Chapter 9 Covalent Bonding: Orbitals

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–2 QUESTION In examining this figure from left to right, you could conclude that…

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–3 QUESTION (continued) 1.Native atomic orbitals have more exact boundaries than hybridized orbitals. 2.During hybridization, orbital number is conserved. 3.The four sp 3 orbitals represent a better way than the native orbitals to obtain 90° bonding angles. 4.The sp 3 orbitals represent a 50–50 combination of s and p orbitals.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–4 ANSWER Choice 2 shows a logical conclusion from the diagram. There are 4 total orbitals on the left (three p orbitals and one s orbital) and 4 total new sp 3 orbitals that can arise from mixing those. Section 9.1: Hybridization and the Localized Electron Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–5 QUESTION The bond angle formed between three carbon atoms in a compound is approximately 120°. Which of the following is the most likely type of hybridization for the middle carbon atom? 1.sp 2.sp 2 3.sp 3 4.This relationship is still something I do not understand.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–6 ANSWER Choice 2 is correct. sp 2 hybridization is most often associated with 120° bond angles. The three orbitals that have sp 2 hybridization will be as far apart as possible, and for three areas that would be 120°. Section 9.1: Hybridization and the Localized Electron Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–7 QUESTION Formaldehyde and methanol both have a carbon to oxygen bond. Yet the oxygen atom in methanol, CH 3 OH, can freely spin around without causing the C to move with it. In formaldehyde, CH 2 O, any spin on the oxygen causes the C to spin with it. What type of hybridization and what consequence is found for the carbon in each of these useful compounds? (Hint: examine the Lewis structures)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–8 QUESTION (continued) 1.Methanol—sp 3 ; formaldehyde—sp; the unhybridized p orbitals get in the way of free rotation for the C to O bond. 2.Methanol—sp 2 ; formaldehyde—sp 2 ; the unhybridized p orbital in methanol forms a new bond with the other three H atoms that allows the bond to be stronger and rotate on its own. 3.Methanol—sp 3 ; formaldehyde—sp 2 ; the unhybridized p orbital of C in formaldehyde forms a bond with an oxygen p orbital to further lock their positions together. 4.I don’t see the connection here.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–9 ANSWER Choice 3 provides the best answer for both the hybridization aspect and its importance to these observations. The unhybridized p orbitals in C and O form a  bond that hinders free rotation of the two atoms. Section 9.1: Hybridization and the Localized Electron Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–10 QUESTION Which of the following would not have d 2 sp 3 hybridization in a compound? 1.S 2.Sc 3.C 4.Ar

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–11 ANSWER Choice 3 provides the correct element choice. To participate in d 2 sp 3 hybridization, an atom must have available d orbitals. C’s valence level (n = 2) only has s and p orbitals. Section 9.1: Hybridization and the Localized Electron Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–12 QUESTION (a) What type of hybridization produces the linear shape of a CO 2 molecule? (b) How many unhybridized p orbitals are involved in the bonding of one CO 2 molecule? (c) What type of hybridization is present in the oxygen atoms? 1.(a) sp; (b) 2; (c) sp 3 2.(a) sp; (b) 4; (c) sp 2 3.(a) sp 3 ; (b) 2; (c) sp 4.(a) sp 2 ; (b) 4; (c) sp 2

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–13 ANSWER Choice 2 provides the correct response to all three questions. Examining the Lewis structure shows that C forms two bonds to each oxygen; therefore it needs two orbitals to bond to each oxygen. sp hybridization provides two linear hybridized orbitals and two unhybridized orbitals, which can then be used in pi bonding. Each oxygen must provide two orbitals to bond to the carbon atom – sp 2 hybridization would provide one hybridized orbital to form a sigma bond, then the unhybridized orbital could form a pi bond. This overlap geometry is possible only with the sp and sp 2 hybridization, respectively for C and O. Section 9.1: Hybridization and the Localized Electron Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–14 QUESTION When comparing the M.O. theory of bonding to the Localized Electron model, which of the following would be an incorrect claim? 1.For a molecule of H 2 ; MO 1 = 1s A + 1s B; MO 2 = 1s A – 1s B. 2.The molecular orbitals (both bonding and anti-bonding) still have a maximum electron occupancy of two just as the localized orbitals. 3.In H 2, the bonding orbital (MO 1 ) is lower in energy than the 1s orbital of hydrogen. 4.Although not used in the molecular bonding, the 1s orbital of hydrogen is present.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–15 ANSWER Choice 4 makes a false statement. Once the MO treatment is made on H 2 ; the previous 1s orbitals of the unbonded hydrogen atoms become new bonding and anti-bonding orbitals, for the entire H 2 molecule. Section 9.2: The Molecular Orbital Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–16 QUESTION The key reason that H 2 is found to be lower in energy than H 2 – and lower in energy than H 2 + is that…

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–17 QUESTION (continued) 1.the extra e – for H 2 – is in the anti-bonding orbital, which detracts from the bond; the lost e – in H 2 + takes away from the bonding orbital. 2.the extra e – for H 2 – causes more repulsion, thus raising its energy. The loss of an e – in H 2 + weakens the bond because losing an e – typically weakens a bond. 3.the bond in H 2 consists of two e – in one orbital; this is very stable. Adding or taking away an e – detracts from the molecule’s stability. 4.Bonding and anti-bonding orbitals still leave me a bit uncertain.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–18 ANSWER Choice 1 states the correct relationship between bonding and antibonding orbitals for H 2 with adding and removing e –. Section 9.2: The Molecular Orbital Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–19 QUESTION Is dilithium real? This was the fuel of choice in the original Star Trek series. Using just the outer level of orbitals for two lithium atoms, construct a molecular orbital diagram and determine the bonding order. 1.B.O. = 0; No, dilithium was only science fiction. 2.B.O. = 1; Captain Kirk was right! It could exist. 3.B.O. = 1; No, B.O. must be > 1 for molecules to actually exist, so Li 2 would not exist. (Captain Kirk should know this.) 4.B.O. cannot be calculated without more information (ask Spock).

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–20 ANSWER Choice 2 shows once again that some science fiction is actually science, however, the warp drive may be another story. The properly constructed MO diagram will show one e – from each Li atom occupying the stable bonding orbital (2s sigma) Subtracting zero from 2, then dividing by 2 yields a bond order = 1. This positive value for the bonding order indicates that Li 2 could exist. Section 9.2: The Molecular Orbital Model

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–21 QUESTION It is known that p atomic orbitals can overlap in two ways during the formation of molecular orbitals. How many pi bonding orbitals are assigned electrons in the bonding of a B 2 – ion? 1.Zero 2.One 3.Two 4.I do not know how to construct the MO diagram and determine the electron arrangement.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–22 ANSWER Choice 3 provides the correct number of orbitals. When the MO diagram is properly prepared for B 2 the two pi p orbitals are lower in energy, therefore no sigma p orbitals are involved. Section 9.3: Bonding in Homonuclear Diatomic Molecules

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–23 QUESTION Using MO diagrams, predict which, if any, of the following would be diamagnetic. 1.O 2 2.O 2 – 3.O None are diamagnetic

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–24 ANSWER Choice 4 correctly predicts that none of these diatomic oxygen species is diamagnetic. The requirement that all electrons must be paired in a diatomic species cannot be met by any of these three. Section 9.3: Bonding in Homonuclear Diatomic Molecules

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–25 QUESTION The bond energy of F 2 is 154 kJ/mol. The bond energy for Cl 2 is 239 kJ/mol. Comparing these values to the bond orders for each molecule allows what conclusion? 1.Cl 2 contains a stronger bond than F 2. It also has a higher bond order. 2.Cl 2 contains a stronger bond than F 2, but because F 2 is such a small highly electronegative element, the molecule of F 2 will have a higher bond order. 3.Cl 2 contains a stronger bond than F 2 ; however, their bond orders are the same. Bond energy does not depend on bond order alone. 4.Cl 2 contains a stronger bond than F 2 ; bond order and bond energy are inversely related.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–26 ANSWER Choice 3 correctly points out that bond order is a general guide to bond strength, but bond order cannot automatically be associated with a specific bond energy. Both diatomic chlorine and fluorine have bond orders of 1. Section 9.3: Bonding in Homonuclear Diatomic Molecules

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–27 QUESTION The carbon monoxide molecule can bond to hemoglobin, causing severe oxygen shortages in humans, at a rate up to 200 times faster than oxygen. What is the bond order and magnetic characteristic of CO? 1.Three; diamagnetic 2.Two; diamagnetic 3.Three; paramagnetic 4.Two; paramagnetic

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 9–28 ANSWER Choice 1 provides the correct bond order and magnetic properties. Carbon atoms bring two p electrons and oxygen has four p electrons. This total of six electrons is then placed into the lowest available molecular orbitals. The arrangement produces filled  2p and both  2p orbitals, so the bond order is 3 and all electrons are paired. Section 9.4: Bonding in Heteronuclear Diatomic Molecules