C101Rec Chaps 13and14Z8e Ques SS 19 PDF

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Page 1 University of Pennsylvania Department of Chemistry Chemistry 101 – Recitation Material Problems for Chemical Bonding Chapter 13: Bonding : General Concepts & Chapter 14: Covalent Bonding: Orbitals

Lewis Octet Structures - Formal Charge - Isomers - Resonance Structures Consider the molecule with the molecular formula NOF. The arrangement of the atoms in this molecule are in a row, i.e., next to each other. (a) Draw all possible Lewis structures for the NOF molecule with the atoms connected in the order shown. Include all resonance structures and assign formal charge to each atom in every structure that you draw. Based upon formal charge considerations, determine which of the drawn structures should contribute the most to the predicted actual structure of the molecule. Explain.

1.

(b) Draw all possible Lewis structures for the ONF isomer of this molecule and for the OFN isomer. Each isomer has the atoms arranged in the order shown. Include all resonance structures and assign formal charge to each atom in every structure that you draw. Based upon formal charge considerations, determine which of the drawn structures should contribute the most to the predicted actual structure of the ONF isomer and of the OFN isomer. Explain. (c) Considering all of the drawn structures in (a) and in (b), determine the most likely isomer and the greatest contributing resonance structure of this isomer. Explain. 2.

Expanded Octet Structures & More Consider BrCl3 - where Br is central atom covalently bound to three (3) chlorine atoms. In the usual

method of determining the Lewis structure, we create only single bonds between each pair of connected atoms and then put in lone pairs as needed to satisfy octets. In this case to do so gives rise to a 26 electron Lewis structure, whereas there are a total of 28 electrons to be distributed. The question is “Where best to put the additional pair of electrons?”. Since Br and Cl can expand their octets (why?), we have three (3) choices: #1. Into one of the three (3) single bonds - thereby creating a double bond between Cl and Br - thus expanding both of their octets to ten electrons. Of course, resonance structures exist. #2. As a lone pair any one of three (3) Cl atoms - thus expanding the octet of only Cl. Of course, resonance structures exist. #3. As a lone pair only on the central Br atom - thus expanding the octet of only Br. There are no resonance structures here (why?). Draw all of the structures indicated by choices #1, #2, and #3 above. Demonstrate - based upon careful formal charge analysis that choice #3 is the best alternative. Of course, we already said that this should be the case, i.e., expand an octet only on an atom that is capable of such and then only if it is a central atom and then only if it leads to better formal charge distribution. This will help you to see that formal charge clearly rationalizes this. (b) Considering octet and expanded octet structures - including resonance - determine the best structure(s) for the sulfite anion, (SO3-2), pay attention to formal charge considerations. For your best structure(s), determine also all S− 0 bond orders. (c) Predict the probable structure of barium sulfite (BaSO3). Be careful!! What should be the geometry of the sulfite anion? 3. Valence Shell Electron Pair Repulsion (VSEPR) Model. For the following species, complete the following (the central atom is underlined): For each of the listed species, determine the following. • Determine the best Lewis structure. • Give the name of the ideal geometry. • Give the name of the observed geometry.. PH3 , CO3-2 , BF3 , SF4 , ClF3 , KrF2 & IF5 .

Page 2 Valence Shell Electron Pair Repulsion (VSEPR) Model & Polar vs Non-Polar Molecules: Adapted from Problem # 3 of Problem Set 6 - in Assignments Folder. (a) For the following species, complete the following (the central atom is underlined): • Draw the best Lewis structure and give the observed geometries. • Determine if the molecule is polar (i.e., has a dipole moment (µ)) or is non-polar (µ = 0).

4.

(b)

PH3 , CO3 − 2 , C2Cl2 (Cl-C-C-Cl, arrangement) , SF4 , ClF3 , KrF2 & IF5 . Consider the PCl2F3 molecule. Draw an isomer of this compound that is found to have no dipole moment, i.e., if it is found to be non-polar, draw it.

5.

Adapted from Problem # 3 of Problem Set 6 - in Assignments Folder. For the following species, complete the following (the central atom is underlined): • Draw the best Lewis structure. • Determine the hybridization of the underlined atom in each structure - and also of each carbon atom in C2 Cl2 . PH3 , CO3 − 2 , C2Cl2 (Cl-C-C-Cl, arrangement) , SF4 , ClF3 , KrF2 & IF5 .

Localized Electron Model & Valence Shell Electron Pair Repulsion (VSEPR) Model: A portion of Problem # 4 from Problem Set # 6 - in Assignments folder - AMENDED. (a) List the hybridization on all non-hydrogen atoms for the structure below:

6.

3

O H C

H

2

S

C

1

O 4

5

C

N

13

N

12

7

C

8

H

C

H C 14

9 10

C

N

11

O

H

15

O 6

Make sure to put in all lone pair (non-bonded) electrons in the above structure, before determining the hybridizations. Follow the numbering scheme above, when listing your answers. (b) How many σ bonds are there in this molecule? How many π bonds are there in this molecule? Now, focus on C (#9) and N(#10). Determine the type of bonds present and carefully explain how each bond arises.

Page 3 7.

Problem # 4 from Problem Set #6 - revisited & AMENDED Is there delocalization in the structure listed below? If so, what atoms are involved and why? If not, explain why not. Follow the numbering scheme above, when listing your answers. Note that all lone pair electrons are present. 3

•• O •• H

1

C

•• 2 S •• •• • •O 5 4 N

C8

H

•• N

12

7

C

H 13

C

H C 14

9

C

10

C

N ••

11

•• O ••

H

15

•• O • 6 • ••

8. The allene problem… Problem #5 - Problem Set #6 Consider the molecule allene, C3H4 , with the structure:

H

H C

C 2

H 1 (a) (b) (c) (d)

9.

C 3

H

Draw the Lewis structure for allene. List the hybridization of all carbon atoms (1, 2, and 3, above). Do the two hydrogen atoms attached to carbon # 1 lie in the same plane as the hydrogen atoms attached to carbon # 3 ? Explain. Is there delocalization of the π electrons in allene?

Comparison of LCAO-MO Theory with VB Theory for Diatomic Molecules: Consider the diatomic species CO and the diatomic anion CO- . Discuss the bonding and magnetic properties of each species according to MO theory and according to VB theory. Do the two theories (MO & VB) predict any conflicting results for either species? Explain how these conflict(s) could serve as a valuable experimental test of the theories. 2

Information: You may also refer to a periodic table - like the one distributed for exams. In order of increasing energy (for a diatomic molecule with its bond axis arbitrarily defined along the z-coordinate axis): * σ1s < σ1s* < σ2s < σ2s* < π2p = π2p < σ2p < π2p * = π2p < σ2p * x

y

z

x

y

z

Page 4 Brief Answers (1a) The structures for the NOF isomer. Structure II = “Best”. Discussion.

+1 •• O

!2

• •• N • ••

-1

+1 •• F

+1 •• O

•• N ••

•• Structure I

••• F •• •

Structure II

(1b) The structures for the ONF isomer. Structure IV = “Best”. Discussion. !1

•• •O • ••

+1 •• F

••

N

•• O ••

••

Structure III

••

••• F• ••

N

Structure IV

The structures for the OFN isomer. Neither is any good. Discussion. !1

+2

•• •O • ••

••

+2

!1

•• N ••

F

•• O ••

Structure V (1c)

•• F

!2

•• • N• ••

Structure VI

The most likely isomer is the ONF isomer - structure IV. Discussion.

(2a) Choice #1 - Cl=Br double bond - and resonance. Non-zero formal charges shown. Discussion.

•• •• Cl ••

!1

+1

••

Br

•• • Cl •• •

•• •• Cl ••

+1

••

Br

•• • Cl •• •

•• •• Cl ••

••

Br

•• Cl •• ••

•• Cl •• ••

•• Cl •• ••

+1

!1

•• ••

Cl ••

!1

(2a) Choice #2 - Extra lone pair on Cl - and resonance. Non-zero formal charges shown. Discussion.

!2

•• ••• Cl • ••

+2

••

Br

•• Cl •• ••

•• • Cl • ••

•• •• Cl ••

+2

••

Br

•• Cl •• •• • •

!2

•• Cl •• ••

•• •• Cl ••

+2

••

Br

•• Cl •• ••

•• • • Cl •• •• !2

Page 5 Choice #3 - Extra lone pair on central Br. All formal charges zero - as shown. Discussion.

0

••

•• Br• •

•• •• Cl ••

Cl ••

••

0

0

•• Cl •• •• 0

(2b) & (2c). We can answer (2b) and (2c) together by focusing on (2c). Discussion. Here we have an ionic compound in which the polyatomic sulfite anion (SO3 −2) is a (polar) covalent −2

+2 − species. Hence the ionic portion is [Ba] (isoelectronic with [Xe]) & SO3 is a polyatomic anion with (polar) covalent bonds between the central S atom and the attached O atoms. The octet and expanded octet structures are depicted below – with non-zero formal charges indicated. Octet structure:

••

!2

! 1 ••O •• +1 •• S !1 !1 •• •••O • • •O• • ••

−

Expanded octet (10 e− ) structures - all resonance structures are equivalent and each has the best formal charge distribution for the circumstances.

!1

!2

•• • •• O •

!1

••

S

••O • •

• ••O •• • !1

!2

•• •• O •• S

• ••O • •!•1

•• O ••

!2

S

•• ••O••

•••O • •• !1

•• !•1 ••O •• •

The set of expanded octet structures above collectively represent the best structure overall. Discussion. 1 2 Thus, the S−O bond in the expanded octet structure of SO3- is 1 3 . Discussion. The central S atom has a SN = 4 & LP = 1. VSEPR predicts an ideal geometry of tetrahedral and a predicted observed geometry of trigonal pyramidal - with a O−S−O bond angle of less than 109.5o due to the lone pair on S.

Page 6 3. Results are tabulated below. Discussion. For the Central Atom (underlined) list the steric # = S.N. , # of pairs of “lone” or non-bonded electrons = L.P. , the Ideal Geometry , and the Observed geometry. The structures are not drawn - but you should draw them for yourself. However, the table below lists the results. Discussion. Abbreviations below: TPlanar (Trigonal Planar), Bent-TPlanar (Bent from Trigonal Planar), Tet (Tetrahedral), TPyr (Trigonal Pyramidal), Bent-Tet (Bent from Tetrahedral ), TB (Trigonal Bipyramidal), Oct (Octahedral), SP (Square Pyramidal). Species PH3

S.N. L.P. 4 1

CO3BF3 SF4

3 3 5

0 0 1

TPlanar TPlanar TB

TPlanar TPlanar See-saw

ClF3

5

2

TB

T-shaped

KrCl2 IF5

5 6

3 1

TB Oct

Linear SP

4.

2

Ideal Geometry Tet

Observed Geometry TPyr

(B only has a sextet)

All detailed solutions to Problem Set Problems are provided in the Assignments folder. For the Central Atom (underlined) list the steric # = S.N. , # of pairs of “lone” or non-bonded electrons = L.P. , the Ideal Geometry , the Observed geometry, and whether the molecule is polar or non-polar. The hybridization of the underlined atom - or the indicated atom - is also listed in the last column. The structures are not drawn - but you should draw them for yourself. However, the table below lists the results. Discussion. Remember, all detailed solutions are also provided in the Assignments folder.

(4a) Discussion. Abbreviations below: TPlanar (Trigonal Planar), Bent-TPlanar (Bent from Trigonal Planar), Tet (Tetrahedral), TPyr (Trigonal Pyramidal), Bent-Tet (Bent from Tetrahedral ), TB (Trigonal Bipyramidal), Oct (Octahedral), SP (Square Pyramidal). Species S.N. L.P. Ideal Geo. Obs'd. Geo. Polar or Non-polar PH3

4

1

Tet

TPyr

Polar

C2Cl2

3 2

0 0

TPlanar Linear (each C)

TPlanar Linear

Non-Polar Non-Polar

SF4

5

1

TB

See-saw

Polar

2 CO3-

ClF3

5

2

TB

T-shaped

Polar

KrCl2 IF5

5 6

3 1

TB Oct

Linear SP

Non-Polar Polar

(4b)

The solution is listed in a bit more detail in the detailed solution in the Assignments folder. Please refer to it. Discussion. The 3 isomers are below, Only #1 is non-polar (µ = 0).

•• • ••Cl •• •• • • F• •• •• F ••

P ••Cl •• ••

Isomer # 1

•• • F• ••

•• •• F •• • •• • •Cl •

•• •• F ••• •• • •F• •• ••Cl ••

P

•• • F• ••

•• •• F ••

P

•• Cl •• ••

••Cl •• ••

•• F •• ••

Isomer # 2

Isomer # 3

Page 7 5. Discussion. Species S.N. PH3 4

-2 3

CO3 C2Cl2

Hybridization 3 sp for P 2

sp for C sp for each C

2

3

3

3

3

3

3

SF4

5

sp d or dsp for S

ClF3

5

sp d or dsp for Cl

KrCl2

5

sp d or dsp for Kr

IF5

6

sp d or d sp for I

3 2

2

3

All detailed solutions to Problem Set Problems are provided in the Assignments folder. (6a) The re-drawn structure - with all lone pair electrons shown - is displayed below: 3

••O •• H

1

C

•• 2 S •• •• • •O 5 4 N

C C

H

•• N

12

7

C

8

13

H

C

H C 14

9 10

N ••

C 11

•• O ••

H

15

••O • 6 • ••

The atom # (from above), Steric Number (S.N.) and atom hybridization are tabulated below. Atom # Atom Name S.N. Hybridization 1

S

4

sp3

2 3

C O

3 3

sp 2 sp 2

4

O

3

sp 2

5 6

N O

3 4

sp 2 sp 3

7 8

N C

3 3

sp 2 sp 2

9

C

3

sp 2

10 11

N C

3 3

sp 2 sp2

12

C

3

sp2

13 14

C C

4 4

sp3 sp3

15

O

4

sp3

Page 8 (6b) There are 21 σ bonds (2 N-H bonds are removed from the original Problem Set 6 structure) and 5 π bonds. The “first” or single bond between a pair of atoms is the σ bond. Any second or third bond - giving rise to an overall double or triple bond, respectively - is a π type bond. We now focus on atoms C(#9) and N(#10). We tabulate again. First, we list the atomic orbitals (A.O.'s) that overlap to form the bond. Finally, we list the bond type (σ or π). We also refer to the numbering scheme on the diagram, as needed, to clarify the answers. Of course, sp, sp2, and sp3 are hybrid A.O.'s and 1s, 2s, 2p, etc., are unhybridized A.O.'s. Note: when we denote an unhybridized “p” A.O. as “px” or “py”, “pz”, this is purely arbitrary. The designation is to insure that it is clear which “variety” of “p” A.O. is being used. It is crucial that the same kind of “p” A.O. 's overlap “side-by-side to form a π bond. Atoms in Bond

A.O.'s overlapping

C (# 9) & N (# 10) C (# 9) & N (# 10)

sp2 on C (# 9) & sp2 on N (# 10)

Bond Type σ π

2px on C (# 9) & 2px on N (# 10)

7. Yes, there is delocalization in this molecule. π-electron delocalization arises from the spreading out of electron density over more than two bonding sites - due to the constructive side-by-side overlap of in-phase (+ with + and − with −) p-type AO's on a string of three atoms or more. This commonly arises - as is th e case here - when we have an uninterrupted sequence of sp2 hybridized atoms. This side-by-side (parallel) alignment of unhybridized 2p atomic orbitals (of the same type - i.e., all np x , or all npy , or all np z) is in a “picket-fence-like” form. The “string” or “ring” of atoms - each have an unhybridized p-type AO that is perpendicular to the plane of sp2 hybridization (the plane of the page, here). These 2p orbitals are aligned side-by-side and in phase so that they all constructively interfere. This allows the electrons in these 2p AO's to spread out or delocalize over the entire sequence of involved atoms - producing a + phase region above the plane of the page and a − phase region below the plane of the page. The atoms that fulfill these criteria are atoms #2 through #12. Prove this for yourself. 2

All of these atoms are sp hybridized - uninterrupted and thus give a planar structure. Hence, all of the unhybridized 2p atomic orbitals are parallel, side-by-side, and in phase.

8.

Remember, this problem (and analysis) is posted as a .pdf in Course Documents in Bb. The solution to this problem is also part of the detailed solutions to Problem Set 6.

9. Discussion. A brief summary of results appears below. Species

MO configuration (ground state)

CO

[KK or[He2]] σ2s σ2s

-

2

*

2

2

* 2

2

* 2

4

Bond Order 2

π2p σ2p 4

2 * 2

2 CO [KK or[He2]] σ2s σ 2s π2p σ2p π 2p (unpaired) * P = paramagnetic & D = diamagnetic

Remember, KK or[He2] = σ1s σ1s

-

# of unpaired e (P or D) *

3.0

0