chapt 1 key - thoburn-chemistry-rmc.org · 2 is a covalent compound with an electronegativity...

1

Answer Key to Chapter 1 Questions Chemical Bonding

P1. How many valence electrons are found on each of the following species? (L1.1)

(a) Na 1s2 2s2 2p6 3s1 – 1 valence electron (b) Ca 1s2 2s2 2p6 3s2 – 2 valence electron

(c) O 1s2 2s2 2p4 – 6 valence electron

O2– 1s2 2s2 2p6 – 8 valence electron

(d) Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 – 17 valence electron

Br+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 – 16 valence electron

P2. Give the electronic configuration of (L1.28, 1.29)

(a) a chlorine atom: Cl 1s2 2s2 2p6 3s2 3p5

(b) a chloride anion: Cl¯ 1s2 2s2 2p6 3s2 3p6

(c) an argon atom: Ar 1s2 2s2 2p6 3s2 3p6

(d) a magnesium cation: Mg2+ 1s2 2s2 2p6 3s0

(e) a silicon atom: Si 1s2 2s2 2p6 3s2 3p2

P3. Name species that satisfy the following criteria: (L1.2)

(a) the singly charged negative ion isoelectronic with neon: F¯

(b) the singly charged positive ion isoelectronic with neon: Na+

(c) the dipositive ion isoelectronic with argon: Mg2+

(d) the neon species that is isoelectronic with neutral fluorine: Ne+

P4. In each of the following sets, specify the one compound that is completely ionic (L1.23)

(a) CCl4 HCl NaAt K2

The electrons in the C–Cl bond are shared. C and Cl make covalent bonds. The electronegativity difference between C (2.6) and Cl (3.2) is small (0.6). CCl4 does not break about into ions when placed in water.

H–Cl is considered a covalent compound, not an ionic compound. The electronegativity difference between H (2.2) and Cl (3.2) is larger (1.0) than in CCl4. It is a covalent compound with some ionic character. In the gas phase is does not dissociate, but in aqueous solution it does dissociate completely.

NaAt is considered to be a ionic compound, mostly because it is made from a group 1 metal and a group 7 non-metal. The electronegative difference between Na (0.9) and At (2.2) is smaller than most ionic compounds (1.3), but we still consider it ionic. One rule of thumb is that when the electronegativity

2

difference is > 2, the compound is ionic. NaAt would probably forms ions when placed in aqueous solution and would probably conduct electricity in solution. This compound is something of a curiosity because there is so little astatine in the universe (<1g in earth’s crust) that one can’t really get enough of it to experiment on. All the astatine isotopes are radioactive with half-lives of less than 8 hours, so it is vanishes almost as quickly as it is made.

K2 is made of group 1 atoms, so you might think it should form an ionic compound, but it doesn’t. That is because the electronegativity difference between the atoms is zero, therefore the electrons must be shared equally as in a covalent bond. It is not possible to do the test to see of K2 forms ions in solution, because K2 reacts (violently) with water.

(b) CS2 CsF HF XeF2 BF3

Carbon disulfide is made up of two covalent bonds, meaning the electrons are shared more or less equally. The electronegativity different between C (2.55) and S (2.58) is very small (0.03) insuring that the electrons are shared equally.

Cesium fluoride is as ionic as it gets, with the largest electronegativity difference possible: F (3.98) - Cs (0.79) = 3.19.

HF is considered covalent despite a largish electronegativity difference: F (3.98) – H (2.20) = 1.78. The H–F bond is considered a polar covalent bond, meaning the electrons are shared, but not equally. There is more electron density around the more electronegative F. It is sometimes described as a covalent molecule with some ionic character. At room temperature it is a gas and thus does not form a crystal lattice.

XeF2 is a covalent compound with an electronegativity difference of F (3.98)– Xe (2.60) = 1.38. The electrons in the bond are shared unequally, with more electron density around F.

BF3 is considered a covalent molecule even though it electronegativity difference is almost 2: F (3.98) – B (2.04) = 1.94. The electrons are shared unequally. BF3 is also a gas and thus does not form a crystal lattice at room temperature. You can’t put it in water to see if ions form because it reacts with water.

P5. Which of the following species has a complete octet? Assume all unshared electrons are explicitly shown. (L1.24)

H C HH

H

C: 4 e¯H: 4 x 1 e¯ = 4 e¯ 8 e¯

H N HH

N: 5 e¯H: 3 x 1 e¯ = 3 e¯ 8 e¯

H C HH

C: 4 e¯H: 3 x 1 e¯ = 3 e¯cation –1 e¯ 6 e¯

H B HH

B: 3 e¯H: 3 x 1 e¯ = 3 e¯ 6 e¯

I: 7 e¯cation: – 1 e¯ 6 e¯

H B HH

B: 3 e¯H: 4 x 1 e¯ = 4 e¯anion: + 1 e¯ 8 e¯

HI

3

P6. Draw Lewis structures for each of the following

CO2 a. Count the total # of electrons 1xC + 2xO = 4 + 2x6 = 16 b. Draw a skeleton connecting the atoms c. Add bonds until C is tetravalent O is divalent d. Add remaining electrons: 8 (follow the octet rule!)

e. Calculate formal charge: none

CH3CH3

a. Count the total # of electrons 2xC + 6xH = 8 + 6 = 14 b. Draw a skeleton connecting the atoms

c. Add bonds until: C already tetravalent & H already monovalent d. Add remaining electrons: none left (follow the octet rule!) e. Calculate formal charge: none

CH3CHCH2

a. Count the total # of electrons 3xC + 6xH = 12 + 6 = 18 b. Draw a skeleton connecting the atoms

c. Add bonds until: C is tetravalent & H is monovalent

d. Add remaining electrons: none left (follow the octet rule!) e. Calculate formal charge: none CH3CN

a. Count the total # of electrons 2xC + 3xH + 1xN = 8 + 3 + 5 = 16 b. Draw a skeleton connecting the atoms

O C O

O C O

O C O

C C HH

HH

H

H

C C HH

CH

H

HH

C C HH

CH

H

HH

C CHH

HN

4

c. Add bonds until: C is tetravalent & N is trivalent

d. Add remaining electrons: 14 used, 2 left, N needs octet (follow the octet rule!)


CH3NH2

a. Count the total # of electrons 1xC + 5xH + 1xN = 4 + 5 + 5 = 14 b. Draw a skeleton connecting the atoms

c. Add bonds until: valency rules already satisfied d. Add remaining electrons: 12 used, 2 left, N needs octet (follow the octet rule!)


CH3Cl a. Count the total # of electrons 1xC + 3xH + 1xCl = 4 + 3 + 7 = 14 b. Draw a skeleton connecting the atoms

c. Add bonds until: valency rules already satisfied: C tetravalent, halogen monovalent d. Add remaining electrons: 8 used, 6 left, Cl needs octet (follow the octet rule!)


CH3CHO acetaldehyde

a. Count the total # of electrons 2xC + 4xH + 1xO = 8 + 4 + 6 = 18 b. Draw a skeleton connecting the atoms

c. Add bonds until: C tetravalent, H monovalent, O divalent

C CHH

HN

C CHH

HN

C N HH

HH

H

C ClHH

H

C ClHH

H

C C OH

HH

H

C N HH

HH

H

5

d. Add remaining electrons: 14 used, 4 left, O needs octet (follow the octet rule!)


C2H4O ethylene oxide


c. Add bonds until: C already tetravalent, H already monovalent, O ready divalent d. Add remaining electrons: 14 used, 4 left, O needs octet (follow the octet rule!)


C2H4O2 acetic acid


c. Add bonds until: C must be tetravalent, H already monovalent, O must be divalent



C2H4O2 acetic acid

a. Count the total # of electrons 2xC + 5xH + 1xCl = 8 + 5 + 7 = 20 b. Draw a skeleton connecting the atoms

c. Add bonds until: C already tetravalent, H & Cl already monovalent

C C OH

HH

H

C C OH

HH

H

C CO

HH H

H

C CO

HH H

H

H C C OOH

HH

H C C OOH

HH

H C C OOH

HH

H C C ClH

H

H

H

6



H2CCO ketene a. Count the total # of electrons 2xC + 2xH + 1xO = 8 + 2 + 6 = 16 b. Draw a skeleton connecting the atoms

c. Add bonds until: C must be tetravalent, O must be divalent, H already monovalent



P7. Give the formal charge on each atom and the net charge in each of following (L1.27)

Formal charge (FC) = # valence electrons – # bonds – # non-bonding electrons

Cl FC = 7 – 4 = +3 Cl FC = 7 – 1 – 6 = 0 C FC = 4 – 2 – 2 = 0 C FC = 4 – 3 – 2 = –1 O FC = 6 – 1 – 6 = –1 O FC = 6 – 1 – 6 = –1 H FC = 1 – 1 – 0 = 0 H FC = 1 – 1 – 0 = 0

P8. Draw Lewis structures with formal charge for the following:

a. Count the total # of electrons 1xN + 4xH = 5 + 4 – 1 (because it is a cation) = 8 b. Draw a skeleton connecting the atoms

c. Add bonds until: H already monovalent, N is usually trivalent, but may be tetravalent if cationic. No

additional bond needed. d. Add remaining electrons: all electrons used (follow the octet rule!)

H C C ClH

H

H

H

H C C OH

H C C OH

H C C OH

Cl OOO

OCl O H

CH H C

H

H

Cl OOO

OCl O H

CH H C

H

H

3+

NH4

H N HH

H

7

e. Calculate formal charge: F.C. = valence e¯ – bonds – # non-bonded e¯ = 5 – 4 – 0 = +1

SO42–

a. Count the total # of electrons 1xS + 4xO = 6 + 24 + 2 (because it is a dianion) = 32 b. Draw a skeleton connecting the atoms

c. Add bonds until: S is usually divalent like O, but it can accommodate 3 or even 4 bonds. Here’s

another exception: O usually has 2 bonds and 2 lone pairs, but it could also have 1 bond and 3 lone pairs and still complete its octet.

d. Add remaining electrons: used 8; add remaining 24 until O has its octet

e. Calculate formal charge: F.C. on S = valence e¯ – bonds – # non-bonded e¯ = 6 – 4 – 0 = +2 F.C. on O = valence e¯ – bonds – # non-bonded e¯ = 6 – 1 – 6 = –1

Because S is in the 3rd row it can expand beyond its octet by taking advantage of low lying

d-orbitals. Thus S can accommodate 8, 10, or even 12 electrons. This gives rise to alternative Lewis structures:

H N HH

H

S OOO

O

S OOO

O

S OOO

O

2+

S OOO

O

HCO2 H3O HPO4²–

H CO

OH O

H

HP OOO

OH

O3 COBH3NH3

H B N HH

H

H

HO

OO C O

8

P9. In the following molecule which is the least polar bond (other than the C–C bond)? Which carbon has the most positive charge? (L1.7)

The C–H bond is the least polar because the electronegativity difference 2.55 – 2.20 = 0.35 is the smallest.

The largest electronegativity difference (3.44 – 2.55 = 0.89) is between oxygen and carbon, thus it is the most polar bond. The C attached to O has the most positive charge.

P10. For which of the following ion does the formal charge give fairly accurate picture of where the charge really is? Hint: draw Lewis structures and think about electronegativity. (L1.8)

(a) No, formal charge does not fit with electronegativity. According to formal charge the N bears a

positive charge. Compared to ammonia H3N:, the N of ammonium is electron poor. It used to have (in ammonia) a lone-pair all to itself, but in ammonium (NH4

+) it has to share that lone pair with a fourth H in a covalent bond. In that sense N has lost electron density.

On the other hand we know N is much more electronegative than H. Therefore, the positive charge should be spread out over the Hs, not the N. Indeed, advanced calculations (i.e. way beyond formal charge calculations) show that there is more positive charge on the Hs.

(b) No, formal charge again does not fit with electronegativity. According to formal charge the O bears a

positive charge. Compared to H2O, the O of hydronium (H3O+) is electron poor. It used to have (in water) two lone-pairs all to itself, but in hydronium it has to share one of those lone pairs with a third H in a covalent bond. In that sense O has lost electron density.

On the other hand we know O is much more electronegative than H. Therefore, the positive charge should be spread out over the Hs, not the O.

(c) In this case formal charge is a good indicate of where the negative charge is. Formally, NH2¯ has one

more lone pair than NH3, fewer shared electrons, and therefore it has “more” electron density and more negative charge. Electronegativity likewise predicts there should be more negative charge on the N.

H3NCH2CO2 (H3C)3NOH3CNO2

C NO

OH

H

HC C

O

ON

H

HH

H

HN OH3CCH3

CH3

H CH

ClCCl

O

H N HH

H

H OH

H

H NH

H CH

H

9

(d) This one is a bit more complicated. Formal charge clearly shows that the positive charge is on the central C. Electronegativity predicts that positive charge ought to be on H because H is electropositive relative to C.

However the electronegativity argument is misleading in this case. More important is that the central C is non-octet. The fact that C is two electrons short of an octet means it is very electron deficient, which is consistent with the positive charge predicted by formal charge.

P11. Predict the approximate geometry (bond angles) in each of the following (L1.10).

(a) H2O (b) BF4¯ (c) H2CO (d) H3C–C≡N P12. (a) Draw a resonance structure for the allyl anion that shows that the two CC bonds are equivalent

(b) How much negative charge is on each CH2? ½ –

(c) Draw a single structure for the allyl anion that shows shared bonds as dashed lines and charges as partial charges. (L1.14)

P13. The Lewis structure for benzene (C6H6) consists of single and double bonds, which in principle have different lengths. Experiments show however that all CC bonds are of equal length. Account for this observation by drawing a second resonance structure for benzene. (L1.15)

The CC bonds are neither single nor double bonds, but rather something in between and all identical.

P14. (a) Draw a second resonance structure for the allyl cation

(b) How much positive charge is on each CH2? +½

(c) Draw a single structure for the allyl cation that shows shared bonds as dashed lines and charges as partial charges.

(d) What is the bond order of each carbon-oxygen bond? (L1.33) 1½ bonds

P15. (a) Draw two more resonance structure for the carbonate (CO32–)

H2C C CH2

HCH2CH2C

H

CH2CH2CH

1/2 1/2

CC

CCC

CH

H

HH

H

HCC

CCC

CH

H

HH

H

HCC

CCC

CH

H

HH

H

H=

H2C C CH2

HCH2CH2C

H

CH2CH2CH

1/2 1/2

10

(b) How much negative charge is on each oxygen? – ⅔

(c) What is the bond order in each of the carbon-oxygen bond? (L1.34)

(2 bonds + 1 bond + 1 bond) / 3 resonance structures = 1 ⅓

P16. Predict the geometry and approximate bond angles for each of the following:

P17. (a) Construct a hybrid orbital picture for water using sp3 hybrid orbitals on oxygen.

Water has 4 “things” (two bonds and two lone pairs) attached for a nominal hybridization of sp3 and a predicted bond angle of 109.5˚.

(b) Predict any departures from tetrahedral geometry that you might expect from the presence of two lone pairs. (L1.21) The non-bonding pair of electrons on O take up more “space” than the covalent bonds, so the lone pairs “push” away from each other. This has the effect “pushing” the two Hs closer together. The experimentally measured is 104.5˚.

P18. (a) Construct a hybrid orbital picture for hydronium (H3O+) using sp3 hybrid orbitals on oxygen.

CO

O OC

O

O

OC

O O

O

C

H

HHH H C N

C CH

H H

H109.5˚ 180˚

180˚~120˚

4 "things" attached therefore sp³

2 "things" attached therefore sp

3 "things" attached therefore sp²

O C ON

H

HHCl

SCl

O ~109.5˚ ~109.5˚ 180˚



2 "things" attached therefore sp

H2O

~109.5˚

H

H

O

~104.5˚

H

H

OHH

H

11

(b) Is the H–O–H bond angle of hydronium smaller or larger than that of water? (L1.22)

The actual H–O–H bond angle in hydronium (H3O+) is 113.4˚. At first this is a bit puzzling. The comparison with water is not the most useful. A better comparison is with ammonia, which is isoelectronic and trigonal pyramidal. The H–N–H bond angle in ammonia is 107.5˚ Why is the bond angle in H3O+ so much larger? O is more electronegative. That means the electrons are more tightly held. In other words the nitrogen lone pair is “big”, the oxygen lone pair is “small”. Since the oxygen lone pair is small, the Hs can open up and take up more space. In other words, while lone pair interaction is highly repulsive, lone pair / bond interactions are not as destabilizing.

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chapt 1 key - thoburn-chemistry-rmc.org · 2 is a covalent compound with an electronegativity...

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