Chemistry 101 -- 2001 Answers to Questions for Assignment #8 and Quiz #8

Problems in McMurry and Fay( 3rd edition) from which these questions were taken. Other question has been made up for Chem 101.7.93, 7.95, 7.107, 7.111 plus 1 additional question not in text Alternatively problems in ( ) refer to McMurry and Fay( 2nd edition) from which these questions were taken. Other question has been made up for Chem 101.7.89, 7.91, 7.103, 7.107 plus 1 additional question not in text

Click Here for correct responses to all versions of Quiz #8

Complete Questions Qn. [3rd edition (2nd edition)]

Q1. [7.93 (7.89)] Use the MO diagram in Figure 7.18a to describe the bonding in N2+, N2

and N2. Which of the three should be stable? What is the bond order of each? Which

contain unpaired electron? Fig 7.18a:

2p

2s

2p

2s

2s

2s*

2p*

2p

2p*

2p

Molecular orbitalsAtomic orbitalson N atom 1(2s and 2p)

Atomic orbitalson N atom 2 (2s and 2p)

bonding

anti-bonding

bonding

bonding

anti-bonding

anti-bonding

.

ANSWER: The MO diagrams for these three molecules are: 

*

*

*

2s

2s

2p

2p

2p

2p

N2

N2 N

2

+ -

 Bond order = (bonding electrons - anti-bonding electrons)/2 * means anti-bonding orbital moleculebonding eanti-bonding e Bond order bond length /N2 unpaired eN2

+ 7 2 2.5 > 1 1N2 8 2 3.0 = 1 0N2

8 3 2.5 > 1 1 All are stable molecules with bond orders of 2.5 or 3. The bond length of the ions is greater than N2 and the bond strength is less. Q2. [7.95 (7.91)] Make a sketch showing the location and geometry of the p orbitals in the nitrite ion, NO2

. Describe the bonding in this ion using a localized valence bond model for the bonding and a delocalized MO model for the bonding.  ANSWER:The electron dot structures for NO2

are: 

. N

OON

O O

 Each resonance form contributes equally since they are both the same except for the position of the electrons. If we averaged over these two forms then we would predict

that each N-O bond is about half way between a N-O single and a N=O double bond. Also on average each oxygen has 2 ½ non bonding pairs on it.

Since the N atom has three independent electron clouds around it the prediction of VSEPR model is the orientation of these clouds is trigonal planar and the hybridization on the N atom is sp2. If we examine the two resonance structures we see that there are two pairs of electrons that are in different places in the two resonance forms i.e. one pair in the double bond and one non-bonding pair on the two O atoms. The fact that these two pairs are in different places suggests that they will be present in PI Molecular orbitals above and below the plane of the molecule. The other electron pairs (the ones that do not appear to move) are in the sigma framework which can be treated using the Valence Bond model.Sigma framework: 

sigma framework

Valence Bond Model

NOO

N

O O

sp2 sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

Each atom has ONE p orbital not involved in the sp2 hybridization. The orbitals are perpendicular to the plane of the molecule. We can use these THREE p orbitals to form THREE pi molecular orbitals which lie above and below the plane of the molecule. Pi MO’s:

O N O O N O O N O

bonding non-bonding antibonding

or

O N O O N O O N O

oxygen atoms

nitrogen atom

anti-bonding

non-bonding

bonding

pi MO's

NO2-

2p2p

 Notice: only the bonding and non-bonding pi orbitals are occupied. We have only four electrons in the pi system and we fill up the possible pi energy levels in the same way that we do atoms, i.e. start at the lowest and use Pauli exclusion principle and Hund’s rule.  The total pi bond order is (2 - 0)/2 = 1 but since the bonding orbital is spread over two N-O bonds the pi bond order per N-O bond is ½ =0.5. The N-O sigma bonds contributes a sigma bond order of 1.0 to each N-O bond so the total bond order per N-O bond is 1.0+0.5 = 1.5 which again predicts the N-O bond is half way between N-O and N=O.  

The pi electrons in the non-bonding orbital spend half their time on one oxygen and half their time on the other oxygen. Adding in the sigma non-bonding pairs we get an average of 2 ½ non-bonding pairs per O atom as was predicted by averaging the resonance forms. 

Q3. [7.107 (7.103)] There are three substances with the formula C2H2Cl2 . Draw electron dot structures for all, and explain how and why they differ.

Answer: Following the procedure for forming electron dot structures for C2H2Cl2 we can get THREE structures (see below) where all the atoms obey the octet rule and the formal charge on all the atoms is zero.

 

These ARE NOT RESONANCE HYBRIDS as the atoms are in different locations and one form cannot be made to look like another form or turned into another form without breaking bonds. The three structure represent three different molecules with slightly different chemical and physical properties. The molecules are said to be isomers of one another since they all contain two C atoms, 2 H atoms and two Cl atoms.

To convert the top structure (1,1 dichloroethene) into one of the others we would have to break and remake two bonds: a C-H bond and a C-Cl bond.

To convert one of the bottom two structures into the other, one end of the molecule would have to be rotated through 180o with respect to the other end. This would require breaking the bond part of the double bond and then remaking it after rotation. Thus, we say that there is no rotation around the double bond, whereas if the C-C bond was only a single bond then one end could be rotated with little resistance, since no bond breaking is required.

has like2C 2 x 4 = 8e 2 x 8 =16e

2 x 1 = 2e 2 x 2 = 4e

total 24e 36epairs 12

18-12=6 bonding pairs

12 - 6 = 6 non-bonding pairs

C

18

2H

H

Ho o

C2H2Cl2

2 x 7 = 14e 2 x 8=16e2Cl

Clo o

CCl

o ooo

using this and octet rule

oo

o o

C

H

Ho o

Clo o

C

Clo o

oo

oo

o o

C

HH

o oCl

o o

CCl

o o

oo o

o

o o

1,1 dichloroethene

cis 1,2 dichloroethene trans 1,2 dichloroethene

Q4. [7.111 (7.107)] In the cyanate ion, OCN , carbon is the central atom.

a) Draw as many resonance structures as you can for OCN , and assign formal charges to the atoms in each.

b) Which resonance structure makes the greatest contribution to the resonance hybrid? Which makes the least contribution? Explain.

c) Is OCN linear or bent? Explain.

d) Which hybrid orbitals are used by the C atom, and how many bonds does the C atom form?

Answer:

a)

o o

o oo o

o ooo

o o o o

oo

oo

oo

+ --o o

CO N CO No o

C NO- 2

I II III

b) Structure III makes the LEAST contribution to the resonance since there is a much larger separation of formal charge compared to I and II. The other two structures contribute roughly equally with structure I probably slightly favored since O is more electronegative than N.

c) and d) In all three resonance structures there are TWO independent groups of electrons around the central C atom, so the orientation around the C is linear. This means that the hybridization around the C atom is sp along the molecular axis with two unhybridized p orbitals at right angles to the molecular axis. This means that the C atom can form two pi bonds.

Note: The actual pi bonding is rather complicated and for a detailed picture requires the use of Molecular Orbital theory involving EIGHT electrons in TWO pi systems stretching over the whole molecule. Question 5 below deals with a similar problem for the CO2 molecule but the symmetry in the CO2 molecule makes it easier to see what is going on so the details are given there.

Q5. (not in text). a) Draw THREE Lewis structures for the CO2 (OCO) molecule which obey the octet rule. Include the formal charge on each atom if not equal to zero. b) Rank these structures in order of decreasing contribution to the overall picture of the molecule. c) Calculate the bond lengths and bond angles in the CO2 molecule. d) EIGHT of the electrons are assigned to pi orbitals and are best described using MO theory. Sketch the sigma framework and indicate the hybrid orbitals that are used to describe the sigma bonding and non-bonding electrons. e) There are two degenerate pi systems spread over the whole molecule. The energy level diagram for these systems is:

oxygen atoms

carbon atom

anti-bonding

non-bonding

bonding

pi MO's

CO2

 Fill in this diagram for the ground state of the CO2 molecule and calculate the total pi bond order, the pi bond order for each CO bond and the total bond order for each CO bond. f) Using the diagram predict what the total bond order for each CO bond will be in the

CO2+ and CO2

ions. Will the CO bond in each of these ions be longer or shorter than in the neutral molecule? ANSWER: a)

o o

o oo o

o ooo

o o o o

oo

oo

oo

+ + --o o

CO O CO Oo o

C OO

  b) The middle structure will contribute much more to a description of the CO2 molecule than either the left or right structures since there is no separation of formal charge.   c) Since the middle structure is the dominant one the prediction is that the two C-O bonds will be double bonds. Since there are only two groups of independent electrons around the central C atom the prediction is that they will be oriented at 180o with respect to one another. Thus the O-C-O bond angle is predicted to be 180o.  d) Although the middle Lewis structure above is estimated to be much more important than either of the other two, when considering how to do the sigma and pi systems, one should at least consider that they might contribute to a small extent. Thus, if we look at all three structures we come to the conclusion that there are four pairs of electrons that do not occupy the same position in all three structures. Moreover, the O atoms should probably be sp hybridized since they are involved in

a triple bond in at least one structure. This means that these four pairs should be treated as pi electrons and the other four pairs be assigned to the sigma system:  Sigma system : two sigma bonds (C-O) and two non-bonding pairs (one on each O). The C atom and each O atom are to be considered sp hybridized. i.e.: 

oo

ooCO O = CO O

sp sp sp sp spsp

  e) This leaves two unhybridized p orbitals (at right angles to the bond axis) on each O atom and on the carbon atom. The pi system can be constructed from the sideways overlap of these p orbitals. Since on each atom they are at right angles only three p orbitals (one on each O and one on the carbon ) can combine to form a pi system. The other three form a SECOND, separate pi system at right angles to the first. This is a similar situation to that which occurs in the triple bond in acetylene where there are two pi bonds at right angles to one another. Three p orbitals will form three pi orbitals so we get two sets of three orbitals with the same energies. These are the ones given in the question handed out. 

oxygen atoms

carbon atom

anti-bonding

non-bonding

bonding

pi MO's

CO2

 

The three pi orbitals are similar to the ones obtained in the NO2 ion or the O3

molecule. i.e.

O C O O C O O C O

bonding non-bonding antibonding

or

O C O O C O O C O

Remember there is a second identical set of orbitals at right angles to these, i.e. in and out of the plane of the paper. The total pi bond order = (4 bonding electrons - 0 antibonding electrons)/ 2 = 2. Since each bonding orbital is spread over two bonds the pi bond order per C-O bond is 2 /2 = 1. The total bond order per C-O bond is 1 sigma bond + one pi bond = 2 per C-O bond. This predicts each C-O bond in CO2 should be close to a double bond, the same prediction as obtained above in part a) to c).   f) Using the above analysis for CO2

+ and CO2 , the only change we have to make is

change the number of electrons that go into the pi system.  molecpi e bondingnon-bond anti-bond unpaired e pi BOpi BO/CO total BO/COCO2 7 4 3 0 1 2 1 2CO2 8 4 4 0 0 2 1 2CO2 9 4 4 1 1 1.5 0.75 1.75 Thus the prediction is that removing one electron to give CO2

+ does not significantly affect the C-O bond length since the electron is removed from the pi non-bonding orbital. When another electron is added to the CO2 molecule to make CO2

+ , it goes into one of the pi-antibonding orbitals. This lowers the total bond order slightly and the prediction is that the bond will lengthen slightly. 

ANSWERS TO QUESTIONS 8 to 10 ON QUIZ #8 To save space I have include answers for all versions of these questions together in tabular form.  8. The total bond order or number of unpaired electrons in the following molecules or ion is (see projectoral for MO energy level diagram): 

  FO NO NF+ NO

  F01 and F03 F01 and F03 F02 and F04 F02 and F04

Quiz Version A, B’, C, D’ A’, B, C’, D A”, B*, C”, D* A*, B”, C*, D”

Total Bond Order 1.5 2.5 2.5  

No. of unpaired electrons

      2

 ANSWER: Just count up the number of valence electrons in each of the two atoms and add or subtract one for the – or + sign. Then fill up energy level diagram in the usual fashion from bottom to top obeying the Pauli exclusion principle and Hunds rule. The filled orbital diagrams for the various molecules considered are given below.  

*

*

*

2s

2s

2p

2p

NO FO

2p

2p

NO NF+

To calculate the bond order use the formula given in the text:Bond order = (no. bonding electrons - no. anti-bonding electrons)/2Remember in the diagrams above the levels with a * as a superscript are anti-bonding. 

9. The hybridization around the central atom in the following molecules or ions ion is:  a) sp3d b) sp2 c) sp d) sp3 e) sp3d2  

OOO

OO O

 

NOO

O

NOO

O

NO O

O

  

NN N NN N NN N 

COO

O

COO

O

CO O

O

2 22

   O atom in O3 N atom in NO3

N atom in N3 C atom in CO3

2

  F01 and F03 F01 and F03 F02 and F04 F02 and F04Quiz Version A, B’, C, D’ A’, B, C’, D A”, B*, C”, D* A*, B”, C*, D”

groups around

central atom

3 3 2 3

hybridization sp2 sp2 sp sp2

 

10. The total number of ALL or JUST THE NON-BONDING PI electrons in the following ions is:  a) 2 b) 3 c) 4 d) 6 e) 8  

  CO32 O3 NO3

NO3

  F01 and F03

F01 and F03

F02 and F04 F02 and F04

Quiz Version A, B’, C, D’

A’, B, C’, D

A”, B*, C”, D*

A*, B”, C*, D”

Total No. of Pi electrons

6 4 6 6

No. of Non-bonding Pi 4 2 4 4 Answer: See diagrams in Question 9. The pi electrons are the ones that do not appear in ALL the resonance structures that make up the resonance hybrid of a molecule. The easiest way to determine this is to write down the structure for the framework using those pairs that appear in the same place in ALL the resonance forms. The pi electrons will be the ones left over. As an example consider the NO3

ion: 

NOO

O

NOO

O

NO O

O

Each O always has one bond (the sigma bond to the N atom) and two non-bonding pairs. The N atom always has the threesigma bonds to the three O atoms. Thus the framework is:

NOO

O

There are 6 non bonding pairs in the frameworkand 3 bonding pairs.

Thus there is ONE bonding pair and TWO non-bondingpairs that do not appear in all structures and these are the pi electrons.