chapter 9 covalent bonding: orbitals. copyright © houghton mifflin company. all rights reserved.crs...

Chapter 9

Covalent Bonding: Covalent Bonding: OrbitalsOrbitals

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 9–2

QUESTION

In examining this figure from left to right, you could conclude that…


QUESTION (continued)1. Native atomic orbitals have more exact boundaries than

hybridized orbitals.2. During hybridization, orbital number is conserved.3. The four sp3 orbitals represent a better way than the native

orbitals to obtain 90° bonding angles.4. The sp3 orbitals represent a 50–50 combination of s and p

orbitals.


ANSWERChoice 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 sp3 orbitals that can arise from mixing those.

Section 9.1: Hybridization and the Localized Electron Model


QUESTIONThe 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. sp2. sp2

3. sp3

4. This relationship is still something I do not understand.


ANSWERChoice 2 is correct. sp2 hybridization is most often associated with 120° bond angles. The three orbitals that have sp2 hybridization will be as far apart as possible, and for three areas that would be 120°.



QUESTIONFormaldehyde and methanol both have a carbon to oxygen bond. Yet the oxygen atom in methanol, CH3OH, can freely spin around without causing the C to move with it . In formaldehyde, CH2O, 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)


QUESTION (continued)1. Methanol—sp3; formaldehyde—sp; the unhybridized p orbitals

get in the way of free rotation for the C to O bond.2. Methanol—sp2; formaldehyde—sp2; 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—sp3; formaldehyde—sp2; 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.


ANSWERChoice 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.



QUESTIONWhich of the following would not have d2sp3 hybridization in a compound?

1. S2. Sc3. C4. Ar


ANSWERChoice 3 provides the correct element choice. To participate in d2sp3 hybridization, an atom must have available d orbitals. C’s valence level (n = 2) only has s and p orbitals.



QUESTION(a) What type of hybridization produces the linear shape of a CO2 molecule? (b) How many unhybridized p orbitals are involved in the bonding of one CO2 molecule? (c) What type of hybridization is present in the oxygen atoms?

1. (a) sp; (b) 2; (c) sp3

2. (a) sp; (b) 4; (c) sp2

3. (a) sp3; (b) 2; (c) sp4. (a) sp2 ; (b) 4; (c) sp2


ANSWERChoice 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 – sp2 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 sp2 hybridization, respectively for C and O.



QUESTIONWhen 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 H2; MO1 = 1s A + 1s B; MO2 = 1s A – 1s B.2. The molecular orbitals (both bonding and anti-bonding) still

have a maximum electron occupancy of two just as thelocalized orbitals.

3. In H2, the bonding orbital (MO1) is lower in energy than the 1sorbital of hydrogen.

4. Although not used in the molecular bonding, the 1s orbital ofhydrogen is present.


ANSWERChoice 4 makes a false statement. Once the MO treatment is made on H2; the previous 1s orbitals of the unbonded hydrogen atoms become new bonding and anti-bonding orbitals, for the entire H2 molecule.

Section 9.2: The Molecular Orbital Model


QUESTION

The key reason that H2 is found to be lower in energy than H2–

and lower in energy than H2+ is that…


QUESTION (continued)1. the extra e– for H2

– is in the anti-bonding orbital, which detracts from the bond; the lost e– in H2

+ takes away from the bonding orbital.

2. the extra e– for H2– causes more repulsion, thus raising its

energy. The loss of an e– in H2+ weakens the bond because

losing an e– typically weakens a bond.3. the bond in H2 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.


ANSWER

Choice 1 states the correct relationship between bonding and antibonding orbitals for H2 with adding and removing e–.



QUESTIONIs 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 Li2 would not exist. (Captain Kirk should know this.)4. B.O. cannot be calculated without more information (ask

Spock).


ANSWERChoice 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 Li2 could exist.



QUESTIONIt 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 B2

– ion?

1. Zero2. One 3. Two4. I do not know how to construct the MO diagram and determine

the electron arrangement.


ANSWERChoice 3 provides the correct number of orbitals. When the MO diagram is properly prepared for B2 the two pi p orbitals are lower in energy, therefore no sigma p orbitals are involved .

Section 9.3: Bonding in Homonuclear Diatomic Molecules


QUESTIONUsing MO diagrams, predict which, if any, of the following would be diamagnetic.

1. O2

2. O2–

3. O2+

4. None are diamagnetic


ANSWERChoice 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.



QUESTIONThe bond energy of F2 is 154 kJ/mol. The bond energy for Cl2 is 239 kJ/mol. Comparing these values to the bond orders for each molecule allows what conclusion?

1. Cl2 contains a stronger bond than F2. It also has a higherbond order.

2. Cl2 contains a stronger bond than F2, but because F2 is such asmall highly electronegative element, the molecule of F2 willhave a higher bond order.

3. Cl2 contains a stronger bond than F2; however, their bondorders are the same. Bond energy does not depend on bond order alone.

4. Cl2 contains a stronger bond than F2; bond order and bondenergy are inversely related.


ANSWERChoice 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.



QUESTIONThe 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; diamagnetic2. Two; diamagnetic3. Three; paramagnetic4. Two; paramagnetic


ANSWERChoice 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

chapter 9 covalent bonding: orbitals. copyright © houghton mifflin company. all rights reserved.crs...

Documents

unhybridized p orbitals

p bond

sp hybridization p

native orbitals

total orbitals

sp2 hybridization

hybridized orbitals

d2sp3 hybridization