Tarique Rahman, Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Here are more questions nishing up Chapter 14. Remember, the HWS could be overloaded when the actual due time arrives Tuesday night. Room Assignments for EXAM 2 A-K L-O P-Z WEL 1.316 WEL 2.122 WEL 1.308 5. stems from sp hybridization of orbitals. 001 (part 1 of 1) 10 points 2 sp hybrid orbitals have 1. linear symmetry. 2. trigonal bipyramidal symmetry. 3. tetrahedral symmetry. 4. trigonal planar symmetry. correct 5. trigonal pyramidal symmetry. Explanation: 002 (part 1 of 1) 10 points In the molecule, C2 H4 , what are the atomic orbitals that participate in forming the sigma bond between the C and H atoms? 1. H: 1s and C: 2p 2. H: sp2 and C: sp2 3. H: 1s and C: sp2 correct 4. H: 1s and C: sp 5. H: 2p and C: sp3 Explanation: The electrons in the H 1s orbital and the
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electrons in the C sp2 hybrid orbital participate in forming the bond. 003 (part 1 of 1) 10 points A sigma bond 1. may exist alone or in conjunction with a pi bond. correct 2. is composed of non-bonding orbitals. 3. always exists in conjunction with a pi bond. 4. is always polar.
Explanation: A sigma, or single bond can exist by itself or with a pi bond, forming a double bond. 004 (part 1 of 1) 10 points In a new compound, it is found that the central carbon atom is sp2 hybridized. This implies that 1. carbon is also involved in a pi bond. correct 2. carbon has a tetrahedral electronic geometry. 3. carbon has four regions of high electron density. 4. carbon has four sigma bonds. 5. carbon has four lone pairs of electrons. Explanation: Carbon forms four bonds. sp2 hybridization enables a C atom to bond to three other atoms, leaving one electron in a p orbital, which will form a pi bond. 005 (part 1 of 4) 10 points Draw the Lewis structure for the following hydrocarbon molecule. The carbons are numbered one to four starting with the far left
Rahman, Tarique Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout carbon as one. CH2 CCHCH3 What is the molecular shape of carbon 3? 1. square planar 3. 109.5 2. bipyramidal 4. 90 , 180 3. trigonal planar correct 5. 90 , 120 , 180 4. angular 6. 180 correct 5. tetrahedral 7. 120 6. trigonal pyramidal 7. linear Explanation: The molecule has the structure C C C C Carbon 3 has bonds to two carbons and one hydrogen or RHED = 3. This is trigonal planar molecular geometry. 006 (part 2 of 4) 10 points What is the hybridization of carbon 4? 1. sp3 d 2. sp3 correct 3. sp 4. sp2 d 5. sp4 6. sp3 d2 7. sp2 Explanation: The molecule has the structure C C C C Carbon 4 has bonds to one carbon and 3 hydrogens or RHED = 4 which is sp3 hybridized. 007 (part 3 of 4) 10 points What are the bond angles of carbon 2? 1. 90 , 120 2. less than 109.5
2
Explanation: The molecule has the structure C C C C Carbon 2 has bonds only to two other carbons (double bonds). This is RHED = 2 or linear electronic geometry. Any linear molecule has 180 degree angles. 008 (part 4 of 4) 10 points Carbon 2 has 1. 2 sigma bonds and 1 pi bonds. 2. no sigma bonds and 4 pi bonds. 3. 4 sigma bonds and no pi bonds. 4. 1 sigma bonds and 3 pi bonds. 5. 2 sigma bonds and 2 pi bonds. correct 6. 3 sigma bonds and 1 pi bond. Explanation: The molecule has the structure C C C C Carbon 2 has bonds only to two other carbons (double bonds). Double bonds are composed of one sigma and one pi bond, therefore two double bonds is two sigma bonds and two pi bonds. 009 (part 1 of 1) 10 points According to molecular orbital theory, which of the following is NOT predicted to exist?
Rahman, Tarique Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout 1. He 2. He2 correct 3. He
2+
3
2p
2p
2p
2p
4. All are predicted to exist 5. He2 2s
2p 2p 2s 2s 011 (part 2 of 2) 10 points What is the magnetism and number of unpaired electrons of B2 ? 1. paramagnetic; 2 correct
Explanation: When the bond order is equal to 0, a molecule is predicted to not exist. bonding e antibonding e B .O . = 2 For He2 , the number of bonding electrons (2) equals the number of antibonding electrons, therefore the bond order is 0 and the molecule is predicted to not exist. 010 (part 1 of 2) 10 points What is the expected bond order for the diatomic species B2 ? 1. 4 3 2. 2 3. 1 correct 4. 1 2
2s
2. paramagnetic; 3 3. diamagnetic 4. paramagnetic; 5. 4 paramagnetic; 1 Explanation: 012 (part 1 of 1) 10 points Which of the following is TRUE about nonbonding orbitals of the molecular orbital theory? 1. Shared electrons are placed in nonbonding orbitals.
5. 2 6. 3
Explanation:
2. Nonbonding orbitals do not overlap other atomic orbitals. correct 3. Nonbonding orbitals are made from the overlap of bonding and antibonding orbitals.
BO =
20 = 1 or 2
BO =
2+22 =1 2
4. Nonbonding orbitals are higher in energy than their corresponding bonding orbitals.
Rahman, Tarique Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout 5. Nonbonding orbitals give added stability to the molecule. Explanation: Nonbonding orbitals maintain the characteristics of the original atomic orbitals. They are not molecular orbitals and do not arise from the overlapping of atomic orbitals. 013 (part 1 of 1) 10 points N2 has a bond order of 3 and O2 has a bond order of 2. Based on this information, choose the response that best completes the following sentence, in the order listed. N2 is (less, more) stable than O2 , and has a (larger, shorter) bond length and a (higher, lower) bond energy. 1. more; shorter; higher correct 2. more; shorter; lower 3. more; longer; higher 4. less; longer; lower 5. less; shorter; lower Explanation: The higher the bond order, the more stable the molecule. Stable bonds have a high bond energy since they are hard to break apart. As bond order increases, the bond length gets shorter. Triple bonds have shorter bond length than double or single bonds. 014 (part 1 of 1) 10 points Which molecule would you predict would be more stable: F2 or F+ ? 2 2 1. F2 2 2. F+ correct 2 3. They would be equally stable. Explanation: 015 (part 1 of 1) 10 points Which of the following is NOT paramagnetic? 1. O+ 2 2. N2+ correct 2 3. O2 4. O 2 5. N2 2
4
Explanation: Only N2 does not possess unpaired electron(s); N2 is diamagnetic. 016 (part 1 of 3) 10 points Consider a hypothetical species HeH. What charge, if any, should be present on this combination of atoms to produce the most stable molecule or ion possible? 1. 1 2. 1 2
3. None of these 4. + 1 2
5. +1 correct Explanation: The helium and hydrogen atoms have only their 1s orbitals available for bonding. Combination of these two will lead to a and a pair of orbitals. The most stable species will be one in which only the orbital is lled, so we need a species that has only two electrons. The charge on this species will be +1. 017 (part 2 of 3) 10 points What is the maximum bond order that such a molecule or ion could have? 1. 2 2. 1 correct
Rahman, Tarique Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout 3. 1 2 1 4. 3 5. 3 Explanation: The maximum bond order will be 1. 018 (part 3 of 3) 10 points If the charge on this species were increased or decreased by one, what would be the eect on the bond order in the molecule? 1 1. increase by 2 1 2. decrease by correct 2 3. increase by 1 4. increase by 2 5. decrease by 1 Explanation: Adding one electron will decrease the bond 1 order by because an antibonding orbital will 2 be populated. Taking an electron away will 1 also decrease the bond energy by because it 2 will remove one bonding electron. He 1s 019 (part 1 of 1) 10 points Draw the molecular orbital diagrams for N2 and CO. Which of the following statements is FALSE? 1. CO and N2 are isoelectric. 2. The atomic orbitals of carbon are lower in energy than the corresponding atomic orbitals of oxygen. correct 1s He 2s 2s 2s 2s
5
3. The diagram of CO is slightly skewed (non-symmetrical). 4. Both molecules have a bond order of 3. 5. Both CO and N2 are diamagnetic.
Explanation: Each N contributes 7 electrons in N2 : 2p 2py 2pz 2p 2p 2py 2pz 2p
1s 1s 1s
1s N2 N N atomic atomic molecular orbital orbital orbital C contributes 6 electrons and O contributes 8 electrons in CO: 2p 2py 2pz 2p 2p 2py 2pz 2p
Rahman, Tarique Homework 7 Due: Oct 16 2007, midnight Inst: Vandenbout
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2s 2s 2s 2s
1s 1s 1s C atomic orbital bond order = 1s CO molecular orbital O atomic orbital
elecbond elecantibond 2 10 4 = =3 2
020 (part 1 of 1) 10 points Which of the following species possesses a delocalized bond? 1. No molecule given here possesses a delocalized bond. 2. H2 O 3. NCl3 4. H2 S 5. NO correct 3 Explanation: Only for NO can resonance structures be 3 drawn. .. .. .O. .O. .O. .. .. .. .. .. .. .. N N N O O O O O O
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