Answer Key, Problem Set 11 Full - Oakton Community College

Chemistry 121 Mines, Fall 2019

PS11-1

Answer Key, Problem Set 11 – Full

1. NT1; 2. NT2; 3. 10.67; 4. 10.66; 5. 10.76(bd); 6. 10.80 in Tro. Add (b) NO+ or NO− [Which has the weaker NO bond? The longer NO bond?]; 7. NT3; 8. 10.81; 9. NT4; 10. NT5; 11. NT6; 12. 11.38; 13. NT7; 14. 11.43 & 11.44; 15. 11.54; 16. 11.56; 17. NT8; 18. 11.63(ad); 19. NT9; 20. NT10; 21. NT11 -------------------------------------------------------

Equivalent Resonance Structures

1. NT1. Draw the Lewis structure for the nitrite ion, NO2-. Include resonance structures.

NOTE: This species is isoelectronic with SO2, but unlike SO2, which can have two double bonds (and 10 electrons around the S), NO2

- cannot have that third resonance structure because N does not have d orbitals and so it cannot have more than 8 electrons around it.

Assessing Relative Importance (Weight) of Nonequivalent Resonance Structures by Assigning Formal Charges

2. NT2. (a) Which of the following statements is/are true?

(i) Each atom in a Lewis structure has a formal charge (ii) Molecules have a formal charge. (iii) Polyatomic ions have a formal charge (iv) LDSs have a formal charge. (v) All of the above are true statements

(b) I have seen year after year at least some students write on an exam that “the best LDS is the one with zero formal

charge”. Why must I take off points on those students’ exams? In other words, what is wrong with this statement?

Answer: This statement is fundamentally flawed because it states that a “LDS” has a formal

charge (e.g., “zero” in this case), when that is not true. As noted in (a), LDS’s don’t have a formal charge. Each atom in an LDS has a formal charge. Please do not mix this up! The “best” LDS is the one that has the most atoms with a formal charge of zero (or closest to zero). That is not the same as what was written in the statement of the problem.

O N O

O N O

v = 5 + 2(6) + 1 = 18 e- (the “+ 1” is because of the -1 charge)

after all 18 electrons are put into the structure, the central N doesn’t have an octet, so……

……a lone pair from one of the O’s is made into a bonding pair (to make a double bond)

But the lone pair could have been donated by the other O atom. That would lead to a second, equally valid Lewis structure. As such, the two structures are called “resonance structures” of NO2

-, and since they are the same in every possible way (except that it is a different O atom making the db), these two structures are called equivalent resonance structures. Each one contributes equally to the description of the bonding in this species.

(1)

(2)

(2)

(3)

O N O

- (3)

-

-

Answer: (i) only! Atoms each “have” a formal charge. Polyatomic “species” (e.g., whole molecules, polyatomic ions, or Lewis Dot Structures of these) do not have a formal charge. They only have an “actual” (overall) charge

Answer Key, Problem Set 11

PS11-2

3. 10.67. How important is this resonance structure [the one with a triple bond to the left O and a single bond to the right

O] to the overall structure of carbon dioxide? Explain.

Answer: Not very important. As shown in class (and below), the formal charges for all the atoms in the three resonance structures for CO2 are as shown below:

*Where both electrons in a lone pair are assigned to the given atom, but only one electron from each covalent bond is assigned to the given atom.

Resonance structures are a way to patch up an inadequate bonding model. When more than one (at least initially, “reasonable”) LDS can be drawn for a species, the actual bonding (i.e., the actual “structure”) is assumed to be a “blending” of all the resonance structures. However, when the resonance structures are not equivalent, each one is (informally) “weighted” before the “blending”. Those that are considered “better” are given more weight and are considered to better represent the actual structure. When assessing “goodness” using formal charges, the ones that have as many atoms with a formal charge of zero as possible are considered to be the “best” (all other things being equal, of course). By that criterion, the best structure for CO2 is the one with two double bonds (all three atoms with FC = 0). The LDS asked about in this problem not only has two atoms with non-zero formal charges, but also has a formal charge of +1 on an O atom, which is particular unfavorable since O is one of the most electronegative atoms (with a value of 3.5 out of 4.0). Thus, this LDS is not considered to be very important in representing the actual structure of CO2.

4.. 10.66. Use formal charges to determine which Lewis structure is better:

*Where both electrons in a lone pair are assigned to the given atom, but only one electron from each covalent bond is assigned to the given atom.

In the left structure, the S has no lone pairs and makes four bonds, thus it is assigned four electrons “formally”. The C has two lone pairs (4 e-‘s) and makes two bonds (2 more e-‘s) for a total of six formally. In the right structure, the S is the one that has two lone pairs and makes two bonds, so it gets six electrons formally. The C makes four bonds (with no lone pairs) and thus has four electrons formally.

v e-‘s if neutral 6 4 6 6 4 6 6 4 6

e-‘s in this structure* 6 6 6 5 4 7 7 4 5

FORMAL CHARGE: 0 0 0 +1 0 -1 -1 0 +1

S C S C

# of v e-‘s a neutral isolated atom has: 6 4 6 4 # of v e-‘s the atom "has" in this

structure* 4 6 6 4

FORMAL CHARGE: +2 -2 0 0

O C O O C O O O C

−1 −1 +1 +1 0 0 0 0 0 “BEST”

H S

H

C H

H

H C

H

S H

H

+2 −2 0 0

The one on the right is better because the structure has more atoms with FC’s that are zero whereas the one on the left has more atoms with large-magnitude formal charges (+2 and -2).


PS11-3

Expanding an Octet by Making Multiple Bonds to Lower Formal Charges in an LDS

5. 10.72(bd). Write a Lewis structure for each ion. Include resonance structures if necessary and assign formal charges

to all atoms. If necessary, expand the octet on the central atom to lower the number of atoms with non-zero formal charges (the authors say “lower formal charge”, but they mean what I wrote).

(b) HSO4-

The initial LDS has all octets where possible, but the formal charges are non-zero for four atoms, and one of those is a +2. Making a double bons “inward” from an outer O atom increases the FC by one on the O and lowers it by one on the (inner) S. Thus, making two double bonds is “best”, yielding all zero FC’s except for one O (which is the best you can do since the overall charge on the ion is -1).

NOTE: There are two equivalent resonance structures to the 2nd one above, so the four structures would be:

Technically, there are three additional resonance structures having just one double bond, but I am not going to show those here. If you included those, then more power to you! ☺

(d) BrO2-

The first LDS has all octets, but no atom has a zero formal charge. Making one double bond “inward” from either (outer) O raises the FC on O by one and lowers the FC on Br by one. That creates the two equivalent LDS’s below, each of which has only one atom with a non-zero formal charge.

O

(terminal)

O (center)

S H

v e-‘s if neutral 6 6 6 1

e-‘s in this structure* 7 6 4 1

FORMAL CHARGE: -1 0 +2 0

O

(term, sb)

O

(term, db) O

(center) S H

v e-‘s if neutral 6 6 6 6 1

e-‘s in this structure* 7 6 6 6 1

FORMAL CHARGE: -1 0 0 0 0

O Br

v e-‘s if neutral 6 7

e-‘s in this structure* 7 6

FORMAL CHARGE: -1 +1

O (db) O (sb) Br

v e-‘s if neutral 6 6 7

e-‘s in this structure* 6 7 7

FORMAL CHARGE: 0 -1 0

v = 1(6) + 4(6) + 1(1) + 1 = 32 e-

O S

O

O H

O

− −1

+2 0 −1

−1

0

O S

O

O H

O

−

−1 0

0

0

0 0

O S

O

O H

O

−

O S

O

O H

O

− −

O S

O

O H

O

−

O S

O

O H

O

v = 1(7) + 2(6) + 1 = 20 e-

− Br O O

−1 −1 +1

− Br O O

− Br O O

−1 -1 0 0 0 0


PS11-4

Predicting Relative Strength and Length of Bonds Using Bond Order Concept

6. 10.80 Which of these compounds has the stronger N-N bond? The shorter N-N bond? H2NNH2 or HNNH

(Hint: You cannot answer these questions, nor will you be given credit for answers, without drawing and considering the LDS’s!)

Answer: HNNH has the stronger and shorter N-N bond because it has a double bond rather than a single bond (see below). The greater the bond order, the stronger and shorter the bond.

H2NNH2 HNNH

(b) NO+ or NO− [Which has the weaker NO bond? The longer NO bond?]

Answer: NO− has the weaker and longer (N-O) bond because it has a double bond whereas NO+ has a triple bond (see below). The smaller the bond order, the weaker and longer the bond.

7. NT3. Based on the three Lewis Structures of OCN- shown in Example 10.8 (p.416) and the data in Table 10.4 (p. 425),

would you expect the C-N bond length to be: (a) greater than 147 pm, (b) between 128 pm and 147 pm, (c) between 116 pm and 128 pm, or (d) less than 116 pm.

Give your reasoning.

Answer: (c), between 116 pm and 128 pm, because the bond order is expected to be between two and three, and so the bond length is expected to be between the length of a C-N double bond and a C-N triple bond (128 pm and 116 pm, respectively [Table 10.4]).

The best Lewis structure is the one with the triple bond, but the second best (not all that far behind) is the one with the double bond (see Example 10.8). That second structure should contribute significantly (although less than the first), making one predict that the bond order should be (at least a little bit) less than three, but still greater than two. The third Lewis structure is so bad (because of the -2 formal charge) that one can consider its contribution essentially negligible (it is the only one with a C-N single bond [bond length 147 pm]).

H N

H

N H

H

v = 4(1) + 2(5) = 14 e- v = 2(1) + 2(5) = 12 e-

H N N H

v = 5 + 6 - 1 = 10 e- v = 5 + 6 + 1 = 12 e-

−

N O +

N O


PS11-5

Using Bond Energies to Calculate an Estimate of H of a Chemical Reaction Equation

8. 10.81. Hydrogenation reactions are used to add hydrogen across double bonds in hydrocarbons and other organic

compounds. Use average bond energies to calculate [an estimate for] Hrxn for the hydrogenation reaction.

(i.e., the H for the thermochemical equation below).

H2C==CH2 + H2 → H3C−−CH3

Answer: -128 kJ/mol

Reasoning:

NOTE: If you do not want to try to figure out the “net” bonds broken and made, you can always just break every bond in every reactant molecule and then make every bond in every product molecule. For this problem, that approach would look like:

Break: Make:

4 C-H (414) 6 C-H (414)

1 C=C (611) 1 C-C (347)

1 H-H (436)

4(414) + 1(611) + 1(436) = +2703 kJ 6(-414) + 1(-347) = -2831 kJ

H = 2703 + (-2831) = -128 kJ (as before, of course)

Relating Bond Strengths to Endo- or Exothermicity of a Chemical Reaction [See Mastering]

Assessing Relative Bond Polarity (or Non-) Using Electronegativities, and Representing Bond Polarity

9. NT4. 10.56 (with added part (e). Determine whether a bond between each pair of atoms would be pure (nonpolar)

covalent, polar covalent, or ionic. Add Part II: For each polar covalent bond, indicate which atom would be partially negative.

(a) C and N Polar covalent. EN = 3.0 – 2.5 = 0.5

(b) N and S Polar covalent EN = 3.0 – 2.5 = 0.5

(c) K and F Ionic EN = 4.0 –0.8 = 3.2

(d) N and N Nonpolar covalent EN = 3.0 – 3.0 = 0

(e) C and H Nonpolar covalent EN = 2.5 – 2.1 = 0.4

H C

H

C H

H

H H

H C

H

C H

H

H H (1) Break one C=C (611 kJ/mol)

and one H-H (436 kJ/mol)

= 611 + 436 = 1047 kJ

(2) Make one C-C (347 kJ/mol) and two C-H (414 kJ/mol)

= -347 + 2(-414) = -1175 kJ

H C

H

C H

H

H H

(1) (2)

H = H1 + H2 = 1047 +(-1175) = -128 kJ

EN values from Figure 10.8: K(0.8), H(2.1); C(2.5); S(2.5); N(3.0); F(4.0)

NOTE: You are required to know the trends in EN values so that if a table is not provid-ed, you can still predict relative polarities of bonds!! I also expect you to memorize the EN values of C, N, O, F, Cl, and H.


PS11-6

10. NT5. 10.102 Draw a Lewis structure for urea, H2NCONH2, one of the compounds responsible for the smell of urine. (The

central carbon atom is bonded to both nitrogen atoms and to the oxygen atom.) Does urea contain polar bonds? Which bond in urea is most polar? Add: Add arrows (with a “+” sign on the back end) to your LDS to indicate the polarity of each polar bond.

Answers: See right and below. Yes, it has polar bonds. In fact every bond is polar! (Below, the more electronegative atom is listed first in each case.)

O-C: EN = 3.5 – 2.5 = 1.0 ← most polar

N-H: EN = 3.0 – 2.1 = 0.9

N-C: EN = 3.0 – 2.5 = 0.5

Using VSEPR to Predict ECG’s, AG’s (MG’s), and Bond Angles in Species with One or More Centers

11. NT6. 11.35(abd). (i) Determine the electron geometry (electron-cloud geometry), molecular geometry (atom

geometry), and idealized bond angles for each molecule. In which cases do you expect deviations from the idealized bond angle? (ii) Add: Sketch each one with wedges and dashes (and atoms and lone pairs, where applicable) as done in the lab handout.

(a) PF3

(b) SBr2

(d) CS2

# electron groups (clouds) 4 (1 lone pair, 3 single bonds)

ECG & idealized bond angles tetrahedral, 109.5°

# lone pairs & AG 1, trigonal pyramidal Deviations from idealized

angle? Yes. A bit less. Lone pair “compresses”

# electron groups (clouds) 4 (2 lone pairs, 2 single bonds)

ECG & idealized bond angles tetrahedral, 109.5°

# lone pairs & AG 2, bent Deviations from idealized

angle? Yes. A bit less. Lone pairs “compress”

# electron groups (clouds) 1 (0 lone pairs, 2 double bonds)

ECG & idealized bond angles linear, 180°

# lone pairs & AG 0, linear Deviations from idealized

angle? No. No lone pairs (& all bonds “same”)

C

O

H

N H H N

H

P F F

F

v = 5 + 3(7) = 26 e-

P

F F

F

S Br Br v = 6 + 2(7) = 20 e-

In these sketches, lone pairs on terminal atoms are not shown—just the central (interior) atoms.

P

Br Br

v = 4 + 2(6) = 16 e- S C S S C S


PS11-7

(e) BrF5

NOTE: You are responsible for all of the VSEPR atom geometries even though I could not “fit” all of them into this problem set. Consider your lab handout to be “fair game” for the exam (and use it and your text for further practice).

12. 11.38. Which species has the smaller bond angle, ClO4- or ClO3

-? Explain.

Answer: ClO3-. Both Lewis structures have 4 electron clouds around the central Cl (see below), but

ClO3- has a lone pair which “pushes harder” on the bonding pairs, making the bond angles a

bit less than the ideal angle of 109.5° predicted for ClO4-, which has no lone pairs. NOTE:

You must show the LDS’s to earn credit for a problem like this since otherwise, how could you know that the number of electron clouds is the same? This is a basic, “show your work” idea!

ClO4- : ClO3

- :

Using VSEPR to Predict ECG’s, AG’s (MG’s), and Bond Angles in Species with One or More Centers

13. NT7. 11.42(bc) Determine the molecular (atom) geometry about each interior atom (center) and make a sketch of the

molecule. NOTE: Use the LDS’s for these that you drew in problem #6 (10.80)! Add: State the HNN bond angle in each.

H2NNH2 HNNH

# electron groups (clouds) 6 (1 lone pairs, 5 single bonds)

ECG & idealized bond angles octahedral, 90 (& 180°)

# lone pairs & AG 1, square pyramidal Deviations from idealized

angle? Yes. A bit less. Lone pair “compresses”

v = 7 + 5(7) = 42 e-

Br

F

F F

F

F F

Br

F

F

F

F

v = 7 + 4(6) + 1 = 32 e- v = 7 + 3(6) + 1 = 26 e- Cl O O

O

- Cl O O

O

- O

H N

H

N H

H

H N N H

Each N: 4 clouds tetrahedral ECG, 109.5° idealized bond angles.

+ 1 lone pair AG = trigonal pyramidal & HNN angle is a bit less than 109°

Each N: 3 clouds trigonal planar ECG, 120° idealized bond angles.

+ 1 lone pair AG = bent & HNN angle is a bit less than 120°


PS11-8

14. 11.43 & 11.44. Each ball-and-stick model shows the electron and molecular geometry of a generic molecule. Explain

what is wrong with each molecular geometry and provide the correct molecular geometry, given the number of lone pairs and bonding groups on the central atom.

11.43: 10:44:

11.43: (a) With four electron clouds (one of them a lone pair), the ECG would be tetrahedral, not “trigonal planar bonding pairs with the lone pair at 90° to all of them” as is shown. Basically, the lone pair would push those three bonding pairs “below” the central atom to make the tetrahedral ECG, opening up those 90° angles to about 109°. The correct AG would be trigonal pyramidal.

(b) With five electron clouds (one of them a lone pair), the ECG would be trigonal bipyramidal, making the four bonded atoms fall into two categories: axial and equatorial. This model shows them all “equivalent” (without any three atoms being linear). The correct AG would be see-saw.

(c) With six electron clouds, the ECG would be octahedral as shown. However, two lone pairs in an octahedral ECG (where all positions are “equivalent”) will get as far apart from each other as possible (180°) making the AG square planar. They will not be at 90° from each other as shown (in a see-saw-like AG). [**Note this may seem contradictory at first to the behavior in a trigonal bipyramidal ECG situation, but it is not. In the latter ECG, there are two non-equivalent positions—axial and equatorial—which is what leads to two lone pairs being at 120° rather than 180° (because that will lead to only two 90° interactions for each lone pair rather than three).

11.44 (a) Four electron clouds (even if two of them are lone pairs) will form a tetrahedral ECG. This model shows the two bonding clouds at 180° and each lone pair at 90° to the bonding clouds (forming a linear AG). The correct AG would be bent (with bond angles a bit less than 109°).

(b) Five electron clouds (three of which are lone pairs) would form a trigonal bipyramidal ECG. The model shows all the clouds in the same plane (I think), with the AG being bent (for sure). In a trigonal bipyramidal ECG, three lone pairs would occupy the equatorial positions (where there are only two 90° interactions instead of three), making the (correct) AG linear.

(c) This model shows five atoms around a central one in a “perfect” AG of trigonal bipyramidal, except with a lone pair (sixth electron cloud) at 180° from one of the equatorial bonding clouds (and bisecting the other two equatorial clouds). Six clouds would actually form an octahedral ECG, and with one of them a lone pair, the correct AG would be square pyramidal.

Valence Bond Theory (VBT): Concept

15. 11.54. The valence electron configurations of several atoms are shown below. How many bonds can each atom make

without hybridization?

Answers: (a) one bond; (b) three bonds; (c) two bonds

Reasoning: In VBT, an atom forms a covalent bond with another by “donating” an unpaired electron (in an orbital, obviously). Without hybridization, you must simply look at the ground-state electron configurations (really, the orbital diagrams, shown below) and see how many unpaired electrons are present to predict how many bonds it could form.


PS11-9

(a) B: (b) N: (c) O:

16. 11.56. Write orbital diagrams to represent the electron configurations—without hybridization—for all the atoms in SF2.

Circle the electrons involved in bonding. Draw a three-dimensional sketch of the molecule and show orbital overlap. What bond angle do you expect from the unhybridized orbitals? How well does valence bond theory agree with the experimentally measured bond angle of 98.2°?

Answer / Reasoning: I find it easiest to analyze these kinds of problems starting from the Lewis

structure (even though it is not necessary). The LDS indicates two single bonds, which means two

bonds. The orbital diagrams indicate that each F atom will use a p orbital to make its bond with S, and that the S will use its two singly occupied p orbitals (which are necessarily perpendicular to one another since all the p orbitals in a given sublevel are perpendicular to one another in an atom) to

make those bonds to the F’s. The result is that the bond angle is expected to be 90° (if one

does not hybridize) because of the two (perpendicular) p orbitals on S. This prediction is 8° lower than the experimentally observed angle of ~98°, which is not great but is not terrible either. Interestingly, the VSEPR / hybridization model would predict something that is about equally “off” but on the high end (a little bit less than 109° which is roughly 10° higher than the actual angle). Thus this molecule is not particularly consistent with either bonding model.

VBT: Hybridization of Atomic Orbitals

17. NT8. 11.58. (a) Write orbital diagrams to represent the electron configuration of carbon before and after sp

hybridization.

Answer:

C:

(b) Rank the following orbitals (all in n = 2 level of a C atom) from lowest to highest energy:

If any orbitals have the same energy, use an equal sign (“=”) rather than a “less than” sign (“<”).

Answer: lowest energy: s < sp < sp2 < p (before hybridization) = p (after hybridization*)

2s 2p 2s 2p 2s 2p

2s 2p

Before hybridization After hybridizing an s and a p to make two sp hybrids (the 2p orbitals remain unhybridized)

sp 2p

v = 6 + 2(7) = 20 e- S F F

Left F:

2s 2p

Right F:

2s 2p

3s 3p

S:

: F(p)—S(p)

: F(p)—S(p)

S F

F

s (before hybridization), p (before hybridization), sp (after sp hybridization), sp2 (after sp2 hybridization), p (leftover, after hybridization)


PS11-10

Reasoning: Hybrid orbitals are a result of mathematical “blending” of pure atomic orbitals. As a result, the properties of each hybrid is intermediary between those of the pure atomic orbitals. That means, for example, that the energy of any hybrid orbital is necessarily intermediary between the energies of the orbitals from which it was “created”. s orbitals are lower in energy than p orbitals (because of shielding and penetration; see PS9). So any hybrid orbitals derived from s and p orbitals (i.e., sp, sp2, or sp3) will all have energies that are higher than the (pure) s and lower than the (pure) p. However, the amount and type of the orbitals used to make the hybrids “matters”—if there are more p orbitals used in one set of hybrids than another, the resulting hybrids will have properties more like “p”. For example, since a pair of sp hybrid orbitals is derived from one s and one p orbital, the energy is basically halfway between the two energies (we say that the sp hybrids have “one half s character” and “one half p character”). But since the three sp2 hybrids in a set are derived from one s and two p orbitals, the hybrids are more like p than like s (we say that the sp2 hybrids are “one third s character” and “two-thirds p character”). As such, the energy of each sp2 hybrid is closer to the energy of the p orbital than the s orbital (e.g., higher in energy than the sp hybrids). The p orbitals after hybridization are exactly the same orbitals as before the hybridization (because they were not involved in the hybridization—they were “leftover”!), so of course their energies must be identical. This can be seen (for the sp hybridization situation) on the graphic on p. 465 in Tro, reproduced here:

Unfortunately, Tro did not properly show this (in a quantitative way) in the graphics for the other hybridization schemes. I have modified the graphic on p. 460 below (for sp2 hybridization) so that it better shows the issue discussed in this problem (I’ve essentially just raised the energy of the sp2 hybrids:

It should also be noted here that the shape of an sp2 orbital will be slightly different than the shape of an sp orbital. The sp, having less “p-character” (only 50%) than an sp2 or sp3 hybrid (which will have 67% and 75% p character respectively) will be less elongated and more “squat” or “stubby” than an sp2 or sp3 hybrid). This has real implications in terms of properties, especially for lone pairs that are in different kinds of orbitals. This will be discussed in an organic chemistry course, should you ever take one.

VBT: Sigma () and Pi () Bonds and Hybridization [Only in Mastering]


PS11-11

VBT: Application of Ideas / Visualization With Sketches / Relation to (Free?) Rotation Around Bonds

18. 11.63(ad). Write a hybridization and bonding scheme for each molecule or ion. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 11.6 and 11.7.

(a) COCl2 (C is the center)

Strategy / Approach:

1) Start from the LDS to figure out how many orbitals are “needed” for bonding. I like to first

determine how many or bonds are made to a center. (Remember, every double bond

equals exactly one bond and one bond. Every triple bond equals one bond and two bonds.)

2) Deal with the bonds first by counting clouds. The number of clouds equals the number of

hybrid orbitals needed for bonding and lone pairs (if present).

3) Then deal with the bonds. Are there any leftover (not hybridized) p orbitals? There had better

be if you have bonds to that center! If there are no bonds to that center, then there should not be any leftover p orbitals (i.e., they should all have been used in the hybridization).

4) Terminal atoms in this textbook are generally never hybridized—there is no need for it. So just use an unhybridized orbital (usually a p orbital for any atom other than H, which uses its s orbital). You can check the orbital diagram just to make sure there is an unpaired electron.

5) To make the sketch, I’d actually draw the wedge and dash structure first so that I get the

“geometry” correct. (Try to get all bonds so that they are in the plane of the paper!) Then draw the sketch with the orbitals next to it (I used to do these “on top of one another” but it gets pretty messy.) For convenience, you may use simple “lines” to connect sideways overlapping p orbitals (rather than showing the overlap the way the text does).

6) Label each bond and lone pair (on the center). Ignore lone pairs on terminal atoms.

7) NOTE: You do not need to write out the orbital diagrams from any atoms in this approach! It should always “work out” for simple molecules.

Execution of Strategy:

There are 3 bonds and one bond made by the center C. Each bond “needs” a distinct orbital on C.

3 clouds around C means 3 hybrid orbitals are needed to make the bonds (there are no lone pairs here)

use the first three orbitals (s, p, p) for hybridization (i.e., C is sp2 hybridized).

That leaves one p orbital (not hybridized) to make the bond to O.

O uses one p orbital for the bond and a different p orbital for the bond (as in Example 11.7).

The Cl atoms each use their singly occupied p orbital for the bond.

The C has a trigonal planar ECG and AG. Put the C=O in the plane of the paper with the Cl’s coming out and going back (wedge and dash) so the sideways-overlapping p orbitals can be in the

plane (to make the bond)

C

O

Cl Cl

v = 4 + 6 + 2(7) = 24 e-

O C

Cl

Cl

: O(p)—C(p)

: O(p)—C(sp2)

: C(sp2)—Cl(p)

: C(sp2)—Cl(p) O C

Cl

Cl


PS11-12

I

I

I

: Central I(sp3d)—Terminal I(p)

I

I

I Lone pairs (3) in sp3d orbitals

: Central I(sp3d)—Terminal I(p)

(d) I3-

Explanation: (see strategy above)

From LDS, only two bonds are made and no bonds.

# of electron groups around central I is 5 use the first five orbitals available (s, p, p, p, d) to

make a set of five sp3d hybrids (i.e., I is sp3d hybridized). Two of these will be used to make the bonds to the terminal I’s; three will be used for the three lone pairs.

Each terminal I atom will just use its singly occupied p orbital to make the bond to the center.

5 clouds means an ECG of trigonal bipyramidal. Use VSEPR to determine that the lone pairs will go into the equatorial positions and make a wedge and dash sketch. Use that (as well as the orbital info noted above) to make the orbital sketch.

19. NT9. 11.94. The compound C3H4 has two double bonds. (a) Describe its bonding and geometry using a valence bond

approach. Add: (b) Sketch the molecule using wedges and dashes as in lab. (Hint: Think about the bonding carefully

(particularly the bonds) before making your sketch. Can all of the atoms in this molecule be in the same plane?) (c) Is this molecule rigid or can it freely rotate around any of its bonds?

C3H4 v: 3(4) + 4(1) = 26 e-

Answers

(a) C1 and C3 are trigonal planar, and C2 is linear. C1 and C3 use sp2 hybrids to make sigma bonds to the two H’s and to C2. C2 uses sp hybrids to make sigma bonds to C1 and C3. C1 and C3 each use their (single) leftover p orbital to make a pi bond to C2. C2 uses one leftover p orbital to make a pi bond with C1 and another p orbital to make a pi bond to C3.

(b) See bottom sketch off to the right. Two of the H’s and all three C’s are in one plane, but the other two H’s are in a plane perpendicular to the first. All 6 atoms cannot be in the same plane because the two pi bonds are in planes perpendicular to one another. I’ve shown the “orbital” sketch to try to show this (although I did not ask you to draw that for this problem)

(c) This molecule is rigid. It cannot rotate freely around either of the C-C bonds because rotation about either one of them would require the breaking of a pi bond, which takes energy (see discussion in Tro, p. 463 and box on “vision” on p. 464)

v = 3(7) +1 = 22 e-

I I I

-

C2

H

C3 H H C1

H

C C

H

H C

H

H

C C C

: C(sp)— C(sp2)

: C(p)—C(p)


PS11-13

Reasoning:

(a) (approach as in prior problem (#18)):

Carbons 1 and 3 each make three and one bond

C2 makes two and two bonds

Carbons 1 and 3 have three clouds sp2 hybrids (p left over for bond)

C2 has two clouds sp hybrids (two p’s left over for two bonds)

Carbons 1 and 3 each have a trigonal planar ECG and AG (3 clouds, no lone pairs)

C2 is linear (ECG and AG; 2 clouds, no lone pairs)

**Since C2 must use two different p orbitals for its two p bonds, those p orbitals (and subsequent p bonds) must be at 90 from one another. That means the H atoms cannot all be in the same plane! They are, in fact, in perpendicular planes to one another. That is why I drew the left two in the plane, and the right two in front of and behind the plane of the paper.**

Combining VSEPR and VBT (Prediction of ECG, AG, Angles, and Hybridization, # of and Bonds from an LDS)

20. NT10. 11.85(c) Amino acids are biological compounds that link together to form proteins, the workhorse molecules

in living organisms. The skeletal structures of several simple amino acids are [cysteine is] shown here. Complete the Lewis structure and determine the geometry and hybridization about each interior atom. [Don’t sketch it!] Also state the value of each of the following bond angles in the molecule:

(i) HOC: a bit less than 109° (ii) OCC: ~120° (iii) CCC: 109.5° (iv) CCS: 109.5°

(v) HNC: a bit less than 109°

NOTE: To “complete the Lewis structure”, I merely added lone pairs in order to get octets around the N, S, and O atoms. This “works” because the multiple bonds were already shown and this is a simple organic molecule. Technically, I used the HONC patterns noted in class (C forms 4 bonds and has no lone pairs; N forms 3 bonds with one lone pair; O (and S, here) forms 2 bonds with two lone pairs). If multiple bonds are not initially shown in a structure, you must be particularly careful not to “just” add lone pairs to get octets!

Atom #

clouds Hybridi-zation

ECG # lone pairs

AG

O, S 4 sp3 tetrahedral 2 bent

C1 3 sp2 trigonal planar

0 trigonal planar

C2, C3 4 sp3 tetrahedral 0 tetrahedral

N 4 sp3 tetrahedral 1 trigonal

pyramidal

H N C2 C1 O H

O H

C3

S

H

H H

H


PS11-14

21. NT11 11.87. The structure of caffeine, present in coffee and many soft drinks, is shown here. How many pi bonds

are present in [one molecule of!] caffeine? How many sigma bonds? Insert the lone pairs in the molecule. What kinds of orbitals do the lone pairs on the N’s occupy?

NOTE: Don’t ignore the condensed notations “H3C-“ and “-CH3”! These groups have bonds that are not explicitly shown! Perhaps expand these to make sure you don’t miss any bonds.

Note: Lone pairs were assigned as in the prior problem (see #20, 11.85(c)).

============================ END OF SET =================================

A couple of extra problems from a prior answer key (may be in Mastering now and may have different chapter/problem numbers)

10.126 Which statement is (generally) true of an endothermic reaction?

(a) Strong bonds break and weak bonds form.

(b) Weak bonds break and strong bonds form.

(c) The bonds that break and those that form are of approximately the same strength.

Answer: (a) Breaking bonds is “energy absorbing” and making them is “energy releasing”, so if the bonds that break are the strong ones, then more energy will be used to break those (strong) bonds than the amount of energy that will be released upon forming the new (weak) bonds.

Note: The savvy student will recognize that this need not always be the case (which is why I added the word “generally” up above) because the amount of bonds made and broken need not be the same (i.e. the argument above implicitly assumes that the number of bonds made equals the number of bonds broken). If more weak bonds are made than the amount of (fewer) strong bonds broken, it is possible that the reaction could be exothermic. In fact, the prior problem in this set (#8, 10.81) is a great example of this!! Both of the bonds that were broken were stronger than the bonds made, but only two of them were broken and three of the weaker bonds were made. The net result was an exothermic process!

11.59. Which hybridization scheme allows the formation of at least one bond? sp3, sp2, sp3d2

Answer: sp2, because it is the only one that leaves a p orbital unhybridized. -bonds are always (in this class) formed by sideways overlapping p-orbitals (which are not hybridized!). Hybrid orbitals are

never used for making -bonds! They are only used for making -bonds or for housing lone pairs!!!

So if all of the p orbitals on an atom are used to make hybrid orbitals, there will not be any -bonds made to that atom. (I think this is easier to see when you draw an orbital diagram. That is why I wrote out the answer to 11.58 above.)

N

C N

C

C C

N

C

N

O

H

C

C

H

H

H

O

C H H

H

H

H

H Answers:

a) 25 sigma bonds (count the “first” bond of a double or triple bond as a sigma)

b) 4 pi bonds (count the second [or third] bond of a double [or triple] bond as a pi bond.

c) Each of the three N’s having only single bonds has its lone pair in an sp3 orbital (because there are four clouds)

The N with the double bond has its lone pair in an sp2 hybrid (because there are three clouds).

Documents

Answer Key, Problem Set 11 Full - Oakton Community College