Chapter 8: Covalent Bonding Solutions for Chapter 8 Questions for Review …academic.udayton.edu/VladimirBenin/123_2004/Problem… · · 2013-12-16Chapter 8: Covalent Bonding 235

Chapter 8: Covalent Bonding 235

Chapter 8: Covalent Bonding

Solutions for Chapter 8

Questions for Review and Thought

Review Questions

1. An ionic bond and a covalent bond differ in the location of the electrons. In ionic bonds, there is no sharing of electrons. The cation is deficient in electrons and the anion has extra electrons. The opposite charges of these ions cause them to be attracted to each other and that is what we call ionic bonding. In contrast, when both atoms want to gain electrons, they achieve this by sharing electrons. The atoms must stay close together to share electrons; hence, a different type of bond – a covalent bond – is formed.

2. To predict the type of bonding, we need to determine what types of elements are involved. Metals are not good at taking electrons from other elements; nonmetals generally are good at taking electrons. If the elements differ in their ability to take electrons, such as metals and nonmetals, electrons will be transferred, so we'll predict ionic bonding. If they are similar, such as nonmetals and nonmetals, they will share, and we'll predict covalent bonding.

(a) Na is a metal and I2 is a nonmetal. The nonmetal can take electrons from the metal, so we predict ionic bonding.

(b) C is a nonmetal and S is a nonmetal. They both want to take electrons, so we predict covalent bonding.

(c) Mg is a metal and Br2 is a nonmetal. The nonmetal can take electrons from the metal, so we predict ionic bonding.

(d) P4 is a nonmetal and Cl2 is a nonmetal. They both want to take electrons, so we predict covalent bonding.

3. If the elements A and X have similar ability to take electrons, such as a nonmetal and a nonmetal, they will share electrons, and we'd find them bonded using a covalent bond.

4. Alkanes have all single bonds between carbons.

C C C

H

H

H

H

H

H

H

H

Alkenes have at least one double bond between two carbons.

C C C

H

H

H H

H

H


Alkynes have at least one triple bond between two carbons.

C C CH

H

H

H

5. (a) It shows several examples of molecules whose central atoms have more than eight electrons. None of them are from Period 2.

(b) The maximum number of electrons around any central atom in the table is twelve, in six pairs.

6. Period 3 elements, such as sulfur, have a set of d orbitals (the 3d orbitals) with similar energy to their valence electrons (in the 3s and 3p orbitals), so there are extra d orbitals available for expanding their number of bonding electrons beyond eight. Period 2 elements, such as oxygen, do not have a set of d orbitals with similar energy to their valence electrons (in the 2s and 2p orbitals), so there are no extra orbitals available for expanding their number of bonding electrons beyond eight.

7. Determine the valence electrons from each atom and find the sum.

NO2 N in Group 5 and O in Group 6 gives: 5 + 2 × (6) = 17 This is odd.

SCl2 S in Group 6 and Cl in Group 7 gives: 6 + 2 × (7) = 20 This is even.

NH3 N in Group 5 and H in Group 1 gives: 5 + 3 × (1) = 8 This is even.

NO3 N in Group 5 and O in Group 6 gives: 5 + 3 × (6) = 23 This is odd.

So, from this list, only NO2 and NO3 have an odd number of electrons.

8. Two resonance structures for NO2–

...... ..

.. ..O N OG ..

.... .... ..O N O

G

Table 8.1 has the bond lengths of single N–O bonds (126 pm) and double N=O bonds (115 pm). Since the two resonance structures are equivalent, and in one the bond is single and in the other the bond is double, we predict that the bond between N and O will be halfway between the length of single and double, or approximately 121 pm.

9. Structures (a) and (b) are resonance structures, because the double bond is moved to alternate oxygen atoms. Structure (c) is equivalent to structure (a). Placing the electrons below the symbol for O does not change what atom they are associated with.

10. Atoms get smaller left to right across a period of the periodic table, so C–F will be the shortest, then C–O, then C–N, then C–C, and C–B will be the longest. We can confirm most of this trend by looking at the bonds lengths given in Table 8.1:

C–F (141 pm), C–O (143 pm), C–N (147 pm), and C–C (154 pm)

11. Looking at Tables 8.1 and 8.2, we find that the single bonded carbons have


C C 154 pm 356 kJ C C 134 pm 598 kJ C C 121 pm 813 kJ

The bonds get shorter and stronger from single to double to triple.

12. In alkynes, the triple bond between two carbons does prevent free rotation about that bond, just like we see with the double bond in alkenes. However, isomers in alkenes are possible when there are two different things bonded to each of those carbons. In alkynes, the triple bonded carbon has only one thing bonded to it. Any changes that are restricted by rotations about the triple bond can be readily achieved by free rotations about the neighboring single bond:

Lewis Structures

13. Define the problem: Write Lewis structures for a list of ions and molecules.

Develop a plan: Following a systematic plan will give you reliable results every time. Trial and error often works for small molecules, but as molecules get more complex, it’s better to follow a procedure than to try to guess where electrons will end up. There are a number of methods. The one described here is the same as that described in the text and it works all the time.

(A) Count the total number of valence electrons. If there is a nonzero charge, adjust the electron count appropriately. Add electrons for negative charges and subtract electrons for positive charges.

(B) Determine which atom is the central atom. This is usually the first element in the formula, and it will have a smaller electronegativity (see Section 8.7) than the rest of the atoms in the formula. There is an exception: H will never be the central atom.

Put the rest of the atoms (called the outer electrons) around the central atom and connect them with single bonds.

(C) Subtract the electrons used for single bonds from the total. Each bond is composed of two electrons.

(D) Complete the octets of all of the outer electrons by adding pairs of dots (called lone pairs) so that each of them ends up with a total of eight electrons, including the two shared with the central atom. There is an exception: H only need two electrons, so never put dots on H.

(E) Subtract the electrons used for lone pairs from the total.

(F) If you still have unused electrons, put all the remaining electrons on the central atom. (This may occasionally result in the central atom having more than eight electrons.)

(G) Check the octet of the central atom.

(i) If it has eight or more electrons, then the structure is complete.

(ii) If it has less than eight electrons, move a lone pair of electrons from one of the outer atoms to make a new shared pair, forming a multiple bond to the central atom. Repeat this procedure until the central atom has an octet.

(H) If the structure is an ion, put it in brackets and designate the net charge in the upper right corner.

X C C Y


Execute the plan:

(a) (A) The atoms Cl and F are both in Group 7A. The + charge removes one electron: 7 + 4 × (7) – 1 = 34 e–

(B) The Cl atom is the central atom with the F atoms around it.

F Cl F

F

F

(C) Connecting four F atoms uses eight electrons: 34 e– – 8 e– = 26 e–

(D) Complete the octets of all the F atoms.

F Cl F

F

F

....

. ...

..

... ... . ...

..

8 eG

inside each circle

..

(E) Completing the octets of the four F atoms uses 24 electrons:

26 e– – 24 e– = 2 e–

(F) Put the last two electrons on the Cl.

F Cl F

F

F

....

. .... .

... ... . ...

....

..

(G) Cl has ten electrons, so the structure is complete.

(H) The structure is an ion, so put it in brackets and add the + charge.


F Cl F

F

F

.. ....

. .... .

... ... . ...

....

+

(b) (A) The Cl atom is in Group 7A and the O atoms are in Group 6A. The – charge adds one electron: 7 + 3 × (6) + 1 = 26 e–

(B) The Cl atom is the central atom with the O atoms around it.

O Cl O

O

(C) Connecting three O atoms uses six electrons: 26 e– – 6 e– = 20 e–

(D) Complete the octets of all the O atoms.

O Cl O

O.. ....

... ... . ...

..

(E) Completing the octets of the three O atoms uses 18 electrons:

20 e– – 18 e– = 2 e–

(F) Put the last two electrons on the Cl.

O Cl O

O.. ....

... ... . ...

....

(G) Cl has eight electrons, so the structure is complete.

(H) The structure is an ion, so put it in brackets and add the – charge.

O Cl O

O.. ....

... ... . ...

....

G

(c) (A) The H atom is in Group 1A, the Cl atom is in Group 7A, and the O atom is in Group 6A: 1 + 7 + 6 = 14 e–

13. (c) (continued)


(B) Contrary to the rule, the H atom in this oxyacid is bonded to the O atom, which in turn is bonded to the Cl atom.

(C) Connecting the H and Cl atoms uses four electrons: 14 e– – 4 e– = 10 e–

(D) H needs no more electrons. Complete the octet of the Cl atom.

(E) Completing the octet of the Cl atom uses six electrons: 10 e– – 6 e– = 4 e–

(F) Put the last four electrons on the O.

(G) O has eight electrons, so the structure is complete.

(d) (A) The S atom is in Group 6A and the O atoms are in Group 6A. The 2– charge adds two electrons: 6 + 3 × (6) + 2 = 26 e–

(B) The S atom is the central atom with the O atoms around it.

O S O

O

(C) Connecting three O atoms uses six electrons: 26 e– – 6 e– = 20 e–


O S O

O.. ....

... ... . ...

..

(E) Completing the octets of the three O atoms uses 18 electrons:

20 e– – 18 e– = 2 e–

(F) Put the last two electrons on the S.

O S O

O.. ....

... ... . ...

.. ..

(G) S has eight electrons, so the structure is complete.

(H) The structure is an ion, so put it in brackets and add the 2– charge.

H O Cl

H O Cl......

H O Cl....

..

.. ..


O S O

O.. ....

... ... . ...

....

2G

Check your answers: All the Period 2 elements have an octet of electrons. The H atoms have two electrons. All the Period 3 elements have eight or more electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.


Develop a plan: Following a systematic plan will give you reliable results every time. Trial and error often works for small molecules, but as molecules get more complex, it’s better to follow a procedure than to try to guess where electrons will end up. There are a number of methods. The one described here is the same as that described in the text and it works all the time.

(A) Count the total number of valence electrons. If there is a nonzero charge, adjust the electron count appropriately. Add electrons for negative charges and subtract electrons for positive charges.

(B) Determine which atom is the central atom. This is usually the first element in the formula, and it will have a smaller electronegativity (see Section 8.7) than the rest of the atoms in the formula. There is an exception: H will never be the central atom.

Put the rest of the atoms (called the outer electrons or lone pair electrons) around the central atom and connect them with single bonds.

(C) Subtract the electrons used for single bonds from the total. Each bond is composed of two electrons.

(D) Complete the octets of all of the outer electrons by adding pairs of dots (called lone pairs) so that each of them ends up with a total of eight electrons, including the two shared with the central atom. There is an exception: H only need two electrons, so never put dots on H.

(E) Subtract the electrons used for lone pairs from the total.

(F) If you still have unused electrons, put all the remaining electrons on the central atom. (This may occasionally result in the central atom having more than eight electrons.)

(G) Check the octet of the central atom.

(i) If it has eight or more electrons, then the structure is complete.

(ii) If it has less than eight electrons, move a lone pair of electrons from one of the outer atoms to make a new shared pair, forming a multiple bond to the central atom. Repeat this procedure until the central atom has an octet.

(H) If the structure is an ion, put it in brackets and designate the net charge in the upper right corner.

Execute the plan:

(a) (A) The atoms Cl and F are both in Group 7A: 7 + 7 = 14 e–

(B) The Cl atom is the central atom with the F atom bonded to it.

Cl F

(C) Connecting the F atom uses two electrons: 14 e– – 2 e– = 12 e–


(D) Complete the octet of the F atom.

Cl F. .....

8 eG

inside the circle

(E) Completing the octet of the F atom uses 6 electrons:

12 e– – 6 e– = 6 e–

(F) Put the last six electrons on the Cl.

Cl F. .....

... ...

(G) Cl has eight electrons, so the structure is complete.

(b) (A) The H atoms are in Group 1A and the Se atom is in Group 6A:

2 × (1) + 6 = 8 e–

(B) The Se atom is the central atom with the H atoms around it.

H Se H

(C) Connecting the H and Se atoms uses four electrons: 8 e– – 4 e– = 4 e–

(D) H atoms need no more electrons.

H Se H

(E) No electrons used in this step. So we still have 4 e–.

(F) Put the last four electrons on the Se.

H Se H....

(G) Se has eight electrons, so the structure is complete.

(c) (A) The B atom is in Group 3A and the F atoms are in Group 7A. The – charge adds one extra electron: 3 + 4 × (7) + 1 = 32 e–

(B) The B atom is the central atom with the F atoms around it.

F B F

F

F

(C) Connecting four F atoms uses eight electrons: 32 e– – 8 e– = 24 e–


(D) Complete the octets of all the F atoms.

F B F

F

F

.. ....

. .... .

... ... . ...

..

(E) Completing the octets of the four F atoms uses 24 electrons:

24 e– – 24 e– = 0 e–

(F) No more electrons are available.

(G) B has eight electrons, so the structure is complete.

(H) The structure is an ion, so put it in brackets and add the – charge.

F B F

F

F

.. ....

. ...

... ... . ...

..

G

..

(d) (A) The P atom is in Group 5A and the O atoms are in Group 6A. The 3– charge adds three extra electrons: 5 + 4 × (6) + 3 = 32 e–

(B) The P atom is the central atom with the O atoms around it.

O P O

O

O

(C) Connecting four O atoms uses eight electrons: 32 e– – 8 e– = 24 e–


O P O

O

O. .... .

... ... . ...

..

.. ....

(E) Completing the octets of the four O atoms uses 24 electrons:


24 e– – 24 e– = 0 e–

(F) No more electrons are available.

(G) P has eight electrons, so the structure is complete.

(H) The structure is an ion, so put it in brackets and add the 3– charge.

O P O

O

O

.. ....

. ...

... ... . ...

..

3G

..

Check your answers: All the Period 2 and 3 elements have an octet of electrons. The H atoms have two electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.

15. Define the problem: Write Lewis structures for a list of molecules.

Develop a plan: Follow the systematic plan given in the answer to Question 15. Use what you learned about organic molecules in Chapter 3 to determine how atoms are bonded in the organic structures.

Execute the plan:

(a) The C atom is in Group 4A, the H atom is in Group 1A, and the halogen atoms are in Group 7A: 4 + 1 + 7 + 2 × (7) = 26 e–. The C atom is the central atom with the other atoms around it. Connecting four atoms uses eight electrons: 26 e– – 8 e– = 18 e–. H needs no more electrons. Complete the octets of the halogen atoms using 18 electrons: 18 e– – 18 e– = 0 e– . No more electrons are available. The central atom, C, has eight electrons in shared pairs, so it needs no more. We get the following structure:

H C F

Cl

F

.. ....

. ...

. ...

..

..

(b) The C atom is in Group 4A, the H atoms are in Group 1A, and the O atom is in Group 6A: 4 + 4 × (1) + 6 = 14 e–. As described in Section 3.1, the structural formula given here tells us that this organic compound has three H atoms and an O atom bonded to C, and another H atom bonded to the O. Connecting six atoms uses ten electrons: 14 e– – 10 e– = 4 e–. For the purposes of this method, let's call C the central atom, since it has the most bonds. H needs no more electrons. Complete the octet of the oxygen atom using four electrons: 4 e– – 4 e– = 0 e– . No more electrons are available. The central atom, C, has eight electrons in shared pairs, so it needs no more. We get the following structure:


H C O

H

H

H....

(c) The C atom is in Group 4A, the H atoms are in Group 1A, and the N atom is in Group 5A: 4 + 5 × (1) + 5

= 14 e–. As described in Section 3.1, the structural formula given here tells us that this organic compound has three H atoms and an N atom bonded to C, and another two H atoms bonded to the N. Connecting seven atoms uses 12 electrons: 14 e– – 12 e– = 2 e–. For the purposes of this method, let's call C the central atom, since it has the most bonds. H needs no more electrons. Complete the octet of the nitrogen atom, using two electrons: 2 e– – 2 e– = 0 e– . No more electrons are available. The central atom, C, has eight electrons in shared pairs, so it needs no more. We get the following structure:

H C N

H

H

H

H

..

Check your answers: All the Period 2 elements have an octet of electrons. The H atoms have two electrons. All valance electrons are accounted for in the structure, either as members of shared pairs or of lone pairs. These answers look right.


Develop a plan: Follow the systematic plan given in the answers to Question 16. Use what you learned about organic molecules in Chapter 3 to determine how atoms are bonded in the organic structures. Execute the plan:

(a) The C atom is in Group 4A, the H atom is in Group 1A, and the Cl atom is in Group 7A: 4 + 3 × (1) + 7 = 14 e–. The C atom is the central atom with the other atoms around it. Connecting four atoms uses eight electrons: 14 e– – 8 e– = 6 e–. H needs no more electrons. Complete the octet of the Cl atom using six electrons: 6 e– – 6 e– = 0 e– . No more electrons are available. The central atom, C, has eight electrons in shared pairs, so it needs no more. We get the following structure:

H C Cl

H

H

. ...

..

(b) The Si atom is in Group 4A and the O atoms are in Group 6A. The 4– charge adds four extra electrons: 4 + 4 × (6) + 4 = 32 e–. The Si atom is the central atom with the O atoms around it. Connecting four O atoms uses eight electrons: 32 e– – 8 e– = 24 e–. Complete the octets of all the O atoms using 24 electrons. 24 e– – 24 e– = 0 e–. No more electrons are available. The central atom, Si, has eight electrons in shared pairs, so it needs no more. The structure is an ion, so put it in brackets and add the 4– charge:


O Si O

O

O

.. ....

... ...

4G. .....

. ...

..

(c) The atoms Cl and F are both in Group 7A. The + charge removes one electron: 7 + 4 × (7) – 1 = 34 e–. The Cl atom is the central atom with the F atoms around it. Connecting four F atoms uses eight electrons: 34 e– – 8 e– = 26 e–. Complete the octets of all the F atoms, using 24 electrons: 26 e– – 24 e– = 2 e– . Put the last two electrons on the Cl. The central atom, Cl, has ten electrons already, so it needs no more. The structure is an ion, so put it in brackets and add the + charge:

F Cl F

F

F

.. ....

. .... .

... ... . ...

....

+

(d) The C atom is in Group 4A and the H atoms are in Group 1A: 2 × (4) + 6 × (1) = 14 e–. As described in Section 3.3, the ethane molecule has three H atoms bonded to each C, and the two carbons bonded to each other. Connecting eight atoms uses 14 electrons: 14 e– – 14 e– = 0 e– No more electrons are available. Each C has eight electrons in shared pairs, so they need no more. We get the following structure:

C C

H

H

H

H

H

H

Check your answers: All the Period 2 elements have an octet of electrons. The H atoms have two electrons. All the Period 3 elements have eight or more electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.

17. Define the problem: Write Lewis structures for a list of molecules.

Develop a plan: Follow the systematic plan given in the answers to Question 15. Use what you learned about organic molecules in Chapter 3 to determine how atoms are bonded in the organic structures.

Execute the plan:

(a) 2 × (1) + 4 + 2 × (6) = 18 e–. Connecting these five atoms with single bonds uses eight electrons: 18 e– – 8 e– = 10 e–. Complete the octets of the oxygen atoms using 10 electrons: 10 e– – 10 e– = 0 e– . At this point, we have the following structure:


.. ....

H C O

O

H....

The central atom, C, has only six electrons in shared pairs, so it needs two more, but there are no more electrons available for lone pairs. Therefore, we must move a lone pair of electrons from one of the O atoms to make a new shared pair, forming a double bond to the C atom. The atomic arrangement given in the problem tells us to put the multiple bond on the top oxygen:

H C O

O

H....

.. ..

The central atom, C, now has eight electrons in shared pairs and this is the proper Lewis Structure.

(b) 2 × (4) + 3 × (1) + 5 = 16 e–. As described in Section 3.1, the structural formula given here tells us that this organic compound has three H atoms and a C atom bonded to C, and another N atom bonded to the C. Connecting six atoms uses ten electrons: 16 e– – 10 e– = 6 e–. For the purposes of this method, let's call the second C the central atom, since the eight bonding electrons on the first carbon satisfy the octet rule for that C atom. H needs no more electrons. Complete the octet of the N atom using six electrons: 6 e– – 6 e– = 0 e– . At this point, we have the following structure:

H C C

H

H

N. .....

The second C atom has only four electrons in shared pairs, so it needs four more, but there are no more electrons available for lone pairs. Therefore, we must move two lone pairs of electrons from the N atom to make two new shared pairs, forming a triple bond to the C atom:

H C C

H

H

N..

The second C atom now has eight electrons in shared pairs and this is the proper Lewis Structure.

(c) 2 × (4) + 3 × (1) + 7 = 18 e–. As described in Section 3.1, the structural formula given here tells us that this organic compound has two H atoms and a C atom bonded to the first C, and an H atom and a Cl atom bonded to the second C. Connecting six atoms uses 10 electrons: 18 e– – 10 e– = 8 e–. For the purposes of this method, let's call the first C the central atom. H needs no more electrons. Complete the octet of the Cl atom and on the second C atom, using eight electrons: 8 e– – 8 e– = 0 e– . At this point, the structure looks like this:


H C C

H

Cl

H. .......

The first C atom has only six electrons in shared pairs, so it needs two more, but there are no more electrons available for lone pairs. Therefore, we must move one lone pairs of electrons from the C atom to make a new shared pair, forming a double bond to the C atom:

H C C

H

Cl

H. .....

The first C atom now has eight electrons in shared pairs and this is the proper Lewis Structure.

Check your answers: All the Period 2 and 3 elements have an octet of electrons. The H atoms have two electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.


Develop a plan: Follow the systematic plan given in the answers to Question 16. Use what you learned about organic molecules in Chapter 3 to determine how atoms are bonded in the organic structures.

Execute the plan:

(a) 2 × (4) + 4 × (7) = 36 e–. This organic molecule looks like ethene with all the H atoms changed to F atoms. The two carbon atoms are bonded to each other, and each has two F atoms bonded to it. Any other arrangement would make F–F bonds. F has the highest electronegativity (see Section 8.7) and won't bond with another F atom if any other element is present. Connecting six atoms with single bonds uses 10 electrons: 36 e– – 10 e– = 26 e–. For the purposes of this method, let's call the first C the central atom. Complete the octets of the F atoms and the second C atom, using 26 electrons: 26 e– – 26 e– = 0 e– . At this point, the structure looks like this:

F C C

F

F

F

... .....

...... ......... ...

The first C atom has only six electrons in shared pairs, so it needs two more, but there are no more electrons available for lone pairs. Therefore, we must move one lone pair of electrons from the C atom to make a new shared pair, forming a double bond to the C atom:

F C C

F

F

F

. ...

..

...... ......

... ...

The first C atom now has eight electrons in shared pairs and this is the proper Lewis Structure.


(b) 3 × (4) + 3 × (1) + 5 = 20 e–. As described in Section 3.1, the structural formula given here tells us that this organic compound has two H atoms and a C atom bonded to the first C atom, an H atom and a C atom bonded to the second C atom, and an N atom bonded to the third C atom. Connecting seven atoms uses 12 electrons: 20 e– – 12 e– = 8 e–. For the purposes of this method, let's call the second and third C atoms "central atoms" so we can start from the ends and work toward the middle. H needs no more electrons. Complete the octet of the N atom and the octet of the first carbon using eight electrons: 8 e– – 8 e– = 0 e– . At this point, we have the following structure:

H C C

H

C

H

N. .......

The second C atom has only six electrons in shared pairs and the third C atom has only four electrons in shared pairs, but there are no more electrons available. Therefore, we must move one lone pair of electrons from the first C atom to make a new shared pair between the first and second C atoms, making a double bond. We must also move two lone pairs of electrons from the N atom to the third C atom to make two new shared pairs, forming a triple bond between the N atom and C atom:

H C C

H

C

H

N..

The second and third C atoms now both have eight electrons in shared pairs and this is the proper Lewis Structure.

Check your answers: All the Period 2 elements have an octet of electrons. The H atoms have two electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.


Develop a plan: Try to follow the systematic plan given in the answers to Question 16. For (a) and (b), refer to Section 8.10 for guidance drawing structures that do not follow the octet rule.

Execute the plan:

(a) 2 × (5) – 1 = 9 e–. An odd number of electrons means that we will not be able to follow the octet rule. Connecting the two nitrogens with single a bond uses 2 electrons: 9 e– – 2 e– = 7 e–. For the purposes of this method, let's call the first N the central atom. Complete the octet of the second N atom: 7 e– – 6 e– = 1 e–. Put the last electron on the first N atom

N N.

. .. ...

The first N atom has only three electrons, so it needs more, but there are no more electrons available for lone pairs. We can’t get to eight, but we can get to seven, if we move two lone pair of electrons from the second N atom to make new shared pair, forming a triple bond to the N atom:

N N ... +


This is the closest to a proper Lewis Structure we can make with an odd number of electrons.

(b) 8 + 7 × (7) + 1 = 58 e–. Use 14 of the electrons to connect the seven F atoms to the central Xe atom. Use 42 more of them to fill the octets of the F atoms. Put the last two on Xe.

Xe

FF F

FF

F F..

.

.

...... ..

. .. .....

..

.

... ..

....

.

. ... . ..

....

This is the closest to a proper Lewis Structure we can make with seven atoms bonded to the same central atom.

(c) The name given here tells us that this organic compound with a two-carbon chain. Tetra means four, and “cyano” refers to –CN. So we’ll put the atoms in this an atomic arrangement:

N C C C C N

C C

N N

6 × (4) + 4 × (5) = 44 e–. Connecting ten atoms uses 18 electrons: For the purposes of this method, let's call the second and third C atoms "central atoms" so we can start from the ends and work toward the middle. Complete the octet of the N atom using 24 electrons: That leaves two electorns. Put them on one of the central C atoms. At this point, we have the following structure:

N C C C C N

C C

N N

...... .

.

. ....

...

..

..

......

The C atoms bonded to N have only four electrons in shared pairs, so we must move two lone pair of electrons from each N to its neighboring C atom to make new shared pairs between the atoms, making a triple bonds. The two “central” C atoms also need to share the last lone pair to complete their octets.

N C C C C N

C C

N N

....

. ...

.. ..

Check your answers: All the Period 2 elements have an octet of electrons (or less as was necessary in the first structure). All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.



Develop a plan: Try to follow the systematic plan given in the answers to Question 16. For (c), refer to Section 8.10 for guidance drawing structures that do not follow the octet rule.

Execute the plan:

(a) 2 × (6) + 2 × (7) = 26 e–. The two sulfur atoms will be bonded to each other with one Cl atom attached to each S atom. We use six electrons to connect the atoms with single bonds, fill the octets of the Cl atoms using 12 more electrons then finish the octets of the S atoms with the remaining eight electrons.

Cl S S Cl ... ..

. ..... .

.... ..

.

.

(b) 1 + 5 + 2 × (6) = 18 e–. Since the structure has no O–O bonds, the atomic arrangement must be H–O–N–O. Six electrons are needed to attach the atoms to each other. Theright O atom needs six more electrons, and the left O atom needs four, leaving one pair of electrons that go on the N atom, producing this structure:

H O N O... ...

.

...

..

The N atom has only six electrons, but there are no more electrons available. Therefore, we must move two lone pairs of electrons from the O atom to the N atom. This can happen one of two ways.

H O N O....

..... .

..H O N O. .....

..

Later in Chapter 8, using formal charges, we will find that the left structure is the better structure.

H O N O....

..... .

(c) 7 + 6 × (7) - 1 = 48 e–. Use 12 of the electrons to connect the six F atoms to the central Cl atom. Use 36 more of them to fill the octets of the F atoms. Put the last two on Xe.

Cl

FF F

F FF

. ... ............

.

.

.

. .. ..

.. ....

....

..

+

This is the closest to a proper Lewis Structure we can make with six atoms bonded to the same central atom.

Check your answers: All the Period 2 elements have an octet of electrons. The H atoms have two electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. The answer to (b) is confirmed at the beginning of Section 8.4.

21. Define the problem: Determine if given Lewis structures are correct and explain what is wrong with the incorrect ones.

Develop a plan: A Lewis structure is correct if all valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs; if all the Period 2 elements have an octet of electrons; if


the H atoms have two electrons; and if all the Period 3 and higher elements have eight or more electrons. So, first count the electrons in the structure and compare that with the total number of valence electrons. If the count is correct, then check the octets of all the elements and check H atoms for two electrons. Only elements in period 3 and higher are allowed to exceed eight electrons.

Execute the plan:

(a) N2

Check electron count: 2 × (5) = 10 e–. The given structure also has 10 electrons (three lone pairs and a double bond), so the total electron count is okay.

Check octets: The first N atom doesn't have an octet. It only has six electrons. The second N atom does have eight electrons. So the octet rule is violated.

This structure is incorrect.

(b) NCl3

Check electron count: 5 + 3 × (7) = 26 e–. The given structure also has 26 electrons (10 lone pairs and three single bonds), so the total electron count is okay.

Check octets: The N atom and all three Cl atoms each have eight electrons, so the octet rule is satisfied.

This structure is correct.

(c) ClO3–

Check electron count: 7 + 3 × (6) + 1 = 26 e–. The given structure has 25 electrons (9 lone pairs, one unpaired electron, and three single bonds), so the total electron count is wrong.


(d) (CH3)2O

Check electron count: 6 + 2 × (4) + 6 × (1) = 20 e–. The given structure also has 20 electrons (two lone pairs and eight single bonds), so the total electron count is okay.

Check octets: The O atom and both C atoms each have eight electrons, so the octet rule is satisfied. The H atoms all have two electrons.


(e) NH4+

Check electron count: 5 + 4 × (1) – 1 = 8 e–. The given structure has 10 electrons (1 lone pair and four single bonds), so the total electron count is wrong.


The incorrect structures are (a) N2, (c) ClO3–, and (e) NH4

+.

Check your answers: We used what we know about Lewis structures to determine the incorrect structures. Let's draw correct Lewis structures for those that were incorrect.


(a) For N2, another pair of electrons from the second N atom must be shared between them for them to both obey the octet rule.

N N....

(c) The correct structure for ClO3– has one more electron on Cl (and brackets with a charge):

O Cl O

O.. ....

... ... . ...

....

G

(e) The correct structure for NH4+

has two fewer electrons on N (and brackets with a charge):

H N H

H

H

+

The corrected structures are different from the incorrect ones, so these answers look right.

22. Define the problem: Determine if given Lewis structures are correct and explain what is wrong with the incorrect ones.

Develop a plan: A Lewis structure is correct if it has the correct type and number of atoms, if all valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs; if all the Period 2 elements have an octet of electrons; if the H atoms have two electrons; and if all the Period 3 and higher elements have eight or more electrons. So, first count the atoms and compare with the formula. Then count the electrons in the structure and compare that with the total number of valence electrons. If the count is correct, then check the octets of all the elements and check H atoms for two electrons. Only elements in period 3 and higher are allowed to exceed eight electrons.

Execute the plan:

(a) OF2

Check electron count: 6 + 2 × (7) = 20 e–. The given structure has 8 electrons (two lone pairs and two single bonds), so the total electron count is wrong.


(b) O2

Check electron count: 2 × (6) = 12 e–. The given structure has 10 electrons (two lone pairs and a triple bond), so the total electron count is wrong.


(c) CCl2O

Check electron count: 4 + 2 × (7) + 6 = 24 e–. The given structure also has 24 electrons (eight lone pairs, two single bonds, and one double bond), so the total electron count is okay.


Check octets, etc.: The O atom, the C atom, and both Cl atoms each have eight electrons, so the octet rule is satisfied.


(d) CH3Cl

Check type and number of atoms: The structure has only two H atoms, not three.

Check electron count: 4 + 3 × (1) + 7 = 14 e–. The given structure has 16 electrons (four lone pairs, and four single bonds), so the total electron count is wrong.

Check octets, etc.: The structure also has one hydrogen atom with too many electrons (4, not 2).


(e) NO2–

Check electron count: 5 + 2 × (6) + 1 = 18 e–. The given structure has 16 electrons (five lone pairs, one single bond, and one double bond), so the total electron count is wrong.


The incorrect structures are (a) OF2, (b) O2, (d) CH3Cl, and (e) NO2–.

Check your answers: We used what we know about Lewis structures to determine the incorrect structures. Let's draw correct Lewis structures for those that were incorrect.

(a) The correct structure for OF2 needs 12 more electrons in the form of three more lone pairs on each F atom:

..F... ... ..O

.

... ..F

....

(b) The correct structure for O2 needs 2 more electrons, two more lone pairs, and one less shared pair:

O O.... .. ..

(d) The correct structure has two fewer electrons and the carbon bonded to the chlorine without an H atom between them. Remember, H atom only shares two electrons:

H.. .... ..

.H

HC .. .

..Cl

(e) The correct structure has two more electrons on N (and brackets with a charge):

..O N O.... ..

.. ..G

The corrected structures are different from the incorrect ones, so these answers look right.

Bonding in Hydrocarbons

23. Define the problem: Write structural formulas for all the branched-chain compounds with a given formula.


Develop a plan: Be systematic. Start with a long chain that has only one methyl branch. Move the methyl around (but don't put it on the end carbon and don't put it on a carbon past the first half of the chain). Then make two methyl branches, and move them around similarly. Continue this process until the main chain is too short for methyl branches. Then make an ethyl branch and move it around (but don't put it on the end carbon or the carbon next to the end carbon and don't put it on a carbon past the first half of the chain). Follow a similar pattern with ethyls as with methyls.

Execute the plan: One methyl branch can go on the second carbon in the chain.

HCCC

H

H

HH

H

H

C H

H

H

This is the only branched hydrocarbon with the formula C4H10, because making a second methyl branch leaves the main chain with only two carbons and we can't put the methyl branches on the end carbon.

Check your answer: The molecule has a branch off of a three-carbon chain. It also has four C atoms and ten H atoms. This answer makes sense.

24. Define the problem: Write structural formulas for all the branched-chain compounds with a given formula.

Develop a plan: Be systematic. Start with a long chain that has only one methyl branch. Move the methyl around (but don't put it on the end carbon and don't put it on a carbon past the first half of the chain). Then make two methyl branches, and move them around similarly. Continue this process until the main chain is too short for methyl branches. Then make an ethyl branch and move it around (but don't put it on the end carbon or the carbon next to the end carbon and don't put it on a carbon past the first half of the chain). Follow a similar pattern with ethyls as with methyls.

Execute the plan: The straight chain isomer of C6H14 has six carbons, so start with a five-carbon chain.

One methyl branch can go on the five-carbon chain in two different ways. The methyl branch can go on the second carbon or on the third carbon.

C C C C

H

C

H C C

H

H

H H

H H

C

H

H

HH

H

C

H

C

H

H

H

H

H

H

H H

C

H

H

HC H

H

H

Two methyl branches can go on the four-carbon chain in two ways. They can both go on the second carbon or one can go on the second carbon and one can go on the third carbon.


HCCCC

H

H

H

H

H

H

H

C H

H

H

C HH

H

Three methyl branches cannot all three be attached to the one middle carbon in a three-carbon chain, so we're done with methyl branches.

We could try to make a four carbon chain with an ethyl branch. However, ethyl branches can't be attached to chain-carbons within two carbons of the end of the chain. (If you do that, the molecules "longest" chain will actually include your branch!) A four-carbon chain isn't long enough and therefore we can't use ethyl branches. So we have found all the branched isomers of the hydrocarbon with the formula C6H14.

Check your answers: The molecules have one or more branches off the three-carbon chain. They also all have six C atoms and 14 H atoms. These answers makes sense.

25. Define the problem: Write structural formulas for all the straight-chain alkenes with a given formula.

Develop a plan: Be systematic. Start with a long chain with one double bond. Move the double bond around (but don't go past the first half of the chain).

Execute the plan: C5H10

The double bond can go between the first and second carbons or between the second and third carbons in the five-carbon chain.

HCCCCC

H

H

H

H

H

HHH

H

HCCCCC

H

H

H

HHH

H

H

H

There are more than two structures that meet these specifications, because the second structure above can have cis- and trans-isomers:

cis-isomer (methyl and trans-isomer (methyl and


ethyl on the same side) ethyl on the opposite sides)

HCC

CC

C

H

H

H

H

HH

H

H

H

HCC

CC

C

H

H

H

H

H

H

H

H

H

Check your answers: The molecules have one straight five-carbon chain, with one double bond. They also have five C atoms and ten H atoms. These answers make sense.

26. Define the problem: Given formulas of hydrocarbons determine if they are alkane, alkene, or alkyne.

Develop a plan: If we assume that each of these straight chain molecules has at most one multiple bond, we can use the formula pattern to determine the class. The formulas of alkanes are CnH2n+2 , for n = 1 and higher. Each multiple bond removes two electrons. The formulas of alkenes are CnH2n, for n = 2 and higher. The formulas of alkynes are CnH2n–2, for n = 2 and higher. Using the number of C atoms, set n. Then figure out 2n+2, 2n, or 2n–2. Compare these numbers to the number of H atoms. Form a conclusion based on the that comparison.

Execute the plan:

(a) C5H8 n = 5, with this n, 2n+2 = 12, 2n = 10, 2n–2 = 8. This hydrocarbon is an alkyne. Note, that it could also be an alkene, if there are two double bonds in the molecule.

(b) C24H50 n = 24, with this n, 2n+2 = 50, 2n = 48, 2n–2 = 46. This hydrocarbon is an alkane.

(c) C7H14 n = 7, with this n, 2n+2 = 16, 2n = 14, 2n–2 = 12. This hydrocarbon is an alkene.

Check your answers: One and only one of the three calculations using n gave a matching number. These answers makes sense.

27. Define the problem: Given formulas of hydrocarbons, determine if they are alkane, alkene, or alkyne.

Develop a plan: If we assume that each of these straight chain molecules has at most one multiple bond, we can use the formula pattern to determine the class. The formulas of alkanes are CnH2n+2 , for n = 1 and higher. Each multiple bond removes two electrons. The formulas of alkenes are CnH2n, for n = 2 and higher. The formulas of alkynes are CnH2n–2, for n = 2 and higher. Using the number of C atoms, set n. Then figure out 2n+2, 2n, or 2n–2. Compare these numbers to the number of H atoms. Form a conclusion based on the that comparison.

Execute the plan:

n 2n+2 (alkane)

2n (alkene)

2n–2 (alkyne)

*predict

(a) C21H44 21 44* 42 40 alkane

(b) C4H6 4 10 8 6* alkyne†


(c) C8H16 8 18 16* 14 alkene

† Note: (b) could also be an alkene, if there are two double bonds in the molecule, i.e., CH2=CHCH=CH2.

Check your answers: One and only one of the three calculations using n gave a number matching the number of H atoms in the molecule. These answers make sense.

28. Define the problem: Given the name of a hydrocarbon, write cis- and trans-isomers for it.

Develop a plan: Use the prefix to determine number of carbons, and the numeral provided in the name to determine where the double bond starts along the chain. Add H atoms to complete the octets. Look at the fragments attached at the double bond put the two larger ones on the same side for the cis-isomer, and on opposite sides for the trans-isomer.

Execute the plan: 2-pentene has a five-carbon chain with the double bond starting at the second carbon.

cis-2-pentene (methyl and trans-2-pentene (methyl and

ethyl on the same side) ethyl on the opposite sides)

HCC

CC

C

H

H

H

H

HH

H

H

H

HCC

CC

C

H

H

H

H

H

H

H

H

H

Check your answers: These are the same isomers discovered when answering Question 25. The structure for cis-2-pentene given in Question 30 is consistent with this structure. These answers makes sense.

29. Define the problem: Given the condensed structural formulas for some organic compounds, determine if cis- and trans-isomers exist.

Develop a plan: In this Question we will first look for a double bond. If there is no double bond, there can be no cis- and trans-isomerism. Identify the fragments attached to each carbon of the double bond. If both carbons have different two fragments, then cis and trans-isomers can exist. If either of the carbons has identical fragments, then there is no cis- and trans-isomerism.

Execute the plan:

(a) Br2CH2 has no double bond, so cis- and trans-isomer do not exist.

(b) CH3CH2CH=CHCH2CH3

CH3CH2

CC

CH2

HH

CH3


CH3CH2 is different from H on both carbons, so cis- and trans-isomer do exist. Shown above is the cis-isomer, and below is the trans-isomer:

CH3CH2

CC

CH2CH3 H

H

(c) CH3CH=CHCH3

CH3

CC

CH3

HH

CH3 is different from H on both carbons, so cis- and trans-isomer do exist. Shown above is the cis-isomer, and below is the trans-isomer:

CH3

CC

CH3 H

H

(d) CH2=CHCH2CH3

CH3CH2

CC

H

HH

On the first carbon, H is not different from H, so cis- and trans-isomer do not exist.

Check your answers: The isomers shown in (b) and (c) are the cis-isomers. Switching the fragments on the left carbon, gives the trans-isomer. If that is done in (d) nothing changes. These answers makes sense.

30. Define the problem: Given some isomers' structures or structural formulas, determine the other cis-or trans-isomers.

Develop a plan: Switching the positions of the fragments on the left carbon, gives the other isomer.

Execute the plan:

(a) Switch the positions of the H and the Cl on the left carbon to make cis-1,2-dichloropropene. (It’s cis- because the two Cl atoms are on the same side.)

CH3CC

ClCl

H

(b) Switch the positions of the H and the CH3 on the left carbon to make trans-2-pentene. (It’s trans- because the ethyl and methyl fragments are opposite sides.)


HCCCC

C

H

H

H

HH

H

H H

H

(c) Switch the positions of the H and the CH3 on the left carbon to make trans-3-hexene. (It’s trans- because the two ethyl fragments are on opposite sides.)

HCCCC

CC

H

H

H

HH

H

H

H

H

H

H

(d) Switch the positions of the H and the CH3 on the left carbon to make cis-2-hexene. (It’s cis- because the methyl and propyl fragments are on the same side.)

H C C CH2 CH2 CH3

CH3

H

Check your answers: We've seen some of these isomers in previous examples. The particular isomers formed can be correlated to the name change. These answers makes sense.

31. Define the problem: Given the condensed structural formulas for some organic compounds, determine if cis- and trans-isomers exist. For those that do, write the structural formula for the isomers and label them. For those that don't, explain why.

Develop a plan: Identify the fragments attached to each carbon of the double bond. If both carbons have different two fragments, then cis and trans-isomers can exist. If either of the carbons has identical fragments, then there is no cis- and trans-isomerism.

Execute the plan:

(a) The fragments on both carbons of CH3CH2BrC=CBrCH2CH3 are ethyl and Br. So, cis- and trans-isomers can exist.

cis-isomer (Br atoms on the same side.)


HCCCCCC

H

H

H

HBrBr

H

H

H

H

H

trans-isomer

(Br atoms on opposite sides.)

HCCCC

CC

H

H

H

HBr

Br

H

H

H

H

H

CH3CH2 is different from Br on both carbons, so cis- and trans-isomers do exist.

(b) The fragments on both carbons of (CH3)2C=C(CH3)2 are all methyl. So, cis- and trans-isomers cannot exist.

CH3

CC

CH3

CH3CH3

Switching the fragments on the left carbon results in no change.

(c) The fragments on both carbons of CH3CH2IC=CICH2CH3 are ethyl and I. So, isomers can exist.

cis-isomer (I atoms on the same side.)

HCCCCIC

H

H

H

HII

H

H

H

H

H

trans-isomer

(I atoms on opposite sides.)

HCCCC

CC

H

H

H

HI

I

H

H

H

H

H


(d) The fragments on the left carbon of CH3ClC=CHCH3 are methyl and Cl. The fragments on the left carbon of CH3ClC=CHCH3 are methyl and H. So, isomers can exist.

cis-isomer (Cl atom and trans-isomer (Cl atom and methyl on the same side.) methyl on opposite sides.)

HCCC

C

H

HH

Cl

H H

H

HCCCC

H

HH

Cl

HH

H

(e) The fragments on the left carbon of (CH3)2C=CHCH3 are both methyl. So, isomers cannot exist.

C C

H

CH3

CH3

CH3

Switching the fragments on the left carbon results in no change.

Check your answers: Appropriate checking has been shown in answering the question. Those with isomers have their isomers shown, and those without are shown not to have isomers. These answers make sense.

Bond Properties

32. It is not possible for oxalic acid to have cis-trans isomerization since the double bonds are only to the oxygen atoms and the C–C bond is a single bond. That means there is free rotation about the C–C bond can occur.

33. It is not possible for 2-methyl propene to have cis-trans isomerization since the left carbon has two methyl groups and the right carbon has two hydrogens.

34. Define the problem: Given a series of pairs of bonds, predict which of the bonds will be shorter.

Develop a plan: Use periodic trends in atomic radii to identify the smaller atoms. The bond with the smaller atoms, will have a shorter bond. In cases where bonds between the same atoms are compared, triple bonds are shorter than double bonds, which are shorter than single bonds.

Execute the plan:

(a) Both bonds have Cl, so compare the sizes of B and Ga. B is smaller than Ga. (It is higher in the same group of the periodic table.) So, B–Cl is shorter than Ga –Cl.

(b) Both bonds have O, so compare the sizes of C and Sn. C is smaller than Sn. (It is higher in the same group of the periodic table.) So, C–O is shorter than Sn–O.


(c) Both bonds have P, so compare the sizes of O and S. O is smaller than S. (It is higher in the same group of the periodic table.) So, P–O is shorter than P–S.

(d) Both bonds have C, so compare the sizes of C and O. O is smaller than C. (It is further to the right in the same period of the periodic table.) So, C=O is shorter than C=C.

Check your answers: Several of these predictions are confirmed in Table 8.1.

35. Define the problem: Given a series of pairs of bonds, predict which of the bonds will be shorter.

Develop a plan: Use periodic trends in atomic radii to identify the smaller atoms. The bond with the smaller atoms, will have a shorter bond. In cases where bonds between the same atoms are compared, triple bonds are shorter than double bonds, which are shorter than single bonds.

Execute the plan:

(a) First, compare the sizes of N and O. O is smaller than N. (It is further to the right in the same period of the periodic table.) Second, compare the sizes of Si and P. P is smaller than Si. (It is further to the right in the same period of the periodic table.) So, P–O is shorter than Si–N.

(b) Both bonds have O, so compare the sizes of Si and C. C is smaller than Si. (It is higher in the same group of the periodic table.) So, C–O is shorter than Si–O.

(c) Both bonds have C, so compare the sizes of F and Br. F is smaller than Br. (It is higher in the same group of the periodic table.) So, C–F is shorter than C–Br.

(d) First compare the length of carbon-carbon triple and double bonds. C≡C is shorter than C=C because triple bonds are shorter than double bonds. Next compare the sizes of N and C. N is smaller than C. (It is further

to the right in the same period of the periodic table.) So, C≡N is shorter than C≡C. Therefore, C≡N is shorter than C=C.

Check your answers: Several of these predictions are confirmed in Table 8.1.

36. Define the problem: Given some bonds and only a periodic table, predict which of the bonds will be strongest.

Develop a plan: Ionic bonds are stronger than polar covalent bonds. Polar covalent bonds are stronger than purely covalent bonds. Use the periodic trend for electronegativity (EN):

smaller larger EN EN

larger EN

smaller EN

The larger the difference in electronegativity, the stronger the bond will be.

Execute the plan: (a) Si–F, (b) P–S, (c) P–O


Looking up these atoms on the periodic table, we find that Si has lower electronegativity than P. (It is further to the left in the same period of the periodic table.) F has higher electronegativity than O. (It is further to the left in the same period of the periodic table.) So the Si–F bond has the largest electronegativity difference; hence, it is the strongest.

Check your answers: These predictions are confirmed in Figure 8.6.

37. Define the problem: Given two chemical formulas, predict which has the shorter nitrogen-nitrogen bond and which has the stronger nitrogen-nitrogen bond.

Develop a plan: First, write resonance structures for the two molecules. In cases where bonds between the same two atoms are compared, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. If more than one resonance structure is possible, average their contributions.

Execute the plan: There is only one Lewis structure for hydrazine, N2H4, with 14 valence electrons.

H N

H

N

H

H....

In hydrazine, the nitrogen-nitrogen bond is a single bond.

There are three plausible resonance structures for laughing gas, N2O, with 12 valence electrons.

In N2O, the average of the three resonance structures indicates that the bond between N atoms is approximately a double bond.

That means the shortest, strongest bond is the bond in N2O.

Check your answers: The Lewis structures obey the octet rule and have the right number of valance electrons. These predictions are upheld in general with values given in Tables 8.1 and 8.2.

38. Define the problem: Given two chemical formulas, predict which has the shorter carbon-oxygen bond.

Develop a plan: First, write Lewis structures for the two molecules. In cases where bonds between the same two atoms are compared, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.

Execute the plan: There is only one plausible Lewis structure for formaldehyde, H2CO, with 12 valence electrons.

H C O...

.

H

In formaldehyde, the carbon-oxygen bond is a double bond.

There is only one plausible Lewis structure for carbon monoxide, CO, with 10 valence electrons.

C O.. ..

..

.. ....

..

.. ....N N O N N O N N O.. .. .. ..


In carbon monoxide, the carbon-oxygen bond is a triple bond.

That means the shorter bond is the bond in CO.

Check your answers: The Lewis structures obey the octet rule and have the right number of valance electrons. These predictions are upheld in general with values given in Table 8.1.

39. Define the problem: Given two chemical formulas, predict which carbon-oxygen bond is harder to break.

Develop a plan: First, write Lewis structures for the two molecules. In cases where bonds between the same two atoms are compared, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.

Execute the plan: There is only one plausible Lewis structure for formaldehyde, H2CO, with 12 valence electrons.

H C O...

.

H

In formaldehyde, the carbon-oxygen bond is a double bond.

There is only one plausible Lewis structure for carbon monoxide, CO, with 10 valence electrons.

C O.. ..

In carbon monoxide, the carbon-oxygen bond is a triple bond.

That means the stronger bond is the bond in CO and this bond will require more energy to break.


40. Define the problem: Given two chemical formulas, predict which has the shorter carbon-oxygen bond.

Develop a plan: First, write resonance structures for the two molecules. In cases where bonds between the same two atoms are compared, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. If more than one resonance structure is possible, average their contribution.

Execute the plan: Consider formate ion first. Two equivalent plausible Lewis structures exist for HCO2–

with 18 valence electrons:

In the formate ion, one resonance structure has a single carbon-oxygen bond and one has a double carbon-oxygen bond. We predict that the carbon-oxygen bond will be halfway between a single and double bond.

There are three plausible equivalent Lewis structures for carbonate ion, CO32–

, with 24 valence electrons:

H C O

O

H C O

O GG.. ......

..

.. .... ....

O C O

O

O C O

O

O C O

O 2G2G2G .. ........

......

........

.. ....

...... ....

.. ....


In the carbonate, one resonance structure has a double carbon-oxygen bond and two have single carbon-oxygen bond. We predict that the carbon-oxygen bond will be closer to a single bond than a double bond.

That means the shortest bond is the bond in HCO2–.


41. Define the problem: Given two chemical formulas, predict which has the shorter nitrogen-oxygen bond.

Develop a plan: First, write resonance structures for the two molecules. In cases where bonds between the same two atoms are compared, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. If more than one resonance structure is possible, average their contributions.

Execute the plan: Three plausible Lewis structures exist for NO2+ with 16 valence electrons.

Looking at the resonance structures for NO2+, we predict that the nitrogen-oxygen bond will be

approximately a double bond.

There are three plausible equivalent Lewis structures for NO3– with 24 valence electrons.

In the nitrate ion, one resonance structure has a double nitrogen-oxygen bond and two have single nitrogen-oxygen bonds. We predict that the nitrogen-oxygen bond will be closer to a single bond than a double bond.

That means the shorter bond is the bond in NO2+

Check your answers: The Lewis structures obey the octet rule and have the right number of valance electrons. These predictions are upheld in general with values given in Table 8.1. Bond Energies and Enthalpy Changes

42. Define the problem: Given a description of a chemical equation for a reaction and a table of bond energies, estimate the standard enthalpy change of the reaction and determine whether the reaction is exothermic or endothermic.

Develop a plan: First, balance the chemical equation. Then use another variation of Hess’s Law to estimate ∆H°. We will break all the bonds in the reactants (by putting their bond energies into the system) and then we will form all the bonds – the opposite of breaking – in the products (by removing their bond energy from the system). That is the logic behind the equation given in Section 8.6.

.. ....

..

.. ....O N O O N O O N O +++.. .. .. ..

O N O

O

O N O

O

O N O

O GGG .. ........

......

........

.. ....

...... ....

.. ....


∆H° = (moles of bond ) × D(bond broken)[ ]∑ − (moles of bond ) × D(bond formed )[ ]∑

Get the balanced equation and use it to describe the moles of each of the reactants and products. Count the moles of bonds of each type that break and form. Set up a specific version of the above equation. Look up the D for each bond in Table 8.2, plug them into the equation and solve for ∆H°.

Execute the plan: The reactants are nitrogen, N2, and hydrogen, H2. The product is ammonia, NH3. The balanced equation is

N2 + 3 H2 2 NH3

N N.. ..H H

H H

H H N H

H

H

..N H

H

H

.. ++

+

+

+N N.. ..

That means we break one mole of N≡N bond and three moles of H–H bonds We form six moles of N–H bonds:

∆H° = (1 mol of N≡N) × DN≡N + (3 mol H–H) × DH–H – (6 mol N–H) × DN–H

Look up the D values in Table 8.2.

∆H° = (1 mol) × (946 kJ/mol) + (3 mol) × (436 kJ/mol) – (6 mol) × (391 kJ/mol)

∆H° = –92 kJ

This reaction is exothermic.

Check your answer: The reaction is twice the formation reaction for NH3. Appendix J tells us that NH3 has

∆H fo

= –46.11 kJ/mol. Twice this value produces a ∆H° = –92.22 kJ/mol. This is very close to the estimate calculated here.

43. Define the problem: Given a description of a chemical equation for a reaction and a table of bond energies, estimate the standard enthalpy change of the reaction and determine whether the reaction is exothermic or endothermic.

Develop a plan: First, balance the chemical equation. Then use another variation of Hess’s Law to estimate ∆H°. We will break all the bonds in the reactants (by putting their bond energies into the system) and then we will form all the bonds – the opposite of breaking – in the products (by removing their bond energy from the system). That is the logic behind the equation given in Section 8.6.

∆H° = (moles of bond) × D(bond broken)[ ]∑ − (moles of bond) ×D(bond formed)[ ]∑

Get the balanced equation and use it to describe the moles of each of the reactants and products. Count the moles of bonds of each type that break and form. Set up a specific version of the above equation. Look up the D for each bond in Table 8.2, plug them into the equation and solve for ∆H°.

Execute the plan: The reactants are 1 mole of carbon monoxide, CO, and molecular oxygen, O2. The


product is carbon dioxide, CO2. The balanced equation is

CO + 12

O2 CO2


At the molecular scale, we must do this reaction twice:

2 CO + O2 2 CO2

C O.. ..+

C O.. ..O O

.

...

.

...+

O C O..

..

.

...

O C O..

..

.

...

+

We can divide the equation by 2 at the mole-level, and break one mole of C≡O bond and half of a mole of O=O bonds. We form two moles C=O bonds:

∆H° = (1 mol of C≡O) × DC≡O + ( 12

mol O=O) × DO=O – (2 mol C=O) × DC=O


∆H° = (1 mol) × (1073 kJ/mol) + ( 12

mol) × (498 kJ/mol) – (2 mol) × (803 kJ/mol)

∆H° = –284 kJ

This reaction is exothermic.

Check your answer: The reaction is easily calculated using the formation reactions and ∆H fo

for CO and

CO2. Appendix J tells us that the ∆H fo

(CO) = –110.525 kJ/mol and ∆H fo

(CO2) = –393.509 kJ/mol. That gives this reaction ∆H° = –282.981 kJ/mol, very close to the estimate.

44. Define the problem: Given a list of molecules, a description of their chemical reactions and a table of bond energies, identify which has the strongest bond and estimate the standard enthalpy change of the reaction and which is the most exothermic.

Develop a plan: The strongest chemical bond has the largest bond energy. Balance the chemical equations for the reactions. Use the equations to describe the moles of each of the reactants and products. Count the moles of bonds of each type that break and form. Set up an equation as described in the answer to Questions 40. Look up the D for each bond in Table 8.2, plug them into the equation, and solve for ∆H°.

Execute the plan: DH–F = 566 kJ/mol, DH–Cl = 431 kJ/mol, DH–Br = 363 kJ/mol, and DH–I = 299 kJ/mol. The largest D is for the H–F bond so the strongest of the four bonds is the one in HF.

Using X to represent the halogen, the chemical reaction looks like this:

H2 + X2 2 HX

That means we break one mole of H–H bonds and one mole of X–X bonds. We form two moles H–X bonds:

∆H° = (1 mol of H–H) × DH–H + (1 mol X–X) × DX–X – (2 mol H–X) × DH–X


When X = F



= –538 kJ

When X = Cl


= –184 kJ

When X = Br


= –97 kJ

When X =I


= –11 kJ The H2 + F2 reaction is the most exothermic.

Check your answer: The reaction is twice the formation reaction for HX. Appendix J tells us that HF has

∆H fo

= –271.1 kJ/mol, HCl has ∆H fo

= –92.307 kJ/mol, and HBr has ∆H fo

= –36.40 kJ/mol. Twice these values produce ∆H° = –542.2 kJ/mol, ∆H° = –184.614 kJ/mol, and ∆H° = –72.80 kJ/mol. These are reasonably close to the estimates calculated here. The bromine value is farther away because Appendix J uses standard state liquid bromine, not gas-phase as assumed in the bond energy numbers. HI would be far off since the standard state of I2 is solid.

Electronegativity and Bond Polarity

45. Bonds are more polar when the electronegativity difference is larger. Look up electronegativity values for the atoms in these pairs.

ENC = 2.5, ENO = 3.5, ENN = 3.0, ENB =2.0,

ENP =2.1, ENS = 2.5, ENH = 2.1, ENI = 2.5

The partial negative (δ–) end of a bond is the atom with the larger electronegativity.

The partial positive (δ+) end of a bond is the atom with the smaller electronegativity.

(a) ∆ENC–O = ENO – ENC = 3.5 – 2.5 = 1.0 more polar

∆ENC–N = ENN – ENC = 3.0 – 2.5 = 0.5

C O δ+ δ-

(b) ∆ENB–O = ENO – ENB = 3.5 – 2.0 = 1.5 more polar

∆ENP–S = ENS – ENP = 2.5 – 2.1 = 0.4


B O δ+ δ-

(c) ∆ENP–H = ENH – ENP = 2.1 – 2.1 = 0.0

∆ENP–N = ENN – ENP = 3.0 – 2.1 = 0.9 more polar

P N δ+ δ-

(d) ∆ENB–H = ENH – ENB = 2.1 – 2.0 = 0.1

∆ENB–I = ENI – ENB = 2.5 – 2.0 = 0.5 more polar

B I δ+ δ-

46. (a) Look up electronegativity values (in Figure 8.6) for the atoms in these bonds.

ENC = 2.5, ENN = 3.0, ENH = 2.1, ENBr =2.8, ENS = 2.5, NO = 3.5

N is more electronegative in C–N,

C is more electronegative in C–H,

Br is more electronegative in C–Br, and

O is more electronegative in S–O.

(b) Bonds are more polar when the electronegativity difference is larger. Find ∆EN to determine which is most polar.

∆ENC–N = ENN – ENC = 3.0 – 2.5 = 0.5

∆ENC–H = ENC – ENH = 2.4 – 2.1 = 0.4

∆ENC–Br = ENBr – ENC = 2.8 – 2.5 = 0.3

∆ENS–O = ENO – ENS = 3.5 – 2.5 = 1.0 most polar

47. Bonds are more polar when the electronegativity difference is larger. Look up electronegativity values for the atoms in these pairs.

ENB = 2.0, ENCl = 3.0, ENO = 3.5, ENF = 4.0,

ENSe =2.4, ENS = 2.5, ENN = 3.0, ENH = 2.1

(a) ∆ENB–Cl = ENCl – ENB = 3.0 – 2.0 = 1.0

B Cl δ+ δ-

∆ENB–O = ENO – ENB = 3.5 – 2.0 = 1.5 B–O is more polar than B–Cl.


B O δ+ δ-

(b) ∆ENO–F = ENF – ENO = 4.0 – 3.5 = 0.5

O F δ+ δ-

∆ENO–Se = ENO – ENSe = 3.5 – 2.4 = 1.1 O–Se is more polar than O–F.

Se O δ+ δ-

(c) ∆ENS–Cl = ENCl – ENS = 3.0 – 2.5 = 0.5

S Cl δ+ δ-

∆ENB–F = ENF – ENB = 4.0 – 2.0 = 2.0 B–F is more polar than S–Cl.

B F δ+ δ-

(d) ∆ENN–H = ENN – ENH = 3.0 – 2.1 = 0.9

H N δ+ δ-

∆ENN–F = ENF – ENN = 4.0 – 3.0 = 1.0 N–F is more polar than N–H.

N F δ+ δ-

48. (a) Bonds are polar if the atoms’ electronegativities are different.

ENH = 2.1, ENN = 3.0, ENC = 2.5, ENO = 3.5

Since all the electronegativities are different, the bonds are all somewhat polar. None of the bonds are nonpolar.

(b) Bonds are more polar when the electronegativity difference is larger.

∆ENN–H = ENN – ENH = 3.0 – 2.1 = 0.9

∆ENC–N = ENN – ENC = 3.0 – 2.5 = 0.5

∆ENC–O = ENO – ENC = 3.5 – 2.5 = 1.0

The largest electronegativity difference is for the C=O bond, so it is the most polar.


C O δ+ δ-

The O atom is the partial negative end of this bond, because it has the larger electronegativity.


49. (a) Bonds are polar if the atoms’ electronegativities are different.

ENH = 2.1, ENC = 2.5, ENO = 3.5

Since all the electronegativities are different, the bonds are all somewhat polar. None of the bonds are nonpolar.

(b) Bonds are more polar when the electronegativity difference is larger.

∆ENC–H = ENC – ENH = 2.5 – 2.1 = 0.4

∆ENC–C = ENC – ENC = 2.5 – 2.5 = 0.0

∆ENC–O = ENO – ENB = 3.5 – 2.0 = 1.5

The largest electronegativity difference is for the C=O bond, so it is the most polar.

C O δ+ δ-

The O atom is the partial negative end of this bond, because it has the larger electronegativity.

Formal Charge

50. The total of the formal charges in a molecule must be zero, because molecules have no net charge. The total of the formal charges in an ion must be its ionic charge.

51. Shared electrons between two atoms are assigned to the more electronegative of the two elements. If non-zero formal charges must be present in a resonance structure, the higher-electronegativity element will have the more negative formal charge in the better structure(s).

52. Define the problem: Given the formulas of molecules or ions, write the correct Lewis structure and assign formal charges to each atom.

Develop a plan: Write the Lewis structures. Then determine the number of lone pair electrons and bonding electrons around each atom. Use the method described in Section 8.9 to determine the formal charges on each atom.

Formal charge = (valence electrons in atom) – [(lone pair electrons) + ( 12

number of bonding

electrons)]

Execute the plan:

(a) Lewis structure for SO3 molecule: 24 electrons total

O S O

O. ...

..

......

....

The formal charges are calculated for each atom using the number of valence electrons, the number of lone pair electrons and the number of bonding electrons. Let's set up a chart for these values:


S =O –O

Valence electrons 6 6 6

Lone pair electrons 0 4 6

Bonding electrons 8 4 2

Formal charge 6 – (0 + 4) = +2 6 – (4 + 2) = 0 6 – (6 + 1) = –1

O S O

O

......

....

. .....

0 G1

G1

2+

There are other resonance structures that could be written for SO3 with the double bond moved to each of the other two O atoms and with more than one double bond to the sulfur atom. (A more involved description of writing and judging the feasibility of resonance structures is found in the answers to Questions 54 - 57.)

(b) Lewis structure for NCCN molecule: 18 electrons total.

N C C N.. ..


C N

Valence electrons 4 5

Lone pair electrons 0 2

Bonding electrons 8 6

Formal charge 4 – (0 + 4) = 0 5 – (2 + 3) = 0

N C C N.. ..0 0 0 0

(c) Lewis structure for NO2– ion: 18 electrons total

O N O...... ..

.. ..G


N =O –O





Formal charge 5 – (2 + 3) = 0 6 – (4 + 2) = 0 6 – (6 + 1) = –1

O N O...... ..

.. ..G

0 0 G1

One other resonance structure could be written for NO2– with the double bond moved to the other O

atom. (A more involved description of writing resonance structures is found in the answers to Questions 54 and 55.)

Check your answers: The sum of the formal charges is zero for the neutral molecules and the ionic charge for the charged ion. The atoms with more bonds have more positive formal charges than those with fewer bonds and more lone pairs. These answers make sense.


Develop a plan: Write the Lewis structures. Then determine the number of lone pair electrons and bonding electrons around each atom. Use the method described in Section 8.9 to determine the formal charges on each atom.

Formal charge = (valence electrons in atom) – [(lone pair electrons) + ( 12

number of bonding

electrons)] Execute the plan:

(a) There are three possible Lewis structures for OCS ion: 16 electrons total; C has lowest electronegativity, so it is the central atom.

First structure:

O C S....

.

...

We need to count number of lone pair electrons and bonding electrons around each atom. Let's set up a chart for the electron count values:

O C S




Formal charge 6 – (4 + 2) = 0 4 – (0 + 4) = 0 6 – (4 + 2) = 0


0 0 0

O C S....

.

...

Second structure:

O C S .... ....

O C S




Formal charge 6 – (6 + 1) = –1 4 – (0 + 4) = 0 6 – (2 + 3) = +1

O C S.... ....

G1 +10

Third structure:

O C S.. ......

O C S




Formal charge 6 – (2 + 3) = +1 4 – (0 + 4) = 0 6 – (6 + 1) = –1

O C S.. ......

+1 0 G1

(b) Lewis structure for HNC molecule: 10 electrons total: H N C ..

H N C




Formal charge 1 – (0 + 1) = 0 5 – (0 + 4) = +1 4 – (2 + 3) = –1

H N C ..0 +1 G1

(c) Lewis structure for CH3– molecule: 8 electrons total:


H C H

H

..

G


C H




Formal charge 4 – (2 + 3) = –1 1 – (0 + 1) = 0

H C H

H

..

0

0 0G1

G



Develop a plan: Write the Lewis structures. Then determine the number of lone pair electrons and bonding electrons around each atom. Use the method described in Section 8.9 and the answers to Question 50 to determine the formal charges on each atom.

Execute the plan:

(a) The Lewis structure for CH3CHO molecule: 18 electrons total.

H C C

H

H

O

H....

Set up the chart:

C– C= H O

Valence electrons 4 4 1 6

Lone pair electrons 0 0 0 4

Bonding electrons 8 8 2 4

Formal charge 4 – (0 + 4) = 0 4 – (0 + 4) = 0 1 – (0 + 1) = 0 6 – (4 + 2) = 0


H C C

H

H

O

H....

0 0 0 0

00

0

(b) There are three possible Lewis structures for N3– ion: 16 electrons total.

First structure:

..

....N N N..

G

Set up the chart:

N= =N=




Formal charge 5 – (4 + 2) = –1 5 – (0 + 4) = +1

......

..N N NG

G1 G1+1

Second structure:

N N N.... ....

G

Set up the chart:

Left-most N Middle N Right-most N




Formal charge 5 – (2 + 3) = 0 5 – (0 + 4) = +1 5 – (6 + 1) = –2

N N N.... .... G

+1 G20


Third structure:

N N N.... ...

. G

Set up the chart:

Left-most N Middle N Right-most N




Formal charge 5 – (6 + 1) = –2 5 – (0 + 4) = +1 5 – (2 + 3) = 0

N N N.. .. .... G

G2 +1 0

(c) Lewis structure for CH3CN molecule: 16 electrons total

H C C

H

H

N..

Set up the chart:

C H N




Formal charge 4 – (0 + 4) = 0 1 – (0 + 1) = 0 5 – (2 + 3) = 0

H C C

H

H

N..0 0 0 0

0

0




Develop a plan: Write the Lewis structures. Then determine the number of lone pair electrons and bonding electrons around each atom. Use the method described in Section 8.9 and the answers to Question 51 to determine the formal charges on each atom.

Execute the plan: (a) Draw the Lewis structures for KrF4 molecule: 36 electrons total.

KrF

F

F

F

.... .

.

......

......

.. ....

....

Set up the chart:

Kr F




Formal charge 8 – (4 + 4) = 0 7 – (6 + 1) = 0

KrF

F

F

F

.... .

.

......

..........

....

..0

0

0

00

(b) Draw the Lewis structure for ClO3– ion: 26 electrons total

O Cl O

O.. ....

.. ..... ..

...

G

.. Set up the chart:

Cl O




Formal charge 7 – (2 + 3) = +2 6 – (6 + 1) = –1

O Cl O

O.. ....

.. ..... ..

...

G

..

G1

G1 G1+2

(c) Lewis structure for SO2Cl2 molecule: 32 electrons total following the octet rule


Cl S O

Cl

O. .... .

... ... . ...

..

.. ....

Set up the chart:

S O Cl




Formal charge 6 – (0 + 4) = +2 6 – (6 + 1) = –1 7 – (6 + 1) = 0

Cl S O

Cl

O. .... .

... ... . ...

..

.. ....

0

0 G1

G1

+2

Because this mo lecule has nonzero formal charges and the central atom is from period 3, we violate the octet rule on the S atom and form two new multiple bonds to the sulfur to reduce the charges to zero:

S O Cl




Formal charge 6 – (0 + 6) = 0 6 – (4 + 2) = 0 7 – (6 + 1) = 0

Cl S O

Cl

O....

... ... . ...

.. ....

0

00

00


56. Given the formula of an ion, write the correct Lewis structure and assign formal charges to each atom. Use the method described in the answer to Question 54 to get the formal charges.

With 16 electrons total, the central N atom needs two multiple bonds:

C N O.. ..... . -

C N O... . . .. .

-

C N O.. .... .. -

I II III


Set up a chart:

C

I II III

N

I II III

O

I II III

Valence electrons 4 4 4 5 5 5 6 6 6

Lone pair electrons 2 4 6 0 0 0 6 4 2

Bonding electrons 6 4 2 8 8 8 2 4 6

Formal charge –1 –2 –3 +1 +1 +1 –1 0 +1

C N O.. .. ... .

-1 +1 -1-

C N O. .. .....-2 +1 0 -

C N O. .. .. .. . --3 +1 +1

I II III

The lowest number formal charge structure is the best Lewis Structure, so the right answer is structure (I).

57. Given the formula of two ions, write the correct Lewis structures and assign formal charges to each atom. Use the method described in the answer to Question 54 to get the formal charges.

The S2O82– molecule has 60 electrons total, and an O–O bond. One Lewis Structure for this ion ends up like

this. (The Period 3 sulfur atom expands its octet t o reduce the formal charge on the S atom. See the answer to Question 63 for more details.)

O S O

O

O ..

....

..

O S O

O

O

..

.

.

2-

...

.

.

.

...

. ....

.

..

..

. .

..

.

Set up a chart:

S O (of S=O) O (of S–O) O (of O–O)

Valence electrons 6 6 6 6

Lone pair electrons 0 4 6 4

Bonding electrons 12 4 2 4

Formal charge 0 0 –1 0

O S O

O

O..

.. ..

..

O S O

O

O

..

.

.

2-

...

.

.

.

...

. ....

.

..

..

. .

..

.

-1 -1

0

0

0

0 0

0

00


The SO42– molecule has 32 electrons total. One Lewis Structure for this ion ends up as show beow. (The

Period 3 sulfur atom expands its octet t o reduce the formal charge on the S atom. See the answer to Question 63 for more details.)

O S O

O

O

.

. .

. .

.

..

...

.

...

.

..

2-

..

Set up a chart:

S O (of S=O) O (of S–O)




Formal charge 0 0 –1

O S O

O

O

.

.

..

.

.

.

. .

.....

.

.

..

..2-

-1

-1

0

0

0

58. Given the atoms in a molecule, write two correct Lewis structures and assign formal charges to each atom. Use the method described in the answer to Question 54 to get the formal charges.

With 18 electrons total, the central N atom needs one multiple bond:

O N Cl...... ..

.. ..

O N Cl .......

.. ...

I II Set up a chart:

O I II

N I II

Cl I II

Valence electrons 6 6 5 5 7 7

Lone pair electrons 6 4 2 2 4 6

Bonding electrons 2 4 6 6 4 2

Formal charge –1 0 0 0 0 +1

....O N Cl. ... ..

-1 0 +1..

O N Cl. .... ...

.. 0 0 0..

I II

The lowest number formal charge structure is the best Lewis Structure, so the right answer is structure (II).


59. Given the atoms in a molecule, write two correct Lewis structures and assign formal charges to each atom. Use the method described in the answer to Question 54 to get the formal charges.

With 18 electrons total, the central N atom needs one multiple bond:

O N F...

... .... ..

O N F ..

.

.. ...

.. ..

I II Set up a chart:

O I II

N I II

Cl I II

Valence electrons 6 6 5 5 7 7

Lone pair electrons 6 4 2 2 4 6

Bonding electrons 2 4 6 6 4 2

Formal charge –1 0 0 0 0 +1

....O N F.. ..

.. -1 0 +1..

O N F ........

.. 0 0 0

..

I II

The lowest number formal charge structure is the best Lewis Structure, so the right answer is structure (II).

Resonance

60. Define the problem: Given the formulas of molecules or ions, write all the resonance structures.

Develop a plan: Write the Lewis structure. Each resonance structure differs only by where the electrons for a multiple bond come from. When two or more atoms with lone pairs are bonded to an atom that needs more electrons, any one of them can supply the electron pair for a multiple bond. To write all resonance structures, systematically and sequentially supply the central atom with needed electrons from each of the possible sources. Separate these different structures with a double-headed arrow to show that they are resonance structures.

Execute the plan:

(a) There are three plausible Lewis structures for nitric acid, HNO3, with 24 valence electrons. They are formed using one lone pair from a different one of the outer O atoms to make the second bond in the double bond to complete the octet of the N atom.

H O N

O

O

H O N

O

OH O N

O

O

.

...

.

...

..

....

....

......

......

......

......

These structures are not equally plausible. (See Question 57 for details.)

(b) There are three plausible Lewis structures for nitrate ion, NO3–, with 24 valence electrons. They are

formed using one lone pair from a different one of the outer O atoms to make the second bond in the double bond to complete the octet of the N atom.


O N

O

O

O N

O

O

O N

O

O....

....

......

......

......

......

.. .... .. ..

.. ....

G G G

Check your answers: The structures drawn all follow the octet rule and have the right number of valence electrons. They differ by which outer atom is double bonded to the N atom. These answers make sense.

61. Define the problem: Given the formulas of molecules or ions, write all the resonance structures.

Develop a plan: Write the Lewis structure. Each resonance structure differs only by where the electrons for multiple bonds come from. When two or mo re atoms with lone pairs are bonded to an atom that needs more electrons, any one of them can supply the electron pair for a multiple bond. To write all resonance structures, systematically and sequentially supply the central atom with needed electrons from each of the possible outer atom sources. Keep in mind that Period 2 elements must have an octet, but Period 3 elements can have 8 or more electrons. Separate these different structures with a double-headed arrow to show that they are resonance structures.

Execute the plan:

(a) There are three plausible Lewis structures for SO3 that follow the octet rule. The molecule has 24 valence electrons. The resonance forms are formed using one lone pair from a different one of the outer O atoms to make the second bond in the double bond to complete the octet of the S atom.

O S

O

O

O S

O

O....

....

......

......

.. .... .. ..

..O S

O

O......

.......

...

There are several other plausible Lewis structures for SO3 that don't follow the octet rule on the S atom from Period 3. They are formed using two and three lone pairs from the outer O atoms to make the more multiple bonds to the S atom. Some of them are here:

O S

O

O

O S

O

O....

....

......

....

.. .... .. .. O S

O

O....

.......

...

O S

O

O

O S

O

O..

..

......

......

.. .... .. ..

..O S

O

O......

......

..


O S

O

O....

....

.. ..

These structures are not equally plausible. (See Question 62 for details.)

(b) There are three plausible Lewis structures for SCN– with 16 valence electrons.

..

....S C N S C NS C N.. .. .. .. ..

.... .... GG G

Check your answers: The Period 2 elements in these structures all follow the octet rule. The Period 3 S atoms have eight or more electrons. All structures have the right number of valence electrons. They differ by which outer atom forms the multiple bonds to the central atom. These answers make sense.

62. Define the problem: Write resonance structures using all single bonds, one, two and three double bonds, and use formal charges to predict the most plausible one.

Develop a plan: Write the Lewis structure. To write all resonance structures, systematically and sequentially supply the central atom with needed electrons from each of the possible sources. Keep in mind that Period 2 elements must have an octet, but Period 3 elements can have 8 or more electrons. Separate these different structures with a double-headed arrow to show that they are resonance structures. To determine the relative plausibility of the structures, determine the formal charges (as described in the answers to Question 50) then use the rules described in Section 8.8:

• Smaller formal charges are more favorable than larger ones.

• Negative formal charges should reside on the more electronegative atoms. Conversely, positive formal charges should reside on the least electronegative atoms.

• Like charges should not be on adjacent atoms.

Execute the plan: BrO4– has 32 electrons. Formal charges for single bonded O atoms are always –1. Formal

charges for double bonded O are always 0. Each time electrons are moved from lone pairs into bonding pairs the positive formal charge on Br goes down: The Lewis structure that follows the octet rule for all the

atoms is the first one:

Since smaller formal charges are more favorable than larger ones, we predict that this fourth resonance structure, with three double bonds and the most zero formal charges, is the most plausible.

O Br O

O

O

... ... . ...

..

G.. ..

.. ....G1

G1

0G1

O Br O

O

O

... ... . ...

..

G.. ....

.. ....

G1G1

G1G1

+2+3

O Br O

O

O

... ... ..

..

G.. ..

....G1

00

O Br O

O

O

... ... ..

..

G.. ..

.. ....G1

00

0+1


Check your answers: The Period 2 O atoms in all these structures follow the octet rule. The Period 3 Br atom has eight or more electrons. All structures have the right number of valence electrons. They differ by which outer atom forms the multiple bonds to the central atom. While it might feel strange to select a resonance structure that does not follow the octet rule as being the most plausible, it is clear from the formal charges that this structure is preferred over the one that does follow the octet rule. These answers make sense.

63. Define the problem: Use formal charges to predict the most plausible resonance structure for each of two molecules.

Develop a plan: Write the Lewis structure. Write the resonance structures as described in the answers to Questions 54 and 55. Keep in mind that Period 2 elements must have an octet, but Period 3 elements can have 8 or more electrons. To determine the relative plausibility of the structures, determine the formal charges (as described in the answers to Questions 50 and 51) then use the rules described in Section 8.8:

• Smaller formal charges are more favorable than larger ones.

• Negative formal charges should reside on the more electronegative atoms. Conversly, positive formal charges should reside on the least electronegative atoms.

• Like charges should not be on adjacent atoms.

You should plan to consider all the possible resonance structures. However, if you see a bonding pattern that violates one of these rules, we may safely avoid permutations with the same bonding pattern.

Execute the plan:

(a) SO3 has 24 electrons. All of the possible resonance structures for SO3 are given in the answer to Question 61. Three of them follow the octet rule for all the atoms. Assign formal charges to all the atoms:

......

O S

O

O

O S

O

O

O S

O

O

.. .... . ..

.

...

.

.

..

.

.

.... ..

.. ..

. ...

....

. ....

G1

G1

G1 G1 G1

G1

0

0

0

+2 +2 +2

Structure 1 Structure 2 Structure 3

There are several other plausible Lewis structures for SO3 that don't follow the octet rule on the S atom from Period 3. They are formed using two and three lone pairs from the outer O atoms to make the more multiple bonds to the S atom. Some of them are here. Assign formal charges to all the atoms:

O S

O

O

O S

O

O

O S

O

O..

. ....

.

...

.

.

.

..

..

....

... .

.... ..

.. .

.

....

..G1G1

G1

0

0 0

0 00

+1 +1 +1

Structure 4 Structure 5 Structure 6


..

..

..O S

O

O.

... .

...

G1

G1

+1

+1

Structure 7

In Structure 7, we see a positive formal charges occurring on adjacent atoms, and one of these positive formal charges is found on a highly electronegative O atom. These two things makes this bonding pattern very unlikely. Therefore we won't bother checking the formal charges of any of the rest of the possible structures that have a triple bond to an O atom.

Structure 8

O S

O

O0

00

0

..

......

....

Since smaller formal charges are more favorable than larger ones, we predict that Structure 8, with three double bonds and all zero formal charges, is the most plausible.

(b) HNO3 has 24 electrons. All of the possible resonance structures for HNO3 are given in the answer to Question 54. Assign formal charges to all the atoms:

.. .. ..

..

.

.

..

...

....

..

..

......

H O N

O

O

H O N

O

O

H O N

O

O

. ..Structure 1 Structure 2 Structure 3

0 0 0 00+1 +1 +1...... +1

G1

G1

G1

G1....0

0

Structures 1 and 2 have the same number of nonzero formal charges. Both have smaller formal charges than Structure 3. Structure 3 has a positive formal charge on the highly electronegative O atom and positive charges on adjacent O and N atoms. Nothing in the rules helps us distinguish between Structures 1 and 2. So we will predict that Structure 1 and Structure 2 are equally plausible.

Check your answers: (a) While it might feel strange to select a resonance structure that does not follow the octet rule as being the most plausible, it is clear from the formal charges that this structure is preferred over the one that does follow the octet rule. (b) The first two structures only differ by which of the terminal O atoms has the double bond, so it makes sense that they are equivalent. These answers make sense.

64. Define the problem: Given the structural formula of an organic molecule, write all the resonance structures.

Develop a plan: Add appropriate lone pairs to the structural formula to complete the Lewis structure. Systematically move electron pairs around the structure. It helps to keep track of the nonzero formal charges, to see how and where the electrons are being moved.

Execute the plan: The five N atoms each need one lone pair to satisfy the octet rule:


CN

CN

C

C

NC

N

NH 2

H

HH

..

..

..

..

..

We can move the three double bonds in the hexagonal ring on the left, as seen in Section 8.11 for the resonance structures of benzene to form a second resonance structure:

CN

CN

C

C

NC

N

NH 2

H

HH

..

..

..

..

..

Both of these resonance structures have all atoms with zero formal charges.

The locations of the lone pairs can be changed, if we rearrange the bonds. This is done systematically by pushing the lone pair of electrons on one atom in to form a double bond, then taking electrons that are part of the neighboring double bond and moving it to form a long pair. The technique is shown here for the –

NH2 lone pair:

CN

CN

C

C

NC

N

NH2

H

HH

..

..

..

..

..G1

+1

CN

CN

C

C

NC

N

NH2

H

HH

..

..

..

..

..

We find that, whenever we start with all-zero-formal-charge structures and move electron pairs in this fashion, we create structures with nonzero formal charges. Because of this, we can conclude that they contribute only slightly to the resonance hybrid. (Note: Structures like this one, with separated formal charges, are described and discussed in much more detail in most organic chemistry courses. They are important for understanding the nature of the reactivity of the molecules.)

Check your answers: Organic molecules based on extended alternating multiple bonds, like benzene, are remarkable stable. All the structures drawn have the same number of electrons and the same number of


multiple bonds. The most plausible structures have all atoms with zero formal charges. These answers make sense.

65. Follow the systematic plan given in the answers to Question 62. One Lewis structures for S2O32– follows

the octet rule. The molecule has 32 valence electrons.

O S O

O

S

.

.

..

. . .

.

.....

. .

...

2-.. ....

Several other plausible Lewis structures for S2O3

2– do not follow the octet rule on the S atom from Period 3, using two lone pairs from the outer atoms to make the more multiple bonds to the S atom:

O S O

O

S

O S O

O

S

O S O

O

S.

.

.

. .

.

2-

.. .

..

.

.

....

...

.

.

...

.

.

.

..

.

2-

.

..

..

.

....

...

.. ..

...

....

.. .

2-..

O S O

O

S

O S O

O

S.O S O

O

S

.

. .

.

.. ...

. .2- .

.

.

...

.

.

.

.

.....

.

..

.

. .

.

.

.

. 2- .

....

...

2-

. .

.

..

..

.

..

... .

.

66. Follow the systematic plan given in the answers to Question 62.

The S2O32– molecule’s resonance structures are determined in questions 65. Look at the resonance forms

determined in the answer to Question 65

Set up a chart for formal charges:

central S O (of S=O) O (of S–O) S (of S=S) S (of S–S)

Valence electrons 6 6 6 6 6

Lone pair electrons 0 4 6 4 6

Bonding electrons 12 4 2 4 2

Formal charge 0 0 –1 0 –1 The following structures show the NON-ZERO formal charges.

O S O

O

S

O S O

O

S

..

.

. .

..

.

....

O S O

O

S

...

.

..

.

. .

. .

. ..

..

..

2- ...

...

..

.

2-

.

...

.

.

.

..

.

.

.

....

..

..

. 2--1-1 -1

-1

-1-1


O S O

O

S

O S O

O

S

.

.

.

. .

.

..

.

.

.

. O S O

O

S

..

..

.2-. ..

. .

..

....

........

.

..

. 2- .

.

.

.

.

.

.

.......

..

2-

.. .

.

-1

-1

-1

-1-1

-1

Because, negative formal charges should reside on the more electronegative atoms, then the first three structures are better than the second three.

Therefore, the most plausible structure for this molecule is a resonance hybrid of the first three structures above.

Exceptions to the Octet Rule

67. Follow the systematic plan given in the answers to Question 16.

(a) BrF5 has 42 valance electrons. Use ten of the electrons to connect the six F atoms to the central Br atom. Use 30 more of them to fill the octets of the F atoms. Put the last two on Br.

Br

FF F

F F

......

.........

...

...

...

..... . ..

(b) IF5 has 42 valance electrons. Use ten of the electrons to connect the five F atoms to the central I atom. Use 30 more of them to fill the octets of the F atoms. Put the last two on I.

I

FF F

F F

......

.........

...

...

...

..... . ..

(b) IBr2– has 22 valance electrons. Use four of the electrons to connect the Br atoms to the central I atom.

Use 12 more of them to fill the octets of the Br atoms. Put the last six on I.

Br I Br ........ ..

.. .. ....

G

Check your answers: All the Period 2 elements have an octet of electrons. All the Period 3 elements have eight or more electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.

68. Define the problem: Write Lewis Structures for a list of ions and molecules.

Develop a plan: Follow the systematic plan given in the answers to Question 15.

Execute the plan:


(a) BrF3 has 28 valance electrons. Use six of the electrons to connect the three F atoms to the central Br atom. Use 18 more of them to fill the octets of F. Put the last four on Br.

Br

F

F F

......

........

... .....

(b) I3– has 22 valance electrons. One I atom is the central atom. Use four of the electrons to connect the

other two I atoms to the central I atom. Use 12 more of them to fill the octets of the I atoms. Put the last six on the central I atom.

I I I.. ....

G........

....

(b) XeF4 has 36 valance electrons. Use eight of them to connect the four F atoms to the central Xe atom. Use 24 more of them to fill the octets of the F atoms. Put the last four on Xe.

XeF F

F F

.. .........

...

...

...

...

... ..

Check your answers: All the Period 2 elements have an octet of electrons. All the Period 3 or greater elements have eight or more electrons. All valance electrons are accounted for in the structures, either as members of shared pairs or of lone pairs. These answers look right.

69. Elements in Periods 3 or higher can form compounds with five or six pairs of valence electrons surrounding their atoms. The Period 2 elements cannot. Using a periodic table, we find that (b) P, (e) Cl, (g) Se, and (h) Sn can, and (a) C, (c) O, (d) F and (f) B cannot.

Aromatic Compounds

70. Benzene does not undergo addition reactions, suggesting that it does not have C=C bonds, which react by addition.

71. “All carbon-to-carbon bond lengths are the same” argues against the presence of C=C bonds in benzene, since the C=C bonds would be shorter than the C–C bonds.

72. Using the structural notation and the naming definitions given in the "Constitutional Isomers of Aromatic Compounds" part of Section 8.11:


Br

Br

Br

Br

Br

Br

ortho-dibromobenzene meta-dibromobenzene para-dibromobenzene

73. Adapting the structural notation given in Section 8.11, anthracene looks like this:

That means it has a formula of C14H10.

74. Using the structural notation and the naming definitions given in the "Constitutional Isomers of Aromatic Compounds" part of Section 8.11:

I

I

I

I

I

I

ortho-diiodobenzene meta-diiodobenzene para-diiodobenzene

General Questions

75. The nitrosyl ion, NO+, has 10 electrons. Using the method described in Problem-Solving Example 8.12.

σ2s σ2s* π2p π2p σ2p π2p* π2p* σ2p*

NO+ (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) ( ) ( ) ( )

As a result of how the electrons fill the MO diagram, the bond order is (8-4)/2 = 3, meaning that the bond between the N and the O is a triple bond. There are no unpaired electrons.

76. Using the method described in Problem-Solving Example 8.12.

(a) CO has 10 electrons.

CC

CC

C

C

CC

CC

CC

CC

H

H

H

H H H

H

H

HH



CO (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) ( ) ( ) ( )

As a result of how the electrons fill the MO diagram, we find four electrons in the 2s molecular orbitals (σ2s and σ2s*) and six electrons in the 2p molecular orbitals (π2p and σ2p). The bond order is (8-4)/2 = 3,

meaning that the bond between the C and the O is a triple bond. There are no unpaired electrons.

(b) F2– has 15 electrons.


F2– (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑ )

As a result of how the electrons fill the MO diagram, we find four electrons in the 2s molecular orbitals (σ2s and σ2s*) and 11 electrons in the 2p molecular orbitals (π2p, σ2p, π2p* and σ2p*). The bond order is

(8-7)/2 = 0.5, meaning that the bond between F and F is a half bond. There is one unpaired electron.

(c) NO– has 12 electrons.


NO– (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑ ) (↑ ) ( )

As a result of how the electrons fill the MO diagram, we find four electrons in the 2s molecular orbitals (σ2s and σ2s*) and eight electrons in the 2p molecular orbitals (π2p and σ2p). The bond order is (8-4)/2 =

2, meaning that the bond between the N and the O is a double bond. There are two unpaired electrons.

77. Using the method described in Problem-Solving Example 8.12.

(a) B2+ has 5 electrons.


B2+ (↑↓) (↑↓) (↑ ) ( ) ( ) ( ) ( ) ( )

As a result of how the electrons fill the MO diagram, the bond order is (3-2)/2 = 0.5, meaning that the bond between B and B is a half bond. There is one unpaired electron.

(b) Li2+ has 1 electron.


Li2– (↑ ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

As a result of how the electrons fill the MO diagram, the bond order is (1-0)/2 = 0.5, meaning that the bond between Li and Li is a half bond. There is one unpaired electron.

(c) O2+ has 11 electrons.


O2+ (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑ ) ( ) ( )


As a result of how the electrons fill the MO diagram, the bond order is (8-3)/2 = 2.5, meaning that the bond between the N and the O is half way between a double bond and triple bond. There is one unpaired electron.

General Questions

78. Use the periodic table and trends in electronegativities to determine which pair is farthest apart. That will mean that their electronegativities are most different and the bond will be most polar.

Here, all the choices have F atom in common so find out which of the other elements is the farthest from F. Si is farthest from F on the periodic table, so (c) Si–F is more polar than these other choices (a) C–F, (b) S–F, (d) O–F

79. Br2 is composed of two Br atoms bonded together. Each of them is the same radius, so that the sum of their radii is the bond length.

Br228 pmBr

114 pm

Each Br atom in a covalent bond is 114 pm. Use that information along with the length of the C–Br bond in CBr4 to determine the C atom's radius:

BrC191 pm

114 pm77 pm

Here we calculate that the C atom's radius is 77 pm. The C atom's radius given in Figure 7.15 is 77 pm. Our calculation here gives an identical result!

80. Yes, it is a good generalization, because elements close together in the periodic table have similar electronegativities and the bonds would be formed by shared electrons (polar covalent). If they are far apart on the periodic table, their electronegativities will likely be very different and the bond more likely to be ionic.

A few exceptions exist, of course. For example, H atom is fairly far away from most of the nonmetals on the periodic table, yet it is also a nonmetal and forms polar covalent bonds.

81. Use the Lewis Structure method outlined in the answers to Question 15.

(a) CS2 has 16 valence electrons. Use four of them to attach the two S atoms to the central C atom. Use 12 of them for the lone pairs on S. Make two double bonds to the central S atom to give the lowest formal charge.

S C S....

....

(b) The structural formula of NH2OH tells us that we have a N atom with two H atoms and an O atom bonded to it. The O has an H atom bonded to it. The molecule has 14 valence electrons. Use eight of them to attach the atoms together using single bonds. Use the remaining six of them for the lone pairs on O and N.

H N

H

O H......


(c) S2Cl2 has a bond between two S atoms with the Cl atoms bonded to them. The molecule has 26 valence electrons. Use six of them to attach the atoms together using single bonds. Use the remaining twenty for the lone pairs on Cl and S.

Cl S S Cl........ ..

.. ....

.

...

(d) Given NS2+. The molecule has 16 valence electrons. With the N atom first in the formula, we might find

that it is the central atom. However, the electronegativity of S atoms are smaller than that of N atoms, so probably the more stable bonding order is N–S–S. Let's do each structure and see.

Use four of the electrons to attach the N and S atoms to the central S atom. Use the remaining 12 electrons to complete the octets of the outer atoms. So, at this point the structure looks like one of these:

N S S.. ....

...... S N S.. ..

.. ......

Remember, S is from period 3, so it can have more than 8 electrons. Hence, the lowest formal charge structure is this one:

N S S.. ......

+

+1

Now, use four of the electrons to attach the two S atoms to the central N atom. Use the remaining 12 electrons to complete the octets of the S atoms. Here we form two necessary double bonds to the Period 2 N atom to complete its octet:

S N S....

.... +

+1

The two structures have the same number of nonzero formal charges, but the first one has the positive charge on the least electronegative atom, so it indeed is more stable than the second structure as we predicted.

NOTE: These are not resonance structures of the same ion. They are two different ions. We did both because the formula alone doesn't tell us the particular bonding arrangement.

82. The resonance structure given in the problem is the only resonance structure with zero formal changes, so it is the dominant form for the molecule.

(a) There are two carbon-carbon bonds, a C=C and a C–C. Double bonds are shorter than single bonds, so C=C is the shortest.

(b) Double bonds are stronger than single bonds, so C=C is the strongest.

(c) The most polar bond is the bond with atoms that have the largest electronegativity difference. The difference between the electronegativities of C (EN = 2.5) and H (EN = 2.1) is 0.4. The difference between the electronegativities of C and N (EN = 3.0) is 0.5. So, the carbon-nitrogen bond is slightly more polar. Nitrogen has the higher electronegativity, so it is the partial negative end of the bond.


83. Use the Lewis Structure method outlined in the answers to Question 15. Use the methods described in the answers to Question 55 to find all the resonance structures.

NO2Cl has 24 valence electrons. We were told that there is no oxygen-oxygen bond. That means either N or Cl could be the central atom. They both have the same electronegativity, but N has fewer valance electrons so it's formal charge as central atom would be lower. N is also first in the formula, so we'll select it as the central atom. Use six electrons to attach the two O atoms and the Cl atom to the central N atom. Use the remaining 18 of them for the lone pairs on the outer atoms. Make one double bond to the central N atom in one of three different ways.

O N

O

Cl

O N

O

Cl

O N

O

Cl

.. ..

. ... ...... .. ..

. ..

. .. ....

...

.

.... ...... ...

...

(Note: The last two are equally plausible, but the first one has two adjacent atoms, N and Cl, with +1 formal charge, so it is not very plausible.)

84. Use the Lewis Structure method outlined in the answers to Question 16.

(a) SCl2 has 20 valence electrons. Use four of them to attach the two Cl atoms to the central S atom. Use the 12 of them for the lone pairs on the Cl atoms. Use the remaining four of them for the lone pairs on the S atom.

Cl S Cl.. .... .

... ...

...

(b) The Cl3+ ion has 20 valence electrons, just as in (a). One of the Cl atoms is the central atom, while the other two Cl atoms are single bonded to the central Cl atom:

Cl Cl Cl.. .... .

... ...

...

+

(c) ClOClO3 has 38 valence electrons. We are told there is a Cl–O–Cl bond, so we will put the last three O

atoms on the second Cl. Use tenelectrons to attach the atoms together using single bonds. Use the remaining 28 of them to complete the octets.

Cl O Cl O

O

O.. ....

.......

..... ..

.... ..

..

The second Cl atom has a +3 formal charge in this Lewis structure, so we can make a lower formal charge resonance structure for this molecule by expanding the octet of the Period 3 Cl atom, as we did for Br atom in Question 56.


Cl O Cl O

O

.. ..

.. ..

.......... ..

..O

Now, all the formal charges are zero.

(d) SOCl2 has 26 valence electrons. Use six of them to attach the O atoms to the central Cl atom. Use the remaining 18 of them complete the octets of the O atoms. Use the remaining 2 to complete the octet on S.

O S

Cl

Cl.....

...

......

.. ....

The S atom has a +1 formal charge in this Lewis structure, so we can make a lower formal charge resonance structure for this molecule by expanding the octet of the Period 3 S atom and making one double bond to the S atom, similarly to what we did in (c) for Cl.

85. The sizes of the atoms and the types of bonds affect the length of the bond, as described in the answers to Question 33. We can also check our predictions using Table 8.1

The H atom, from Period 1, is very small compared to the others. Atoms of Period 2 elements O and C, are smaller than the Period 3 element (Cl). The smaller of the two Period 2 elements is O, since it further to the right in the same period. The Period 3 Cl atom is the largest. Double bonds are shorter than single bonds between atoms of the same type.

So, we’ll predict that O–H is shortest; then the double bonds, O=O then O=C; and then the single bonds O–O, then O–Cl.

The predictions are confirmed in Table 8.1:

O–H (94 pm) < O=O (112 pm) < O=C (122 pm) < O–O (132 pm) < O–Cl (165 pm)

86. The strongest chemical bond has the largest bond energy. Look up the bond energies for the bonds in

Table 8.2: DO–H = 467 kJ/mol, DO=O = 498 kJ/mol, DO=C = 695 to 803 kJ/mol, DO–O = 146 kJ/mol, and DO–

Cl = 205 kJ/mol. The order of increasing strength: O–O < Cl–O < O–H < O=O < O=C

87. We will again follow the systematic plan given in the answers to Question 15.

ClF3 has 28 valance electrons. Use six of the electrons to connect the three F atoms to the central Cl atom. Use 18 more of them to fill the octets of F. Put the last four on Cl.

O S Cl

Cl

...... ..

.. .... ....


F Cl

F

F........

....

......

.. ..

88. Given formulas of hydrocarbons, determine if they are alkane, aromatic, or neither. We can try what was done in the answers to Question 26.

n 2n+2

(alkane)

2n

(alkene)

2n–2

(alkyne)

C6H? 6 ? = 14 ? = 12 ? = 10

C8H? 8 ? = 18 ? = 16 ? = 14

C10H? 10 ? = 22 ? = 20 ? = 18

It’s not clear how (a) and (b) fit, yet, so save them for later. We find the answers to (c) - (f) in this chart:

(c) C6H12 has the formula of an alkene, so this is neither an alkane nor aromatic (though it could be a cyclic alkane).

(d) C6H14 is an alkane.

(e) C6H18 is an alkane.

(f) C6H10 is an alkyne, so this is neither an alkane nor aromatic.

Let’s draw some structures that fit the other two formulas as aromatic compounds with the help of Section 8.11:

(a) Xylene is an aromatic compound given in Section 8.11, and its formula is C8H10 (para-xylene shown here).

CC

CC

C

C

CH3

H

H

CH3

H

H


(b) Naphthalene is an aromatic compound given in Section 8.11, and its formula is C10H8.

Applying Concepts

89. The number of valence electrons in ClO3– is 7 + 3 × (6) + 1 = 26 e

–. It looks like the student subtracted an

electron from the total rather than adding, since the given structure has 7 + 3 × (6) – 1 = 24 e–.

90. The number of valence electrons in SF5+ is 6 + 5 × (7) – 1 = 42 e–. It looks like the student forgot to subtract

one electron for the positive charge, since the given structure has 6 + 5 × (7) = 41 e–. 91. Resonance structures must only be different by where the electrons are. The bonding arrangement of the

atoms (what atom is bonded to what atom) must be the same from structure to structure. In the first structure, the O atom has only one N atom bonded to it. In the second structure the O atom has an N atom and an O atom bonded to it. These bonding arrangements differ; therefore these structures represent different molecules, not resonance structures of one molecule.

92. Resonance structures must only be different by where the electrons are. The bonding arrangement of the atoms (what atom is bonded to what atom) must be the same from structure to structure. In the first structure, the C atom has only one S atom bonded to it. In the second structure the C atom has an N atom and an S atom bonded to it. These bonding arrangements differ; therefore these structures represent different molecules, not resonance structures of one molecule.

93. Group 3A elements would form four bonds to obey the octet rule. They have only three valence electrons, which means they still need five more, so they must pair the three they have and find an atom that provides two more to complete the octet. For example: BF4

–

..

F B F

F

F

.. . ...

..

.. ...

...

......

G

Group 4A elements would form four bonds to obey the octet rule. They have four valance electrons, which means they still need four more, so they must pair the four they have with the electrons of other atoms For example: CF4

CC

CC

C

C

CC

CC

H

H

H

H H

H

H

H


..

F C F

F

F

.. . ...

..

.. ...

...

......

Group 5A elements would form three bonds to obey the octet rule. They have five valance electrons, which means they still need three more, so they must pair three of the five they have with the electrons of other atoms For example: NF3

..

N F

F

F

.. . ...

..

..

..

......

Group 6A elements would form two bonds to obey the octet rule. They have six valance electrons, which means they still need two more, so they must pair two of the six they have with the electrons of other atoms For example: OF2

Group 7A elements would form one bond to obey the octet rule. They have seven valance electrons,

which means they still need one more, so they must pair one of the seven they have with the electron of another atom. For example: F2

94. In this fictional universe, a nonet is nine electrons. We will assume that the number of electrons in the atom called “O” in the other universe is still six, the number of electrons in the atom called “H” is still one, and the number of electrons in the atom called “F” is still seven.

The 6 electrons in the O atom would need three more electrons to make a total of nine. That means it would combine with three H atoms to make this molecule:

H H

H

O .....

....

The 7 electrons in the F atom would need two more electrons to make a total of nine. That means it would combine with two H atoms to make this molecule:

H

H

F .....

....

95. P4 has 20 valence electrons. The best bonding arrangement for a Group 5 element is to have three bonding pairs and one lone pair (that gives it a formal charge of zero). We were told each P atom is bonded to three

F O F....

..

.. .... ....

F F ........

..

..


other P atoms. That uses 12 electrons and gives each P six electrons towards its octet. Use the remaining eight electrons to complete the octet of each P atom by adding one a lone pair to each:

P P

P

P

.. ..

..

..

96. Using the periodic trend for electronegativity (EN)

S ClAs Se Br

smaller larger EN EN

larger EN

smaller EN

We certainly know that Cl’s EN is the largest, so we’ll assign it the value of 3.0. We certainly know that As’s EN is the lowest, so we’ll assign it value of 2.1 The trend shows that EN’s of S and Br are both larger than that of Se, so we’ll assign them both the value of 2.5. That leaves the EN value of 2.4 for Se. With the values of 2.5 and 2.4 so close together, the assignment of those values are uncertain.

97. Define the problem: Given the percent mass of the elements in a compound, the density of the gas and the volume of one mole of gas (also called the molar volume), determine the empirical formula, molar mass, and two possible Lewis structures for the molecule.

Develop a plan: We start out using the method similar to that used to answer Question 76 in Chapter 3. Choose a convenient mass sample of acetylene, such as 100.0 g. Using the given mass percents, determine the mass of C and H in the sample. Using molar masses of the elements, determine the moles of each element in the sample. Find the whole number mole ratio of the elements C and H to determine the subscripts in the empirical formula. Find the mass of 1 mol of the empirical formula. Now we need to use the density and the molar volume as unit factors to determine the molar mass of the compound. Then we divide the molar mass of the compound by the calculated empirical mass to get a whole number. Multiply all the subscripts in the empirical formula by this whole number to get the molecular formula. Finally write two Lewis structures for various isomers of this compound.

Execute the plan:

A 100.0 gram sample will have 85.7 grams of C and 14.3 grams of H.

Find moles of C and H in the sample:

85.7 gC×

1mol C12.011 g C

= 7.14 mol C 14.3 g H ×

1 mol H1.0079 g H

= 14.2 mol H


Mole ratio =

7.14 mol C14.2 mol H

= 0.503 =1 mol C2 mol H

The empirical formula is CH2, so the molecular formula is (CH2)n.

Mass of 1 mol CH2 = 12.011 g + 2×(1.0079 g) = 14.027 g/mol

Molar mass of (CH2)n =

1.87 g (CH2)n1 L (CH2)n gas

×22.4 L (CH2)n gas1 mol (CH2)n gas

= 41.9 g/ mol

n =

mass of 1 mol of moleculemass of 1 mol of CH

=41.9 g

14.027 g= 2.99 ≈ 3

Molecular Formula is (CH2)3 = C3H6

Two Lewis Structures are:

HCCC

H

HHH

H

C

C C

H H

H

H

H

H

Check your answer: The moles of H in the sample are nearly twice the moles of C, so the empirical formula makes sense. The molar mass is about three times larger than the mass of one mole of the empirical formula, so the molecular formula C3H6 makes sense. Two possible Lewis structures could be written for the resulting formula. That is a satisfying result.

98. All of these compounds require the central atom to have more than eight electrons around it in order to have bonds to five atoms. Three of the four of the central atoms are in Periods 3 or higher, making it possible to expand their octets. The Period 2 N atom cannot have more than eight electrons, so NF5 is least likely to exist.

99. Write an appropriate balanced chemical equation, and use Hess’s Law to solve for its enthalpy as done in answers to Questions 40 - 42 and in Chapter 6.

(a) If we find ∆H° for the reaction that breaks all four of the bonds in gas-phase methane, then we can divide it by four to get the average.

CH4(g) C(g) + 4 H(g)

∆H° = (1 mol) × ∆H fo (C(g)) + (4 mol) × ∆H f

o (H(g)) – (1 mol) × ∆H fo (CH4(g))

= (1 mol) × (716.7 kJ/mol) + (4 mol) × (218.0 kJ/mol)

– (1 mol) × (–74.81 kJ/mol) = 1663.5 kJ

Average Bond Energy =

1663.5 kJ4 bonds

= 415.88 kJ /bond

(b) CH4(g) C(g) + 2 H2(g)


There are four C–H bonds in methane broken, and two H–H bonds formed.

∆H° = (4 mol of C–H) × DC–H – (2 mol H–H) × DH–H


∆H° = (4 mol of C–H) × (416 kJ/mol) – (2 mol H–H) × (436 kJ/mol) = 792 kJ

(c) To draw figures like Figure 8.4 for these reactions, we need to break C–H bonds and form (in some cases) H–H bonds.

CH3(g) C(g) + H2(g) + H(g)

+ + +

+ +E

3 C–H bonds

broken: 3(416 kJ) = 1248 kJ

1 H–H bond formed: 436 kJ

²H° = 812 kJ

CH2(g) C(g) + H2(g)

+ +

+E 2 C GH bonds

broken: 2(416 kJ) = 8 32 kJ

1 HGH bond formed: 436 kJ

ÆH¡ = 396 kJ

CH(g) C(g) + H(g)


+

E1C–H bond

broken: 416 kJ

²H° = 416 kJ

Average Bond Energy in gas-phase CH3:

If we find ∆H° for the reaction that breaks all three of the bonds in gas-phase CH3, then we can divide it by three to get the average.

CH3(g) → C(g) + 3 H(g)

∆H° = (1 mol) × ∆H fo (C(g)) + (3 mol) × ∆H f

o (H(g)) – (1 mol) × ∆H fo (CH3(g))


– (1 mol) × (146.0 kJ/mol) = 1224.7 kJ


1224.7 kJ3 bonds

= 408.23 kJ /bond

Average Bond Energy in gas-phase CH2:

If we find ∆H° for the reaction that breaks both of the bonds in gas-phase CH2, then we can divide it by two to get the average.

CH2(g) C(g) + 2 H(g)

∆H° = (1 mol) × ∆H fo (C(g)) + (2 mol) × ∆H f

o (H(g)) – (1 mol) × ∆H fo (CH2(g))


– (1 mol) × (392.5 kJ/mol) = 760.2 kJ


760.2 kJ2 bonds

= 380.1 kJ /bond

If we find ∆H° for the reaction that breaks both of the bonds in gas-phase CH, then we can divide it by two to get the average.

CH(g) C(g) + H(g)

∆H° = (1 mol) × ∆H fo (C(g)) + (1 mol) × ∆H f

o (H(g)) – (1 mol) × ∆H fo (CH(g))


– (1 mol) × (596.3 kJ/mol) = 338.4 kJ



338.4 kJ1 bond

= 338.4 kJ /bond

Average bond energy varies depending on the stability of the molecule whose bonds are being broken. Fewer H atoms on the C atom make the molecule less stable. It is easier (i.e., it takes less energy) to break a bond in an unstable molecule than in stable molecule.

100. (a) FNO has 18 electrons. The best Lewis Structure for this molecule has a double bond on the oxygen, since the formal charges are all zero for this resonance structure.

F N O.. .... . .. ...

(b) Single bonds are longer than double bonds so, N–F is the longest bond.

(c) N=O is the the shortest bond.

(d) Both bonds include N, and the periodic trend for electronegativity indicates that ENF is greater than

ENO, so the most polar bond is N–F.

101. (a) ClNO has 18 electrons. The best Lewis Structure for this molecule has a double bond on the oxygen, since the formal charges are all zero for this resonance structure.

Cl N O. .. .. ....

...

(b) Single bonds are longer than double bonds so, N–Cl is the longest bond.

(c) N=O is the the shortest bond.

(d) Both bonds include N, and the periodic trend for electronegativity indicates that ENO is greater than

ENCl, so the most polar bond is N=O.

102. (a) Bond strength increases from single to double to triple, so the weakest bond will be a single bond. Table 8.2 gives the various bond energies for single bonds. The C–C bond energy is 356 kJ/mol. The C–O bond energy is 336 kJ/mol. Therefore, the C–O bond is the weakest carbon-containing bond.

(b) The triple C≡N bond is the strongest carbon-containing bond.

(c) The electronegativity of O atom is greater than that of any other element in this structure, so the C–O bond is the most polar.

103. (a) Bond strength increases from single to double to triple, so the triple C≡N bond is the strongest carbon-containing bond.

(b) The double C=C bond will be the second strongest.

(c) The electronegativity of N atom is greater than that of any other element in this structure, so the C≡N bond is the most polar.

More Challenging Questions

104. (a) B4Cl4 cannot follow the octet rule; there are too few electrons. The structure has 40 valence

electrons. We must use 8 electrons to attach the Cl atoms to the B atoms and we must use 24 electrons to fill the octets of the Cl atoms. If the B atoms cluster together and use a total of four


electrons to bond with each other, then all the electrons could be used. This structure is misleading because the lines between B and B do not represent two electron bonds. (At the time of production of this textbook, the three-dimensional ball-and-stick model of the B4Cl4 molecule is provided at

http://www.chem.ox.ac.uk/inorganicchemistry3/index.html, as are several of the others in this section of problems .)

B B

B

B

Cl

Cl

Cl

Cl

..... .

.. .. ..

......

......

(b) This molecule is an acid so we will attach the H atoms to the O atom as we have seen with other acids. The molecule has 24 electrons. The best Lewis Structure would look like this:

H O N N O H....

........

105. N4O has a nitrosyl, which we have learned is NO. We also saw three nitrogens bonded together in the

answer to Question 54(b), so lets line up the atoms in a row. This produced two resonance structures:

N N N N O... .

.. . .... .

N N N N O....

... . . ...

Upon determining the formal charge, we find the following non-zero charge assignments:

N N N N O... .

.. . .... .

-1 +1

N N N N O....

.... ....

+1 -1

A resonance structure is favored when the negative formal charge is on the more electronegative atom, so the second of these two structures is the preferred structure.

106. Many different and creative guesses could result from attempting to write Lewis Structures of this molecule. There is limited information provided, so several structures will be equally good at answering the question. The N atoms must follow the octet rule. It is also not likely that the N atoms will bond to each other or to Cl atoms, since their electronegativity is relatively high. (At the time of production of this textbook, the three-dimensional ball-and-stick model of the (Cl2PN)3 and (Cl2PN)4 molecules are provided

at http://www.chem.ox.ac.uk/inorganicchemistry3/index.html, as are several of the others in this section of problems.)

(a) There are 72 valence electrons in (Cl2PN)3.


PN

PN

P

N

Cl Cl

Cl

Cl

Cl

Cl ..... .

.....

.

.

....

.

.

.

.

.

.

. .

..

..

......

......

.

(b) There are 96 valence electrons in (Cl2PN)4.

N

P

N P

N

P

NP

Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl

... .

.....

. ..

.

..

..

.. .

..

.

.

.

.

....

...

. ....

..

.. .

.. . ... ..

.

. .

.

.

107. (a) There are 30 valance electrons in the diothiocyanogen molecule, (NCS)2. Cyano- suggests that the

structure has C atom bonded to N. The N atom has the highest electronegativity, so make it the terminal atom. Using these as a guide, the molecule’s Lewis Structure turns out to look like this:

N C S S C N....

..

... ...

(b) There are 30 valance electrons in the tetrahydrofuran molecule, C4H8O. Again, guessing can really

produce some creative answers, including double bonds, This molecule is a ring structure. The Lewis Structure looks like this:

CC

CO

C

H

H

H

H H

H

H

H....

108. The molecule has 56 electrons. If we imagine an SO3 molecule bonding with a sulfuric acid molecule, by

having one of the O atoms on H2SO4 attached to the S atom of SO3, then the molecule would look like

this.


O S O S O

O

O

O

OH

H

.

....

....

....

.. . ...

....

.

...

.

..

This structure has two atoms with non-zero formal charge (the left-most O and the right S), so it is likely that the H atom will shift to the other side of the molecule during the reaction to form the more stable (all zero-formal-charge) molecule.

H O S O S O H

O

O

O

O

..

.

.

..

... .. .

..

..

....

....

... .

109. (a) With so few atoms, it is clear this molecule, S2N2, must be a cyclic compound. The N atoms must

satisfy the octet rule. The S atom does as well, in this structure with 22 valence electrons.

N N

S

S

. .. .....

....

(b) Using the answer above to pattern an answer for S4N4 with 44 valence electrons:

S

N

N

S

N

NS S

....

.... .

.

.

.

....

.

.

. ..

.. .

110. (a) Having done hyponitrous acid in the answer to Questions 104 (b) makes hyponitrite anion fairly easy. Remove the H atoms from the acid and add the charge for this 24 balance electron structure:

O N N O. .... . ..

...

...

.. 2-

(b) This molecule has 72 electrons. We will follow the same theories applied in the answer to 106. The S atoms must be interspersed in the structure, so make it an alternating-atom cyclic compound. Make sure that the O atoms satisfy the octet rule.


SO

SO

S

O

O O

O

O

O

O

....

.

.. .

.

.

...

.

.

..

.

.

..

..

. ..

.

.

.

.

.

.

.

.

.

.

111. (a) S2F10 can really go together one way. The F atoms will not bond to each other with such high

electronegativity, so the S atoms be central. Notice that the S atoms have an expanded octet; however, F must follow the octet rule. This structure has 82 valence electrons.

F S S F

F

F

F

FFF

F F.. ... . .

..

. .. .. ... ... .. ..

.... ...... ...... ..... ..

......

.

.. ....

(b) This molecule has 70 electrons. Xenon can and often does have a large number of electrons. The organization of the formula suggests that SO2F is a component part of the molecule. A plausible Lewis Structure is:

....

N

Xe

F

S S

F

O

O

F

O

O

.. .. ..

.......

.

.

..

. ..

..

..

.

.

..

. ..... . .

....

112. The C and N atoms in this molecule must all follow the octet rule. Make a straight chain, and add as many multiple bonds to fulfill the octet rule for both period two atoms.

H C C C N..

113. This is a WEIRD one. The valence electron count is only 16, so the extreme multiple bonding is necessary. The larger size of the Si atom probably relieves some of the strain.

Si

CC

C


Seven transitions of this molecule have been detected toward the evolved carbon star IRC+10216 using the NRAO 12m radio-telescope. The molecule is rhomboidal in shape, planar, and highly polar

114. The aldehyde functional group is present, along with the thio meaning replace an O atom with a S atom. The valence electron count is 30. All atoms must satisfy the octet rule.

H C C C S H

O H

H

H

H

..

.

. ..

..

115. Doing some searching on the web, suggests that the skunk odor comes from thio-ethers. So, these two structures are various structural isomers of thio-ethers.

S

C C

H

O C

H

H

HH

H

.. ... ...

H

C C

S

O C

H

H

H

H H

.. ..

....

116. Tetraphosphrous trisulfide is P4S3, with 38 valance electrons.

P

P S

P

SP

S

.....

.

.. .

.

..

....

..

..

117. (a) The formula is C17H19NO3. Remember to count the unsubstituted corners in saturated rings as

“CH2” and in unsaturated rings like benzene as “CH”.

(b) The shortest carbon-to-carbon bond in this molecule is the benzene carbon-to-carbon bond.

(c) The shortest carbon-to-oxygen bond in this molecule is the carbon-to-oxygen double bond.

(d) The strongest carbon-to-carbon bond in this molecule is the benzene carbon-to-carbon bond.

(e) The most polar bond in this molecule is the C–O double bond.

118. (a) This four carbon chain as has an amine group and an organic acid. Gamma is the forth letter of the alphabet, so put the amine on the last carbon in the chain. The structure has 42 electrons.

H N C C C C O H

OH

H

H

H

H

HH

.. ... ....

...

119.


σ2p*

π2p* π2p

*

Eπ2p π2p

σ2p

σ2s*

σ2s

The switch affects the molecules that have six (B2) and eight (C2) electrons, since a degenerate pair is

replaced with a single orbital in the energy order.

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