2. determine the arrangement ofelectron pairs using the

AP Chem — Norato/SunCh.9classwork

When using Localized Electron Model to describe bonding in a molecule, follow thesethree steps,

1. Draw the Lewis structure2. Determine the arrangement of electron pairs using the VSEPR model.3. Specify the hybrid orbitals needed to accommodate the electron pairs.

Use Localized Electron Model to describe bonding in the following molecules.

1. CO

2. CO2

3. BFZI

4. XeF2

5. C3H4

6. 13

7. N2

Drawing VESPR Structures

Name: Date:__________ Period:________

Determine the geometry of each of the following structures or ions.

1. Co2

2. SO

3. BrF3

4. Xe04

5. 1C12+

Arrangement of Valence Pairs(Bonding & Lone)

ElectonicTotal RegionsStructure

2 Linear

3 Trigonal Planar4 Tetrahedral

TrigonalBypyramidal

6 Octahedral

* An electronic region is the sharedelectrons between 2 atoms. One regionmay include one, two or three pairs ofelectrons.

Predicted Molecular Structure (VSEPR)Total Sets of Electronic Regions

Bond Angles 180°

2 3 4 5 6

0

1200

Linear

Q_4l —o

Trigonal Planar0

109.5° 120°/90° / 90°

I

Tetrahedral Trigonal Bipyramidãi Octahedrãf0 0a

o——-oa

2

Bent Trigonal Pyramidal Seesaw Square Pyramidal0;

DOAE /.{q0

Ci)

9ci0)

0

0..•0

Cl)

Linear Bent

3

T-shape

a

Square Planar

•3 Al3Linear

4

Linear T-shape

- ,--, F.

5

Linear Linear

- -

Linear

-4

:t

I

•

-

t—%

‘w.

-‘F

Ch.9Covalent Bonding

9.1 Hybridization

Problems in BondingExample: methane, CH4

i. If kept separate, the bondingbetween valence orbitals should bevaried:

• one type for is from H and 2s from C• one type for is from H and 2p from Cbut all the bonds are equal(bond energies!)

2. 2p orbitals are perpendicular but C-Hbonds are at 109.5° angles

I

Problems in BondingExample: methane, CH4

H

H—C-—H

H

(a) (b)

py

y/ê’x

P2:

Solutions for BondingExample: methane, CH4

> carbon must accept a new set oforbitals used for bonding other thanits 2s and 2p

> all 4 must be the same> must form tetrahedral shape (109.5°)

hybridization - mixing of normalorbitals to for special bonding orbitals

2

sp3 hybridization of CH4

four new sp3 orbitals are created fromone 2s and three 2p orbitalsidentical in shape:• one large lobe• one small lobetetrahedral arrangement (1O95°)

> When a set of equal tetrahedrallyarranged orbitals is needed, the atomadopts a set of four sp3 orbitals


HIIyhndizal,on

E

Hybridization

—2s

Orbitaisinafree C atom

sp3

Orbitals in C inthe CH4 molecule

3


P3

sp2 hybridization of C2H4 /14c=c

> three new sp2 orbitals are created fromone 2s and two 2p orbitals

>identicalin shape:• one large lobe• one small lobe

> trigonal planar arrangement (120°)> When a set of equal trigonal planar-

arranged orbitals is needed, the atomadopts a set of three sp2 orbitals

4

sp2 hybridization of C2H4

2p

E

Hybridization

— 2s

2p

Orbitals in an isolatedcarbon atom

— — — sp2

Carbon orbitalsin ethylene

r—Sp2 /sp2 —ï-—


FH1s

—H1

-sp2

H1- sp2

Jsp2—

sp2

5


> What happens to the other 2p orbitalleft out of the hybridization in each C?

> they share an electron pair> occupies the space above and below

the sharing between sp2 orbitals> forms pi (rc) bond

sigmaband


> the sp2 orbitals are used to shareelectrons too

> shared in the area centered on a directline between 2 atoms

> form a sigma (a) bond

1ti++ —aitbond

+____.___

6

crJ N C C n

x\

/

./\ \ \

CV C rD C,,

Ctr C,,

C (Th C C

,)

C,,

C,)

CC C

,,

Tj

C

C C Cii

C,,

sp hybridization of CO2two new sp orbitals are created fromone 2s and one 2p orbitals

> identical in shape:• one large lobe• one small lobe

> linear arrangement (1800)

> When a set of equal linearly-arrangedorbitals is needed, the atom adopts aset of two sp orbitals

:o=cd:

sp hybridization of CO2

;514’ HAI 4

E

Hybridization

2s

2p

Orbitals in afree C atom

—sp —sp

Orbitals ip the sphybridized C in CO2

8


sp2

C

sp2

0sp

sp2

sp2


each C?

What happens to the other two 2porbitals left out of the hybridization in

they share electron pairsextra 2p orbital from sp2

with the onehybridized 0

spsp2-1

p

9


pi bond(1 pair ofelectrons)

sigma bond(1 pair of electrons)

sp3dhybridization of PCi5:C1:..I4,C1:

:C1—..

:C1:> five new sp3d orbilals are created from

one 3s, three 3p, and one 3d orbitals> trigonal bipyramidal arrangement> When a set of equal

trigonal bipyramidal-arranged orbitalsis needed, the atomadopts a set offive sp3dorbitals

dsP3r?

dsp

10

sp3d2 hybridization of SF6 :iJ:•_*

%b••> six new sp3d2 orbitals are created :F F:from one 3s, three 3p, and two 3dorbitals

> octahedral arrangeme>Whena set of

equal octahedrally-arranged orbitalsis needed, the atomadopts a set of sixsp3d2orbitals

Rq

?

11

(b)

1:wo pi

Example 1 Describe the bonding inammonia using the LE model

5P3

IHF’3

steric #: 4PG:tetrahedral

is

MS: trigonalpyramidal

LH N H

is

three sigma bonds, no pi bonds

H

•M M••IN

ExampleDescribe the bondingusing the LE

2in

modelN2

lone pair sigma bond lone pair

N N 4

SI, SI, SI, SI,

one sigmabond

bonds

(dl

12

r..__.-‘._•• 1” Example 31:1 I 1:1L J Describe the bonding in

steric #: 13 using the LE model

> PG: trigonal bipyramidal> MS: linear

requires 5: sp3d for central I• 3 hold lone pairs• 2 make sigma bonds with outside I’s sp3

> requires 4: sp3 for outside I’s• 3 hold lone pairs• 1 makes sigma bond with central I’s sp3d

Describe the bonding in Ex p1 4XeF4 usmg the LE model••

>steric#:6> PG: octahedral> MS: square planar> requires 6: sp3d2 for central Xe

• 2 hold lone pairs• 4 make sigma bonds with outside F’s sp3

> requires 4: sp3 for outside F’s• 3 hold lone pairs• 1 makes sigma bond with central Xe’ssp3d2

13

AP Chemistry — NoratoChapter 8 BondingStudy Guide

Example 91 A sp3 Hybridization

Describe the bonding in the water molecule using the localized electron model.

Strategy

We must establish the VSEPR structure. This is done by drawing the Lewis structure and determiningthe number of effective electron pairs around the central atom.

Solution

The Lewis structure ofH20 is (see Sections 8.10 and 8.13)

H\/H

Four effective electron pairs around th oxygen atom indicate a tetrahedral basis leading to aV-shaped structure. Therefore the oxygen is sp3 hybridized. Two sp orbitals are occupied with lselectrons from hydrogen. The other two sp3 orbitals are occupied by lone pairs.

In molecules or ions with a trigonal planar configuration (3 effective electron pairs around an atom), sp2hybridization occurs. With this hybridization, an s orbital and twop orbitals are used. That leaves anunhybridized (unchanged)p orbital perpendicular to the sp2 plane.

Look at jgire 9.12 in your textbook.

• Bonds formed from the overlap of orbitals in the plane between two atoms are called sigma (a) bonds.• Bonds formed by the overlap ofunhybridizedp orbitals (above and below the centerp1ane) are called

p1 (sr) bonds.• A single bond is a a bond.• A double bond consists of one a and one r bond.• A triple bond consists of one a bond and two t bonds.

Example 9.1 B Sigma and P1 Bonds

How many a bonds are there in the commercial inseôticide, “Sevin,” shown below? How many itbonds?

OH H

ii I IH O—C—N—C---—H

I I IH H H

I II IH

I IH H

Solution

There are 27 a bonds (single bonds and one of the bonds in a double bond).There are 6 it bonds (the other bond in a double bond).

AP Chemistry — Norato

Chapter 8 Bonding

Study Guide

Your textbook goes over a variety of different hybridization schemes, all of which center on the idea thatthe hybridization of the orbitals of an atom depends on the total number of effective electron pairsaround it. In order to determine the hybridization of an atom, it is essential that you can figure out its VSEPRstructure.

The following table summarizes effective electron pairs around an atom with hybridization. Rememberthat, for purposes of the VSEPR model, double and triple bonds count as only one effective electron pair.

Effective ElectronPairs Around an Atom Arrangement Hybridization

2 linear sp3 trigonal planar sp24 tetrahedral sp35 trigonal bipyramid sp3d6 octahedral sp3d2

Keep in mind that a ligand can have a hybridization different from that of the central atom. Each atom in amolecule must be considered separately based on the Lewis and VSEPR structures of that molecule.

Example 9.1 C Practice with Hybrid Orbitals

Give the hybridization, and predict the geometry of each of the central atoms in the followingmolecules or ions.

+a. tF2b. OSF4 (sulfur is the central atom)c. SiF62’d. HCCH (work with both carbons)

Solution

a. Lewis structure: : F—I—F:

All three atoms have 4 effective electron pairs. They are all sp3 hybridized. The VSEPRstructure has a tetrahedral basis. Because the central atom (iodine) has two bonding pairs, it willtake on a V-shape, but the bond angle will be smaller than 109.5°.

b. Lewis structure: : 0:Il-F:

Sulfur has 5 effective electron pairs. The VSEPR structure is a trigonal bipyramid. Sulfur is sp3dhybridized, Each of the fluorines has 4 effective electron pairs.

c. Lewis structure: : F:f__F:

Silicon has 6 effective electron pairs. The VSEPR structure is octahedral. Silicon is sp3d?hybridized. Each of the fluorines has 4 effective electron pairs.

AP Chemistry — NorateChapter 8 BondingStudy Guide

d. Lewis structure: H—CC—H

Each carbon has 2 effective electron pairs. The VSEPR structure is linear. Each carbon is sphybridized. Hydrogen atoms bond using1s orbitals. The orbitals are unhybridized.

Example 9J D Summing it All up

Answer the following questions regarding aspartame (Nutrasweetm1).

a. How many bonds are in the molecule?b. How many t bonds?c. What is the hybridization on carbon”a”? Carbon ‘9,”?cL What is the hybridization on nitrogen “a”? Oxygen “a”?e. What is the CbOliHa bond angle?

So’ution

You must remember to complete the octets in this “shorthand” Lewis structure. Lone pairs are often“assumed,” so always be on the lookout.

a. 39abondsb. 67tbondsc. sp3, sp2 (3 effective electron pairs)d. sp3 (remember the “assumed” lone pair to complete the octet), sp2 (3 effective electron pairs,.

including the two to complete the octet)e. Ob is sp3 hybridized (complete the octet!); therefore, the angle is based on a tetrahedron, with two

lone pairs compressing the C—O—H bond angle to 104.5°.

H

H

H—N\

H 0

AP Chemistry — NoratcChapter 8 BondingStudyGuide

QgIi:i

(chemistry 6th eci page 372/7th ed. page 352)

As the number of bonds between two atoms increases, the bond

grows shorter and stronger.

EXAMR.E: Arrange the following molecules in order of decreasing

C-C bond length: C2H4,C2H2,C2H6.

In order of decreasing C—C bond length: C2H€, C2H4,C2H2.

First, draw the Lewis structure for each molecule. The C—C triple

bond is th shortest, followed by the double bond, and the single

bond.

Molecular Shapes and Bond Angles

sp

ElectronPair No. of No. of

No. of Geometry Bonding LoneElectron (Bond Electron Electro Molecular HybridPairs Angle) Pairs n Pairs Geometry Formula 2D Structure Orbital

2 Linear 2 0 Linear BeHz(180°)

3 Trigonal 3 0 frigonal C032 sp2planar lanar(120°)

3 Trigonal 3 1 3ent N02 sp2planar(120°)

4 Tetrahedral 4 0 fetrahedra CH4(109.5°)

sp3

AP Chemistry — Noratc..Chapter 8 BondingStudy Guide

retrahedral 4(900 and>1200)

Seesawunsyrnmetricaltetrahedron

Electron 2D Structure Hybrid

Pair No. of No. of Orbital

No. of Geometry Bonding Lone

Electron (Bond Electron Electro Molecular

Pairs ngle) Pairs n Pairs Geometry Formula

4 rtrahedral 3 1 rrigonal NTh sp3

(>109.5°) pyramidal I

y____

4 retahedral 2 2 Bent HzO sp3

(>>109.5°)

@—py.

5 Tri011al 5 0 rngonal PCIs spd

midal midal(90°, 120°)

5 1 SF4 sp3d


,

Electron HybridPair No. of No.of OrbitalNo, of Geometry Bonding LoneElectron (Bond Electron Electron MolecularPairs ng1e) Pairs Pairs Geometry Formula 2D Structure

5 tetrahedral 3 2 r’-shaped BrF3 sp3d(900 and

5 tetrahedral 2 3 Linear 1C12 sp3d(180°)

6 Octahedral 6 6 Octahe- SF6 sp3d2(90°) dral

6 Octahedral 5 1 Square BrFs sp3d2(90°) pyramidal

24

AP Chemistry — Norato’Chapter 8 BondingStudy Guide

ElectronPair No. of No. of

No. of Geometry Bonding LoneElectron (Bond Electron Electron MolecularPairs Angle) Pairs Pairs Geometry Formula 2D Structure

6 Octahedral 4 2 Square ICW sp3d2(900) planar

EXAMPLE: Predict the molecular structure of the carbon dioxidemolecule. Is this molecule expected to have a dipole moment?

First we must draw the Lewis structure for the CO2 molecule.

O=c=O

In this structure for C02, there are two effective pairs around thecentral atom (each double bond is counted as one effective pair).There are two atoms attached to the central atom,

According to the table, a linear arrangement is required.

Each C—O bond is poiar, but, since the molecule is linear, thedipoles cancel out and the molecule is nonpolar.

EXAMPLE: Draw the Lewis structure for PF5 and identify themolecular geometry.

PF5 has 40 valence electrons. Connecting 5 F atoms to P. thecentral atom, uses 10 electrons in five single bonds. The 30remaining electrons can be used to satisfy the octet rule for eachF atom. There are 5 electron pairs around P and 5 atomsattached to P. resulting in a trigonal bipyramidal shape.

EXAMPLE: Draw the Lewis structure for CH3OH.

H

This molecule has more than one central atom, the carbon andthe oxygen. Hydrogen cannot be in the middle since it can form

25


only one bond. Draw the correct Lewis structure and then usingthe rules for VSEPF{, assign the molecular geometry around eachcentral atom. The carbon atom has four electron pairs around itand four atoms attached resulting in tetrahedral geometry.Oxygen also has four electron pairs around it, but only has twoatoms attached. This leads to bent geometry around oxygen.

EXAMPLE: Arrange the following molecules in order of increasingbond angles: H2S, CC14,NF3.

First, draw the correct Lewis structure for each molecule andthen assign an approximate bond angle for each.

Each of the four molecules has four electron pairs around thecentral atom, resulting in a tetrahedral arrangement of theelectron pairs about the cetitral atom and approximately a 109.5bond angle. HS has two lone pairs on the central atom; NF3 hasone lone pair on the. central atom. These lone pairs repel morethan the bonded electron pairs and will cause the bond angle tobe smaller than the expected 109,5°. The more lone pairs, thesmaller the bond angle.

In order of increasing bond angles, the answer is H2S <NP3 <

CC’4.

ExercisesSection 8.1

i.fl Indicate whether the bonds between the following would be primarily covalent, polar covalent, or ionic:a. ()—f’{ C. H-Clb. Cs—Cl d. Br—Br

2. Calculate the energy of interaction for KCI if the internuclear distance is 0.3 14 nm.

3. Calculate the energy of interaction between Ag and Bf if the internuclear distance of AgBr is 0.120 rim(in Id/mole).

Section 8.2

4. )sing a periodic table, order the following from lowest to highest electronegativity:a. Fr, Mg, Rb c. P, As, Ga, 0b. B, Al, C, N d. Cl, 5, P

5.) Using the periodic chart of elements, place the following in order from the lowest to the highestelectronegativity:

F, Nb, N, Si, Rb, Ca, Pt

6. Using Figure 8.3 in your textbook, calculate the difference in electronegativity () for each of thefollowing bonds:

Cl-Cl c. Fe—C) ) S-Hb. K-Br d. H—C)


7. Place the following in order of increasing polarity:

NaBr, ‘2, H20, MnO2, CN

8. Which of the following molecules contain polar covalent bonds? List in order of increasing bondpolarity. (Use Fig. 8.3 in your text).

03, Pg, NO, C02, CH4, H2S

9. How will the charge be distributed on each of the following molecules: HF, NO, CO. and HCI?

10. Why is it that BeF2 is ionic, and BeCI2 is öovalent?

Section 8.3

II. Determine the orientation of the dipole of the following, if any.

a. AICI3 (planar with aluminum atom at the center)b. CH3F (tetrahedral with carbon at the center)c. N20 (linear with N-N-O structure)d. AgCI.4 (planar molecule, silver atom at center, Ag--Cl bonds 900 apart)

12. Which of the molecules in problem 11 contain one or more polar bonds?

13. Which of the following molecules would you expect to have a dipole moment of zero? Describe thedipole orientation of the other two molecules.

a, KI c. H2Se (bent structure)b. CF4 (tetrahedral structure)

Section 8.4

14. Determine the most stable ion for each of the following atoms, and indicate which element they wouldbe isoelectronic with if they lost or gained electrons:

a. 0 c. I e. Nab. Be d. Te

15. List four ions that are isoelectronic with argon and have charges from —2 to +2. Arrange these in orderof increasing ionic radius.

16. List four ions that are isoelectronic to Kr. Arrange these in order of increasing radius.

17. Determine the formula of the binary compound formed from the following sets of atoms.

a. Caand0b. KandClc. RbandSd. BaandP V

18. Predict formulas for the following binary ionic compounds.

a. MgandN V

b. NaandF V

c. CaandSd. SrandTe

V7

AP Chemistry — Norato,Chapter 8 BondingStudy Guide

19. Using shorthand notation, list the core electron configurations for the ions rn the compounds inproblem 18.

20. Place the following in an order of increasing ionic size. Use the shorthand notation to list the coreelectron configuration for each of the ions.

a. Ba2,Te2,Cs, fb. Cs, S2. 2— K

21. Place the following in an order of increasing ionic size. Use shorthand notation to list the core electronconfiguration for each of the ions.

a. C1, F, Sr, Ca2 b. Na, Mg2,Li, Be2

Section 8.8

22. Using the bond energy values listed in Table 8.4 of your text, calculate the All for the followingreactions:

a. 2H2(g) + 02(g) — 2H20(g)b. 2C2H6g) + 702(g) -+ 4C02(g) + 6H20(g)c. HCN(g) + 2H2(g) - CH3NN2(g)

23, Use bond energy values from 1L4 jny Jcipck to calculate AN for the following reactions:a. H-CC-H(g) + H2(g) -* CH2CH2(g)b. 2CH4(g) + 302(g) — 2C0(g) + 4H20(g)c. N2(g) + 2H2(g) - N2H4(g)(M12-NH2)

24. Compare the values obtained in parts “a” and “b” of Problem 23 to AN values calculated from AR° datain Appendix 4 in your textbook.

25. Calculate the enthalpy of reaction Aff for the following reaction. Use the enthalpies of formationfound in Appendix 4 of your textbook.

H2(g) + C2H4(g) - C2H6(g)

Section 8.10

26. Draw Lewis dot structures for the following atoms, ions, or molecules:a. Sr d. Ga g. NH2b. 3r e, GaC14’ h. CSe2c. ICN f. P’

27. Draw Lewis structures for the following:a. C. P e.Crb. C d. P5’’

28. Draw Lewis- structures for the following:a. AsF3 c. H30’ e. NHb. 03 d. BH. f. 0


Section 8.11

29. Draw Lewis dot structures for the following:

a, BCI3 (i’) Br03b. AsF5 () S2F10 (contains a S—S bond)

30. Draw Lewis dot structures for the following:

a. SbCI3AIF6

C. PCI5

Section 8.12

Assign formal charges to each of the labeled atoms, a-e.e

• 2—•Br. c1=C—0:

0C—Br d :S—O: : ci:a b c:):

Draw the remaining resonance forms forN204.

/0

N—N

How many reasonable resonance structures are there for carbon monoxide, CO. What are the formalcharges?

Hydrazin, N2H4,is used as a propellant on the Space Shuttle. Draw all reasonable structures forN2H4,and assign formal charges.

Draw a Lewis structure and any resonance forms of benzene, C6H6. (Benzene consists of a ring of sixcarbon atoms with one hydrogen bonded to each carbon.)

Section 8.13

Predict the structure of each of the following molecules or ions:

a. SeF6 C. CIF e. CF3CI (carbon is central atom)b. N20 d. C10

Predict the structure of HCN.

Using the VSEPR model, determine the molecular geometry for each of the following molecules:

a. SCI4 c. IF4 e. TlCl2b. H2Se d. SnClj

Which of the molecules or ions in Problem 36 contain polar covalent bonds? e polar?

2


SectIon 9,1 9What geometry do the following hybrid bonds possess?a. sp C. sp3 e. sp3d2b. sp2 d. sp3d

Predict the type of hybrid orbital that the central atoms of each of the following compounds display:a. S1H4 d. NC13 f. SiF6b. H30 e. AsH3 g. CH3C. PC15

Predict the geometries of the following compounds:

a. SF2 b. SF4 c. XeF2 d. XeF4 e. IF5 f. CIF3

Predict the geometry about the indicated atom, and identify the hybridization of each atom.a. the two carbon atoms and the nitrogen atom of glycine

H o

H—C—Cat \

0

b. the carbon atom in CF2CI2c. the phosphorous atom in PCi5d. the nitrogen atom in NH2

Section 9.2

() Determine the number of sigma and pi bonds in each of the following:

HHO I1 H

H)c=—c—oH b. H—CN c. HCC—C<

HH H

The structure of urea is

H O.HI II I

H—N—C—N—H

a. How many a bonds are there?b. How many it bonds?c. What is the hybridization at the carbon?d. How are the nitrogen atoms hybridized?e. What is the N—C—N bond angle expected to be?f. How many lone pairs of electrons are there?

30

AP Chemistry — NoratoChapter 8 BondingStudy Guid

Answers to Exercisesa. polar covalent c. polar covalentb. ionic d. covalent

2. —736 x io J

3. E —1160 Id/mole

4. a. Fr,Rb,Mg c. Ga,As,P,Ob. A1,B,C,N d. P,S,C1

5. Rb<Ca<Nb<Si<Pt<N<F0.8 1.0 1.6 1.8 2.2 3.0 4.0

6. a. 0 c. 1.7 e. 0.4b, 2.0 d. 1.4

7.12<CW<H20<NaBr<Mn020 0.5 1.4 1:9 2.0

8. 03, P8 < H2S, CH4 < NO < CO2

9. HF;NO;CO;HC1+—+--. +—+—

10. The difference in the electronegativity between Be and F is higher than the one between Be and Cl.3.0— 1.5 1.5 (covalent bond)4.0— 1.5 2.5 (ionic)

11. a. no dipole c. negative toward 0b. negative toward F d. no dipole

12. all

13. b. The opposing bond polarities in a tetrahedral structure cancel out. Thus CF4 has no dipole moment.KI is a binary ionic compound that has a negative dipole toward I. Selenium will have a partial negativecharge as its electronegativity is greater than that of hydrogen. Thus the resulting dipole moment ofH2Se would be orientated as shown:

H H

14. a. O2, isoelectronic with neonb. Be2, isoelectronic with heliumc. F, isoelectronic with xenond. Te2, isoelectronIc with xenone. Na4, isoelectronic with neon

15. Ca24,K4, CF. S2

16. Sr < Rb4 < Br <Se2

17. a. CaO c. Rb2Sb. KCI d. Ba3P2

31

18. a, Mg3N2b.NaF

c. CaSd. SrTe

19. a. [Ne], [Ne]b. [Ne], [Ne]

20. a. Ba2’’<Cs<F<Te2;b. K < Cs <O2’ <

21. a. Ca2<Sr2<F”<CFb. Be2 < Li < Mg2< Na

c. [Ar], [Ar]d. [Kr], [Xe]

all can be written as [Xe]K” [Ar], Cs [Xe], O2 = [Ne], S2” = [Ar]

Ca2” [Ar], Sr2 = [Kr], F = [Ne]. CV [Ar]Be2 = [He], Li” = [He], Mg2”= [Ne], Na = [Ne]

22. a. —509kJ b. —2881kJ C. —lS8kJ

23. a. —169kJ

24. a. 6kJ difference

25. iH = —136.7 Id/mole

26. a. •Sr.

b.. —1091 kJ

b. 7kJdifference

d,.Ga.

C, I +81 Id

g. [H——Hj’

b. [:Br:] ci: —

:C1—Ga—C1:

ci:

h. SeC=Se

c. : I—CN:

27. a. [H]

b. • C.

13—f. :P:j

c. : P.

d. [P]5

[:ci:J

28. a. : F—As—F:

: F:

c. H—O—H +

H

e. H +

H—N—H

H

b. 0=0—0: d. H

H—B—H

H

f. 00

AP Chemistry — Noratc,Chapter 8 BondingStudy Guide


29. a. : Cl—B—Cl:

ci:

b. : F.L_..F:As/\

:F: :F:

c. :O—Br—O:

0:

:F: :F: :F: :F:\/ \/d : F—S S—F:

../\ /\..:F: F: :F: :F:

30. a. : Cl—Sb—Cl:

31. a. 0b. 0

Ci:

.. .. 3—b. :F::F:\/

F—Al—F:.. / \ ..:F::F:

C. —1d. +1

:ci:..c. :ci.ja:

:c’a:

e. +1£0

0\

____

N—N 4/ \

o 0

0\ /0

33. : oc: is the only reasonable structure. The carbon has a formal charge of—i, and oxygen has a +1formal charge. It is possible to draw a structure such as o=c: in wbich both C and 0 have a zeroformal charge, but the carbon would be electron deficient (lacks an octet).

34. The only reasonable structure has an N—N single bond. All formal charges =0.

H H\ ../

N—N/.. \

HHEach nitrogen has a formal charge of+1 and the two hydrogens that are single have formal chargesof—i.

H

35.

II

H

36. a. octahedralb. linear

H

H H

I IIH H

H

c. see-sawd. linear

e. tetrahedral

37. HCN is linear (H—CN:)

32. 0 0/

N—N -40\

33

AP Chemistry — NoratoChapter 8 BondingSwdy Guide

38. a. see-saw c. square planar e. linearb. bent d. trigonal bipyramidal

39. All contain polar covalent bonds; b, c, d, and e are polar. Negative is toward: 0 in b; equatorial fluorineinc;Oind;fluorineine,

1. a, linear c. tetrahedral e. octahedralb. trigonal planar d. trigonal bipyramidal

2. a. sp3 c, dsp3 e. sp3 g. sp2b. sp3 d. sp3 f, d2sp3

3, a, angular (like water) c, linear e, square pyramidalb. see-saw ci square planar f. T-shaped

4. a. carbon a-sp3,tetrahedral; carbon b-sp2,trigonal planar; nitrogen-sp3,based on tetrahedrontrigonal pyramid

b. sp-tetrahedralc. dsp3-trigonal bipyramidd. sp3-based on tetrahedron-bent’

5, a. 8sigrna,2pib. Ssigrna,2pic. 7sigma,3pi

6. a. 7 c sp2 e. 1200b. l d.sp3 f.4

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2. determine the arrangement ofelectron pairs using the

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