Chapter 8 Basic Concepts of Chemical Bonding

© 2012 Pearson Education, Inc.

Chapter 8

Basic Concepts of Chemical Bonding

Basic Concepts of Chemical

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Chemical Bonds

• Three basic types of bonds Ionic

• Electrostatic attraction between ions.

Covalent• Sharing of electrons.

Metallic• Metal atoms bonded to

several other atoms.


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A. MetallicB. IonicC. Covalent


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Lewis Symbols

• G.N. Lewis pioneered the use of chemical symbols surrounded with dots to symbolize the valence electrons around an atom.

• When forming compounds, atoms tend to add or subtract electrons until they are surrounded by eight valence electrons (the octet rule).


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A. Yes, all three are correct.B. No, the first two are different Lewis structures

because of differing placement of unpaired electron.

C. No, because the Lewis structure on the right has only five valence electrons and elemental Cl has seven valence electrons.


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Energetics of Ionic BondingAs we saw in the last chapter, it takes 496 kJ/mol to remove an electron from sodium.

We get 349 kJ/mol back by giving an electron to chlorine.


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Energetics of Ionic Bonding

But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!


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A. Yes, the relative sizes of the reactants are similar.B. No, the physical properties of the reactants are different.C. Yes, the metals are alkali metals and the nonmetals are elemental

halogens.D. No, the metals and nonmetals are each from different families of

elements.


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A. No, because we must know the names of each element assigned to a sphere to make a conclusion.B. Yes, because Na and Cl are in the same period. Atomic size generally increases from left to right in a period—a cation is smaller than its neutral atom, and an anion is greater than its neutral atom.C. No, because we need actual ion sizes to make a conclusion.D. Yes, because Na and Cl are in the same period. Atomic size generally decreases from left to right in a period —a cation is larger than its neutral atom, and an anion is smaller than its neutral atom.


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A. No pair of an alkali metal with a halogen gives a sum of I1 and EA that is negative, exothermic.

B. Na and FC. K and ClD. Rb and I


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• There must be a third piece to the puzzle.• What is as yet unaccounted for is the

electrostatic attraction between the newly formed sodium cation and chloride anion.


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Lattice Energy

• This third piece of the puzzle is the lattice energy:The energy required to completely separate a mole

of a solid ionic compound into its gaseous ions.• The energy associated with electrostatic

interactions is governed by Coulomb’s law:

Eel = Q1Q2

d


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Lattice Energy

• Lattice energy, then, increases with the charge on the ions.

• It also increases with decreasing size of ions.


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Sample Exercise 8.1 Magnitudes of Lattice EnergiesWithout consulting Table 8.2, arrange the ionic compounds NaF, CsI, and CaO in order of increasing lattice energy.

Practice ExerciseWhich substance do you expect to have the greatest lattice energy, MgF2, CaF2, or ZrO2?


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A. Lattice energy range of KF cannot be estimated unless we know ion sizes exactly and the type of crystal structure.B. Between 691 kJ/mol and 788 kJ/mol estimated from the range between the lattice energies of NaCl and RbCl because KCl is between these two lattice energies.C. Between 910 kJ/mol and 788 kJ/mol estimated from the range between the lattice energies of NaF and NaCl because a fluoride ion is present in the pair.D. Between 701 kJ/mol and 910 kJ./mole estimated by recognizing that the distance between adjacent ions of opposite charge should be larger than that in NaF and smaller than in KCl.


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By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.


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Sample Exercise 8.2 Charges on IonsPredict the ion generally formed by (a) Sr, (b) S, (c) Al.

Practice ExercisePredict the charges on the ions formed when magnesium reacts with nitrogen.


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A. Each calcium atom loses one electron and each fluorine atom gains two electrons.

B. Each calcium atom loses two electrons and each fluorine atom gains one electron.

C. Each calcium atom gains one electron and each fluorine atom loses two electrons.

D. Each calcium atom gains two electrons and each fluorine atom loses one electron.


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A. FeB. CoC. PdD. Rh


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Covalent Bonding

• In covalent bonds, atoms share electrons.

• There are several electrostatic interactions in these bonds:Attractions between electrons

and nuclei,Repulsions between

electrons,Repulsions between nuclei.


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Repulsions: Electron-Electron Nuclei-Nuclei Attractions: Electron-NucleiA. Unaffected Decrease DecreaseB. Unaffected Decrease IncreaseC. Increase Increase IncreaseD. Increase Increase Decrease


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A. Stronger, because a H-H covalent bond in H2+ has

one less electron than in H2.B. Stronger, because a H-H covalent bond in H2

+ has one more electron than in H2.

C. Weaker, because a H-H covalent bond in H2+ has

one less electron than in H2.D. Weaker, because a H-H covalent bond in H2

+ has one more electron than in H2.


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Polar Covalent Bonds

• Though atoms often form compounds by sharing electrons, the electrons are not always shared equally.

• Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does.

• Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.


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Electronegativity• Electronegativity is the ability of atoms in a molecule

to attract electrons to themselves in a bond.• On the periodic chart, electronegativity increases as

you go……from left to right across a row.…from the bottom to the top of a column.


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A. EN and EA are the same; they both measure the same characteristic.

B. EA measures the energy released when an isolated atom/ion gains an electron; EN measures the ability of an atom in a molecule to attract electrons to itself.

C. EN values of neutral atoms are just the negative of EA values of neutral atoms.

D. EN measures the energy released when an isolated atom/ion gains an electron; EA measures the ability measures the ability of an atom in a molecule to attract electrons to itself.


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A. IncreasingB.DecreasingC.constant


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The greater the difference in electronegativity, the more polar the bond.

• \


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A. Nonpolar, because both S and O are in the same family.B. Polar covalent, because the difference in electronegativity

values is 1.0.C. Ionic, because the difference in electronegativity values is –

1.0.D. Ionic, because the difference in electronegativity values is

>0.9.


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A. Polarity of HBr and HI bonds are insufficiently polar to cause significant excess electron density (shown by red shading) on the halogen atoms.B. Polarity of HBr and HI bonds are sufficiently polar to cause excess electron density (shown by dark green shading) on the halogen atoms.C. Polarity of HBr and HI bonds are insufficiently polar to cause significant excess electron density (shown by yellow shading) on the halogen atoms.D. Polarity of HBr and HI bonds are sufficiently polar to be represented by significant yellow shading between the bonded atoms


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A. OsO4

B. Cr2O3

C. Melting points are not useful in distinguishing between types of compounds.

D. More information about types of bonds within structures is needed to answer this question.


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Sample Exercise 8.4 Bond PolarityIn each case, which bond is more polar: (a) B—Cl or C—Cl, (b) P—F or P—Cl ? Indicate in each case which atom has the partial negative charge.

Practice ExerciseWhich of the following bonds is most polar: S—Cl, S—Br, Se—Cl, or Se—Br ?


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• When two atoms share electrons unequally, a bond dipole results.

• The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:

= Qr• It is measured in debyes (D).


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A. Increase, because r increasesB. Decrease, because r decreasesC. Stays the same because the values of Q and r change in opposite but equal directions.


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Sample Exercise 8.5 Dipole Moments of Diatomic MoleculesThe bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were and 1+ and 1–, respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms leads to this dipole moment?

Practice ExerciseThe dipole moment of chlorine monofluoride, ClF(g), is 0.88 D. The bond length of the molecule is 1.63 Å.(a) Which atom is expected to have the partial negative charge? (b) What is the charge on that atom in units of e?


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A. H—C has the same bond dipole moment as H—I, because the magnitude of Q and r in are about the same for both.

B. H—C has the larger bond dipole moment, because Q and r are larger in H—C than in H—I.

C. H—I has the smaller bond dipole moment, because Q is about the same for both, but r is larger in H—C than in H—I.

D. H—I has the larger bond dipole moment, because Q is about the same for both, but r is larger in H—I than in H—C.

Q r


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A. IFB. ClF C. They have the same dipole moment.D. Neither has a dipole moment; they are both

nonpolar.


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Lewis Structures

Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.


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Sample Exercise 8.3 Lewis Structure of a CompoundGiven the Lewis symbols for nitrogen and fluorine in Table 8.1, predict the formula of the stable binary compound (a compound composed of two elements) formed when nitrogen reacts with fluorine and draw its Lewis structure.

Practice ExerciseCompare the Lewis symbol for neon with the Lewis structure for methane, CH4. In what important way arethe electron arrangements about neon and carbon alike? In what important way are they different?Answer: Both atoms have an octet of electrons. However, the electrons about neon are unshared electron pairs, whereas those about carbon are shared with four hydrogen atoms.


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Writing Lewis Structures1. Find the sum of valence electrons

of all atoms in the polyatomic ion or molecule. If it is an anion, add one

electron for each negative charge.

If it is a cation, subtract one electron for each positive charge.

2. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.

3. Fill the octets of the outer atoms.4. Fill the octet of the central atom.


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Sample Exercise 8.6 Drawing a Lewis Structure

Practice Exercise(a) How many valence electrons should appear in the Lewis structure for

CH2Cl2?(b) Draw the Lewis structure.


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Writing Lewis Structures

5. If you run out of electrons before the central atom has an octet…

…form multiple bonds until it does.


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A. Single covalent bondB. Double covalent bondC. Triple covalent bond


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Sample Exercise 8.7 Lewis Structure with a Multiple Bond

Practice ExerciseDraw the Lewis structure for (a) NO+ ion, (b) C2H4.


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Sample Exercise 8.8 Lewis Structure for a Polyatomic Ion

Practice ExerciseDraw the Lewis structure for (a) ClO2

–, (b) PO43–

Draw the Lewis structure for the BrO3– ion.


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• Then assign formal charges.For each atom, count the electrons in lone pairs and

half the electrons it shares with other atoms.Subtract that from the number of valence electrons for

that atom: the difference is its formal charge.


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• The best Lewis structure……is the one with the fewest charges.…puts a negative charge on the most

electronegative atom.


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Sample Exercise 8.9 Lewis Structures and Formal ChargesThree possible Lewis structures for the thiocyanate ion, NCS– , are

(a) Determine the formal charges in each structure. (b) Based on the formal charges, which Lewis structure is the dominant one?

Practice ExerciseThe cyanate ion, NCO, has three possible Lewis structures. (a) Draw these three structures and assign formal charges in each. (b) Which Lewis structure is dominant?


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A. The structure actually represents an ion.B. The F atom in the structure must have four covalent bonds

attached to it.C. There must be another F atom in the structure carrying a –1

formal charge, since F is the most electronegative element and it should carry a negative formal charge.

D. There must be a better Lewis structure, since F is the most electronegative element and it should carry a negative formal charge.


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Resonance

This is the Lewis structure we would draw for ozone, O3.


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Resonance

• But this is at odds with the true, observed structure of ozone, in which……both O—O bonds

are the same length.…both outer

oxygens have a charge of −1/2.


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A. The colors of the two O atoms are the same.B. The substance is bent.C. The substance has a bond angle of 117°.D. The two O—O bond lengths are equivalent.


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Resonance

• One Lewis structure cannot accurately depict a molecule like ozone.

• We use multiple structures, or resonance structures, to describe the molecule.


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Resonance

Just as green is a synthesis of blue and yellow…

…ozone is a synthesis of these two resonance structures.


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A. Yes, The map shows the same electron densities on the left and right sections of the ozone molecule, indicating that resonance results in two equal O—O bonds.B. No. The map show unequal electron densities on the upper and section sections of the ozone molecule, indicating that resonance does not exist.


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A. Yes, because in each of the two resonance forms there is one O=O bond and one O—O bond, giving an overall average of 1.5 bonds per O—O bond.

B. No, because a half bond cannot exist in a bonding situation.


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A. One-and-a-fifth bondsB. One-and-a-fourth bondsC One-and-a-third bonds D. One-and-a-half bonds


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Sample Exercise 8.10 Resonance StructuresWhich is predicted to have the shorter sulfur–oxygen bonds, SO3or SO3

2– ?

Practice ExerciseDraw two equivalent resonance structures for the formate ion, HCO2

–.


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Resonance• In truth, the electrons that form the second C—O

bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon.

• They are not localized; they are delocalized.


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Resonance

• The organic compound benzene, C6H6, has two resonance structures.

• It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.


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A. The C—C bonds in benzene are weaker than a normal C—C single bond.

B. The C—C bonds in benzene are stronger than a C=C bond.C. Two resonance forms for benzene exist resulting in 1 ½ C—C

bonds with the dashed lines representing a “half of a bond”.D. Three resonance forms for benzen exist resulting in 1 ⅓ C—C

bonds with the dashed line representing a “third of a bond”.


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A. Yes, because it has multiple Lewis resonance structures.B. Yes, because it has three C=C bonds that can be moved throughout

the structure.C. No, because the carbon chain is linear.D. No, because we cannot write other reasonable Lewis structures.


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Exceptions to the Octet Rule

• There are three types of ions or molecules that do not follow the octet rule:ions or molecules with an odd number of

electrons,ions or molecules with less than an octet,ions or molecules with more than eight

valence electrons (an expanded octet).


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Odd Number of Electrons

Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.


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A. The first NO structure, because all atoms have zero formal charge.

B. The first NO structure, because N and O have equal but opposite formal charges.

C. The second NO structure, because all atoms have zero formal charge.

D. The second NO structure, because N should not have a positive formal charge.


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Fewer Than Eight Electrons

• Consider BF3:Giving boron a filled octet places a negative charge

on the boron and a positive charge on fluorine.This would not be an accurate picture of the

distribution of electrons in BF3.


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Fewer Than Eight ElectronsThe lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.

Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.


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More Than Eight Electrons• The only way PCl5 can exist is if phosphorus

has 10 electrons around it.• It is allowed to expand the octet of atoms on

the third row or below.Presumably d orbitals in these atoms participate in

bonding.


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More Than Eight ElectronsEven though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygen atoms.

•This eliminates the charge on the phosphorus and the charge on one of the oxygens.•The lesson is: When the central atom is on the third row or below and expanding its octet eliminates some formal charges, do so.


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Sample Exercise 8.11 Lewis Structure for an Ion with More than an Octet of Electrons

Draw the Lewis structure for ICI4–.

Practice Exercise(a) Which of the following atoms is never found with more than an octet of valence electrons around it: S, C, P, Br? (b) Draw the Lewis structure for XeF2.


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Covalent Bond Strength

• Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.

• This is the bond enthalpy.• The bond enthalpy for a Cl—Cl bond, D(Cl—

Cl), is measured to be 242 kJ/mol.


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Average Bond Enthalpies• Table 8.4 lists the average bond

enthalpies for many different types of bonds.

• Average bond enthalpies are positive, because bond breaking is an endothermic process.

• Note: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.


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Enthalpies of Reaction• Yet another way to estimate H for a

reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.

• In other words, Hrxn = (bond enthalpies of bonds broken) −

(bond enthalpies of bonds formed)

CH4(g) + Cl2(g)

CH3Cl(g) + HCl(g)

In this example, one C—H bond and one Cl—Cl bond are broken; one C—Cl and one H—Cl bond are formed.


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Enthalpies of Reaction

So,H = [D(C—H) + D(Cl—Cl)] − [D(C—Cl) + D(H—Cl)]

= [(413 kJ) + (242 kJ)] − [(328 kJ) + (431 kJ)]= (655 kJ) − (759 kJ)= −104 kJ


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Sample Exercise 8.12 Using Average Bond EnthalpiesUsing data from Table 8.4, estimate H for the reaction

Practice ExerciseUsing Table 8.4, estimate H for the reaction


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A. The enthalpy of atomization / 7 bonds broken = a good estimate of D(CC).

B. The enthalpy of atomization – 6 [ D(CH)] = a good estimate of D(CC).

C. The enthalpy of atomization + 6 [ D(CH)] = a good estimate of D(CC).

D. The enthalpy of atomization / 7 bonds broken – 6[D(CH)] = a good estimate of D(CC).


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A. O2 is more reactive, because the O=O bond enthalpy is less than that of the O—O bond enthalpy in hydrogen peroxide.

B. O2 is more reactive, because the O=O bond enthalpy is greater than that of the O—O bond enthalpy in hydrogen peroxide.

C. H2O2 is more reactive, because the O—O bond enthalpy is less than that of the O=O bond enthalpy in O2.

D. H2O2 is more reactive, because the O—O bond enthalpy is greater than that of the O=O bond enthalpy in O2.


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A. Exothermic because ΔHrxn shows its arrow pointing downward.B. Exothermic because ΔH1 shows its arrow pointing upward.C. Endothermic because ΔHrxn shows its arrow pointing upward.D. Endothermic because ΔH1 shows its arrow pointing downward.


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Bond Enthalpy and Bond Length

• We can also measure an average bond length for different bond types.

• As the number of bonds between two atoms increases, the bond length decreases.


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A. A negative slope means the N—N bond strength increases to the right.

B. A negative slope means the stronger the N—N bond the longer the bond.

C. A negative slope means that as the N—N bond gets longer it gets weaker.

D. A negative slope means that when a N—N breaks it is an exothermic event.

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Chapter 8 Basic Concepts of Chemical Bonding