Version 038 – Exam 2 – Mccord – (53130) 1 This Print-Out Should

Version038–Exam2–McCord–(53130) 1

This print-out should have 31 questions. 3. sp3 Multiple-choice questions may continue on the next column or page – ﬁnd all choices 4. Pure pz orbitals are used in bonding. before answering. 5. sp3d McCord CH301tt Explanation: The structure is

b b b This exam is only for b O McCord’s Tues/Thur CH301 class. C If Quest works properly, you should be able H3C CH3 to see your score for this exam by 11:30 PM tonight. Do realize however that Quest has 003 10.0 points been taking up to 8 hours to get through the Which of the molecules

b b

b grading, so your score might appear bit by bit. b O I will post an announcement on our web page I) and/or Quest when I feel like all bubblesheets H H

have been graded. b b

b b

b b Cl PLEASE carefully bubble in your UTEID and Version Number! We av- II)

b b B

b b

erage about 20-30 students per exam that b

b b b

Cl Clb b misbubble this information. Get it right! b

b b b b b b

b III) b Cl Cl b b b b 001 10.0 points contains polar covalent bonds but is NOT Choose the species that is incorrectly matched itself a polar molecule? with electronic geometry about the central atom. 1. III only

1. CF4 : tetrahedral 2. II and III only

2. BeBr2 : linear 3. I and II only

3. H2O : tetrahedral 4. All ﬁt the criteria.

4. PF3 : pyramidal correct 5. None ﬁt the criteria.

5. NH3 : tetrahedral 6. II only correct Explanation: 7. I and III only 002 10.0 points 8. I only Identify the hybrid orbitals used by the underlined atom in acetone (CH3COCH3). Explanation: I) Water has polar covalent bonds and is 1. sp2 correct itself polar. II) Boron trichloride contains polar covalent 2. sp bonds but is not polar overall due to symmetry. Version038–Exam2–McCord–(53130) 2 III) Chlorine gas has nonpolar bonds only. b b

b H C S b 004 10.0 points 10. b b

b b Which of the following is the correct Lewis b H b formula for thioformaldehyde (CH2S)? b b

b b b b Explanation:

b b b H C S b The Lewis formula for thioformaldehyde 1. b b b b

b b b b

b b H b b

b b S (CH2S) is b b b b b b H C

b b b H C S b H 2. b b b b

b b b H b b b 005 10.0 points The molecule AX3 is polar and obeys the b b octet rule; therefore, the central atom A H C S 3.

b b b H b 1. has no lone pairs of electrons.

b b 2. b b has two lone pairs of electrons. b b S C 4. b 3. correct b has one lone pair of electrons.

H b H b b b 4. has four lone pairs of electrons.

b b

b S b 5. has three lone pairs of electrons. b b

b Explanation: 5. b H C

b b b The molecule AX3 could obey the octet b

b H b rule in one of two ways. The molecule could b b contain two A−X single bonds and one A−X double bond, yielding the structure H C S X 6.

b b A X b H b b b X

b b This molecule would have a trigonal planar

b S b geometry. It is symmetrical and therefore 7. correct H C non-polar, so this is not the correct structure for our molecule. H The second possible dot structure for AX3 is ·· X A X 8. HS C H X b b b b This molecule obeys the octet rule and has b H C S b a pyramidal molecular geometry. Assuming 9. b b

b b there is an electronegativity diﬀerence be- b H b b b tween A and X, this molecule would be polar. Version038–Exam2–McCord–(53130) 3 This structure has one lone pair on A. what is the interaction energy between a zinc − ion (Zn2+) and a sulﬁde ion (S2 ) in a hy- 006 10.0 points pothetical structure in which the inter-ionic Which of the following compounds would be distances are the same as that of NaCl? expected to have the shortest bonds? 1. 1.140 kJ/mol − 1. NO3 2. 1140 kJ/mol 2. CCl4 3. 3040 kJ/mol correct − 3. NO2 correct 4. 760 kJ/mol 4. NCl3 5. 1520 kJ/mol Explanation: triple bonds < double bonds < single Explanation: − bonds. The more bonds, the shorter the bond qZn =+2C qS = 2C − length. Resonance structures have bonds qNa =+1C qCl = 1C somewhere between single and double bonds. VNaCl = 760 kJ/mol The more bonding atoms sharing the 2 elec- If r is the distance between the two ions (the trons, the weaker the bond and the longer the sum of the ionic radii), the energy interaction − − between ions is given by Coulomb’s Law: bond length. NO2 and NO3 are both reso- − nance structures, but NO2 has fewer bonding q1 q2 atoms. Therefore it has stronger bonds, and V = shorter bond lengths. 4 π ǫ0 r q1 q2 r = 007 10.0 points 4 π ǫ0 V A double bond consists of

1. two pi bonds. rNaCl = rZnS qNa qCl qZn q = S 2. one sigma bond containing four elec- 4 π ǫ0 VNaCl 4 π ǫ0 VZnS qZn qS VNaCl trons. VZnS = qNa qCl 3. one sigma and one pi bond. correct (+2 C)(−2 C)(−760 kJ/mol) = (+1 C)(−1 C) 4. two sigma bonds. = −3040 kJ/mol ,

5. two beta bonds. the energy released by the interaction.

6. an alpha and a beta bond. 009 10.0 points Consider the following pairs A and B

Explanation: b b b b b b

b b b The sigma bonds overlap along a line drawn A1) N N O A2) b N N O b b b b b b between the two nuclei, and the pi bonds − b b − 3 b b 3

b b b b overlap perpendicular to this line. b b b b  O   O 

b b b b b b b b

b B1) O P O b 2 O P O b b B ) 008 10.0 points  b b   b b b b     

If the interaction energy between a sodium ion  b b   b b   b O b   b O b  and a chloride ion in table salt is 760 kJ/mol,  b b   b b  Version038–Exam2–McCord–(53130) 4 of Lewis structures. onal bipyramidal. Select the one from each pair that is likely to make the dominant contribution to a reso- 011 10.0 points nance hybrid. Consider the species − a) I2, b) O3, c)I3 , d) CS2, e) CO. 1. A1 and B1 Which of the species is/are polar?

2. A2 and B1 correct 1. c) and e) only

3. A1 and B2 2. b) and e) only correct

4. A2 and B2 3. e) only Explanation:

b b b b b b 4. b) and c) only b b b A1) N N O A2) b N N O b b b b b b −1 +1 0 0 +1 −1 Explanation: Of the species listed, only O3 and CO are The second structure is preferred because it polar. CO is polar due to the difference in places the negative formal charge on the most electronegativity between O and C; O3 is po- electronegative atom. − b b − 3 b b 3 lar because it has 3 RHED and one lone pair b b − b b − b O b 1 b O b 1 on the central atom. This lone pair is an  0   0  b b b b b b − b b 0 b 1 area where negative charge is concentrated, 1 O P O b 2 O P O B ) b b b b B )    b b b b      so this results in the molecule having an over- −1 0  b b   b b  2 b b all dipole moment. In the other species, I and  −1 O   −1 b O b   b b   b b  CS2 are both linear and in the case of CS2, The first structure is preferred for the same the two opposing dipoles of the C-S bonds reason. − will cancel. Finally I3 has 5 RHED and three lone pairs on the central atom but they are ar- ◦ 010 10.0 points ranged at 120 so their effects cancel and the ion is nonpolar. On the basis of valence shell electron pair repulsion theory, the electronic geometry of 012 10.0 points XeF2 is Which of the following has bond angles ◦ slightly less than 109.5 ? 1. trigonal bipyramidal. correct 2− 1. CS3 2. linear. 2. COCl2 3. octahedral. 3. AsF3 correct 4. square planar. 4. COS 5. trigonal planar. Explanation: 5. SO2

b b b

b b b b

b F Explanation: b

b b b

Xe b b b b

b AsF3 has four regions of electron density,

b F b b one of which is a lone pair, which repels the For the central atom Xe, HED = 5, lone other regions, making those bond angles less ◦ pairs = 3, and the electronic geometry is trig- than 109.5 . Version038–Exam2–McCord–(53130) 5

013 10.0 points 3. 67%; 33% Which bond is not found in ethylene (C2H4)? 4. 33%; 67% 1. π2p−2p 5. 50%; 50% 2. All of these bonds are found in C2H4. Explanation:

3. σ1s−sp2 1 s = = 25% 4 4. σsp2−2p correct 3 p = = 75% 4 5. σsp2−sp2 016 10.0 points Explanation: + Consider the species H2CCH .

014 10.0 points +

→ What hybridization would you expect for P H → a b  − in PO ? 3  C C H      1. spd H 

3 What hybridization is at a and b, respec- 2. sp tively? 3. p 1. sp3; sp3 4. sp3d 2. sp; sp 5. sp3d2 3. sp3; sp 6. sp 4. sp2; sp2 7. sp2 correct 5. sp2; sp correct 8. s2p2 6. sp2; sp3 Explanation: Explanation: This is a resonance structure where P +

shares eight electrons with three oxygen 2

→ atoms. There are no lone pairs of electrons. H sp→ sp  Therefore there are 3 regions of HED and the 2  C C H  hybridization is sp .     H  015 10.0 points The sp3 hybridization has what percent s 017 10.0 points character and what percent p character? A molecule with the formula AY4 has eight 1. 25%; 75% correct shared electrons and has the trigonal bipyramidal electronic geometry. The central atom 2. 75%; 25% is “A”. Pick the true statement. Version038–Exam2–McCord–(53130) 6 4. slightly less than 109.5 degrees 1. It is sp3 hybridized. 5. 180 degrees 2. It has square planar molecular geometry. 6. slightly less than 120 degrees correct

3. It exhibits resonance. Explanation: N is the central atom. One O atom makes 4. It has one lone pair of electrons on A. a single bond to N and the other makes a correct double bond to N. This means there are TWO diﬀerent resonance structures. Additionally 5. It has two single bonds and two double there is a lone pair on the N atom. bonds. There are 3 RHED around N so the electronic geometry is trigonal planar. As one 6. It is an anion. RHED is a lone pair, the molecular geometry is bent and the O-N-O bond angle is a little 7. It is a non-polar molecule. less than 120 degrees.

Explanation: 020 10.0 points

b b Assume that in each pair of compounds be- The Lewis structure is Y A Y low the type of crystal structure is similar. Consider the strength of the attractive force Y Y bonding the ions together in each compound. 018 10.0 points In which case would the relative strengths be Which of the following compounds contains a ranked correctly? σ bond formed from the overlap of 1s and sp3 atomic orbits? 1. LiCl > LiBr correct

1. CHF3 correct 2. NaI > LiI

2. C2H2 3. MgCl2 < CaBr2

3. CF4 4. KF < KCl Explanation: 4. C2F6 The strength of the coulombic interaction 5. C2H4 between the two ions is directly proportional to their charges, and inversely proportional to Explanation: their separation. The separation of the ions is determined by their radii. Therefore the 019 10.0 points strongest interactions will be between small Predict the bond angle in the nitrite ion, ions and with a high charge. − NO2 . 021 10.0 points 1. 60 degrees A molecule has one lone pair of electrons on the central atom and three atoms bonded to 2. 90 degrees the central atom. The central atom follows the octet rule. What is its electronic arrange- 3. slightly less than 180 degrees ment and its hybridization? Version038–Exam2–McCord–(53130) 7

1. pyramidal; sp3 shell. Each of the ﬂuorines will have a true octet. Thus 2. tetrahedral; sp2 N =12+5(8)=52 3. pyramidal; sp2 A =7+5(7)=42 S = N − A = 10 . 4. angular; sp3 10 electrons are shared in the 5 single bonds 5. trigonal planar; sp2 to the ﬂuorines. In addition to those bonding electron pairs, there is also one lone pair 6. trigonal planar; sp3 of electrons on the chlorine. (“Left over” electrons are placed on the central atom.) 7. tetrahedral; sp3 correct 023 10.0 points Explanation: Which substance has nonpolar covalent One lone pair plus 3 bonded pairs equals bonds? 4 electronic regions. This means that the central atom is tetrahedral and hybridization 1. O2 correct is sp3. 2. NO2 022 10.0 points The compound ClF5 (where Cl is the central 3. CO atom) has an electronic geometry of octahedral. What would be the quantities for the 4. NaCl Lewis dot formula? N (needed electrons), A Explanation: (available electrons) and S (shared electrons)

1. N = 48; A = 42; S = 12 024 10.0 points Write the electron-dot notation for the ele- 2. N = 52; A = 42; S = 12 ment P.

3. N = 48; A = 42; S = 10 1. None of these

b b

b b 4. N = 48; A = 42; S =8 2. b P b b b

b b

b 5. N = 50; A = 42; S = 10 3. b P b b

b b 6. N = 52; A = 42; S =8 4. P b

7. N = 50; A = 42; S =8 5. P b

b b

b 8. N = 50; A = 42; S = 12 6. b P b b b

b 9. N = 52; A = 42; S = 10 correct 7. P b

b b Explanation: 8. P b b

The number of electrons needed in this case b b

b must be adjusted for an expanded octet for 9. b P correct the central atom, chlorine, which will have b 6 pairs of electrons (12 total) for its valence Explanation: Version038–Exam2–McCord–(53130) 8 According to its electron conﬁguration 1. 16; 4 ([Ne] 3s2 3p3), P has 2 + 3 = 5 valence electrons. The electron-dot notation for P is 2. 14; 5 therefore b b

b b P 3. 14; 4 b 025 10.0 points 4. 14; 6 What total number of valence electrons should appear in the dot formula for the chlo- 5. 16; 6 correct − rate ion ClO3 ? 6. 15; 5 1. 32 7. 15; 6 2. 28 8. 15; 4 3. 30 9. 16; 5 4. 24 Explanation: 5. 26 correct The total number of electrons equals its Explanation: atomic number plus one extra electron as is indicated by the charge: 15 + 1 = 16 electrons Chlorine has 7 valence electrons and oxy- total. The valence electrons are the outer gen 6. The overall −1 charge indicates that − most s and p electrons. P has 6 electrons in there is 1 extra electron. The total number of its outer shell. valence electrons is − − 1 × 7 e (from Cl) + 3 × 6 e (from O) − − +1 e (from −1 charge) = 26 e 028 10.0 points How many non-bonding electron pairs are 026 10.0 points − on the central atom in SF ? Which of the following species are radicals? 5 1. 3 1. NO2 correct 2. 0 2. CH2O 3. 2 3. HCN 4. 4 4. HNO2 5. 1 correct 5. N2O Explanation: − Explanation: NO2 has 17 valence e . The total number of valence electrons is

027 10.0 points − The phosphorous anion (P ) has how many − − S 1 × 6 e = 6 e total electrons and how many valence elec- × − − F 5 7 e− = 35 e− trons? −1 charge +1 e = 1 e − 42 e Version038–Exam2–McCord–(53130) 9 − 1. None of these

b b

b b b b

b  b 

b b

F b b F b  b  2. correct

b b S b and  b  is the Lewis struc-

b  b 

b b b

F b F b

 b 

b b

b  b F   b b  3.   ture. 4. 029 10.0 points Which of the following statements concerning valence bond theory is/are true? 5. I) Hybridizing one 2s orbital with three 2p 3 orbitals would produce four sp orbitals. Explanation: II) When a 2s orbital is hybridized with a The term ‘sigma’ implies the entity has single 2p orbital, there will be a single symmetry of rotation of any angle around a unhybridized 2p orbital left to form a π central axis; like viewing a cylinder down the bond. long axis from one of the circular ends. III) Among the types of hybrid orbitals A sigma-antibonding molecular orbital is we have learned about, the minimum formed when two atomic orbitals overlap in amount of s-character for a hybrid or- such a way that they are out of phase. In th bital is 1/6 . H2 the atomic orbitals which combine to form the antibonding sigma molecular orbital are 1. II only both 1 s: ∗ H1 s H1 s H2 σ 2. I only + → 3. I, III correct 031 10.0 points After constructing the Lewis dot formula for 4. II, III − CCl3 , give its electronic arrangement and 5. I, II molecular shape.

6. I, II, III 1. trigonal bipyramidal, T-shaped

7. III only 2. trigonal bipyramidal, seesaw 3. Explanation: tetrahedral, tetrahedral Statement II is false because there will be 4. tetrahedral, trigonal pyramidal correct a pair of unhybridized p orbitals available for π bonding. Statements I is true because hy- 5. trigonal planar, trigonal planar bridizing any number of atomic orbitals al- ways results in an equal number of hybrid 3 2 6. tetrahedral, angular orbitals. Statement III is true because sp d hybrid orbitals have 1/6th s-character. Explanation:

030 10.0 points N = 8(4) = 32 Identify the antibonding orbital corre- sponding to the σ-bond in H2. A =1+4+3(7)=26 S = 32 − 26=6 Version038–Exam2–McCord–(53130) 10 ativity increases going from left to right and 032 10.0 points upward in the periodic table, so the most electronegative element is O, meaning the H O Which of the following ions has a tetrahe- bond is most polar. dral molecular (actual) geometry?

+ 1. NH4 correct

+ 2. H2F

+ 3. H3O

2− 4. CO3

− 5. NO3 Explanation: H

H N H

H For the central atom N, HED = 4 and lone pairs = 0, so the electronic and molecular geometries are also tetrahedral. H

N H H H

033 10.0 points Which covalent bond is the most polar?

1. H S

2. H H

3. H O correct

4. H N Explanation: In each case a pair of electrons is being shared between H and one other element. The greater the diﬀerence in electronegativity between the two elements (H and O, N, H and S), the more polar the bond. Electroneg-