Chapter 10 Chemical Bonding II

Copyright 2011 Pearson Education, Inc.Tro: Chemistry: A Molecular Approach, 2/e

Chapter 10Chemical Bonding II

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro


Taste• The taste of a food depends on the interaction

between the food molecules and taste cells on your tongue

• The main factors that affect this interaction are the shape of the molecule and charge distribution within the molecule

• The food molecule must fit snugly into the active site of specialized proteins on the surface of taste cells

• When this happens, changes in the protein structure cause a nerve signal to transmit

2Tro: Chemistry: A Molecular Approach, 2/e


Sugar & Artificial Sweeteners• Sugar molecules fit into the active site of taste cell

receptors called Tlr3 receptor proteins• When the sugar molecule (the key) enters the

active site (the lock), the different subunits of the T1r3 protein split apart

• This split causes ion channels in the cell membrane to open, resulting in nerve signal transmission

• Artificial sweeteners also fit into the Tlr3 receptor, sometimes binding to it even stronger than sugarmaking them “sweeter” than sugar



Structure Determines Properties!• Properties of molecular substances depend on

the structure of the molecule• The structure includes many factors, such as:

the skeletal arrangement of the atomsthe kind of bonding between the atoms

ionic, polar covalent, or covalentthe shape of the molecule

• Bonding theory should allow you to predict the shapes of molecules



Molecular Geometry• Molecules are 3-dimensional objects• We often describe the shape of a molecule

with terms that relate to geometric figures• These geometric figures have characteristic

“corners” that indicate the positions of the surrounding atoms around a central atom in the center of the geometric figure

• The geometric figures also have characteristic angles that we call bond angles



Lewis Theory PredictsElectron Groups

• Lewis theory predicts there are regions of electrons in an atom

• Some regions result from placing shared pairs of valence electrons between bonding nuclei

• Other regions result from placing unshared valence electrons on a single nuclei



Using Lewis Theory to PredictMolecular Shapes

• Lewis theory says that these regions of electron groups should repel each otherbecause they are regions of negative charge

• This idea can then be extended to predict the shapes of molecules the position of atoms surrounding a central atom will be

determined by where the bonding electron groups are the positions of the electron groups will be determined

by trying to minimize repulsions between them



VSEPR Theory

• Electron groups around the central atom will be most stable when they are as far apart as possible – we call this valence shell electron pair repulsion theorybecause electrons are negatively charged, they

should be most stable when they are separated as much as possible

• The resulting geometric arrangement will allow us to predict the shapes and bond angles in the molecule



Electron Groups• The Lewis structure predicts the number of valence

electron pairs around the central atom(s) • Each lone pair of electrons constitutes one electron

group on a central atom• Each bond constitutes one electron group on a

central atom regardless of whether it is single, double, or triple

O N O ••••••

•••••• there are three electron groups on N

three lone pairone single bondone double bond



Electron Group Geometry• There are five basic arrangements of electron

groups around a central atombased on a maximum of six bonding electron groups

though there may be more than six on very large atoms, it is very rare

• Each of these five basic arrangements results in five different basic electron geometries in order for the molecular shape and bond angles to be

a “perfect” geometric figure, all the electron groups must be bonds and all the bonds must be equivalent

• For molecules that exhibit resonance, it doesn’t matter which resonance form you use – the electron geometry will be the same



Linear Electron Geometry• When there are two electron groups around the

central atom, they will occupy positions on opposite sides of the central atom

• This results in the electron groups taking a linear geometry

• The bond angle is 180°



Linear Geometry



Trigonal Planar Electron Geometry• When there are three electron groups around

the central atom, they will occupy positions in the shape of a triangle around the central atom

• This results in the electron groups taking a trigonal planar geometry

• The bond angle is 120°



Trigonal Geometry



Tetrahedral Electron Geometry• When there are four electron groups around the

central atom, they will occupy positions in the shape of a tetrahedron around the central atom

• This results in the electron groups taking a tetrahedral geometry

• The bond angle is 109.5°



Tetrahedral Geometry



Trigonal Bipyramidal Electron Geometry• When there are five electron groups around the central

atom, they will occupy positions in the shape of two tetrahedra that are base-to-base with the central atom in the center of the shared bases

• This results in the electron groups taking a trigonal bipyramidal geometry

• The positions above and below the central atom are called the axial positions

• The positions in the same base plane as the central atom are called the equatorial positions

• The bond angle between equatorial positions is 120°• The bond angle between axial and equatorial positions is 90°



Trigonal Bipyramid



Trigonal Bipyramidal Geometry



Octahedral Electron Geometry• When there are six electron groups around the central

atom, they will occupy positions in the shape of two square-base pyramids that are base-to-base with the central atom in the center of the shared bases

• This results in the electron groups taking an octahedral geometry it is called octahedral because the geometric figure has

eight sides• All positions are equivalent• The bond angle is 90°



Octahedral Geometry



Molecular Geometry• The actual geometry of the molecule may be

different from the electron geometry• When the electron groups are attached to

atoms of different size, or when the bonding to one atom is different than the bonding to another, this will affect the molecular geometry around the central atom

• Lone pairs also affect the molecular geometrythey occupy space on the central atom, but are not

“seen” as points on the molecular geometry



Not Quite Perfect Geometry

Because the bonds and atom sizes are not identical in formaldehyde, the observed angles are slightly different from ideal



The Effect of Lone Pairs

• Lone pair groups “occupy more space” on the central atombecause their electron density is exclusively on the

central atom rather than shared like bonding electron groups

• Relative sizes of repulsive force interactions isLone Pair – Lone Pair > Lone Pair – Bonding Pair > Bonding Pair – Bonding Pair • This affects the bond angles, making the bonding

pair – bonding pair angles smaller than expected



Effect of Lone Pairs

The bonding electrons are shared by two atoms, so some of the negative charge is removed from the central atom

The nonbonding electrons are localized on the central atom, so area of negative charge takes more space



Bond Angle Distortion from Lone Pairs



Bent Molecular Geometry:Derivative of Trigonal Planar Electron Geometry• When there are three electron groups around

the central atom, and one of them is a lone pair, the resulting shape of the molecule is called a trigonal planar — bent shape

• The bond angle is less than 120°because the lone pair takes up more space



Pyramidal & Bent Molecular Geometries:Derivatives of Tetrahedral Electron Geometry• When there are four electron groups around the

central atom, and one is a lone pair, the result is called a pyramidal shapebecause it is a triangular-base pyramid with the

central atom at the apex• When there are four electron groups around the

central atom, and two are lone pairs, the result is called a tetrahedral—bent shape it is planar it looks similar to the trigonal planar—bent shape,

except the angles are smaller• For both shapes, the bond angle is less than

109.5°33Tro: Chemistry: A Molecular Approach, 2/e


Methane



Pyramidal Shape



Tetrahedral–Bent Shape



Derivatives of theTrigonal Bipyramidal Electron Geometry

• When there are five electron groups around the central atom, and some are lone pairs, they will occupy the equatorial positions because there is more room

• When there are five electron groups around the central atom, and one is a lone pair, the result is called the seesaw shape aka distorted tetrahedron

• When there are five electron groups around the central atom, and two are lone pairs, the result is called the T-shaped

• When there are five electron groups around the central atom, and three are lone pairs, the result is a linear shape

• The bond angles between equatorial positions are less than 120°

• The bond angles between axial and equatorial positions are less than 90° linear = 180° axial–to–axial



Replacing Atoms with Lone Pairsin the Trigonal Bipyramid System



Seesaw Shape



T–Shape



Linear Shape



Derivatives of theOctahedral Geometry

• When there are six electron groups around the central atom, and some are lone pairs, each even number lone pair will take a position opposite the previous lone pair

• When there are six electron groups around the central atom, and one is a lone pair, the result is called a square pyramid shape the bond angles between axial and equatorial positions is

less than 90°• When there are six electron groups around the

central atom, and two are lone pairs, the result is called a square planar shape the bond angles between equatorial positions is 90°



Square Pyramidal Shape



Square Planar Shape



Predicting the Shapes Around Central Atoms

1. Draw the Lewis structure2. Determine the number of electron groups

around the central atom3. Classify each electron group as bonding or

lone pair, and count each type remember, multiple bonds count as one group

4. Use Table 10.1 to determine the shape and bond angles



Example 10.2: Predict the geometry and bond angles of PCl3

1. Draw the Lewis structure

a) 26 valence electrons2. Determine the

Number of electron groups around central atom

a) four electron groups around P



Example 10.2: Predict the geometry and bond angles of PCl3

3. Classify the electron groupsa) three bonding groupsb) one lone pair

4. Use Table 10.1 to determine the shape and bond angles

a) four electron groups around P = tetrahedral electron geometry

b) three bonding + one lone pair = trigonal pyramidal molecular geometry

c) trigonal pyramidal = bond angles less than 109.5°



Practice – Predict the molecular geometry and bond angles in SiF5

−



Practice – Predict the molecular geometry and bond angles in SiF5

─

Si = 4e─

F5 = 5(7e─) = 35e─

(─) = 1e─

total = 40e─

5 electron groups on Si

5 bonding groups0 lone pairs

Shape = trigonal bipyramid

Bond anglesFeq–Si–Feq = 120°Feq–Si–Fax = 90°

Si least electronegative

Si is central atom



Practice – Predict the molecular geometry and bond angles in ClO2F



Practice – Predict the molecular geometry and bond angles in ClO2F

Cl = 7e─

O2 = 2(6e─) = 12e─

F = 7e─

Total = 26e─

4 electron groups on Cl

3 bonding groups1 lone pair

Shape = trigonal pyramidal

Bond anglesO–Cl–O < 109.5°O–Cl–F < 109.5°

Cl least electronegative

Cl is central atom



Representing 3-Dimensional Shapes on a 2-Dimensional Surface

• One of the problems with drawing molecules is trying to show their dimensionality

• By convention, the central atom is put in the plane of the paper

• Put as many other atoms as possible in the same plane and indicate with a straight line

• For atoms in front of the plane, use a solid wedge• For atoms behind the plane, use a hashed wedge



SF6

S

F

FF

F F

F



H O

| ||

H C C O H|

H

Multiple Central Atoms• Many molecules have larger structures with many

interior atoms • We can think of them as having multiple central

atoms• When this occurs, we describe the shape around

each central atom in sequence shape around left C is tetrahedral

shape around center C is trigonal planar

shape around right O is tetrahedral-bent



Describing the Geometryof Methanol



Describing the Geometryof Glycine



Practice – Predict the molecular geometries in H3BO3



3 electron groups on B

B has3 bonding groups0 pone pairs

4 electron groups on O

O has2 bonding groups2 lone pairs

Practice – Predict the molecular geometries in H3BO3

B = 3e─

O3 = 3(6e─) = 18e─

H3 = 3(1e─) = 3e─

Total = 24e─Shape on B = trigonal planar

B least electronegative

B Is Central Atom

oxyacid, so H attached to O

Shape on O = tetrahedral bent



Polarity of Molecules


• For a molecule to be polar it must1. have polar bonds

electronegativity difference - theory bond dipole moments - measured

2. have an unsymmetrical shape vector addition

• Polarity affects the intermolecular forces of attraction

therefore boiling points and solubilities like dissolves like

• Nonbonding pairs affect molecular polarity, strong pull in its direction


Molecule Polarity

The H─Cl bond is polar. The bonding electrons are pulled toward the Cl end of the molecule. The net result is a polar molecule.



Vector Addition



Molecule Polarity

The O─C bond is polar. The bonding electrons are pulled equally toward both O ends of the molecule. The net result is a nonpolar molecule.



Molecule Polarity

The H─O bond is polar. Both sets of bonding electrons are pulled toward the O end of the molecule. The net result is a polar molecule.



Predicting Polarity of Molecules

1. Draw the Lewis structure and determine the molecular geometry

2. Determine whether the bonds in the molecule are polar

a) if there are not polar bonds, the molecule is nonpolar

3. Determine whether the polar bonds add together to give a net dipole moment



Example 10.5: Predict whether NH3 is a polar molecule

1. Draw the Lewis structure and determine the molecular geometry

a) eight valence electronsb) three bonding + one lone

pair = trigonal pyramidal molecular geometry




2. Determine if the bonds are polar

a) electronegativity difference

b) if the bonds are not polar, we can stop here and declare the molecule will be nonpolar

ENN = 3.0ENH = 2.13.0 − 2.1 = 0.9therefore the bonds are polar covalent




3) Determine whether the polar bonds add together to give a net dipole moment

a) vector additionb) generally, asymmetric

shapes result in uncompensated polarities and a net dipole moment

The H─N bond is polar. All the sets of bonding electrons are pulled toward the N end of the molecule. The net result is a polar molecule.



Practice – Decide whether the following molecules are polar

ENO = 3.5N = 3.0Cl = 3.0S = 2.5



Practice – Decide whether the following molecules Are polar

polarnonpolar

1. polar bonds, N-O2. asymmetrical shape 1. polar bonds, all S-O

2. symmetrical shape

TrigonalBent Trigonal

Planar2.5



Molecular Polarity Affects

Solubility in Water• Polar molecules are attracted

to other polar molecules• Because water is a polar

molecule, other polar molecules dissolve well in waterand ionic compounds as well

• Some molecules have both polar and nonpolar parts



Problems with Lewis Theory• Lewis theory generally predicts trends in

properties, but does not give good numerical predictionse.g. bond strength and bond length

• Lewis theory gives good first approximations of the bond angles in molecules, but usually cannot be used to get the actual angle

• Lewis theory cannot write one correct structure for many molecules where resonance is important

• Lewis theory often does not predict the correct magnetic behavior of moleculese.g. O2 is paramagnetic, though the Lewis structure

predicts it is diamagnetic77Tro: Chemistry: A Molecular Approach, 2/e


Valence Bond Theory

• Linus Pauling and others applied the principles of quantum mechanics to molecules

• They reasoned that bonds between atoms would occur when the orbitals on those atoms interacted to make a bond

• The kind of interaction depends on whether the orbitals align along the axis between the nuclei, or outside the axis



Orbital Interaction• As two atoms approached, the half-filled

valence atomic orbitals on each atom would interact to form molecular orbitalsmolecular orbtials are regions of high probability of

finding the shared electrons in the molecule• The molecular orbitals would be more stable

than the separate atomic orbitals because they would contain paired electrons shared by both atomsthe potential energy is lowered when the molecular

orbitals contain a total of two paired electrons compared to separate one electron atomic orbitals



Orbital Diagram for the Formation of H2S

+

↑

H

1s ↑↓ H─S bond

↑

H

1sS↑ ↑ ↑↓↑↓

3s 3p

↑↓ H─S bond

Predicts bond angle = 90°Actual bond angle = 92°



Valence Bond Theory – Hybridization• One of the issues that arises is that the number

of partially filled or empty atomic orbitals did not predict the number of bonds or orientation of bonds C = 2s22px

12py12pz

0 would predict two or three bonds that are 90° apart, rather than four bonds that are 109.5° apart

• To adjust for these inconsistencies, it was postulated that the valence atomic orbitals could hybridize before bonding took placeone hybridization of C is to mix all the 2s and 2p

orbitals to get four orbitals that point at the corners of a tetrahedron



Unhybridized C Orbitals Predict the Wrong Bonding & Geometry



Valence Bond TheoryMain Concepts

1. The valence electrons of the atoms in a molecule reside in quantum-mechanical atomic orbitals. The orbitals can be the standard s, p, d, and f orbitals, or they may be hybrid combinations of these.

2. A chemical bond results when these atomic orbitals interact and there is a total of two electrons in the new molecular orbitala) the electrons must be spin paired

3. The shape of the molecule is determined by the geometry of the interacting orbitals



Hybridization• Some atoms hybridize their orbitals to

maximize bonding• more bonds = more full orbitals = more stability

• Hybridizing is mixing different types of orbitals in the valence shell to make a new set of degenerate orbitalssp, sp2, sp3, sp3d, sp3d2

• Same type of atom can have different types of hybridizationC = sp, sp2, sp3



Hybrid Orbitals• The number of standard atomic orbitals

combined = the number of hybrid orbitals formed combining a 2s with a 2p gives two 2sp hybrid

orbitalsH cannot hybridize!!

its valence shell only has one orbital• The number and type of standard atomic orbitals

combined determines the shape of the hybrid orbitals

• The particular kind of hybridization that occurs is the one that yields the lowest overall energy for the molecule



Carbon Hybridizations

Unhybridized

2s 2p

sp hybridized

2sp

sp2 hybridized

2p

sp3 hybridized

2p

2sp2

2sp3



sp3 Hybridization• Atom with four electron groups around it

tetrahedral geometry109.5° angles between hybrid orbitals

• Atom uses hybrid orbitals for all bonds and lone pairs



Orbital Diagram of the sp3 Hybridization of C



sp3 Hybridized AtomsOrbital Diagrams

Unhybridized atom

2s 2p

sp3 hybridized atom

2sp3

C

2s 2p

2sp3

N

• Place electrons into hybrid and unhybridized valence orbitals

as if all the orbitals have equal energy• Lone pairs generally occupy hybrid orbitals



Practice – Draw the orbital diagram for the sp3 hybridization of each atom

Unhybridized atom

3s 3p Cl

2s 2p O

sp3 hybridized atom

3sp3

2sp3



Bonding with Valence Bond Theory

• According to valence bond theory, bonding takes place between atoms when their atomic or hybrid orbitals interact“overlap”

• To interact, the orbitals must either be aligned along the axis between the atoms, or

• The orbitals must be parallel to each other and perpendicular to the interatomic axis



Methane Formation with sp3 C



Ammonia Formation with sp3 N



Types of Bonds• A sigma (s) bond results when the interacting

atomic orbitals point along the axis connecting the two bonding nucleieither standard atomic orbitals or hybrids

s–to–s, p–to–p, hybrid–to–hybrid, s–to–hybrid, etc.• A pi (p) bond results when the bonding atomic

orbitals are parallel to each other and perpendicular to the axis connecting the two bonding nucleibetween unhybridized parallel p orbitals

• The interaction between parallel orbitals is not as strong as between orbitals that point at each other; therefore s bonds are stronger than p bonds



Orbital Diagrams of Bonding

• “Overlap” between a hybrid orbital on one atom with a hybrid or nonhybridized orbital on another atom results in a s bond

• “Overlap” between unhybridized p orbitals on bonded atoms results in a p bond



CH3NH2 Orbital Diagram

sp3 C sp3 N

1s H

s

sss s s

1s H 1s H 1s H 1s H

H C

H

H

N H

H

··ss

ss

s

s



Formaldehyde, CH2O Orbital Diagram

sp2 C

sp2 O

1s H

s

ss

1s H

p C p Op



sp2

• Atom with three electron groups around it trigonal planar system

C = trigonal planar N = trigonal bent O = “linear”

120° bond angles flat

• Atom uses hybrid orbitals for s bonds and lone pairs, uses nonhybridized p orbital for p bond



sp2 Hybridized AtomsOrbital Diagrams

Unhybridized atom

2s 2p

sp2 hybridized atom

2sp2

2p

C3 s1 p

2s 2p

2sp2

2p

N2 s1 p



Practice – Draw the orbital diagram for the sp2 hybridization of each atom. How many s

and p bonds would you expect each to form?

Unhybridized atom

2s 2p

B3 s0 p

2s 2p

O1 s1 p

sp2 hybridized atom

2sp2

2p

2sp2

2p



Hybrid orbitals overlap to form a s bond.Unhybridized p orbitals overlap to form a p bond.



CH2NH Orbital Diagram

sp2 C

sp2 N

1s H

s

ss s

1s H 1s H

p C p Np

105

C N

H

H H

･･

Tro: Chemistry: A Molecular Approach, 2/e


Bond Rotation

• Because the orbitals that form the s bond point along the internuclear axis, rotation around that bond does not require breaking the interaction between the orbitals

• But the orbitals that form the p bond interact above and below the internuclear axis, so rotation around the axis requires the breaking of the interaction between the orbitals



sp• Atom with two electron groups

linear shape180° bond angle

• Atom uses hybrid orbitals for s bonds or lone pairs, uses nonhybridized p orbitals for p bonds

p

ps



sp Hybridized AtomsOrbital Diagrams

Unhybridized atom

2s 2p

sp hybridized atom

2sp

2p

C2s2p

2s 2p

2sp

2p

N1s2p



HCN Orbital Diagram

sp C

sp N

1s H

s

s

p C p N2p



sp3d• Atom with five electron groups

around ittrigonal bipyramid electron

geometrySeesaw, T–Shape, Linear120° & 90° bond angles

• Use empty d orbitals from valence shell

• d orbitals can be used to make p bonds



sp3d Hybridized AtomsOrbital Diagrams

Unhybridized atom

3s 3p

sp3d hybridized atom

3sp3d

S3s 3p

3sp3d

P

3d

3d

(non-hybridizing d orbitals not shown)



SOF4 Orbital Diagram

sp3d S

sp2 O

2p F

s

s

d S p Op

2p F

s

2p F

s

2p F

s



sp3d2

• Atom with six electron groups around itoctahedral electron geometrySquare Pyramid, Square Planar90° bond angles

• Use empty d orbitals from valence shell to form hybrid

• d orbitals can be used to make p bonds



sp3d2 Hybridized AtomsOrbital Diagrams

Unhybridized atom sp3d2 hybridized atom

S3s 3p 3sp3d2

↑↓

3d

↑↓ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑

I5s 5p↑↓

5d

↑↓ ↑↓ ↑

5sp3d2

↑↓ ↑ ↑ ↑ ↑ ↑

(non-hybridizing d orbitals not shown)



Predicting Hybridization and Bonding Scheme

1. Start by drawing the Lewis structure2. Use VSEPR Theory to predict the electron group

geometry around each central atom3. Use Table 10.3 to select the hybridization

scheme that matches the electron group geometry

4. Sketch the atomic and hybrid orbitals on the atoms in the molecule, showing overlap of the appropriate orbitals

5. Label the bonds as s or p122Tro: Chemistry: A Molecular Approach, 2/e


Example 10.7: Predict the hybridization and bonding scheme for CH3CHO

123

Draw the Lewis structure

Predict the electron group geometry around inside atoms

C1 = 4 electron areas C1= tetrahedral C2 = 3 electron areas C2 = trigonal planar




124

Determine the hybridization of the interior atoms

C1 = tetrahedral C1 = sp3

C2 = trigonal planar C2 = sp2

Sketch the molecule and orbitals



Label the bonds




Practice – Predict the hybridization of all the atoms in H3BO3

H = can’t hybridizeB = 3 electron groups = sp2

O = 4 electron groups = sp3



Practice – Predict the hybridization and bonding scheme of all the atoms in NClO

N = 3 electron groups = sp2

O = 3 electron groups = sp2

Cl = 4 electron groups = sp3

••O N Cl ••••

••••••

127

Cl

O Ns:Nsp2─Clp

Cl

O N

s:Osp2─Nsp2

p:Op─Np

↑↓

↑↓↑↓

↑↓

↑↓

↑↓



Problems with Valence Bond Theory• VB theory predicts many properties better than

Lewis theorybonding schemes, bond strengths, bond lengths,

bond rigidity• However, there are still many properties of

molecules it doesn’t predict perfectlymagnetic behavior of O2

• In addition, VB theory presumes the electrons are localized in orbitals on the atoms in the molecule – it doesn’t account for delocalization



Molecular Orbital Theory• In MO theory, we apply Schrödinger’s wave

equation to the molecule to calculate a set of molecular orbitals in practice, the equation solution is estimated we start with good guesses from our experience as to

what the orbital should look like then test and tweak the estimate until the energy of the

orbital is minimized• In this treatment, the electrons belong to the

whole molecule – so the orbitals belong to the whole moleculedelocalization



LCAO

• The simplest guess starts with the atomic orbitals of the atoms adding together to make molecular orbitals – this is called the Linear Combination of Atomic Orbitals methodweighted sum

• Because the orbitals are wave functions, the waves can combine either constructively or destructively



Molecular Orbitals• When the wave functions combine constructively,

the resulting molecular orbital has less energy than the original atomic orbitals – it is called a Bonding Molecular Orbitals, pmost of the electron density between the nuclei

• When the wave functions combine destructively, the resulting molecular orbital has more energy than the original atomic orbitals – it is called an Antibonding Molecular Orbitals*, p*most of the electron density outside the nuclei nodes between nuclei



Interaction of 1s Orbitals



Molecular Orbital Theory

• Electrons in bonding MOs are stabilizinglower energy than the atomic orbitals

• Electrons in antibonding MOs are destabilizinghigher in energy than atomic orbitalselectron density located outside the

internuclear axiselectrons in antibonding orbitals cancel

stability gained by electrons in bonding orbitals



Energy Comparisons of Atomic Orbitals to Molecular Orbitals



MO and Properties• Bond Order = difference between number of

electrons in bonding and antibonding orbitalsonly need to consider valence electronsmay be a fractionhigher bond order = stronger and shorter bonds if bond order = 0, then bond is unstable compared to

individual atoms and no bond will form• A substance will be paramagnetic if its MO

diagram has unpaired electrons if all electrons paired it is diamagnetic



1s 1s

s*

s

Hydrogen AtomicOrbital

Hydrogen AtomicOrbital

Dihydrogen, H2 MolecularOrbitals

Because more electrons are in bonding orbitals than are in antibonding orbitals,

net bonding interaction136Tro: Chemistry: A Molecular Approach, 2/e


H2

s* Antibonding MOLUMO

s bonding MOHOMO



1s 1s

s*

s

Helium AtomicOrbital

Helium AtomicOrbital

Dihelium, He2 MolecularOrbitals

Because there are as many electrons in antibonding orbitals as in bonding orbitals,

there is no net bonding interaction

BO = ½(2-2) = 0



1s 1s

s*

s

Lithium AtomicOrbitals

Lithium AtomicOrbitals

Dilithium, Li2 MolecularOrbitals

Because more electrons are in bonding orbitals than are in antibonding orbitals, there is a

net bonding interaction

2s 2s

s*

s Any fill energy level will generate filled bonding and antibonding MO’s;therefore only need to consider valence shell

BO = ½(4-2) = 1



s bonding MOHOMO

s* Antibonding MOLUMO

Li2



Interaction of p Orbitals



O2

• Dioxygen is paramagnetic• Paramagnetic material has unpaired electrons• Neither Lewis theory nor valence bond theory

predict this result



O2 as Described by Lewis and VB Theory



OxygenAtomicOrbitals

OxygenAtomicOrbitals

2s 2s

2p 2p

s*

s

s*

s

p*

pBecause more electrons are in bonding orbitals than are in antibonding orbitals, there is a net bonding interaction

Because there are unpaired electrons in the

antibonding orbitals, O2 is predicted to be

paramagnetic

O2 MO’s

BO = ½(8 be – 4 abe)BO = 2



N has 5 valence electrons

2 N = 10e−

(−) = 1e−

total = 11e−

Example 10.10: Draw a molecular orbital diagram of N2

− ion and predict its bond order and magnetic properties

147

Write a MO diagram for N2

− using N2 as a base Count the number of valence electrons and assign these to the MOs following the aufbau principle, Pauli principle & Hund’s rule

↑↓ s2s

s*2s

p2p

s2p

p*2p

s*2p

↑↓

↑ ↑↓ ↓

↑↓

↑



Example 10.10: Draw a molecular orbital diagram of N2

− ion and predict its bond order and magnetic properties

148

Calculate the bond order by taking the number of bonding electrons and subtracting the number of antibonding electrons, then dividing by 2Determine whether the ion is paramagnetic or diamagnetic

↑↓ s2s

s*2s

p2p

s2p

p*2p

s*2p

↑↓

↑ ↑↓ ↓

↑↓

↑

BO = ½(8 be – 3 abe)BO = 2.5

Because this is lower than the bond order

in N2, the bond should be weaker

Because there are unpaired electrons, this ion is paramagnetic



Practice – Draw a molecular orbital diagram of C2+

and predict its bond order and magnetic properties


Copyright 2011 Pearson Education, Inc.Tro: Chemistry: A Molecular Approach, 2/e 150

↑↓ s2s

s*2s

p2p

s2p

p*2p

s*2p

↑↓

↑ ↑↓BO = ½(5 be – 2 abe)

BO = 1.5

Because there are unpaired electrons, this ion is paramagnetic

Practice – Draw a molecular orbital diagram of C2+

and predict its bond order and magnetic properties

C has 4 valence electrons

2 C = 8e−

(+) = −1e−

total = 7e−



Heteronuclear Diatomic Molecules & Ions

• When the combining atomic orbitals are identical and equal energy, the contribution of each atomic orbital to the molecular orbital is equal

• When the combining atomic orbitals are different types and energies, the atomic orbital closest in energy to the molecular orbital contributes more to the molecular orbital



Heteronuclear Diatomic Molecules & Ions

• The more electronegative an atom is, the lower in energy are its orbitals

• Lower energy atomic orbitals contribute more to the bonding MOs

• Higher energy atomic orbitals contribute more to the antibonding MOs

• Nonbonding MOs remain localized on the atom donating its atomic orbitals



NO

a free radical

s2s Bonding MOshows more

electron density near O because it is mostly O’s

2s atomic orbital153

BO = ½(6 be – 1 abe)BO = 2.5



HF



Polyatomic Molecules

• When many atoms are combined together, the atomic orbitals of all the atoms are combined to make a set of molecular orbitals, which are delocalized over the entire molecule

• Gives results that better match real molecule properties than either Lewis or valence bond theories



Ozone, O3

156

MO Theory: Delocalized p bonding orbital of O3


Chapter 10 Chemical Bonding II

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Transcript of Chapter 10 Chemical Bonding II