the next page for a chart of shape and name info you should also study. In addition to knowing the names, bond angles, etc. you should also know the hybridization type for each category. Below you will find a quick listing. #RHED hybridization 2 sp 3 sp 2 4 sp 3 5 sp 3 d 6 sp 3 d 2 RHED = region of high electron density Note: other options exist beyond those in this table... but we ll not be testing all the way up to a 16 coordinate system that employs all of the s, p, d and f orbitals from a single atom!

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