2. determine the arrangement ofelectron pairs using the
TRANSCRIPT
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
sp3 hybridization of CH4
HIIyhndizal,on
E
Hybridization
—2s
Orbitaisinafree C atom
sp3
Orbitals in C inthe CH4 molecule
3
sp3 hybridization of CH4
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-—
sp2 hybridization of C2H4
FH1s
—H1
-sp2
H1- sp2
Jsp2—
sp2
5
sp2 hybridization of C2H4
> 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
sp2 hybridization of C2H4
> 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
sp hybridization of CO2
sp2
C
sp2
0sp
sp2
sp2
sp hybridization of CO2
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
sp hybridization of CO2
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
AP Chemistry — NoratoChapter 8 BondingStudy Guide
,
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
AP Chemistry — NoratoChapter 8 BondingStudy Guide
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)
AP Chemistry — NoratoChapter 8 BondingStudy Guide
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
AP Chemistry — NoratoChapter 8 BondingStudy Guide
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
AP Chemistry — NoratoChapter 8 BondingStudy Guide
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
AP Chemistry — NoratoChapter 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
34