Catherine MacGowan

Armstrong Atlantic State University

Chapter 9

Chemical Bonding I:The Lewis Model

© 2013 Pearson Education, Inc.

Lecture Presentation

It is a measure of the ability of an atom in a bond to attract It is a measure of the ability of an atom in a bond to attract electrons to itself.electrons to itself.

This attraction or pulling of electrons causes a separation of This attraction or pulling of electrons causes a separation of charge within the bond.charge within the bond. Dipole moment is formed.Dipole moment is formed.

Symbol: Symbol:

The greater the difference, the more The greater the difference, the more POLARPOLAR the bond. the bond.

What Is Electronegativity ()?

Electronegativity and Bond PolarityIf the difference in electronegativity between bonded atoms is:

– Zero (0), then the bond is pure covalent.• Equal sharing of electrons between atoms

– 0.1 to 0.4, then the bond is nonpolar covalent.

– 0.5 to 1.9, then the bond is polar covalent.

– ≥2.0, then the bond is ionic.• Transfer of electrons between atoms

100%

0 0.4 2.0 4.0

4% 51%

Percent ionic character

Electronegativity difference

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Chapter |Periodic Trends: Electronegativity

4

• Periodicity of the Elements • 1• 4

Increasing electronegativity

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Electronegativity and Bond Types

Illustration of Main Types of Intramolecular Bonds

Ionic Bonding• Occurs between a metal and a nonmetal element.Occurs between a metal and a nonmetal element.

• Transfer of electron(s) from one element to another.Transfer of electron(s) from one element to another.• Cation/anion formationCation/anion formation

• Strength of bond can be calculated using Coulomb’s law.Strength of bond can be calculated using Coulomb’s law.

• E E = = kQ1Q2kQ1Q2 RR RR

• Ionic bonding results in the attraction of the between the two Ionic bonding results in the attraction of the between the two charges (cation + anion).charges (cation + anion).

Q2Q1

What Is Covalent Bonding?• The bond arises from the mutual attraction of two nuclei for the same electrons.The bond arises from the mutual attraction of two nuclei for the same electrons.

– Electron sharingElectron sharing results. results.

• The bond is a balance between: The bond is a balance between: – Attractive forcesAttractive forces

• (p+ to e-) (p+ to e-) – Repulsive forces Repulsive forces

• (e- to e-) (e- to e-) • (p+ to p+)(p+ to p+)

• Bond can result from a “head-to-head”Bond can result from a “head-to-head” overlapoverlap of atomic orbitals on neighboring of atomic orbitals on neighboring atoms.atoms.

• Occurs between nonmetal elementsOccurs between nonmetal elements

Sigma Bond Formation by Orbital Overlap

Measured by the energy required to break a bond

Bond Bond dissociation enthalpy (kJ/mol) H - H 436 C - C 346 C=C 602 C = C 835 N = N 945

• The The GREATERGREATER the number of bonds (bond order), the the number of bonds (bond order), the HIGHERHIGHER the bond the bond strength and the strength and the SHORTERSHORTER the bond. the bond.

Bond Energies: StrengthBond Energies: Strength

• Decide on the central atom; never H.Therefore, N is central.

• Count the number of valence electrons each atom is bringing.H = 1 but you have 3 H atomsso, 3 x 1 = 3 electrons

N = 5 Total = 3 + 5 = 8 electrons

• Eight valence electrons need to be accounted for in the Lewis structure such that H has 2 electrons or 1 bonding pair and N has a total of 8 valence electrons surrounding it.– 3 BOND PAIRS and 1 LONE PAIR1 LONE PAIR

H••NN

H

H

H NN H

H

NH3: Ammonia’s Lewis Dot Structure

• Central atom = B (6 electrons surrounding max.)– The B atom has a share in only 6 electrons or 3 pairs.– B atom in many molecules is electron deficient

• Total valence electrons in molecule = 24

• B atom is an exception to octet “rule.” F••

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F

F

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Boron Trifluoride, BF3

Why?Why?According to quantum mechanics (quantum numbers) elements in the 3rd period According to quantum mechanics (quantum numbers) elements in the 3rd period can have more than 8 valence electrons if center because “3can have more than 8 valence electrons if center because “3dd” predicted by the ” predicted by the quantum number (quantum number (ll))

Step 1Step 1: : Center atom is SCenter atom is S

Step 2Step 2: : Count number valence electronsCount number valence electrons F: 7 x 4 = 28F: 7 x 4 = 28 S: 6 x 1 = 6S: 6 x 1 = 6Total valence electrons = 34Total valence electrons = 34

Step 3Step 3: : Make bonds Make bonds F only single bondsF only single bonds S must have 8 valence electronsS must have 8 valence electrons

BUT can have > 8 valence electrons if center atomBUT can have > 8 valence electrons if center atom

*Usually nonmetal elements of periods 3 and greater*Usually nonmetal elements of periods 3 and greater

F

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SF4: Exception to “Octet Rule”

Central atom = CTotal valence electrons = 16

Place lone pairs on outer atoms.

NOTENOTE: For carbon and oxygen to have 8 valence electrons, need to form double bond

O OC

••O OC

•• ••

••••••

Carbon Dioxide, CO2

Central atom is oxygen.

O – O – O

Total valence electrons for this molecule is 18.

• Oxygen needs to form a double bond so that it has 8 electrons around it.

• This leads to the following structures; both satisfy the octet guidelines.

• These equivalent structures are called These equivalent structures are called RESONANCE STRUCTURESRESONANCE STRUCTURES. . • The true electronic structure is aThe true electronic structure is a HYBRIDHYBRID of the two. of the two.

Ozone, O3

• Total number of electrons: 24– N atom brings in 5 electrons– O atom brings in 6 each

• You have 3 oxygen atoms so 6 x 3 = 18 electrons– Anion = 1 electron

• The N atom and O atom must have 8 electrons surrounding each of them.

• Three legitimate Lewis dot structures for the nitrate ion can be drawn.Three legitimate Lewis dot structures for the nitrate ion can be drawn.– ResonanceResonance

Nitrate ion, NO3-

CO

SeOF2

NO2−

H3PO4

SO32−

P2H4

Practice Problems: Lewis Structures

CO (10 e-) : c = o:

SeOF2 (26 e-)

NO2− (18 e-)

Answers:

H3PO4 (32 e-)

SO32− (26 e-)

P2H4 (14 e-)

Answers:

H3PO4 (32 e-)

SO32− (26 e-)

P2H4 (14 e-)

Answers:

Problem:Predict the most stable structure: ONC- or OCN- or NOC-.

. . . . . . . . . . . .. . . . . . . . . . . . : O :: N :: C : : O :: N :: C : : O :: C :: N : : O :: C :: N : : N :: O :: C : : N :: O :: C :

(A) (B) (C)

Formal charge: (# valence electrons) - (# of bonds) - (# nonbonding electrons)(# valence electrons) - (# of bonds) - (# nonbonding electrons)

(A) (B) (C)

O 6 – 2 - 4 = 0 6 – 2 - 4 = 0 6 – 4 - 0 = +2

N 5 – 4 - 0 = +1 5 – 2 - 4 = -1 5 – 2 - 4 = -1

C 4 – 2 - 4 = -2 4 - 2 - 0 = 0 4 – 2 - 4 = -2

NOTE: Structure B has the combination with the lowest formal charge, and it has the NOTE: Structure B has the combination with the lowest formal charge, and it has the negative formal charge on one of the more electronegative atoms. negative formal charge on one of the more electronegative atoms.

Sigma Bond Formation by Orbital Overlap

Measured by the energy required to break a bond

Bond Bond dissociation enthalpy (kJ/mol) H - H 436 C - C 346 C=C 602 C = C 835 N = N 945

• The The GREATERGREATER the number of bonds (bond order), the the number of bonds (bond order), the HIGHERHIGHER the bond the bond strength and the strength and the SHORTERSHORTER the bond. the bond.

Bond Energies: StrengthBond Energies: Strength

• Net energy is equal to the ∆Hrxn.

∆Hrxn can be calculated using the following equation:

• Breaking bonds REQUIRES energy (use + value of bond energies).• Making bonds RELEASES energy (use the - value of bond energies).

• Breakage of weak bonds and the formation of stronger bonds: exothermic reaction occurs.

• Breakage of strong bonds and the formation of weak bonds: endothermic reaction occurs.

Using Bond Dissociation EnthalpiesUsing Bond Dissociation Enthalpies

Problem: Find the Hrxn for CH4(g) + Cl2(g) HCl(g) + CH3Cl(g)

Average Bond EnergiesGiven: H – H 436 kJ/mol Cl - Cl 242 kJ/mol

H – Cl 432 kJ/mol H – C 414 kJ/molCl – Cl 339 kJ/mol

∆Hrxn = energy required to break bonds + energy evolved when bonds are made

∆Hrxn = {(4 mole x 414 kJ/mol) + (1 mol x (242 kJ/mol)} +

{(1mol x - 432 kJ/mol) + (3 mol x - 414 kJ/mol) + (1 mole x -339 kJ/mol )}

= 1898 kJ + (-2013 kJ) = - 115 kJ

This reaction is exothermic!