Chemical Bonding - Mr. Mello's AP & HONORS CHEMISTRYseekonkchemistry.weebly.com/.../prsi.bonding.pdf · • METALLIC BONDING ... bonds are neither 100% ionic or 100% covalent and

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Chemical Bonding

Chapter 8Each person should get out a small

WHITE BOARD

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Salt vs Sugar• NaCl• Ionic (metal+non)• Electrolyte• High MP (801ºC)

• C6H12O6

• Molecular (non+non)• Nonelectrolyte• Lower MP (185ºC)

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Who gets with who?• Metals + nonmetals

• IONIC

• Nonmetals + nonmetals• MOLECULAR

• Metals + metals• METALLIC BONDING (Alloys if a mixture.)

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Valence Electrons• Lewis Structures

– (aka: Lewis Symbols or Dot Diagrams or Lewis Dot Diagrams)

• Phosphorus– Electron configuration: [Ne] 3s23p3

– Lewis Structure

– Shows only the valence electrons

P

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Electronic Nirvana• The octet rule - 8 valence electrons• The reasons that atoms come to the party to

bond and form molecules is to obtain an octet– H only acquires a “duet” 2 electrons

• Using Lewis Structures to account for all the electrons and attempt to achieve “electronic nirvana.” i.e stable.

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Drawing Lewis Structures1. Sum the valence electrons.

– for polyatomic ions assign an extra electron for each negative charge.

– Subtract and electron for each positive charge

2. Connect all atoms with single bonds3. Complete the octet of all atoms.4. If there are not enough electrons to go

around, use multiple bonds (one at a time).5. Put excess electrons on the central atom

– This is called an expanded octet.

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Rules to guide you when deciding what connects to what.

• Sometimes the order listed may help.• Central atom is often written first.• H’s and F’s are terminal (on outside).• C’s love to hook together and do not like to be

terminal. – Avoid unshared pairs on C’s.

• O’s do not bond together.– Except in O2 and in peroxide molecules

• Avoid rings of 3 or less atoms.

Lewis Structures1. H2

2. O2

3. N2

4. CH4

5. NF3

6. H2S

7. CH2O

8. (CH3)2CO

9. SO2

10.CO28

Formal ChargeChoosing between two

legitimate structures

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You might be asked to decide between more than one possible structure?

• First be sure you have followed the guidelines above that would eliminate some possible structures.

• Chemists have invented a tool called formal charge that will provide information to help eliminate some structures.

C OO＝＝−−− − ∣C OO ≡∣−

−−

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Applying Formal Charge• Count the number of valence electrons “assigned” to each atom.

– Unshared pairs belong entirely to the atom on which they reside.– Shared pairs are divided evenly between the atoms on which they are

shared: half “assigned” to one atom and half belong to the other atom.• Subtract the number of valence electrons that are “assigned” to each atom

in the molecule from the number of valence electrons that each atom has as an isolated atom (FORMULA: valence electrons in the un-bonded atom – electrons that are “assigned” to the atom in the molecule)– If the number of electrons that are “assigned” to the atom in the molecule

is one more than the number of valence electrons in the elemental atom, the formal charge is -1 (two more, the formal charge is −2)

– If the number of electrons that are “assigned” to the atom in the molecule is one less than the number of valence electrons in the elemental atom, the formal charge is +1 (two less, the formal charge is +2)

– If the number of electrons that are “assigned” to the atom in the molecule is equal to the number of valence electrons in the elemental atom, the formal charge is 0

• The formula with the least amount of formal charge will be prefered.

C OO＝＝−−− −

0 0 0∣C OO ≡∣−

−−-1 +10

Charges, Formal and oxidation #• Something to remember about formal

charge and oxidation number.

• Formal charge is the charge on an atom if the bonds were 100% covalent.

• Oxidation number is the charge on an atom if the bonds were 100% ionic.

• Of course, bonds are neither 100% ionic or 100% covalent and so oxidation numbers and formal charges are not "real" charges, just "constructs" that help us do chemistry.

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Using formal charge to evaluate the validity of structures

✓The sum of the formal charges must equal the charge of the species

✓Molecules have a charge of 0✓A polyatomic ion’s charge should equal the formal

charge.

• Preferred structures have the lowest amount formal charge‣ Atoms without any charge (or just the charge of the ion)

is ideal‣ Lower charges are preferred‣ Two charges of −1 and +1 would be better than −2 and 0

• All else being equal, any negative charge that must exist is best placed on the most electronegative atom.

• Ultimately it may be necessary to have other information to determine the “actual” structure.

Lewis Structures1. NO3−

2. NO2−

3. CO32−

4. SO42−

5. SO32−

6. ClO3−

7. ClO−

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Expanded Octet• Elements with three or more energy levels

have unused d-orbitals which can be accessed to put electrons into for bonding to allow for expanded octets.

• The maximum number of domains we will study will be 6– 7 and 8 do exist, but they are quite rare − and

go well beyond the scope of this course.

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Lewis Structures

1. PCl5

2. SF4

3. ClF3

4. XeF2

5. SF6

6. BrF5

7. XeF4

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Bond LengthSingle.Double.Triple

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Bonds − Length & Strength• Single

• Overlap of one pair of electrons - Longest = weaker

• Double• Overlap of two pair of electrons - Shorter = stronger

• Triple• Overlap of three pair of electrons - Shortest = strongest• more on why it is the shortest in the next chapter.

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Length (A) / strength (kJ/mol) Length (A) / strength (kJ/mol)

614839305615891

3587451070

160418945201607

204498

PolarityUnfair Sharing of Electrons

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Bond Polarity• Electronegativity

– A measure of the ability of an atom (that is bonded to another atom) to attract electrons to itself.

– Increases up and to the right on the periodic chart.

• To determine polarity of bond– compare electronegativity of the two atoms.– Treat each bond individually.

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Determining Polarity & Symbolizing Polarity

δ+ δ- or

2.1 - 4.0 = 1.9, very polarF is the negative end.

H F

H F

H F

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Polarity is a Continuum

• Nonpolar covalent

• Polar covalent

• Ionic

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Nonpolar - Polar - Ionic• Using shape and color as a visual

representation of the three types of bonds.• Polarity is measured as a dipole moment

measure in debye units.0 D 1.92 D

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Hydrogen-halide series.• Larger halide atoms mean smaller electronegativity.• Difference in electronegativity decreases.• Therefore polarity decreases and measured dipole

moment is smaller.

C6H14 (nonpolar) H2O (polar)• C6H14 has polar bonds symmetrically arranged,

and the molecule ends up nonpolar.• H2O has polar bonds not symmetrically

arranged, and the molecule ends up polar.

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The effect of a charged rod

• The stream of polar water coming from a buret is bent by the charged rod.

• Nonpolar hexane is not bent.

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Of the 3 isomers that exist for C2H2F2 how many are polar?1. 1 structure2. 2 structures3. all 3 structures4. none are polar

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Of the 3 isomers that exist for C2H2F2 how many are polar?1. 1 structure2. 2 structures3. all 3 structures4. none are polar

cis-dichloroethenecis- trans-

1,1 dichloroethenetrans dichloroethene

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Which of these structures is most polar?

1. Ortho2. Meta3. Para4. Ortho and meta are equally polar5. All three structures are equally polar6. None of the structures are polar

ortho parameta

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Which of these structures is most polar?

1. Ortho2. Meta3. Para4. Ortho and meta are equally polar5. All three structures are polar• This is a planar molecule, and the

molecule will be most polar when the polar C-Cl bond is oriented in the same direction.

ortho parameta

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Select the non-polar bond(s) shown below. Select all that apply.

1. C−F2. C−H3. C−O4. C−C5. C=C

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Select the least polar bond(s) shown below1. C-F2. C-H this bond is considered nonpolar.

However it is ever SO slightly polar.3. C-O4. C-C5. C=C• Whether single or double, the polarity or

lack there of is calculate in the same way.

IsomersSame Chemical Formula

Different Arrangement of atoms

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Isomers• Isomers - Molecules with the same chemical

formula, but different arrangement of atoms.• Draw a Lewis structure for C2H6

• Draw a Lewis structure for C2H4F2

– Draw isomers and name these isomers– Free rotation around single

• Draw a Lewis structure for C2H2F2

– Draw isomers– Let’s name these isomers

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Isomers• Sketch and build the molecule C4H10

• Sketch and build the molecule C4H8

• Draw as many isomers as you can for this molecule.

Draw a Lewis Structure for C2H4F2

1. Draw as many isomers as you can

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Isomers are:1. same chemical structure, different

number of neutrons.2. same chemical structure, different

arrangement of electrons.3. same chemical formula, but different

arrangement of atoms.4. same chemical formula, different

number of electrons.

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Isomers are:1. same chemical structure, different

number of neutrons.2. same chemical structure, different

arrangement of electrons.3. same chemical formula, but different

arrangement of atoms.4. same chemical formula, different

number of electrons.

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Draw the structure C2H4F2Select the number of different isomers that can exist for C2H4F2

1. 1 structure2. 2 structures3. 3 structures4. 4 structures5. 5 structures

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Select the number of different isomers that can exist for C2H4F2


This structure is the same as the one above it because of free rotation around the single bond.

same

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Draw the structure C2H2F2



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cis-1,2 dichloroethenecis- trans-

1,1 dichloroethenetrans 1,2 dichloroethene

ResonanceSame arrangement of atoms

Movement of electrons

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Resonance• Two or more Lewis Structures in which;

– The connection and placement of the atoms is the same

– But the arrangement of the electrons is different• In contrast with an isomer which is a

difference arrangement of atoms

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Carboxylic Acids: stable conjugate• After the H+ falls off, the C=O and C−O bonds can exhibit

resonance✓ Resonance is the relocation electrons in an available nearby

location resulting in an “averaging” of the structures.

• Resonance allows the electrons to “spread out,” distributing the charge and stabilizing the conjugate base.

H

Resonance − we draw two (or three) separate Lewis structures, but really, the molecule is a blend of the various structures we can draw.

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Resonance in Ozone: O3• Draw a Lewis structure for ozone.• Re-sketch the molecule with electrons in a different

arrangement - This is called a resonance structure.• You wouldn’t ordinarily write two different versions of this, but

chemists noticed that the two bonds in ozone were the same length AND that they both were shorter than a single bond, but longer than a double bond.

• Chemists think that molecule does not actually flip between the two structures, but is an average of the two.

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Resonance in N2O• Draw a Lewis structure for N2O

– Draw two isomers• N−N−O• N−O−N

– Can you draw resonance structures for these?

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Resonance in N2O• Draw four or five Lewis structures for N2O to

represent two different isomers.– and draw at least two resonance structures for

each isomer– How do we decide which is more likely to occur

• Formal Charge

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Resonance in N2O• Draw a Lewis structure for N2O

– Put one of the N’s in the middle.– Putting the O in the middle leads to lousy formal

charge• Draw two other resonance structures.

• Draw an isomers of the structure above and it’s resonance structure.

−＝− ＝−O NN− −

−− ≡ ∣O NN∣

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Using Formal Charge on N2O• Apply formal charges to the 3 resonance

structures of N2O.

• Remember the rule:– If you must choose between two equal amounts

of formal charge, the negative formal charges should reside on the more electronegative atoms.

+1+1 00 -1-2-1 +1+1

−＝− ＝−O NN− −

−− ≡ ∣O NN∣-1 +2 +2-1 -2 0

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Resonance in Benzene: C6H6

• Draw a Lewis Structure C6H6 with 30 valence electrons - make a ring structure.

• Resonance

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Resonance in Benzene: C6H6

• C6H6 with 30 valence electrons - make a ring structure - double bonds will be necessary.

• The molecule does not flip between the two structures, but is an average of the two.

• Bond length studies tell us that all 6 C-C bonds have the same length - intermediate between single and double.

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Representing Benzene: C6H6

• This is an alternative way of representing carbon structures.

• At each corner the appropriate number of hydrogens are attached.

• The circle structure on the right emphasizes the delocalization of the electrons.

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Which structure(s) below exhibit(s) resonance structures: A) CO2 B) NO2

− C) SO2

1. B2. C3. A & C only4. B & C only5. A, B, & C

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Which structure(s) below exhibit(s) resonance structures: A) CO2 B) NO2

− C) SO2 1. B2. C3. A & C only4. B & C only5. A, B, & C

While you can write a resonance structure for CO2, with a triple bond and a single bond, the formal charge is very poor, thus the resonance structure is not valid.

−−O−− −−＝O−S

＝ −− −O−−O− − S＝ −− −O−−O− − N

−−O−− −−＝O−N−

−

O −−＝CO− −＝O −≡C−O− −

−

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The End

Exceptions to the Octet RuleOther than the

Expanded Octet

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Exceptions to the Octet Rule• Odd number of electrons.• Sketch a Lewis Structure for

– NO, NO2

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Odd Number of ElectronsFree Radicals

• Free radicals are very unstable and react quickly with other compounds, trying to capture the needed electron to gain stability.

• Generally, free radicals attack the nearest stable molecule, "stealing" its electron.

• When the "attacked" molecule loses its electron, it becomes a free radical itself, beginning a chain reaction.

• Once the process is started, it can cascade, finally resulting in the disruption of a living cell.

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Exceptions to the Octet Rule• Odd number of electrons.

– NO, NO2

• Less than an octet.– BF3

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Less Than an Octet• Sketch a Lewis Structure for BF3

• BF3 gas is very reactive because of its incomplete octet.

• It reacts with molecules such as water and ammonia with their available unshared pair of electrons.BF3 + NH3 → H3NBF3

• In H3NBF3 the boron would now have an octet.

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Exceptions to the Octet Rule• Odd number of electrons.

– NO, NO2

• Less than an octet.– BF3

• More than an octet.– Expanded octet on a central atom– only atoms that have unused “d” orbitals– Period 3 and beyond

Silly Chemistry Jokes

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Early chemists describe the first dirt molecule.

HybridizationSince one bonding model

may not be enough!

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Why another model?• VSEPR was useful to predict the 3-D shape of

molecules, with little explanation as to why the bond exists.

• We have said that a bond is an overlap of orbitals.• In polyatomic molecules we need to blend the

concept of overlap of orbitals with the observed geometry.

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Hybrid Orbitals• Atomic orbitals have particular shapes.• When an atom is about to become part of a

molecule, that atom’s atomic orbitals must morph into a new set of orbitals with different shapes and orientations than the original atomic orbitals. These new morphed orbitals are called hybrid orbitals.

• The orientation of hybrid orbitals is the same orientation of the shapes you learned from the VSEPR model.

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Summary of Hybrid OrbitalsAtomic orbitals that morph together

To make hybrid orbitals

The resulting shape of the hybrid orbitals

s + p(px py or pz)

Two equivalent sp orbitals

Linear

s + p + p Three equivalent sp2 orbitals

Trigonal planar

s + p + p + p Four equivalent sp3 orbitals

Tetrahedral

s + p + p + p + d

Five equivalent sp3d orbitals

Trigonal bipyramidal

s + p + p + p + d + d

Six equivalent sp3d2 orbitals

Octahedral

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sp hybrid

s p

sp sp

sp sp

Linear

next....

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sp2 hybrids p

psp2 sp2

sp2

sp2sp2

sp2

next....

Trigonal Planar

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sp3 hybrid

sp3

sp3sp3

sp3

p

p

p

s

Tetrahedral

next....

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Bonding is Overlap of OrbitalsNH3

• So when N gets ready to bond in the ammonia molecule, its 5 valence electrons in the s and p orbitals are morphed into 4 orbitals called sp3 hybrid orbitals.

• One of the sp3 orbitals will have 2 electrons and the other three sp3 orbitals each have 1 electron.

• Hydrogen only has 1 electron in a single s orbital, so there is no morphing to be done.

• Three of the hybrid sp3 orbitals in the N overlap with the single atomic s orbital in each of the three H’s to make the three bonds.

sp3sp3

sp3

sp3

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So What about Multiple Bonds?- Double and Triple Bonds -Sigma (σ) and Pi (π) Bonds

• Overlap region that occurs directly between two nuclei on the internuclear axis is called a sigma bond - these are single bonds.

• Overlap of two p orbitals oriented perpendicular to the internuclear axis is called a pi bond - the other part(s) of the double and triple bonds.

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Pi (π) Bonds - The “other” part of Double and Triple Bonds

• Side to side overlap of the p orbitals.

• Above and below the internuclear axis.

• Total overlap covers less area causing the pi (part) bond to be weaker.

• The “hot dog bun” bonds.

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Draw a Lewis structure for C2H4

• Let’s look at what’s going on around each CHow many domains around one of the C’s?

3 domainsWhat is the domain shape?

Trigonal planarWhat is the name of the hybrid orbitals?

The trigonal planar geometry around each carbon indicates sp2 hybridorbitals.

CCH

CH

H H

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Hybridization in C2H4

• The view below looks down on the molecule. The unhybridized p orbitals are not showing. They would be projecting above and below this picture.

• The sp2 hybrids on the carbons are overlapping with the s orbitals of the hydrogens.

CCH

CH

H H

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Hybridization in C2H4• Let’s consider the unhybridized p orbitals on the

carbon again and look at the side view.• Overlap of those p orbitals completes the pi bond

(the double bond)Note that the pi bond has two parts (one part above the

sigma bond and one part below the sigma bond).Only 1 part is labeled.

Cπ

σσσ

σ σC

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Sigma (σ) and Pi (π) Bonds in C2H4

• Look at your Lewis structure of C2H4 • Indicate the Sigma (σ) and Pi (π) Bonds

CH

CH

H H

Between the C’s 1 σ bond and one π bond

1 σ bond in each C-H bond

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Hybridization in CH2O• Sketch Lewis Structure CH2O• What is the # of domains around the C and the O?

3 domains around each• What does this tell us about the hybridization around the C

and O? sp2 hybridization on both C and O

• This leaves one p orbital unhybridized on each atom. This unhybridized p overlaps to make the π bond

C OH

Hsp2

π p

p

sp2sp2 sp2

sp2sp2

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Draw a Lewis Structure for CO2• What is # number of domains around the C?

2 domains

• What does this tell us about the hybridization around the C?sp hybridization on the C

• This leaves two p orbitals unhybridized on the C. These unhybridized p’s overlap to make the π bonds

C OO

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More about hybridization in CO2• What are the number of domains around one of the O’s

3 domains

• What would be the hybridization orbitals? sp2 hybridization.

• This would leave one unhybridized p orbital around the oxygen.

C OO

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Hybridization Overlap in CO2• The view below looks down on the molecule. The

unhybridized p orbitals are not showing. They would be projecting above and below this picture.

• The carbon’s sp hybrid orbitals overlap with the oxygen’s sp2 hybrid orbitals.

• Note how the oxygen orbitals are oriented in opposite planes.

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Bonding in CO2

• Put the sigma bonds together with the pi bonds for the entire molecule.

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Draw a Lewis structure for C6H6• What’s the number of domains and their shape around each C?

3, trigonal planar• What’s the hybridization around each C?

Trigonal planar geometry around each carbon indicates sp2 hybridization• This leaves a single p orbital on each carbon projecting above and below

the planar ring.• The p orbitals overlap for pi bonding and because of the alternating

double bonds you can write resonance structures• These electrons are said to be “delocalized.”

tip this ring flat

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Draw a Lewis structure for N2• The presence of a triple bond tells us that there are how

many unhybridized p orbitals? Two p orbitals on each N must be unhybridized.

• What is the hybridization of each N? 2 p’s unhybridized indicates sp hybridization

• The last diagram shows the sigma overlap of the sp hybrid orbitals to make up one third of the triple bond.

N N

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Bonding in N2

• Pi orbitals overlap above and below the internuclear axis.

• The “hot dog bun” bonds !

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Draw a Lewis structure for PF5

• How many domains and the shape around the P? 5 domains, trigonal bipyramid.

• What is the hybridization of around the P? sp3d hybridization

• How many domains and the shape around each F? 4 domains, tetrahedral shape

• What is the hybridization around each F?

• sp3 hybridization

FP

F

F FF

sp3d

sp3dsp3d

sp3dsp3d

sp3sp3

sp3

sp3

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Geometry, Hybridization & σ, π bonds

• Name the electron domain geometry. tetrahedral

• Name the molecular geometry (shape). bent

• Give the bond angle <<109.5º

• Name the hybridization orbitals. sp3

• List the number of σ and π bonds. 2 σ bonds, no π bonds

Around the left O Around the Cl• Name the electron domain geometry.

tetrahedral• Name the molecular geometry (shape).

Trigonal pyramid• Give the bond angle

<109.5º• Name the hybridization orbitals.

sp3


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Geometry, Hybridization & σ π bonds


• Name the molecular geometry (shape). tetrahedral

• Give the bond angle 109.5º





• Give the bond angle <<109.5º



Around the C Around the O

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• Name the electron domain geometry. linear

• Name the molecular geometry (shape). linear

• Give the bond angle 180º

• Name the hybridization orbitals. sp

• List the number of σ and π

bonds. 2 σ bonds, 2 π bonds

Around the left C

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• Name the electron domain geometry. Trigonal planar

• Name the molecular geometry (shape). trigonal planar

• Give the bond angle (6) >120º


• List the number of σ and π bonds. 3 σ bonds, 1 π bonds



• Give the bond angle (7) <<109.5º



Around the left C Around the right O

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Bonding and Geometry Review

Chapter 8 & 9True / False

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The Lewis structure of N2 is:

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The Lewis structure of N2 is:

• False• There is a triple bond and one unshared pair

of electrons on each N for a total of 10 (not 12) valence electrons in the molecule.

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The product in the reaction:BF3(g) + NH3(g) → F3BNH3(g)contains a single covalent bond between boron and nitrogen

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The product in the reaction:BF3(g) + NH3(g) → F3B:NH3(g)contains a single covalent bond between boron and nitrogen• True• This bond forms because nitrogen can donate

its two non-bonded electrons to form what is sometimes called a coordinate covalent bond. Ultimately it’s a bond - made of two shared electrons - just like any other bond.

• This is a Lewis acid/base reaction.• LA - accepts electrons, LB - donates electrons

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The bond between Ca and O is covalent in character.

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The bond between Ca and O is covalent in nature.

• False: • It is ionic. Oxides of metals are always ionic

as it is made of a metal and a nonmetal.

100

For the following hydrogen halides, bond polarity increases in the order listed below. HCl < HBr < HI

101

For the following hydrogen halides, bond polarity increases in the order listed below. HCl < HBr < HI

• False.• Polarity increases with electronegativity

on the halogen. Thus the order is HI < HBr < HCl

102

Review:For the following hydrogen halides, acidity increases in the order listed below. HCl < HBr < HI

103

Review:For the following hydrogen halides, acidity increases in the order listed below. HCl < HBr < HI• True.• The Acidity increase is NOT due to the

polarity of the bond, but instead associated with the length of the bond.

• While they are all strong acids with very large Ka values HI is the strongest.

104

In MgCl2 there is both an ionic Mg-Cl bond and a covalent Cl-Cl bond.

105

In MgCl2 there is both an ionic Mg-Cl bond and a covalent Cl-Cl bond.

• False:• Mg is ionic bonded to both Cl’s which are not

bonded to each other.

106

The lattice energy of MgO is greater than that of NaCl.

107

The lattice energy of MgO is greater than that of NaCl.

• True. • Lattice energy increases with increasing ionic

charge. (The most important of the two factors.) The charge on Mg and O is 2 (not 1 as in Na and Cl) thus the lattice energy in MgO is greater.

• The second factor is that lattice energy increases with decreasing size of the ions. For this compound the size of O2− is significantly smaller than Cl− (In addition, Mg2+ is slightly smaller than Na+) making MgO an even tighter lattice.

108

In H3CCH3 there exists both sigma and pi bonds.

109

In H3CCH3 there exists a both sigma and pi bonds.• False.• There are no double bonds in this

molecule, thus there are no pi bonds.

CH C

H

H H

H

H

110

In HCCH, acetylene, there exists two pi bonds.

111

In HCCH, acetylene, there exists two pi bonds.

• True.• There is a triple bond between the two

C’s and and a triple bond always contains two pi bonds.

C CH H

112

Fe can form a 2+ ion because it loses its two 4s valence electrons.

113

Fe can form a 2+ ion when it loses its two 4s valence electrons.• True. • Fe 4s2 3d6

• ⊗ ⊗ ∅ ∅ ∅ ∅• The 4s electrons are furthest away from the

nucleus and get ripped off by other atoms.• One of the 3d6 electron is also quite willing

to leave and thus Fe often forms 3+ ion as well. Fe+3 O ∅ ∅ ∅ ∅ ∅

114

In the BF3 molecule, the differing electronegativities of B and F cause the bonds and the molecule to be polar.

115

In the BF3 molecule, the differing electronegativities of B and F and cause the bonds and the molecule to be polar.

• False• The B-F bonds are polar, but because the

molecule is trigonal planar and all three bonds are the same, the polarity cancels out.

116

There are three isomers each for the following molecules: C2H4F2 and C2H2F2

117

There are three isomers each for the following molecules:C2H2F2 and C2H4F2

• False there are only two for C2H4F4 because there IS rotation around a single bond, but there ARE three for C2H2F2 because there is NO rotation around a double bond

cis-1,2 dichloroethenecis- trans-

1,1 dichloroethenetrans 1,2 dichloroethene

CF CH

H H

HF

CH CH

H H

FF

118

The dissociation of Cl2 to form two Cl atoms in the gas phase is endothermic, and this energy is termed bond enthalpy.

119

The dissociation of Cl2 to form two Cl atoms in the gas phase is endothermic, and this energy is termed bond enthalpy.

• True• It is called bond enthalpy and it is

always endothermic. Bond breaking is always endothermic and bond forming is exothermic.

120

The H2Se and H2O are analogous molecules. The bonds in both are polar.Electronegativity values:H=2.2, O=3.5 Se=2.4

121

The bonds in H2Se are analogous to the bonds in H2O and are polar.Electronegativity values:H=2.2, O=3.5 Se=2.4• False.• They are analogous molecules (comparable in

certain respects, in particular their valence electrons and Lewis structures), but the H2Se bonds are nonpolar.

122

The reaction of calcium with Cl to form CaCl2 illustrates the octet rule.

123

The reaction of calcium with Cl to form CaCl2 illustrates the octet rule.

• True.• The calcium achieves an octet by losing

its two outer electrons. The chlorines complete their octets by each gaining one electron.

• All ionic bonds form in order to satisfy their octets

124

The oxidation state (or number) of iodine in ICl is -1.

125

The oxidation state (or number) of iodine in ICl is -1.• False.• Because Cl is more electronegative, it

carries a -1 oxidation and the I is +1

126

One of the three resonance structures of the sulfite ion is:

127

One of the three resonance structures of the sulfite ion is:

• False.• There are not enough electrons in this structure.

There should be 4(6) + 2 = 26, 13 lines.• This is a resonance structure for SO3, the

molecule, not SO3-2 the polyatomic ion. The sulfite

ion does not have a double bond and thus does not have resonance structures

128

FeCl3 contains a metallic bond.

129

FeCl3 contains a metallic bond.• False.• FeCl3 has ionic bonds.• Metallic bonds are the name given to bonds

between metal atoms - a subject we will take up later in Chap 23.

130

CO2 has polar bonds and is a polar molecule.Electronegativity values:C=2.5 O=3.5

131

CO2 has polar bonds and is a polar molecule.Electronegativity values:C=2.5 O=3.5• False.• The C=O bond is polar, but they are

symmetrically oriented causing the polarity to cancel out and the molecule to be nonpolar.

132

The bond enthalpy of a CN triple bond is larger than that for a CN double bond.

133

The bond enthalpy of a CN triple bond is larger than that for a CN double bond.• True.• The enthalpy for a triple bond is always

higher than for an analogous double bond.

134

A sigma bond can exist between hybridized orbitals and atomic orbitals.

135

A sigma bond can exist between hybridized orbitals and atomic orbitals.• True.• Remember that a bond is overlap of any

two orbitals, which can be atomic or hybrid orbitals.

136

According to the VSEPR model the H2S molecule is linear.

137

According to the VSEPR model the H2S molecule is linear.

• False.• There are 4 domains around the central S. It is

bent due to the two unshared pairs on the S atom.

138

Hybridization of the s and p orbitals named sp3 hybrids in carbon in CH4 is responsible for its tetrahedral structure.

139

Hybridization of the s and p orbitals named sp3 hybrids in carbon in CH4 is responsible for its tetrahedral structure.

• True.

140

Benzene, C6H6 is an example of a molecule in which electrons are delocalized over the s orbitals of carbon.

141

Benzene, C6H6 is an example of a molecule in which electrons are delocalized over the s orbitals of carbon.

• False.• It is the p orbitals that contain the

delocalized electrons.

142

In NH3 the electron domain geometry is tetrahedral.

143

In NH3 the electron domain geometry is tetrahedral.

• True.• The electron domain geometry is tetrahedral.• The molecular geometry is trigonal pyramidal

due to the unshared pair on the N.

144

• All three of the AB3 molecules below contain unshared pairs on the central A atom.

145

• All three of the AB3 molecules below contain unshared pairs.

• False.• Molecule i is trigonal planar.• Trigonal planar never has unshared pairs.

146

Molecules ii and iii each contain only one nonbonded pair of electrons.

147

Molecules ii and iii each contain one nonbonded pair of electrons.

• False.• Structure ii does have 1 unshared pair causing

trigonal pyramid• But structure iii is a (laying sideways) trigonal

bipyramid with two nonbonded pairs on iii causing the t-shape.

148

The End

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Chemical Bonding - Mr. Mello's AP & HONORS CHEMISTRYseekonkchemistry.weebly.com/.../prsi.bonding.pdf · • METALLIC BONDING ... bonds are neither 100% ionic or 100% covalent and