08b Bonding 6.pdf

(1)

Chapters 8 and 9

Part 2 representing molecules

How Can We Predict the Shape of a Molecule?

Most molecules are not flat as implied by the Lewis structures.

Most molecules have three-dimensional shapes.

The shape contributes to the properties of the molecule or ion.

Diamond Graphite

Thalidomide: Anti-Nausea Drug given to

Pregnant Women

Helped

Alleviate

nausea

Caused

Babies to

Be born

Without

arms

and legs

Valence Shell Electron Pair Repulsion

(VSEPR) Theory: Used to Predict Shape

•Valence electron groups around the central atom repel each

other and consequently are arranged as far apart as possible to minimize repulsions.

•An electron group may be a •single bond

•double bond •triple bond •lone pair of electrons

•There are five basic shapes based on the number of electron groups around the central atom.

A is the central atom surrounded by

x

B atoms.

x

can have integer values of 2 to 6.

AB

_x

Valence Shell

Electron Pair Repulsion

(VSEPR) Theory: Used to Predict Shape

Lewis Structure Vocabulary

A lone pair (single electron group)

A bonding pair of electrons

(2)

Lewis Structure Vocabulary

Lone pairs, single bonds, double bonds, and triple bonds, are all considered as one electron group each for VSEPR

2 double bonds

2 electron groups (on central atom)

1 single bond

1 double bond

1 lone pair

3 single bonds

1 lone pair

Electron-Domain Geometry and Molecular Geometry

The basis of the VSEPR model is that electrons repel each other.

Electrons will arrange themselves to be as far apart as possible.

Arrangements minimize repulsive interactions.

2 electron domains Linear

3 electron domains Trigonal planar

Electron-Domain Geometry and Molecular Geometry

5 electron domains Trigonal bipyramidal

6 electron domains Octahedral 4 electron domains Tetrahedral

Electron Group Geometry,

Molecular Geometry, Bond Angle

The electron domain geometryis the arrangement of electron domains around the central atom.

The molecular geometryis the arrangement of bonded atoms.

In an ABx molecule, a bond angleis the angle between two adjacent A-B bonds.

Trigonal bipyramidal Octahedral Tetrahedral

Linear Trigonal planar

180°

120°

109.5° 90°

120°

90°

2 Electron Groups BeCl

2

1

2

Linear

3 Electron Groups BF

3

1

2

3

(3)

4 Electron Groups CH

4

Tetrahedral

1

2

3

4

5 Electron Groups PF

5

Trigonal Bipyramidal

1

2 3

4

5

AB

5

molecules contain two different bond

angles between adjacent bonds

Trigonal bipyramidal Axial positions; perpendicular to the

trigonal plane

Equatorial positions; three bonds arranged

in a trigonal plane. ₁₂₀_°

90°

6 Electron Groups SF

6

Octahedral

When the central atom in an AB

_x

molecule bears one

or more lone pairs, the electron-domain geometry

and the molecular geometry are no longer the same.

O O O

••

=

− ••

••

Steps to Determine the Electron Group and

Molecular Geometries

Step 1:

Draw the Lewis structure.

Step 2:

Count the number of e

–

groups on the central atom.

Step 3:

Determine the electron-group geometry.

(4)

3 e

–

groups, 1 lone pair

SO

2

1

2 3

Bent

4 e

–

groups, 1 lone pair

NH

3

1

2

3

4

Trigonal Pyramid

4 e

–

groups, 2 lone pair

H

2

O

1

2

3

4

Bent

5 e

–

groups, 1 lone pair

SF

4

See-Saw

5 e

–

groups, 2 lone pair

ClF

3

T-Shape

5 e

–

groups, 3 lone pair

XeF

2

(5)

6 e

–

groups, 1 lone pair

BrF

5

Square Pyramid

6 e

–

groups, 2 lone pair

XeF

4

Square Planar

2

electron groups

0

lone pairs

Linear

3

electron groups

0

lone pairs

3

electron groups

1

lone pair

Trigonal

Planar

Bent

4

electron groups

0

lone pairs

Tetrahedral

4

electron groups

1

lone pair

4

electron groups

2

lone pairs

Trigonal

(6)

5

electron groups

0

lone pairs

5

electron groups

1

lone pair

Trigonal

Bipyramidal

See-saw

5

electron groups

2

lone pairs

5

electron groups

3

lone pairs

T-shape

Linear

6

electron groups

0

lone pairs

Octahedral

6

electron groups

1

lone pair

6

electron groups

2

lone pairs

Square

Pyramidal

Square

Planar

Some electron domains are better than others at

repelling neighboring domains.

•Lone pairs of electrons (unshared pairs) take up more space than bonded pairs of electrons.

•Multiple bonds repel more strongly than single bonds.

•Consequently, there is an ordering of repulsions of electrons around the central atom.

bp–bp < bond pair–lone pair < lp–lp

least repulsive most repulsive

In a Bonding Pair of

Electrons the Electrons

are Shared by Two

Nuclei

In a Lone Pair, Both

Electrons Must Be Close

(7)

Tetrahedral

e– group geometry Trigonal pyramid shape 4 e– groups

1 lone pair

Tetrahedral e– group geometry

4 e– groups

2 lone pairs Bent shape

The Bond Angles In

CH

₄

, NH

₃

, and H

₂

O

Possible Electron Pair Arrangements

for XeF

₄

Three Possible Arrangements of the

Electron Pairs in the I

₃-

Ion

2 Electron Groups

Electron Pairs Molecular Geometry

Example

Total Bonding Lone

2 2 0 Linear F Be F

O C O

3 Electron Groups

Electron Pairs Example

3 0

3

2 1

Molecular Geometry

Trigonal Planar

Bent

B F

F F

S

(8)

4 Electron Groups

Electron Pairs Example

4 0

4 3 1

2 2

Molecular Geometry Tetrahedral Trigonal Pyramidal Bent N H H H C H H H H O H H

5 Electron Groups

Electron Pairs Example

5 0

5

4 1

Trigonal Bipyramidal

See-saw

5 Electron Groups

Electron Pairs Example

3 2

5

2 3

T-shaped

Linear

6 Electron Groups

Electron Pairs Example

6 0

6 5 1

4 2

Molecular Geometry Octahedral Square Pyramidal Square Planar

The geometry of more complex molecules can be

determined by treating them as though they have

multiple central atoms.

Central C atom

No. of electron domains: 4 Electron-domain geometry: tetrahedral Molecular geometry: tetrahedral

Central O atom

No. of electron domains: 4 Electron-domain geometry: tetrahedral Molecular geometry: bent

Predict the geometry around each carbon

atom in the amino acid alanine.

N C C

C O O H H H H H H H

The carbons in red are

tetrahedral

(4 bonding pairs, 0 lone pairs).

(9)

Molecular polarity is one of the most important

consequences of molecular geometry.

A diatomic molecule is polar

when the electronegativities of the two atoms are different.

H− F

••

H− F

••

δ+ δ−

Molecules 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

bonds

in H

₂

O are

polar

and the

molecule

is

polar

.

Both Shape and Bond Polarity Determine

Molecular Polarity

Two conditions that must be true for a molecule to be polar:

1. There must be at least one polar bond present. 2. The polar bonds, if there are more than one, must be

arranged so that their dipole moments do NOT CANCEL one another.

Molecular Geometry and Polarity

The polarity of a molecule made up of three or more atoms depends on:

(1) the polarity of the individual bonds

(2) the molecular geometry

The bonds in CO2 are polar but the molecule is nonpolar. Carbon dioxide, CO2

Molecular Geometry and Polarity

The polarity of a molecule made up of three or more atoms depends on:

(1) the polarity of the individual bonds

(2) the molecular geometry

The bonds in BF3 are polar but the molecule is nonpolar. Boron trifluoride, BF3

Molecular Geometry and Polarity

Dipole moments can be used to distinguish between structural isomers.

trans-dichloroethylene

nonpolar

cis-dichloroethylene

(10)

The Classification of a

Molecule

as Polar or Nonpolar

Depends on:

The polarity of the individual bonds The overall shape of the molecule

Polar molecules contain:

polar bonds AND

asymmetrical charge distribution

Nonpolar molecules contain:

nonpolar bonds OR

polar bonds in which the resulting charges are distributed symmetrically throughout the molecule

Which Shapes are Symmetric?

Linear is symmetric when the two outer atoms are the same.

Trigonal planar is symmetric when the three outer atoms are the same.

Tetrahedral is symmetric when the four outer atoms are the same.

1. Identical atoms attached to central atom and

2. No lone pairs on central atom

Which Shapes are Symmetric?

Trigonal bipyramidal is symmetric when:

1. The 5 outer atoms are the same or 2. The equatorial atoms are the same and the axial atoms are the same

Octahedral is symmetric when: 1. The 6 outer atoms are the same or 2. Two opposite outer atoms are the

same

Square planar is symmetric when 1. The 4 outer atoms are the same or 2. 2 opposite outer atoms are the same

Which Shapes are Asymmetric?

Bent and Trigonal

Pyramid are always

asymmetric

.

1. Non-identical atoms attached to central atom that

cancel bond dipoles or

2. Lone pairs on central atom not symmetrically

arranged

See-saw, T-shape, and Square-pyramid

are always

asymmetric

.

Polar Bonds & Polar Molecules

O O

C O d

-d+

nonpolar bonds doesn’t matter nonpolar molecule

C O d

-d+

O d

-polar bonds asymmetric polar molecule

(11)

Is the trigonal planar molecule boron trichloride

(BCl

₃

) polar? Explain.

B (EN = 2.0), Cl (EN = 3.0)

so the bond polarization is toward the Cl

B

Cl Cl

Cl d–

d+

d–

This compound has polar bonds, but it is symmetrical.

So it is not polar.

SO

₃

: Polar or Nonpolar? Why?

CH

₄

: Polar or Nonpolar? Why?

H

₂

S: Polar or Nonpolar? Why?

Covalent Bonding Models

VSEPR Geometry

Valence-Bond Theory

Hybridization of Orbitals

Molecular Orbital Theory

Bond Order, Magnetism

The Valence Bond Model

Pictures a molecule as a collection of atoms bound

together by sharing e

-

between hybrid atomic orbitals.

•

Draw the Lewis structure

(s).

•Represents the arrangement of valence e-_.

•

Determine the arrangement of electron groups

.

•VSEPRtheory

•

Specify the necessary hybrid orbitals

.

(12)

Valence Bond Theory: H

₂

The H−H bond in H2 forms when the singly occupied 1s orbitals of the two H atoms overlap:

Valence Bond Theory: F

₂

The F−F bond in F2 forms when the singly occupied 2p orbitals of the two F atoms overlap:

Valence Bond Theory: HF

The H−F bond in HF forms when the singly occupied 1s orbital on the H atom overlaps with the singly occupied 2p orbital of the F atom:

Worked Example 7.4

Strategy The ground-state electron configuration of Se is [Ar]4s2₃_d10₄_p4_{. Its} orbital diagram (showing only the 4p orbitals) is

Hydrogen selenide (H2Se) is a foul-smelling gas that can cause eye and respiratory tract inflammation. The H−Se−H bond angle in H2Se is approximately 92°. Use valence bond theory to describe the bonding in this molecule.

Solution Two of the 4p orbitals are singly occupied and therefore available for bonding. The bonds in H2Se form as the result of the overlap of a hydrogen 1s

orbital with each of these orbitals on the Se atom.

Think About It Because the 4p orbitals on the Se atom are all mutually perpendicular, we should expect the angles between bonds formed by their overlap to be approximately 90°.

Hybridization or mixing of atomic orbitals can account

for observed bond angles in molecules that could not

be described by the direct overlap of atomic orbitals.

Hybridization of Atomic Orbitals: BeCl

₂

Lewis theory and VSEPR theory predict Cl−Be−Cl bond angle of 180°

A ground state beryllium atom can not form two bonds; there are no unpaired electrons.

(13)

Hybridization of Atomic Orbitals: BeCl

₂

An excited state configuration for Be has two unpaired electrons and can form two bonds.

The two bonds formed would not be equivalent.

Hybridization of s and p Orbitals: BeCl

₂

Experimentally the bond in BeCl2 bonds are identical in length and strength.

Mixing of one s orbital and one p orbital to yield two sp orbitals.

Hybridization of s and p Orbitals: BeCl

₂

The 2s orbital on Be and one of the 2p orbitals on Be combine to form two sp hybrid orbitals.

Like any two electron domains, the hybrid orbitals on Be are 180° apart.

Hybridization of s and p Orbitals: BeCl

₂

The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a Cl atom.

The energy required to form an excited state Be atom is more than compensated for by the energy given off when a bond forms.

Whenever a set of

equivalent

linear

atomic orbitals is

required by an atom,

the valence bond model assumes that the

atom adopts a set of

2 sp orbitals

and

that the atom becomes

sp hybridized

.

Hybridization of s and p Orbitals: BF

3

Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3

(14)

Hybridization of s and p Orbitals: BF

3

Step 2:

Count the number of electron domains on the central atom.

This is the # of hybrid orbitals necessary to account for the molecule’s geometry.

(This is also the # of atomic orbitals that must undergo hybridization.)

Three electron domains Three hybrid orbitals required

₃

Step 3:

Draw the ground-state orbital diagram for the central atom.

Step 4:

Maximize the number of unpaired valence electrons by promotion.

Hybridization of s and p Orbitals: BF

₃

Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3

Step 5: Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.

Step 6: Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.

₃

Mixing of ones orbital and twop orbitals to yield threesp2_orbitals.

Hybridization of s and p Orbitals: BF

₃

sp2_{hybrid orbitals on boron overlap with 2}_p_{orbitals on fluorine.}

Whenever a set of

equivalent

trigonal planar

atomic orbitals

is required by an atom,

the valence bond model assumes that the

atom adopts a set of

3 sp

2

orbitals

and

(15)

Hybridization of s and p Orbitals: CH

₄

Determine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.

Step 1: Draw the Lewis structure:

Hybridization of s and p Orbitals: CH

₄

Step 2:

The number of electron groups on the central atom is the number of hybrid orbitals necessary to account for the molecule’s geometry.

Four electron domains Four hybrid orbitals required

₄

Step 3:

Step 4:

Maximize the number of unpaired electrons by promotion.

Hybridization of s and p Orbitals: CH

₄

Step 5:

Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.

Step 6:

Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.

₄

Mixing of ones orbital and threep orbitals to yield foursp3_orbitals

Hybridization of s and p Orbitals: CH

₄

(16)

The 2s and 2

p

Atomic Orbitals of Carbon

Combine to Form 4 sp

3

Orbitals

Hybridization:

Mixing of atomic orbitals to form special orbitals for bonding.

2

s

+ 3 2

p



4 sp

3

An Energy-Level Diagram Showing the

Formation of Four

sp

3

Orbitals

•What is important?

•The total number of e-_{and their arrangement.}

•For the C atom in CH4, we have 8 electrons which will be accommodated by 4 equivalent sp3_{orbitals which point} toward the corners of a tetrahedron.

Tetrahedral Set of Four

sp

3

Orbitals

The 4 new sp

3

hybrid atomic orbitals on C are used to share

e

-

pairs with the 1

s

orbitals from the 4 H atoms.

Whenever a set of equivalent

tetrahedral

atomic orbitals is required

by an atom, the valence bond model

assumes that the atom adopts a set

of

4 sp

3

orbitals

and that the atom

becomes

sp

3

hybridized

.

Hybridization of s, p and d Orbitals: PCl

5

Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.

Step 1: Draw the Lewis structure:

Hybridization of s, p and d Orbitals: PCl

₅

Step 2:

The number of electron domains on the central atom is the number of hybrid orbitals necessary to account for the molecule’s geometry.

(17)

₅

Step 3:

Step 4:

Maximize the number of unpaired electrons by promotion.

Hybridization of s, p and d Orbitals: PCl

5

Step 5:

Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.

Step 6:

Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.

5

Hybrid orbitals on phosphorus overlap with 3p orbitals on chlorine.

Whenever a set of equivalent

trigonal bipyramidal

atomic orbitals is

required by an atom, the valence bond

model assumes that the atom adopts a

set of

5 sp

3

d orbitals

and that the atom

becomes

sp

3

d hybridized

.

101

Valence Bond (VB) Theory

Number of

Electron

Groups

Electronic

Arrangement

Hybridization

2

Linear

s + p



2 sp

3

Trigonal planar

s + 2 p



3 sp

2

4

Tetrahedral

s + 3 p



4 sp

3

5

Trigonal

bipyramidal

s + 3 p + d



5 sp

3

d

6

Octahedral

s + 3 p + 2 d



6 sp

3

d

2

Worked Example 7.5

Strategy Starting with the Lewis structure, determine the number and type of hybrid orbitals necessary to rationalize the bonding in NH3.

The ground-state electron configuration of the N atom is [He]2s2₂_p3_{. Its valence} orbital diagram is

(18)

Worked Example 7.5 (cont.)

Solution Although the N atom has the three unpaired electrons needed to form three N−H bonds, we would expect bond angles of ~90° (not 107°) to form from the overlap of the three mutually perpendicular 2p orbitals. Hybridization, therefore, is necessary to explain the bonding in NH3. Although we often need to promote an electron to maximize the number of unpaired electrons, no promotion is necessary for the nitrogen in NH3. We already have the three unpaired electrons necessary, and the promotion of an electron from the 2s orbital to one of the 2p

orbitals would not result in any additional unpaired electrons. Furthermore, there are no empty d orbitals in the second shell.

Worked Example 7.5 (cont.)

Solution According to the Lewis structure, there are four electron domains on the central atom (three bonds and a lone pair of electrons). Four electron domains on the central atom require four hybrid orbitals, and four hybrid orbitals require the hybridization of four atomic orbitals: one s and three p. This corresponds to

sp3_{hybridization. Because the atomic orbitals involved in the hybridization} contain a total of five electrons, we place five electrons in the resulting hybrid orbitals. This means that one of the hybrid orbitals will contain a lone pair of electrons:

Solution

Each N−H bond is formed by overlap between an sp3_{hybrid orbital on the N} atom and the 1s atomic orbital on an H atom. Because there are four electron domains on the central atom, we expect them to be arranged in a tetrahedron. In addition, because one of the electron domains is a lone pair, we expect the H−N−H bond angles to be slightly smaller than the ideal tetrahedral bond angle of 109.5°.

Solution This analysis agrees with the experimentally observed geometry and bond angles of 107° in NH3.

Describe bonding in NH

3

using valence bond theory (hybrid orbitals)

1.

Draw Lewis structure.

2. Determine

arrangement of e-

groups using the

VSEPR model.

3. Determine the hybrid

atomic orbitals needed

to describe bonding in

the molecule.

Tetrahedral e– group arrangement, 4 sp3_{hybrid orbitals on N.} 3 of the sp3 _{hybrid orbitals overlap with the 1s atomic orbital of H, and}

the 4th_sp3_{hybrid orbital holds the lone pair of e–.}

Hybridization in Molecules

Containing Multiple Bonds: C

2

H

4

Valence bond theory and hybridization can be used to describe the bonding in molecules

containing double and triple bonds.

Each carbon has three electron groups: 2 single bonds

1 double bond

ethylene (C2H4)

Expect sp2_{hybridization}

Hybridization in Molecules

₂

H

₄

Maximize unpaired electrons on carbon by promotion:

(19)

Hybridization in Molecules

2

H

4

Three equivalent sp2 hybrid orbitals

explain three bonds around carbon Hybridization scheme for the carbon atoms in ethylene:

One unhybridized atomic 2p

orbital gives rise to multiple bonds

ethylene (C2H4)

Hybridization in Molecules

₂

H

₄

A sigma (σ) bond forms when sp2_{hybrid orbitals} on the C atoms overlap.

In a sigma bond, the shared electron density lies directly along the internuclear axis.

Hybridization in Molecules

₂

H

₄

The ethylene molecule contains five sigma bonds:

1 between the two carbon atoms (sp2_and_sp2_overlap) 4 between the C and H atoms (sp2_{and 1}_s_overlap)

Hybridization in Molecules

₂

H

₄

The remaining unhybridized p orbital

is perpendicular to the plane in which the atoms of the molecule lie.

The unhybridized p orbitals

overlap in a sideways fashion

to form a pi (π) bond.

Worked Example 7.6

Strategy Remember, a single bond is composed of a sigma bond, whereas a double bond is usually composed of one sigma bond and one pi bond. Thalidomide (C13H10N2O4) is a sedative and antiemetic that was widely prescribed during the 1950s, although not in the United States, for pregnant women suffering from morning sickness. Its use was largely discontinued when it was determined to be responsible for thousands of devastating birth defects. Determine the number of carbon-carbon sigma bonds and the total number of pi bonds in thalidomide.

Solution Thalidomide contains 12 carbon-carbon sigma bonds and a total of seven pi bonds (three in carbon-carbon double bonds and four in carbon-oxygen double bonds.

Think About It The Lewis structure given to thalidomide is one of two possible resonance structures. Draw the other resonance structure, and count sigma and pi bonds again. Make sure you get the same answer.

Hybridization in Molecules

Containing Multiple Bonds

Sigma bonds exhibit free rotation around the bond axis.

(20)

Hybridization in Molecules Containing Multiple Bonds

Pi bonds restrict free rotation around the bond axis.

There are two isomers of 1,2-dichloroethylene

cis-1,2-dichloroethylene trans-1,2-dichloroethylene

Hybridization in Molecules

₂

H

₂

The acetylene molecule is linear with sp hybridized carbons.

Promotion of an electron maximizes the number of unpaired electrons:

acetylene (C2H2)

Hybridization in Molecules

₂

H

₂

acetylene (C2H2)

The 2s orbital and one of the 2p orbitals then mix to form two sp hybrid orbitals:

3 sigma bonds:

1 between the two carbon atoms (sp and sp) 2 between the C and H atoms (sp and 1s) 2 pi bonds

2 between the two carbon atoms (2p and 2p)

Hybridization in Molecules

2

H

2

acetylene (C2H2)

Two equivalent sp hybrid orbitals

give rise to 2 sigma bonds

Two unhybridized atomic 2p orbitals gives rise to 2 pi bonds

Hybridization in Molecules

2

H

2

Formation of the C−C sigma bond in acetylene:

acetylene (C2H2)

3 sigma bonds:

1 between the two carbon atoms (sp and sp) 2 between the C and H atoms (sp and 1s)

Hybridization in Molecules

2

H

2 Formation of the C−C pi bond in acetylene:

acetylene (C2H2)

2 pi bonds

(21)

Worked Example 7.7

Strategy The Lewis structure of formaldehyde is

The C and O atoms each have three electron domains around them. [Carbon has two single bonds (C−H) and a double bond (C=O); oxygen has a double bond (O=C) and two lone pairs.]

In addition to its use in aqueous solution as a preservative for laboratory specimens, formaldehyde gas is used as an antibacterial fumigant. Use hybridization to explain the bonding in formaldehyde (CH2O).

Worked Example 7.7 (cont.)

Solution Three electron domains correspond to sp2_{hybridization. For carbon,} promotion of an electron from the 2s orbital to the empty 2p orbital is necessary to maximize the number of unpaired electrons. For oxygen, no promotion is necessary. Each undergoes hybridization to produce sp2_{hybrid orbitals; and each} is left with a singly occupied, unhybridized p orbital.

Solution

A sigma bond is formed between the C and O atoms by the overlap of one of the

sp2_{hybrid orbitals from each of them. Two more sigma bonds form between the} C atom and the H atoms by the overlap of carbon’s remaining sp2_{hybrid orbitals} with the 1s orbital on each H atom. Finally, the remaining p orbitals on C and O overlap to form a pi bond.

Think About It Our analysis describes the formation of both a sigma bond and a pi bond between the C and O atoms. This corresponds correctly to the double bond predicted by the Lewis structure.

What orbitals do the C atoms in C

2

H

4

employ

to achieve a trigonal planar arrangement

with bond angles of 120°?

s

+ 2

p



3 sp

2

When An

s

and Two

p

Orbitals Combine to Form a Set of

Three sp

2

Orbitals, One p Orbital Remains Unchanged and is

Perpendicular to the Plane of the Hybrid Orbitals

(22)

How can

sp

2

hybrid orbitals account

for the bonds in C

2

H

4

?

sigma (

s

) bonds

The 3 sp2_{hybrid orbitals on each C atom are used to share e}- _{with the H} 1s atomic orbital. In each of these bonds, the e-_{pair is shared in an area}

centered on a line running between the atoms. This type of covalent bond is called a sigma (s) bond.

How can we explain the double bond

between the C atoms in C

₂

H

₄

?

sigma (s) bond

pi (p) bond

pi (p) bond In the sigma bond, the e-_pair

occupies the space between the C atoms. The 2nd_bond

results from sharing an e-_pair

in the space above and below the sigma bond. This type of bond can be formed using the 2p orbital perpendicular to the sp2_{hybrid orbitals on each C}

atom. These parallel p orbitals share an e-_{pair, which}

occupies the space above and below a line joining the C atoms, to form a pi bond.

Sigma (

s

) bonds

are formed from orbitals whose

lobes point toward each other.

Pi (

p

) bonds

result

from parallel orbitals.

A

double bond

consists of

one sigma bond

and

one pi bond

.

Orbitals Used to Form the Bonds of C

2

H

4

When an atom is surrounded by

3 e

-

groups

, a set of

sp

2

hybrid orbitals is required to account for the

resulting

trigonal planar

molecular geometry.

s bond p bond

s bond

s bond s bond

s bond

What orbitals do the C atoms in CO

₂

employ to achieve a linear

arrangement with bond angles of

180°?

When 1-

s

Orbital and 1-

p

Orbital are

Hybridized, a Set of 2

sp

Orbitals

Orientated at 180

°

is Formed

2 e-_groups_{around an atom}

will always require sp hybridization

(23)

Orbitals of an

sp

Hybridized C - Atom

Orbital Energy-Level Diagram for

Formation of

sp

Hybrid Orbitals on C

Orbitals for an

sp

2

Hybridized O - Atom

Sigma (s) Bonds in the CO

2

Molecule

Orbitals Used to Form the Bonds in CO

₂

Use the Valence Bond Theory

(hybrid orbitals)

to describe the bonding

in a N

₂

molecule.

Strategy

1.

Draw lewis structure.

2.

Determine the arrangement of e

-

pairs using the

VSEPR model.

(24)

Bonding in N

₂

What orbitals does the P atom in PCl

₅

employ to

achieve a trigonal bipyramidal arrangement

with bond angles of 90° and 120°?

When an atom is surrounded by 5 e

-

groups, a set of

sp

3

d

hybrid orbitals is required to account for the

resulting trigonal bipyramidal shape.

Orbitals Used to Form the Bonds in PCl

₅

What orbitals does the S atom in SF

6

employ

to achieve an octahedral arrangement with

bond angles of 90°?

How is the Xenon Atom