chapter 9 chemical bonding - university of central...

Chapter 9

Chemical Bonding

Chapter Goals• Understand the difference between ionic and

covalent bonds.• Draw Lewis electron dot structures for small

molecules and ions.• Use the valence shell electron-pair repulsion

theory (VSEPR) to predict the shapes of simple molecules and ions and to understand the structure of more complex molecules.

• Use electronegativity to predict the charge distribution in molecules and ions and to define the polarity of bonds.

• Predict the polarity of molecules• Understand the properties of covalent bonds

and their influence on molecular structure.

Introduction: Valence Electrons * Attractive forces that hold atoms together incompounds are called chemical bonds. * The electrons involved in bonding are usually those in the outermost (valence) shell. The (inner) core electrons are not involved in chemical behavior.

Valence electrons:For the main groups (representative) elements they are the outer most s and p electrons. The # of valence electrons is equal to the group number.

* For the transition elements they are the ns and (n−1)d electrons.

Lewis Dot Symbols for Atoms* Lewis dot formulas or Lewis dot symbols are a

convenient bookkeeping method for tracking valence electrons.

Valence electrons are those electrons that are transferred or involved in chemical bonding. They are chemically important.

Symbol considered to have 4 sides. Two dots per sidemaximum.

Li Be B C N O F Ne.... .. ..

..HeH

.

.. . .

.. ....

...

..

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

.

Lewis Dot Symbols for Atoms

Elements that are in the same periodic group have the same Lewis dot structures.

Li & Na. .

N & P.. ..

..

. ..

. F & Cl...

.... .... ..

.

Formation of BondsWhen a chemical reaction occurs, the valence electrons of the atoms are reorganized so that net attractive forces −chemical bonds− occur between atoms.Chemical bonds are classified into two types:Ionic bonding results from electrostatic

attractions between ions, which are formed by the transfer of one or more electrons from one atom to another.

Covalent bonding results from sharing one or more electron pairs between two atoms.

Comparison of Ionic and Covalent Compounds

Melting point comparisonIonic compounds are usually solids with

high melting pointsTypically > 400oC

Covalent compounds are gases, liquids, or solids with low melting pointsTypically < 300oC

Solubility in polar solvents (such as water)Ionic compounds are generally solubleCovalent compounds are generally

insoluble


Solubility in nonpolar solventsIonic compounds are generally insolubleCovalent compounds are generally soluble

Conductivity in molten solids and liquidsIonic compounds generally conduct electricity

They contain mobile ions

Covalent compounds generally do not conduct electricity


Conductivity in aqueous solutionsIonic compounds generally conduct electricityThey contain mobile ionsCovalent compounds are poor conductors of

electricityFormation of Compounds

Ionic compounds are formed between elements with large differences in electronegativityOften a metal and a nonmetal

Covalent compounds are formed between elements with similar electronegativitiesUsually two or more nonmetals

Ionic BondingFormation of Ionic CompoundsAn ion is an atom or a group of atoms possessing a net electrical charge.Ions come in two basic types:positive (+) ions or cations

These atoms have lost 1 or more electrons.

negative (−) ions or anionsThese atoms have gained 1 or more

electrons.That applies to binary compounds

Formation of Ionic CompoundsIonic bonds are formed by the attraction of

cations for anions usually to form solids.Commonly, metals react with nonmetals to form ionic compounds.The formation of NaCl is one example of an ionic compound formation.

Formation of Ionic CompoundsReaction of Group IA Metals with Group VIIA Nonmetals

gas solidyellow silver

F Li 2 nonmetalVIIA metalIA

2(g)(s)


point melting C842an with gas solid

solid whiteyellow silver

LiF 2 F Li 2 nometalVIIA metalIA

o

(s)2(g)(s)


The underlying reason for the formation of LiFlies in the electron configurations of Li and F.

1s 2s 2pLi loses one electronF gains one electron

These atoms form ions with these configurations.

Li+ same configuration as [He] 1s2

F− same configuration as [Ne]

Formation of Ionic CompoundsReaction of Group IA Metals with Group VIIANonmetals*We can also use Lewis dot formulas to represent the neutral atoms and the ions they form.*electrons are transferred until the metal loses all its valence electrons and the nonmetal has an octet (Octet Rule)

Li + F.....

.. . Li+ F[ ]..

.... ..

Formation of Ionic CompoundsFor the reaction of IA metals with VIA

nonmetals, a good example is the reaction of lithium with oxygen.

The reaction equation is:

-2s22(g)(s) OLi O2/1Li 2

Formation of Ionic CompoundsDraw the electronic configurations for Li,

O, and their appropriate ions.

2s 2p 2s 2p Li [He] Li+

O [He] O2- octet rule

Draw the Lewis dot formula representation of this reaction.

Formation of Ionic CompoundsDraw the electronic representation of the Ca and N reaction.

4s 4p 4s 4p Ca [Ar] Ca2+

2s 2p 2s 2p N [He] N3-

Other IIA and VA elements behave similarly. Symbolically, this reaction can be represented as:

3 M(s) + 2 X(g) M32+ X2

3-

M can be the IIA elements Mg to Ba.X can be the VA elements N to As.

Ca N+ ..... . .3 2 N

3-.. ......2 [ ]Ca

2+3

Formation of Ionic Compounds

Simple Binary Ionic Compounds TableReacting Groups Compound General Formula Example

IA + VIIA MX NaFIIA + VIIA MX2 BaCl2IIIA + VIIA MX3 AlF3

IA + VIA M2X Na2OIIA + VIA MX BaOIIIA + VIA M2X3 Al2S3

Ion Attraction and Lattice Energy

Ionic Bond

• electrostatic attraction between oppositely charged ions

• non-directional

• strength directly proportional to charges of ions and inversely proportional to distance between ion centers

Coulombic Ion Attraction and Lattice Energy

E QcQa

dQc QaE = C ────── is negative because of charges

d1

E = ionic bond strength C (a constant) = ——40

Qc = charge on cation (is positive)

Qa = charge on anion (is negative)

d = distance between centers of ions

E QcQa

d

NaCl Na+ Cl–

CaCl2 Ca2+ Cl–

CaS Ca2+ S–2

Al2S3 Al3+ S2–

Product of charges

−1

−6

−2

−4

Qc x Qa is the heaviest factor. Product is much more negative for Al2S3. The least negative product is NaCl’s

E QcQa

d

F–Na+ Na+

Na+Na+

Cl–

Br–I–

shortest d d

longest dd

NaF has the most negative E and NaI the least 0.095 nm 0.136 nm

0.095 nm 0.216 nm

E − 4.33

E − 3.22

Lattice Energy (Hlattice )the energy of formation of 1 mol of solid crystalline ionic compound when ions in the gas phase combine. It is always negative.Among several compounds, the one with themost negative Hlattice is said to have the highest lattice energy (absolute value), and highest TfusionIt is also said to be the most ionic.

Mg2+(g) + 2F–

(g) MgF2(s) Hlattice = −2910 kJ/mol

for NaF, Hlattice = −911 kJ/molfor KF, Hlattice = −815 kJ/molWHY? charge of Mg2+ (2) > Na+= K+, # of F− (1)

d, radius of Mg2+ < Na+ < K+ (see chapt. 8)

Lattice EnergySmall ions with high ionic charges have more negative lattice energies (higher lattice energies). Large ions with small ionic charges have less negative lattice energies.

Use this information, plus the periodicity rules fromChapter 8, to arrange these compounds in order ofdecreasing values of lattice energy KCl, Al2O3, CaOIonic radii: 1.33 1.81 1.06 1.40 0.57 1.40 Å

Hlattice K+Cl− > Ca2+O2− > 2Al3+3O2−least negative most negative

|Hlattice| (absolute values) Al2O3 > CaO > KCl

Lattice EnergyArrange the following compounds in order of decreasing value of lattice energy (least negative tomost) CaO, MgO, SrO, and BaOThe anion (O2−) is common to the four oxides. In the equation Qc Qa

E = C ─────d

Qc and Qa are the same (+2 and −2). Then, the difference is made by d. The four cations are in the same group, so d will be for Ba2+>Sr2+>Ca2+>Mg2+.E is negative and inversely proportional to d:the least negative the most negativeBa2+O2− > Sr2+O2− > Ca2+O2− > Mg2+O2− (highest E)

About Lattice EnergyAccording to its definition, lattice energy is negative for any compound. Please look at the examples (kJ/mol are the units)

KBr SrCl2 MgO Elat -671 > -2127 > -3795

Elat 671 < 2127 < 3795

where Elat means absolute value (magnitude) of lattice energy.The two rows in the table can be read as: MgO has the most negative lattice energy or the highest magnitude of lattice energy (highest absolute value).KBr has the least negative lattice energy or the lowest magnitude of lattice energy (lowest absolute value).

Ionic Bond Formation• Born-Haber CycleBreaks formation of an ionic compound from its elements into a series of theoretical steps and considers the energetics of each

Example: consider formation of NaClNa(s) + 1/2Cl2(g) NaCl(s) H° = –410.9 kJ

The Born-Haber Cycle is a level diagram that applies Hess’s law (chapter 6): H of reaction = ∑ H of all theoretical steps

E(kJ)

0

–410.9

107.7

229.4

376.4

725.4

Na(s) + 1/2Cl2(g)

Na(g) + 1/2Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g) + e–

Na+(g) + Cl–(g)

NaCl(s)

Hlattice

Hsub, Na

Hatom,Cl

IENa

EACl

H°f, NaCl

The Born-Haber Cycle of NaCl

The Born-Haber Cycle of NaClH°f = Hsublimation, Na + Hatom, Cl + IE1, Na + EACl +

Hlattice, NaCl (the two in blue are negative)

H°f = 107.7kJ + 121.7kJ + 496.0kJ + (–349.0kJ) + (–787.3 kJ) (all values are given here).

H°f = –410.9 kJ

If unknown, Hlattice can be calculated

Hlattice = H°f − Hsublimation,Na − Hatom,Cl − IE1, Na− EACl

Practice – Given the information below, determine the lattice energy of MgCl2

Mg(s) Mg(g) H1°f = +147.1 kJ/mol2{½ Cl2(g) Cl(g)} 2H2°f = 2(+122 kJ/mol)Mg(g) Mg+(g) H3°f = +738 kJ/molMg+(g) Mg2+(g) H4°f = +1450 kJ/mol2{Cl(g) Cl−(g)} 2H5°f = 2(−349 kJ/mol)Mg2+(g) + 2 Cl−(g) MgCl2(s) H° lattice energy = ? kJ/molMg(s) + Cl2(g) MgCl2(s) H6°f = −641 kJ/mol

The 2 in front of the 1/2Cl2 dissociation is due to the fact we need 2 moles of Cl atoms. The 2 also multiplies the respective H (122 kJ). The same 2 is in front of the Cl Electron Affinity (-349 kJ). Contrary to Na, with only 1 e- lost, Mg loses 2e-. Due to this, we have to write the two ionization equations for Mg, IE1 = 738 kJ, IE2 = 1450 kJ.

Covalent BondingCovalent bonds are formed when atoms share electrons.If the atoms share 2 electrons (a pair) a single covalent bond is formed.If the atoms share 4 electrons (two pairs) a doublecovalent bond is formed.If the atoms share 6 (three pairs) electrons a triple covalent bond is formed.

The attraction between the nuclei and the electrons is electrostatic in nature. Directional.The atoms have a lower potential energy when bound.

Formation of Covalent BondsThis figure shows the potential energy of an H2 molecule as a function of the distance between the two H atoms.

Sharing two electronsBond is formed, E < 0 Free atoms,

Energy = 0

Unstable, because of high energy

Bond energy

Formation of Covalent BondsRepresentation of the formation of an H2 molecule from H atoms

Formation of Covalent BondsWe can use Lewis dot formulas to show covalent bond formation. H2 molecule formation representation.

duet rule

HCl molecule formation

octet rule (for Cl)

+H. H . H H.. or H2

H Cl H Cl+... ..

.. ....

..

... or HCl

Formation of Covalent BondsHomonuclear diatomic molecules

Heteronuclear diatomic molecules: hydrogen halides

H HorH H.. F F.. .. ....

..

.. .. F F.. .... ..

.. ..or

N N········ ·· N N·· ··or

or ··H F··

··H F..

·· ···· or ··H Cl··

··H Cl..

·· ····

or ··H Br··

··H Br..

·· ····

+

H H HHy y

x

F

F

Writing Lewis Formulas:The Octet RuleThe octet rule states that representative elements usually attain stable noble gas electron configurations in most of their compounds.Lewis dot formulas are based on the octet rule.We need to distinguish between bonding (or shared) electrons and nonbonding (or unshared or lone pairs of) electrons.

The Octet RuleN - A = S rule N A S

Simple mathematical relationship to help us write Lewis dot formulas.

N = number of electrons needed to achieve a noble gas configuration.N usually has a value of 8 for representative elements. N has a value of 2 for H atoms.

A = number of electrons available in valence shells of the atoms.

A is equal to the periodic group number for each element. A is equal to 8 for the noble gases.

S = number of electrons shared in bonds (in bonding pairs.)A-S = number of electrons in unshared, lone, pairs.

The Octet RuleFor ions we must adjust the number of electrons available, A.

Add one e− to A for each negative charge.Subtract one e− from A for each positive charge.

The central atom in a molecule or polyatomic ion is determined by:The atom that requires the largest number of e-sto complete its octet. It goes in the center. H is never central atom, it shares two e-s only.

For two atoms in the same periodic group, the lesselectronegative element goes in the center (theone towards the bottom of the group.)

Lewis Structures of Covalent Compounds and Polyatomic IonsDrawing Lewis structures by this method, use the following as a guide:a) Draw skeletal Lewis structure.

b) Draw the Lewis electron dot structure for each atom. (Use the method in which the electrons are spread to all four sides of an imaginary square before being paired.) For the sake of keeping the drawing as neat as possible, direct single electrons on adjacent atoms towards each other.

c) Draw a line from a single unpaired electron on the central atom to a single unpaired electron on the surrounding atom. This constitutes the formation of a covalent bond.) Continue doing this until each atom has an octet (exceptions are H, Be, B, Al, elements on rows 3, 4, 5, and 6.) No electrons should be left unpaired (only in rare cases will a species contain an unpaired electron.) For those atoms that can have more than an octet, if all of its single electrons are used in a covalent bond, and there are surrounding atoms with electrons still to be paired, then lone electron pairs are used in bonding. The electron pair(s) being shared must be placed between the two atoms forming the bond.

d) For polyatomic ions,-1 charge, add 1 electron to the most electronegative atom-2 charge, add 2 electrons (one to each of the most electronegative atoms)-3charge, add 3electrons (one to each of the most electronegative atoms)+1 charge, remove 1 electron from the least electronegative atom.

Lewis Structures of Covalent Compounds and Polyatomic Ionse) Coordinate Covalent or Dative Bond Formation (a bond formed when

both of the electrons in a bond are supplied by the same atom): If all the single electrons of one atom (atom 1) are used in binding and the adjacent atom (atom 2) has single electrons which need to be shared, then the electrons on atom 2 (the one still having single electrons) are paired and atom 1 donates a pair of electrons to atom 2 thus forming a coordinate covalent bond. A coordinate covalent bond is represented by . The arrowhead is pointed to the atom to which the electron pair has been donated.

Exceptions to Octet:a) Some atoms have less than an octet (Be, B, and Al are metals, but

they can form covalent compounds.) | |

H─ ─Be─ ─B─ ─Al─

b) Central atoms of elements of rows 3, 4, 5, and 6 of the periodic table, can have more than an octet due to the availability of low laying d-orbitals. These elements can have up to 18 electrons surrounding them.

A Quick-N-Dirty Method for Drawing the Lewis Structure of Covalent Compounds and IonsThe method will be illustrated by drawing the Lewis structure of SO3.1) Find the total number of valence electrons by adding the number of valence electrons 1S = 1(6) = 6from each atom in the formula. 3O = 3(6) = 18(For polyatomic ions add (anions) or subtract 24(cations) the charge to or from the totalnumber of valence electrons.) O2) Draw a single bond from the central atom | to each of the atoms surrounding it. O ─S─ O

3) Subtract the electrons used so far in the 3 bonds x 2 e− = 6 e−structure from the total number of electrons. 24 − 6 = 18 e−Each bond contains two electrons.

4) The remaining electrons are spread, as ..pairs, to the surrounding atoms first. Each :O:surrounding atom (except H) will receive .. | ..enough electron pairs to have an octet. If :O ─S─ O:any electrons are left over, give them to the central atom. •• ••

A Quick-N-Dirty Method for Drawing the Lewis Structure of Covalent Compounds and Ions5) If the central atom does not have an octet, then one or two single bonds are converted to double bonds or one double bond is converted to triple bond. If possible, make two double bonds before making one triple bond.

.. ..:O: :O:

.. | .. | ..:O ─S─ O: :O=S ─ O:•• •• •• ••

Remember exceptions to the octet rule (H, Be, B, Al, and elements from rows 3, 4, 5, and 6).

Lewis Structures of Covalent Compounds and Polyatomic IonsExample: Write Lewis dot and dash formulas for hydrogen cyanide, HCN.A = 1 (H) + 4 (C) + 5 (N) = 10 H−C−NS (so far) = 2 bonds x 2 = 4electrons to be spread = 10 − 4 = 6

..We put the 6e- over N, H−C−N:

¨

But, to fulfill the octet of C and N, H:C:::N:

or H−CN: with four BP and one LP over N

Lewis Structures of Covalent Compounds and Polyatomic Ions

Example: Write Lewis dot and dash formulas for the sulfite ion, SO3

2−. O−S−OA = 6(S) + 3 x 6 (O) + 2 (- charge) = 26 |S (so far) = 3 bonds x 2 = 6 O

.. .. ..e-s to be spread = 26 − 6 = 20 :O−S−O:

¨ | ¨with an overall 2− charge :O:

¨

Thus this polyatomic ion has 3 BP and 10 LP.


Other examples:CO2, NO2

+, NO3−, HNO3, SO4

2−, H2SO4,

PO43−, H3PO4

Cl2O7, CH3CH2OH (ethanol), CH3COOH (acetic acid)

PCl3, PCl5, SF6, SOCl2, IF5, IF4+


Other examples:BBr3, BBr4

−

less than octet6 e- on B

AsF5 more than octet10 valence e- on As

B··. Br··

··

··.

BBr Br

Br

········

····

····

····

····

Br B

Br

Br····

··

·· ····

·· ····

or

As··..

. F····

··.

··

AsF

F F

F F

····

··

·· ······

······

····

·· ··or

····

····

··

·· ······

······

···· AsF

F F

FF

······ ··

··

··


Isoelectronic Species:Molecules and/or ions having the same number of valence electrons and the same Lewis structures.

[:NO:]+ [:NN:] [:CO:] [:CN:]−

They all have 3 BP and 2 LP

Resonance

Example: Write Lewis dot and dash formulasfor sulfur trioxide, SO3.N = 8 (S) + 3 x 8 (O) = 32A = 6 (S) + 3 x 6 (O) = 24S = 8A-S = 16 4BP, 8 LP

orO S O

O··

····

····

··

····

····

·· ·· O S

OO

·· ···· ·· ····

····

ResonanceThere are three possible structures for SO3. The double bond can be placed in one of three places.

When two or more Lewis formulas are necessary to show the bonding in a molecule, we must use equivalent resonance structures to show the molecule’s structure.

Double-headed arrows are used to indicate resonance formulas.

O S

OO

·· ···· ·· ····

····

OS

O

O·· ···· ·· ··

··

······

O S

OO·· ···· ·· ··

····

ResonanceResonance is a flawed method of representing molecules.• There are no single or double bonds in

SO3.• In fact, all of the bonds in SO3 are

equivalent.The best Lewis formula of SO3 that can be drawn is:

SO O

O

CO2

C2 O 12 e–

4 e–

16 e–

2 bonds – 4 e–

12 e–

– 12 e–

0 e–

O O C

O O C

O O C

O O C

Three Resonance structures for CO2

ResonanceOther examples:O3, C6H6 (benzene), NO3

− (nitrate ion)

Molecules with odd number of electronsNO has 11 valence electrons

.. .. NO

NO2 has 17 valence electrons

:O ─NO ON─O:

They are members of a family of substances called Free Radicals: they have an unpaired e−.Very reactive: e.g. dimerization of NO2 to N2O4.

2 NO2(g) N2O4(g)

Central Atoms with Single-Bond Pairs and Lone PairsValence Shell Electron-Pair Repulsion theory (VSEPR Theory):Regions of high electron density around the centralatom are arranged as far apart as possible to minimize repulsions.

Lone pairs of electrons require more volume than shared pairs. Hence,

_Lone pair to lone pair is the strongest repulsion.

_Lone pair to bonding pair is intermediate repulsion.

_Bonding pair to bonding pair is the weakest repulsion.

Mnemonic for repulsion strengthslp/lp > lp/bp > bp/bp

VSEPR TheoryThere are five basic molecular shapes based on the number of regions of high electron density (r.h.e.d) around the central atom.

Several modifications of these five basic shapes will also be examined.

VSEPR Theory

1. Two regions of high electron density (BP + LP) around the central atom.

2. Three regions of high electron density (BP + LP) around the central atom.

VSEPR Theory

3. Four regions of high electron density (BP + LP) around the central atom.

VSEPR Theory

4. Five regions of high electron density (BP + LP) around the central atom.

VSEPR Theory

5. Six regions of high electron density (BP + LP) around the central atom.

VSEPR Theory

Frequently, we will describe two geometries for each molecule or ion.Electronic geometry is determined by the locations of regions of high electron density around the central atom(s).Molecular geometry determined by the arrangement of atoms around the central atom(s).

Electron pairs are not used in the molecular geometry determination, just the positions of the atoms in the molecule are used.

VSEPR Theory

An example of a molecule that has the same electronic and molecular geometries is methane - CH4.Electronic and molecular geometries are tetrahedral.

H

CHH

H

VSEPR Theory

An example of a molecule that has different electronic and molecular geometries is water- H2O.Electronic geometry is tetrahedral.Molecular geometry is bent or angular.

Predicting Molecular Geometries and Bond AnglesWrite the Lewis structure.

Count the number of bonding pairs and lone pairsaround the central atom (regions of high electron Density = groups of electrons).Treat a double or triple bond as one region h.e.d.

Use Table provided to predict geometry.When lone pairs are present, bond angles are slightly less than those given (except, linear = 180° and squared planar = 90°.) When only two atoms are present, it is inappropriate to speak of bond angles.

Predicting Molecular Geometries and Bond Angles2 BP CO2, CS2, BeCl2 linear

3 BP BF3, CO32− trigonal planar

2 BP 1 LP SO2, NOCl bent

4 BP CH4, NH4+, BF4

− tetrahedral

3 BP 1 LP NH3, H3O+ trigonal pyramidal

2 BP 2 LP H2O, NH2− V-shaped or bent

5 BP PCl5, AsF5 trigonal bipyramid

6 BP SF6, PCl6− octahedral

Charge Distribution in Covalent Bonds and Molecules

The way the electrons are distributed in the molecule is called its charge distribution.Formal Charges on Atoms:an accounting tool for electron ownership= (# valence e– in free atom) – (# e– in lone pairs on atom) – ½(# bonded e– on atom)

electronsFormal charge = FC =

group number of atom − [LPE + ½(BE)]FC = group number − LPElectr − BP

Formal Charge (FC) and Best structure

Best structure has• zero FC on all atoms• lowest FC possible• negative FC on most electronegative

atoms and positive FC on least electronegative atoms

The most electronegative elements are at thetop right of periodic table (except for the noble gases.)

For example, consider thiocyanate ion SCN–

N CS

C SN

S N

C

For example, consider thiocyanate ion, SCN–

What is the best structure?

N CS

C SN

FCN = 5 – 4 – 2 = –1

S N

C

For example, consider thiocyanate ionSCN–

N CS

C SN

FCN = 5 – 4 – 2 = –1

FCC = 4 – 0 – 4 = 0

S N

C

FCS = 6 – 4 – 2 = 0


N CS

C SN

FCN = 5 – 4 – 2 = –1

FCC = 4 – 4 – 2 = –2

FCS = 6 – 0 – 4 = +2

–1 0 0S N

C


N CS

C SN

S N

C

FCN = 5 – 0 – 4 = +1

FCC = 4 – 4 – 2 = –2

FCS = 6 – 4 – 2 = 0

–1 0 0 –1–2 +2


N CS

C SN

S N

C

–1 0 0 –1–2 +2 +1 –20


N CS

C SN

S N

C

–1 0 0 –1–2 +2 +1 –20

best structure because lowest FC andnegative FC on most electronegative atom

Sum of formal charges = −1 + 0 + 0 = −1,

that is the charge of the ion.

Formal Charge (FC)H N= 5 − 8/2 = +1 H = 1 − 2/2 = 0

NH4+ H─N─H O

..H SO4

2− O─S─O: ..

O.. .. S = 6 − 8/2 = 6 −4 =+ 2

CO2 OCO O = 6 − (6 + 2/2) = −1.. ..

C = 4 − 8/2 = 0 O = 6 − (4 + 4/2) = 0

-1 +1 0 0 +1 -1 .. .. .. ..

O3 (ozone) :O─OO OO─O:.. .. .. .. .. ..

CO2

SeOF2

NO2−

all 0

Se = +1

H3PO4

SO32−

P2H4

Practice – Identify Structures with Better or Equal Resonance Forms and Draw Them

P = +1rest 0

S = +1

all 0

79

Practice – Identify Structures with Better or Equal FC Resonance Forms and Draw Them

none

−1

+1

none

H3PO4

SO32−

P2H4

80

CO2

SeOF2

NO2−

Bond Polarity and Electronegativity Polar and Nonpolar Covalent BondsCovalent bonds in which the electrons are shared equally are designated as nonpolarcovalent bonds.Nonpolar or pure covalent bonds have a symmetrical charge distribution. To be pure covalent, the two atoms involved in the bond must be the same element toshare equally.

H HorH H.. N N········ ·· N N·· ··or

+

H H HHy y

x

F

F

Pure Covalent Bonds

Bond Polarity and Electronegativity Polar and Nonpolar Covalent BondsCovalent bonds in which the electrons are not shared equally are designated as polarcovalent bonds.Polar covalent bonds have an asymmetrical charge distribution.To be a polar covalent bond the two atoms involved in the bond must have different electronegativities.

y y

Br

F

Polar Covalent Bonds

Br F

ElectronegativityElectronegativity is a measure of the relative tendency of an atom to attract electrons to itself when chemically combined with another element. Electronegativity is measured on the Pauling scale. Fluorine is the most electronegative element. Cesium and francium are the least electronegative elements.For the representative elements, electronegativities usually increase from left to right across periods and decrease from top to bottom within groups.

Electronegativity

ElectronegativityExample: Arrange these elements based on their electronegativity.

Se, Ge, Br, As (period # 4)

Ge < As < Se < Br

Arrange these elements based on their electronegativity.

Be, Mg, Ca, Ba (group IIA)

Ba < Ca < Mg < Be

Electronegativity, polarity, and Ionic bonds A non-polar bond is formed when the difference of electronegativity is between 0.0 – 0.4

A polar bond is formed when the difference of electronegativity is 0.5 – 1.8 (partial charge)

Ionic bonds form when the difference of electronegativity is greater than 1.8 (net charge)

Cl is more electronegativeno charge + −

Se−H (0.3) C−Cl (0.5) K+ and Br− (1.9)non-polar polar ionic

Bonding Continuumpureionic

purecovalent

polarcovalent

CsF (3.3) H2 (0)BrF (1.3)

E’neg

01.8 0.4

nonpolarcovalentionic

Molecular Polarity

The higher the diff. of electronegativity the more polar the bond is

bondpolar slightly 0.4 Difference

2.5 2.1 ativitiesElectronegI H

0.4

bondpolar very 1.9 Difference

4.0 2.1 ativitiesElectronegF H

1.9

Polar and Nonpolar Covalent BondsPolar molecules can be attracted by magnetic and electric fields.

Dipole MomentsMolecules whose centers of positive and negative charge do not coincide, have an asymmetric charge distribution, and are polar. These molecules have a dipole moment.The dipole moment has the symbol ; is the product of the distance, d, separating charges of equal magnitude and opposite sign, and the magnitude of the charge, .It is indicated by crossed arrow pointing from positive end to negative end of dipole = d

Dipole MomentsMolecules that have a small separation of charge have a small Molecules that have a large separation of charge have a large For example, HF and HI (longer d):

Ratio

5.0E’neg 1.9 0.4 5.0

units Debye0.38 units Debye1.91 I- H F- H

--

Dipole MomentsThere are some nonpolar molecules that havepolar bonds.There are two conditions that must be true fora molecule to be polar.1. There must be at least one polar bond present or one lone pair of electrons.2. The polar bonds, if there are more than one, and lone pairs must be arranged so that their dipole moments do not cancel one another.

Polarity of MoleculesMolecules in which dipole moments of the bonds do not cancel are polar moleculestotal 0

Molecules that do not contain polar bonds or in which all dipole moments cancel are non-polar molecules

total = 0

CO2 vs H2O

CO O OH H

+ –

+ +

––

nonpolar polar

total 0total = 0

Dipole Moments of Polyatomic moleculesSome models

BeCl2 BF3

CH4 NH3 H2O

H

CHH

H

H

CHH

H

H

CHH

H

H

CHH

H

1 l o n e p a i r

p y r a m i d a l

NH H

H

. .

Dipole Moments of Polyatomic moleculesSome modelsBeCl2

BF3

H2O

Dipole Moments of Polyatomic moleculesSome examples to work…CCl4 ( = 0) CHCl3 ( = 1.04 D)

No net dipole Net dipole

CH2Cl2 ( = 1.60 D) CH3Cl ( = 1.92 D)

NH3 ( = 1.47 D)

CH4 … CCl4 Polar or Not?

Only CH4 and CCl4 are NOT polar. These are theonly two molecules that are “symmetrical.”

Bonding OrderOrder of a bond is the number of bonding electron pairs shared by two atoms in a molecule. There are bond orders of 1 (single), 2 (double), 3 (triple), and fractional (1.5, 2.5, ..)

Average # of shared pairs linking atoms Bond order = ────────────────────────

# of links in the molecule or ion (no L.P.)

H – Cl O – O = O O = C = O H – CN1 1 2 2 2 1 3

For the molecule (around the central atom):A.B.O. 1/1 = 1 3/2 = 1.5 4/2 = 2 4/2 = 2

Bond Strength and Bond Lengthbond length

(shortest) triple < double < single

bond strengthsingle < double < triple (strongest)

Bond Length (Å) Strength (kJ/mol)

C – C 1.54 348

C = C 1.34 614 (not 2x)

C C 1.20 839 (not 3x)

Bond Energies• chemical reactions involve breaking bonds in

reactant molecules and making new bonds to create the products

• the H°reaction can be calculated by comparing the cost of breaking old bonds to the profit from making new bonds

• the amount of energy it takes to break one mole of a bond in a compound is called the bond energy– in the gas state– homolytically – each atom gets ½ of

bonding electrons

Trends in Bond Energies• the more electrons two atoms share, the

stronger the covalent bond– C≡C (837 kJ) > C=C (611 kJ) > C−C (347 kJ)– C≡N (891 kJ) > C=N (615 kJ) > C−N (305 kJ)

• the shorter the covalent bond, the stronger the bond– Br−F (237 kJ) > Br−Cl (218 kJ) > Br−Br (193 kJ)– bonds get weaker down the column of

periodic table.

Using Bond Energies to Estimate H°rxn

• the actual bond energy depends on the surrounding atoms and other factors

• we often use average bond energies to estimate the Hrxn– works best when all reactants and

products in gas state• bond breaking is endothermic, H(breaking) = +

• bond making is exothermic, H(making) = −

Hrxn = ∑ (H(bonds broken)) + ∑ (H(bonds all >0 all <0 formed))

Estimate the Enthalpy of the Following Reaction

H H + O O H O O H

Estimate the Enthalpy of the Following Reaction

H2(g) + O2(g) H2O2(g)reaction involves breaking 1mol H-H and 1 mol

O=O and making 2 mol H-O and 1 mol O-Obonds broken (energy cost)(+436 kJ) + (+498 kJ) = +934 kJbonds made (energy released)2(−464 kJ) + (−142 kJ) = -1070Hrxn = (+934 kJ) + (-1070. kJ) = -136 kJ(Appendix H°f = -136.3 kJ/mol)

H H + O O H O O H

Calculating the Hcombustion of acetic acid(g)CH3COOH(g) + 2O2(g) 2H2O(g) + 2CO2 (g)

H OH

HCC=O + 2 O=O 2HOH + 2 O=C=O

CHcomb = 3EC-H + EC-C + 2EO=O + EC=O + EC-O + EO-H

4EC=O(* in CO2) 4EO-H

Hcmb = 3×414 + 347 + 2×498 + 736 + 360 4x799 3x464

Hcomb = 907 kJ

Bond Lengths• the distance between the

nuclei of bonded atoms is called the bond length

• because the actual bond length depends on the other atoms around the bond we often use theaverage bond length– averaged for similar

bonds from many compounds

Trends in Bond Lengths

• the more electrons two atoms share, the shorter the covalent bond– C≡C (120 pm) < C=C (134 pm) < C−C (154 pm)– C≡N (116 pm) < C=N (128 pm) < C−N (147 pm)

• decreases from left to right across period– C−C (154 pm) > C−N (147 pm) > C−O (143 pm)

• increases down the column– F−F (144 pm) < Cl−Cl (198 pm) < Br−Br (228

pm) • in general, as bonds get longer, they also get

weaker

Bond Lengths

chapter 9 chemical bonding - university of central...

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