AP Chemistry 3: Chemical Bonding Name __________________________A. Bonding (8.1 to 8.4)

1. why bond?a. attached (bonded) state is a lower energy stateb. complete valence shell = lower energy state

1. metals lose valence electrons (+ ion)2. nonmetals with 5-7 valence electrons gain

electrons to complete valence shell (– ion)3 nonmetals shares valence e- with nonmetals

a. 1-3 valence electrons share 1 for 1 doubling valence number

b. 4-7 valence electrons share to fill s and p orbitals (8 electrons = octet rule)

c. Lewis symbols1. chemical symbol + dots for valence electrons2. Na•, •Mg•, etc.

d. three major types of bonds1. ionic bond: electrostatic attraction between

cations and anions2. covalent bond: shared electrons between

non-metal atoms3. metallic bond: metal atoms collectively share

valence electrons2. ionic bonding

a. metal and nonmetal: Na(s) + ½ Cl2(g) NaCl(s)b. cations and anions: Na+(aq) + Cl-(aq) NaCl(s) c. cation listed first in formula and name (number of

ions is not indicated in name)d. formula represents simplest ratio of ions to make

a neutral compound—not a moleculee. bond strength—energy needed to break the bond:

E Q1Q2/r, where Q1 and Q2 are ion charges and r is distance between ions

3. covalent bondinga. bonding atoms' orbitals overlap, which maximizes

attraction between nuclei and bonding electronsb. atoms share 2, 4 or 6 electrons

1. 2 (single), 4 (double), 6 (triple) bond2. multiple bonds reduce bond distance

a. bond distance < sum of atomic radiib. shorter bond distance = stronger bond

c. polar bond when electrons are not shared equally1. electronegativity

a. measures atom's attraction for bonding electron pair (higher # = stronger)

b. relative scale where period 2 elements are 1.0 (Li) to 4.0 (F), with 0.5 intervals1. noble gases are excluded2. trend:

a. increase across periodb. decrease down groups

2. bond polaritya. electronegativity difference between

atoms result in uneven sharing of electrons partially positive side, +, and a partial negative side, –

b. notation

c. measured as dipole moment3. bond strength increases with polarity

d. naming binary molecules 1. two types of nonmetals covalently bonded2. + atom is written first with element name 3. second element is given –ide ending4. prefix used to indicate number of atoms

a. 1—mono, 2—di, 3—tri, 4—tetra, etc.b. mono never used for first elementc. example: CO2 is carbon dioxide

5. common names: NH3 (ammonia), H2O2 (hydrogen peroxide) and H2O (water)

B. Lewis Structures (8.5 to 8.7) 1. shows the atoms in a molecule with their bonding and

non-bonding electron pairsa. bonding electrons (– single, = double, triple)b. lone (non-bonding or unshared) electron pair (••)

2. drawing Lewis structures with one central atom count the total number of valence electrons

(subtract charge for ions)CO2: 4 + 2(6) = 16IF2

–: 7 + 2(7) + 1 = 22 draw a skeleton structure

o first element in formula is central, except Ho single bonds to other atoms (max. 4)

O–C–O [F–I–F]–

(ions are bracketed) place electrons around each atom

o 8 total electronso except H, Be and B or when total number of

electrons is an odd number

.. .. .. .. .. ..:O – C – O: [:F – I – F:]–

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

count Lewis structure electrons (including bonding electrons)o if equal to valence electrons, stopo if valence e- < Lewis e-, add additional bonds to

reduces # of electrons by 2'so if valence e- > Lewis e-, add 2 or 4 electrons to

central atom (3rd period or higher); called expanded octet

.. .. .. .... ..O = C = O [:F – I – F:]–

.. .. .. .. ..added bonds expanded octet

3. when more than one Lewis structure is possible use formal charge to decide which is more likelya. each atom is assigned its lone electrons plus half

the bonding electronsb. formal charge = valence e- – assigned e- c. preferred structure

1. atoms have formal charges closest to zero2. negative formal charge reside on the more

electronegative atom (upper right most on the periodic table)

d. example: NCS-

[:::N–CS:]- [::N=C=S::]- [:NC–S:::]-

valence e- assigned e-formal

5 4 67 4 5-2 0 +1

5 4 66 4 6-1 0 0

5 4 65 4 70 0 -1

[::N=C=S::]- is preferred because formal charges are closest to zero and negative charge is on the nitrogen (higher electronegativity)

e. technique can produce erroneous structures (experiments are required to determine actual structure)

C. VSEPR Model (9.1 to 9.3)1. rules

a. maximum separation between electron pairsb. atom positions define molecular geometryc. lone electron pairs squeeze bond angle

(actual angle < ideal angle)ElectronDomains

Domain Geometry – : Molecular

GeometryBond Angle

2 2 0 180o

3

3

2

0

1

120o

4

4

3

2

0

1

2

109.5o

5

5

4

3

2

0

1

2

3

90o

120o

6

6

5

4

0

1

2

90o

2. polar moleculesa. lone electron pairs distort symmetry except for

sp3d-linear and sp3d2-square planarb. different perimeter atomsc. polar interactions increase water solubility,

increase melting and boiling temperatures, decrease evaporation (volatility)

D. Valence-Bond Theory (9.4 to 9.5)1. explains electron domain geometries in terms of

electron orbitals2. atomic orbitals from bonding atoms merge, which

allows single electrons from each atomic orbital to occupy overlapping area and simultaneously attract both nuclei (i.e. H–H: overlap of 1s orbitals)

3. more complex molecules require a fusion of s and p orbitals into equivalent (hybrid) orbitalsa. explains why covalent bonds around an atom are

all the same even if electrons were originally in different shaped (s, p and d) atomic orbitals

b. 1 s + 1 p form 2 sp hybrids (2 electron domains)

ground state excited state hybridized state

s p s p sp p

c. 1 s + 2 p form 3 sp2 hybrids (3 electron domains)

ground state excited state hybridized state

s p s p sp2 p

d. 1 s + 3 p form 4 sp3 hybrids (4 electron domains)

ground state excited state hybridized state

s p s p sp3

e. expanded octet hybridization1. 1 s + 3 p + 1 d = 5 sp3d hybrids (5 domains)2. 1 s + 3 p + 2 d = 6 sp3d2 hybrids (6 domains)

4. not all valence electrons enter hybrid orbitalsa. one electron pair per bond enters a hybrid orbital

1. sigma bond ()2. electrons located between bonding atoms

b. lone pairs of electrons enter hybrid orbitalc. remaining bonding pairs of electrons from multiple

bonds remain in pure p orbitals1. pi bond ()2. electrons located above/below bonding atoms

d. example: ::O=C=O::

p p sp2 p p sp2

sp2 sp sp sp2

p p sp2

sp2 p p

e. bond electrons can spread out across entire molecule (delocalized)1. NO3

- has one bond, which is shared evenly and simultaneously between 3 O's

2. multiple Lewis structures show all possible locations for bonds = resonance forms

3. bond order = sigma bond + share of bonds(each N–O bond has bond order = 1 1/3)

E. Simple Organic Molecules—Hydrocarbons (25.1 to 25.6) 1. general properties

a. contain C and Hb. nonpolar, flammable (fuels)

2. formulas and namesa. number of carbons in parent chain1 2 3 4 5

meth eth prop but pent6 7 8 9 10

hex hept oct non decb. bond between carbons

1. alkanes (all single bonds) end in “ane”2. alkenes (1 or more double bonds)

1 double end in “ene”, 2 double end in "diene"3. alkynes (1 or more triple bonds) end in “yne”4. cyclical

a. 3 to 6 carbon ring with single bonds between carbons: prefix "cyclo"

b. 6 carbon ring with 3 shared bonds: benzene (called aromatic hydrocarbon)

c. branches1. C-branches—“yl”2. benzene branch—"phenyl"3. location of branches

a. number of the parent carbonb. lowest number possiblec. dash: # – word, comma: #, # d. number of branches (2—di, 3—tri, etc.)

3. condensed structural formulaa. hydrogens are written after the carbon b. branches are in parentheses after hydrogensc. example: 4-ethyl-2-methyl-1-hexene

CH2C(CH3)CH2CH(C2H5)CH2CH3

d. semi-condensed (shows branches and bonds) CH3 C2H5 | |

CH2=C–CH2–CH–CH2–CH3

4. functional groupsa. dramatically modify properties of hydrocarbonb. haloalkanes: halogen replaces one or more H

1. reduces reactivity (flammability)2. named as a branch with an “o” ending

c. oxygen containing groups1. hydroxyl group (C–OH)

a. water solubleb. alcohols (-ol ending)c. acids (-anoic acid ending)

2. carbonyl group (C=O)a. aldehydes (-al ending)b. ketones (-one ending)c. esters (-oate ending )

3. ethers have C–O–C (-yl -yl ether ending)4. increases polarity: C–OH > C=O > C–O–C

d. amines1. replace H in ammonia with hydrocarbon

group = amine (CH3NH2 = methylamine)2. when NH2 branches off hydrocarbon = amino

CH3CH(NH2)CH2CH3 (2-aminobutane)3. weak bases (neutralize acids—absorb H+)

5. isomerism a. structural isomers: same molecular formula,

different structure and name1. move double/triple bond position 2. move branch3. form cycloalkane from alkene

b. geometric isomers: same molecular formula and structure, but different spatial arrangement around the >C=C<, where carbons can't rotate 1. x>C=C<x : cis2. x>C=C<x : trans

Experiments1. Isomerism Lab—Make all the isomers for the given formula using the molecular model kits and information provided.

a. C5H12 Draw the structure for each isomer named.

pentane 2-methybutane 2,2-dimethylpropaneb. C4H8 Write the names for each structure drawn.

C = C – C – C

C \

C = C \

C

C C\ /C = C

C|

C = C – CC – C| |

C – C

c. C5H10 Draw the structure and name each isomer.

2. Aspirin Synthesis Lab (Wear Goggles)—Synthesize aspirin from acetic acid and salicylic acid and determine its purity.a. In order to increase the yield of aspirin and to speed up the reaction, acetic anhydride is used, which is made by removing

water from two acetic acid molecules (dehydration synthesis). Highlight the H and OH that are removed from the two acetic acid molecules.

CH3–C–OH + HO–C–CH3 CH3–C–O–C–CH3 + H2O || || || || O O O O acetic acid acetic acid acetic anhydride waterb. Aspirin is an ester that is produced by dehydration synthesis. Highlight the H and OH from the carboxyl group (-COOH)

on acetic acid and the hydroxyl group (-OH) on salicylic acid.CH3-C-OH + HO-C6H4-COOH CH3-C-O-C6H4COOH + H2O

|| || O O acetic acid salicylic acid aspirin waterAdd 2.0 g of salicylic acid, 5.0 mL of acetic anhydride and 5 drops of 85 % H3PO4 (catalyst) to a 125-ml Erlenmeyer flask. Place the flask in a 600-mL beaker half filled with 75oC water and clamp in place. Heat for 15 minutes (stir the contents occasionally with a stirring rod). Slowly add 2 ml of water to the flask to decompose any excess acetic anhydride. When the contents stop smelling like vinegar, remove the flask from the water bath and add 20 ml of water. Put the flask in an ice bath for 5 minutes to hasten crystallization and increase the yield. Collect the aspirin by filtering the cold mixture. To test the purity of the aspirin, add 0.10 g aspirin to 5 mL of 95% ethanol in a 50-mL beaker and dissolve. Add 5 mL of 0.025 M Fe(NO3)3 in 0.5 M HCl and 40 ml of distilled water and stir. Fill a cuvette with the solution and measure the absorbance of the solution at 525 nm.c. Record the absorbance.Salicylic acid forms a magenta complex with Fe3+ and the spectrophotometer measures its color intensity (absorbance). The concentration of salicylic acid, an impurity, is directly proportional to the absorbance, A, according to Beer's law (A = abc), where a (molar absorptivity) = 1000 mol/(L•cm) and b (cuvette path length) = 1.0 cm.d. Calculate c, the moles of salicylic acid per liter.

e. Calculate the moles of salicylic acid in 50 mL.

f. Calculate the moles of aspirin, C9H8O4, in 0.10 g.

g. Calculate the mole percent salicylic acid in the aspirin.

h. Perform the following on the chemical equation for the synthesis of aspirin below.(1) Cross out the bond on the acetic anhydride molecule that is broken during the reaction.(2) Highlight the hydrogen atom on salicylic acid's hydroxyl group and draw a line showing where it attaches to acetic

anhydride to form acetic acid.(3) Draw a line that bonds the remaining half of the acetic anhydride molecule to the salicylic acid molecule.(4) Highlight the catalyst for the reaction.

acetic anhydride salicylic acid aspirin acetic acid

Practice ProblemsA. Bonding

1. Consider the main group elements (1-2, 13-18).a. Record the number of valence electrons.b. Draw the Lewis dot structure for element "X" c. Record the ionic charge when forming ionic bondd. Record the total number of electrons surrounding the

atom when forming covalent bond(s)1 2 13 14 15 16 17 18

a

b X X X X X X X X

cd

2. Illustrated below are four ions—A+, B+, C- and D-—showing their relative ionic radii.

A+ B+ C– D–

What combinations of ions are impossible?

What would have the greatest lattice energy?

What would have the least lattice energy?3. The table lists the ionic radius (x 10-10 m) of common ions.

Li+ (0.68) Be2+ (0.31) O2- (1.40) F- (1.33)Na+ (0.97) Mg2+(0.66) Al3+ (0.51) S2- (1.84) Cl- (1.81)a. Use the above information to estimate the relative

lattice energy for each ionic bond. E Q1Q2/d (Q1 and Q2 = ionic charge and d (rcation + ranion)

Ionic Bond Relative Lattice Energy

LiF

MgO

NaCl

Al2S3

b. Ionic compounds melt when the temperature is high enough to break the ionic bond. Rank the above compounds in order of lowest to highest melting point.

c. What is the relative lattice energy for the strongest ionic bond formed from the ions listed in the table?

4. Use the electronegativity values to answer the questions. H 2.1Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.2K 0.8 Ca 1.0 Ga 1.6 Ge 1.8 As 2.0 Se 2.4 Br 2.8

Rb 0.8 Sr 1.0 In 1.7 Sn 1.8 Sb 1.9 Te 2.1 I 2.5a. What is the range of electronegativities?

metals metalloids nonmetals

b. Rank the following bonds from most polar (1) to least polar (6). Place + next to the atom with the lower electronegativity.

N–S O–S F–S

P–S S–S Cl–S

5. Consider the following data for hydrogen halides.

H-Halide

Electronegativity difference

Dipole moment

Bond Strength(kJ/mol)

Bond Length

(Å)HF 1.9 1.82 436 0.92HCl 0.9 1.08 431 1.27HBr 0.7 0.82 366 1.41HI 0.4 0.44 299 1.61

Indicate whether the following correlate directly or inversely.

Direct Inverse

Electronegativity difference & dipole momentElectronegativity difference & bond strengthDipole moment & bond strengthBond length & bond strength

6. Consider the following data for C-C bonds.C-C bond Single Double Triple

Bond Strength (kJ/mol) 348 614 839How does bond strength correlate with the number of shared electrons?

7. Complete the chart with the formula or name of the binary molecule.

Formula NameN2O5

carbon tetrachloride

CO2

nitrogen monoxide

OF2

hydrogen bromide

HBr(aq)

B. Lewis Structures8. Record the number of covalent bonds typically

formed by the main group elements.1 2 13 14 15 16 17 18

9. In the Lewis structure shown below, A, D, E, Q, X and Z represent non-metal elements in the first two rows of the periodic table. Identify the elements.

A D E Q X Z

10. Draw the Lewis structure for the diatomic molecules.

H2 N2 O2 F2

11. There are three ways to draw Lewis structures for NCO–

a. Calculate the formal charge for each versionStructure [:::N–CO:]– [::N=C=O::]– [:NC–O:::]–

FormalCharge

b. Which is the preferred structure? Give two reasons.

12. Draw Lewis structures for the following molecules where the first atom listed is the central atom, unless indicated.CH4 NH3

CO2 CH2O

POCl3 (lowest formal charge)

CNO– (lowest formal charge)

SCN– (C is central atom) BCl3

SF6 BrF3

H2O (O is the central atom)

HF (F is the central atom)

SO3 N2O (lowest formal charge)

IF5 AsF5

SF4 SO2Cl2 (lowest formal charge)

RnCl2 IF4–

XeF4 NO2

C. VSEPR Model13. Make the following ideal molecules using clay and tooth

picks. Complete the table for each ideal molecule.

(–) (• •) DomainGeometry

BondAngl

e

Molecular Geometry Polar?

2 03 03 14 03 14 25 04 13 22 36 05 14 2

14. Based on the Lewis structures from question 11, determine electron domain geometry, molecular geometry, polarities and bond angle (indicate if it is less than the ideal angle "<").Molecul

e Domain Geo Molecular Geo Polarity Angle

CH4

NH3

CO2

CH2O

POCl3

CNO-

SCN-

BCl3

SF6

BrF3

H2O

HF

SO3

N2O

IF5

AsF5

SF4

SO2Cl2

RnCl2

IF4-

XeF4

NO2

D. Valence-Bond Theory15. Based on the Lewis structures from question 11, determine

the number of bonds, the number of bonds, number of lone electron pairs, hybridization around the central atom, and the bond order for the perimeter atoms.

Molecule

Bonds

BondsLone Pairs Hybridization Bond

OrderCH4

NH3

CO2

CH2OPOCl3CNO-

SCN-

BCl3SF6

BrF3

H2OHFSO3

N2OIF5

AsF5

SF4

SO2Cl2RnCl2

IF4-

XeF4

NO2

16. Label the hybridization for each carbon atom.

17. Draw the resonance structures for the following molecules.Molecule Resonance Structures

CO32-

SO2

E. Simple Organic Molecules—Hydrocarbons 18. Draw the structure formulas, including hydrogen, and

names for the first six members of the alkane series.

19. Draw semi-condensed structural formulas for the following organic compounds.

ethane benzene

1,3-pentadiene 3-ethylbutyne

diethylamine ethanoic acid (acetic acid)

2,2-dimethylpropane methylbenzene

ethanol (ethyl alcohol) 2-propanol (isopropyl alcohol)

propanone (acetone) diethyl ether (ether)

methanal (formaldehyde) methyl propanoate20. Name the following organic molecules from their semi-

condensed structural formulas.

CH3 CH3 | |CH3 – C – CH2 – CH – CH3 |

CH3|

C2H5

H2C–CH2

/ \HC C – CH2Cl

Cl-CH CH2

\ /H2C–CH2

CH3 – CH2 – C – OH || O

CH3 – O – C – CH2 – CH3 || O

F F F| | |

CH2 – CH – CH2

..CH3 – N – CH3|

CH3

C2H5|CH3 – CH2 – C – CH2 – CH3|

C2H5

CH3 CH3

\ /C = C/ \

H H

CH3 – C – CH2 – CH3 || O

HO – CH2 – CH2 – CH3 CH3 – O – C3H7

CH3 – CH2 – CH2 – CH || O

OH|

CH3 – C – CH3| OH

CH3 – CH2 – CH – CH3 | NH2

H2N – CH2 – C – OH || O

21. Draw the structures and name six isomers of C4H10O, which include 4 alcohols and 2 ethers.

SummaryBonding

Bonds are classified into three broad groups: ionic bonds are the result of electrostatic forces between cations and anions; covalent bonds form when electrons are shared between non-metal atoms; and metallic bonds, which bind metal cations with mutually shared valence electrons.

Bonds involve the interaction of valence electrons, which are represented by electron-dot symbols, called Lewis symbols. The tendencies of atoms to gain, lose, or share their valence electrons often follow the octet rule, which can be viewed as an attempt by atoms to achieve a noble gas electron configuration.

The strength of the electrostatic attractions between ions is measured by the lattice energy, which increases with ionic charge and decreases with distance between ions.

Electronegativity measures the ability of an atom to attract electrons in a covalent bond. Electronegativity generally increases from left to right in the periodic table and decreases down a column. The difference in atoms' electronegativities is used to determine the polarity of a covalent bond; the greater the difference, the more polar the bond. A polar molecule has a positive side (+) and a negative side (–). The separation of charge produces a dipole, the magnitude of which is given by the dipole moment. Polar bonds are stronger and shorter than non-polar bonds.

Bond strength and length is also affected by the number of shared electrons. Sharing of one pair of electrons produces a single bond; whereas the sharing of two or three pairs of electrons produces double or triple bonds, respectively. Multiple bonds are stronger and shorter than single bonds.

The procedures used for naming two-element, binary, molecular compounds follow the rules below.

1. The lower electronegative element is written first in the formula and named as an element.

2. The name of the second element is given an –ide ending.

3. Prefixes are used to indicate the number of atoms of each element; mono is not used with the first.

Lewis StructuresElectron distribution in molecules is shown with Lewis

structures, which indicate how many valence electrons are involved in forming bonds and how many remain as unshared electron pairs. If we know which atoms are connected to one another, we can draw Lewis structures for molecules and ions by a simple procedure, where eight electrons are placed around each atom. When there are too few valence electrons, then it will be necessary to add double or triple bonds. When there are too many valence electrons (and the central atom has at least third energy level electrons), then it will be necessary to place additional electrons (up to 10 or 12) around the central atom; forming an expanded octet. When the total number of valence electrons is an odd number, then it will be necessary to place seven electrons around the atom with the odd number of valence electrons.

When there are multiple valid Lewis structures for a molecule or ion, we can determine which is most likely by assigning a formal charge to each atom, which is the sum of half the bonding electrons and all the unshared electrons. Most acceptable Lewis structures will have low formal charges with any negative formal charge on the more electronegative atom.

VSEPR Model The valence-shell electron-pair repulsion (VSEPR) model

rationalizes molecular geometries based on the repulsions between electron domains, which are regions about a central atom where electrons are likely to be found. Pairs of electrons, bonding and non-bonding, create domains around an atom, which are as far apart as possible. Electron domains from non-bonding pairs exert slightly greater repulsions, which leads to smaller bond angles than idealized values. The arrangement of electron domains around a central atom is called the electron domain geometry; the arrangement of atoms is called the molecular geometry.

Certain molecular shapes have cancelling bond dipoles, producing a nonpolar molecule, which is one whose dipole moment is zero. In other shapes the bond dipoles do not cancel and the molecule is polar (a nonzero dipole moment). In general non-bonding pairs of electrons around the central atom produce polar molecules.

Valence-Bond TheoryValence-bond theory is an extension of Lewis's notion of

electron-pair bonds. In valence-bond theory, covalent bonds are formed when atomic orbitals on neighboring atoms overlap. The bonding electrons occupy the overlap region and are attracted to both nuclei simultaneously, which bonds the atoms together.

To extend valence-bond theory to polyatomic molecules, s, p, and sometimes d orbitals are blended to form hybrid orbitals, which overlap with orbitals on another atom to make a bond. Hybrid orbitals also hold non-bonding pairs of electrons. A particular mode of hybridization can be associated with each of the five common electron-domain geometries (linear = sp; trigonal planar = sp2; tetrahedral = sp3; trigonal bipyramidal = sp3d; and octahedral = sp3d2).

Covalent bonds formed between hybridized electrons are called sigma () bonds, where the electron density lies along the line connecting the atoms. Bonds that form between non-hybridized p orbitals are called pi () bonds. A double bond consists of one bond and one bond and a triple bond consists of one and two bonds.

Sometimes a bond can be placed in more than one location. In such situations, we describe the molecule by using two or more resonance structures. The molecule is envisioned as a blend of these multiple resonance structures and the bonds are delocalized; that is, spread among several atoms. The bond order value represents the actual bond strength and is sum of the bond plus a share of the bond(s).

HydrocarbonsCarbon molecules (except CO, CO2) are called organic.

Hydrocarbons are organic molecules that contain mostly carbon and hydrogen. The four groups of hydrocarbons are alkanes, alkenes, alkynes, and aromatic. The naming of hydrocarbons is based on the longest continuous chain of carbon atoms in the structure. The locations of alkyl groups, which branch off the chain, are specified by numbering along the carbon chain. Ring structures have the prefix cyclo. The names of alkenes and alkynes are based on the longest continuous chain of carbon atoms that contains the multiple bond, and the location of the multiple bond is specified by a numerical prefix.

The chemistry of an organic compound is dominated by the presence of the functional group.

Isomers are substances that possess the same molecular formula, but differ in the arrangements of atoms. In structural isomers the bonding arrangements differ. Different isomers are given different names. Alkenes exhibit not only structural isomerism but geometric isomerism (cis-trans) as well. In geometric isomers the bonds are the same, but the molecules have different geometries. Geometric isomerism is possible in alkenes because rotation about the C=C bond is restricted.

Practice Multiple ChoiceBriefly explain why the answer is correct in the space provided.1. Types of hybridization exhibited by the C atoms in

propene, CH3CHCH2 include which of the following? I. sp II. sp2 III. sp3

(A) I only (B) II only (C) III only (D) II and III

2. Which molecule contains 1 sigma () and 2 pi () bonds?(A) H2 (B) F2 (C) N2 (D) O2

3. Which molecule has the shortest bond length?(A) N2 (B) O2 (C) Cl2 (D) Br2

4. Which molecule has only one unshared pair of valence electrons?(A) Cl2 (B) NH3 (C) H2O2 (D) N2

5. The electron pairs in a molecule where the central atom exhibits sp3d2 hybrid orbitals are directed toward the corners of (A) a tetrahedron (B) a square pyramid(C) a trigonal bipyramid (D) an octahedron

6. The SbCl5 molecule has trigonal bipyramid structure. Therefore, the hybridization of Sb orbitals should be(A) sp2 (B) sp3 (C) dsp2 (D) dsp3

7. For which molecule are resonance structures necessary to describe the bonding satisfactorily?(A) H2S (B) SO2 (C) CO2 (D) OF2

8. Which molecules have planar configurations? I. BCl3 II. CHCl3 III. NCl3

(A) I only (B) II only (C) III only (D) II and III

9. CCl4, CO2, PCl3, PCl5, SF6

Which does NOT describe any of the molecules above?(A) Linear (B) Octahedral (C) Square planar (D) Tetrahedral

10. The geometry of the SO3 molecule is best described as (A) trigonal planar (B) trigonal pyramidal(C) square pyramidal (D) bent

11. Pi () bonding occurs in each of the following EXCEPT(A) CO2 (B) C2H4 (C) CN- (D) CH4

12. According to the VSEPR model, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the (A) increasing strength of the bonds (B) decreasing size of the central atom (C) increasing electronegativity of the central atom (D) increasing number of unshared pairs of

electrons

Questions 13-15 refer to the following molecules. (A) PH3 (B) H2O (C) CH4 (D) C2H4

13. The molecule with only one double bond

14. The molecule with the largest dipole moment

15. The molecule that has trigonal pyramidal geometry

16. Which molecule has a zero dipole moment?(A) HCN (B) NH3 (C) SO2 (D) PF5

17. Which molecule has the largest dipole moment? (A) CO (B) CO2 (C) O2 (D) HF

18. Which molecule has a dipole moment of zero? (A) C6H6 (benzene) (B) NO(C) SO2 (D) NH3

19. Which pair of atoms should form the most polar bond?(A) F and B (B) C and O(C) F and O (D)N and F

20. Which pair of ions should have the highest lattice energy?(A) Na+ and Br- (B) Li+ and F-

(C) Cs+ and F- (D) Li+ and O2-

21. Which compound has the greatest lattice energy?(A)BaO (B) MgO (C) CaS (D)MgS

22. Which molecule has the weakest bond?(A) CO (B) O2 (C) Cl2 (D) N2

23. How are the bonding pairs arranged in the best Lewis structure for ozone, O3?(A) O–O–O (B) O=O–O (C) OO–O (D)

O=O=O

24. Which species has the shortest bond length?(A) CN- (B) O2 (C) SO2 (D) SO3

25. Which species has a valid non-octet Lewis structure?(A) GeCl4 (B) SiF4 (C) NH4+ (D) SeCl4

26. The Lewis structure for SeS2 with zero formal charge has (A) 2 bonding pairs and 7 nonbonding pairs of electrons.(B) 2 bonding pairs and 6 nonbonding pairs of electrons.(C) 3 bonding pairs and 6 nonbonding pairs of electrons.(D)4 bonding pairs and 5 nonbonding pairs of

electrons.

27. Which molecular shape cannot exhibit geometric isomerism?(A) tetrahedron (B) square planar(C) trigonal bipyramid (D) octahedron

28. Which molecule is NOT polar?(A) H2O (B) CO2 (C) NO2 (D) SO2

29. Which species has sp2 hybridization for the central atom?(A) C2H2 (B) SO32- (C) O3 (D) BrI3

30. In which species is the F-X-F bond angle the smallest?

(A) NF3 (B) BF3 (C) CF4 (D) BrF3

31. For ClF3, the electron domain geometry of Cl and the molecular geometry are, respectively, (A) trigonal planar and trigonal planar. (B) trigonal planar and trigonal bipyramidal.(C) trigonal bipyramidal and trigonal planar.(D) trigonal bipyramidal and T -shaped.

32. The size of the H-N-H bond angles of the following species increases in which order? (A) NH3 < NH4

+ < NH2- (B) NH3 < NH2

- < NH4+

(C) NH2- < NH3 < NH4+ (D) NH2- < NH4+ < NH3

33. What is the molecular geometry and polarity of BF3?(A) trigonal pyramidal and polar(B) trigonal pyramidal and nonpolar(C) trigonal planar and polar(D) trigonal planar and nonpolar

34. Which set does not contain a linear species?(A) CO2, SO2, NO2 (B) H2O, HCN, BeI2 (C) OCN-, C2H2, OF2 (D) H2S, CIO2-, NH2-

35. The hybrid orbitals of nitrogen in N2O4 are(A) sp (B) sp2 (C) sp3 (D) sp3d

36. How many sigma and how many pi bonds are inCH2=CH–CH2–C–CH3? ||

O (A) 5 sigma and 2 pi. (B) 8 sigma and 4 pi. (C) 11 sigma and 2 pi. (D) 13 sigma and 2 pi.

37. What is the best estimate of the H-O-H bond angle in H3O+?(A) 109.5o (B) 107o (C) 90o (D) 120o

38. Which of the following pairs of compounds are isomers? (A) CH3–CH2–CH2–CH3 and CH3–CH(CH3)–CH3 (B) CH3–CH(CH3)–CH3 and CH3–CH3–C=CH2

(C) CH3–O–CH3 and CH3–CO–CH3 (D) CH3–OH and CH3–CH2–OH

39. CH3–CH2–CH2BrWhich of the following structural formulas represents an isomer of the compound that has the structural formula represented above? (A) CH2Br–CH2–CH2 (B) CH3–CHBr–CH3

(C) CH2Br–CH2–CH2Br (D) CH3–CH2–CH2–CH2Br

Practice Free Response1. Rank the oxides in order of greatest lattice energies (1) to

least (3), without looking up any values. Explain.BaO CaO MgO

2. There are several oxides of nitrogen; among the more common are N2O, NO2 and NO3

-.a. Draw the Lewis structures of these

molecules.

N2O NO2 NO3-

b. Which of these molecules "violate" the octet rule? Explain.

c. Draw resonance structures of N2O.

d. For each resonance structure from question 2c, calculate the formal charge and evaluate which structure is most likely. Explain.

e. Which side of the N–O bond is +? Explain.

f. Rank the strength of the N–O bond in order of strongest (1) to weakest (3). Explain your answer.

N2O NO2 NO3-

3. Complete the chart for SeF2, SeF4, and SeF6.SeF2 SeF4 SeF6

Lewis Structure

Se- HybridizationDomain GeometryMolecular GeometryIdeal Bond AnglePolarity

4. Consider the ion SF3+.

a. Draw a Lewis structure.

b. Identify the type of hybridization exhibited by sulfur.

c. Identify the electron-domain and molecular geometries.

Electron-domain geometry Molecular geometry d. Predict whether the F-S-F bond angle is equal to,

greater than or less than 109.5°. Explain

5. Consider the ion SF5-

a. Draw a Lewis structure.

b. Identify the type of hybridization exhibited by sulfur.

c. Identify the electron-domain and molecular geometries.Electron domain geometry Molecular geometry

6. Two Lewis structures can be drawn for the OPF3 molecule.Structure 1

..: O :

.. | ..: F – P – F :

.. | ..: F :

..

Structure 2

: O :.. || ..

: F – P – F :.. | ..

: F :..

Which Lewis structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.

7. a. Draw the condensed structural formula and name four structural isomers of C4H9Cl.

Formula Name

b. Draw the structural formula and name two geometric isomers of C2H2Cl2.

Formula Name

8. Compounds of Xe and F form molecules where the hybridization of Xe is sp3d and sp3d2. Write the formula and draw the Lewis structure for the two molecules.

sp3d sp3d2