announcements chapter 10 the shapes of molecules · the shapes of molecules chemical bonds and the...

AnnouncementsChem 7 Final Exam Wednesday, Oct 10 1:30-3:30AMChapter 1-1270 or 75 multiple choice questions

Exam III (Chapter 7-10)Wednesday, October 3, 2012Time: 6:00PM - 7:30PMSEC A 214A and 215A

Please note the following topics will be excluded from assessment. The page numbers refer to the 2nd Edition of the textbook.

1. Numerical problems involving the Rydberg equation (Chapter 7)2. Spectral analysis in the laboratory (Chapter 7 p. 226-227)3. Trends among the transition elements (Chapter 8 p. 261)4. Trends in electron affinity (Chapter 8 pp. 265-266)5. Pseudo-noble gas configuration (Chapter 8 p. 269)6. Lattice energy (Chapter 9 pp. 283-285)7. IR spectroscopy (Chapter 9 p. 292)8. Numerical problems involving electronegativity (Chapter 9 p. 296)9. Electronegativity and oxidation number (Chapter 9 p. 297)10. Section 11.3: MO theory and electron delocalization (Chapter 11 p.343)11. All sections in chapter 12 except 12.3 (types of intermolecular forces).

Chapter 10The Shapes of Molecules

Chemical bonds and the chemistry of an element is related to the number of valence electrons are in the outer shell (highest value of n quantum number) of the atom.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

LOOK FOR Group #

• For B group elements, the valence electrons are in the highest value ns orbital and the (n-1)d orbitals.

Lewis dot structures are used to depict valence electrons and bonding between atoms.

• A chemical symbol represents the nucleus and all core e-.

• A single dot around the symbol represents one valence e-.

+1 +2 +3 -1-2-3

1

GOAL: Draw and connect Lewis structures with geometry of a molecule (VSEPRT) and in Chapter 11 connect the geometry with how orbitals bond (VBT).

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

1 2 3 4

2 3 4

VB Theory

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

OctahedralLinearTrigonalPyramidal Tetrahedral Trigonal

Bipyramidal

sp3d2sp sp2 sp3 sp3dValance Bond Theory

Valence Bond theory explains how bonding occurs between atoms using “hybridized orbitals”.

Must Learn How To Draw Lewis Structures1) Put least electronegative element as the central atom. C,S,P and N are often central atoms. H and halogens are often bonded to central atoms.

2) Sum all valence e- from each atom in the molecule (careful with ions add or subtract e-). Use the Group numbers!

3) Place bonds to central atoms using 2 e- per bond.

4) Place an octet of electrons (octet rule) around bonded atoms remembering that H--bonds have no octet. Know the incomplete and expanded octet exceptions.

5) Place remaining electrons around central atom which should have an octet if period 2 or less, but could be more than octet if period 3 or higher.

6) Helpful Rules of Thumb: H forms 1-bond, C forms 4-bonds, N forms 3-bonds, O forms 2-bonds.

Electronegativity is an element’s inherent ability to draw electrons to itself when chemically bonded to another atom in a molecule (relative to Li).

F, O, N, Cl, Br, C are highly electronegative with F being the most electronegative

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F --> N is central atom!

Step 3 - Write Lewis structrue with N central and three bonds and rest non-bonding octet electrons around the central atom.

Step 2 - Count valence electrons

F N F

F

octet

octet

octet

octet

N 5e-

F 7e- x 3 = 21e!Total 26e!

An electron group (domain) is either a pair of bonding electrons or a pair of non-bonding electrons surrounding a central atom. Multiple bonds only count as 1-group or domain.We count and “code” the bonding/non-bonding information into shorthand called AXnEm classification.

AX2E0 = AX2A = Central Atom

X = # of BondedDomains

E = # Non-Bonded Domains

shorthand

F N F

F

4 electron groups3 bonding1 non-bonding AX3E1

1) Incomplete Octet - Occurs with Be, B and Al as central atoms.

2) Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens.

3) Odd-number electrons highly reactive species called radicals that have an odd number of electrons (uneven).

There are three major exceptions to the octet rule.

BF3

BeH2

AlCl3

1. Incomplete Octet - no “octet” around central atom. Occurs with Be, B and Al as central atoms.

Draw Lewis structures for the following

Incomplete Octet: Occurs With Group 2A (Be) and 3A (Boron and Aluminum)

BF3B – 3e-

3F – 3x7e-

24e-

F B F

F

Be – 2e-

2H – 2x1e-

4e-

BeH2 H HBe

Cl Al Cl

Cl

AlCl3Al – 3e-

3Cl – 3x7e-

24e-


AX3

AX3

AX2

SF6Phosphorous trichloride

[ICl4]-1PCl5

2. Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens.


SF6

S – 6e-

6F – 42e-

48e- S

F

F

F

FF

F

••

•• ••

••

P

Cl

ClCl

••••

••••

••

••

Phosphorous trichloride PCl3

P

Cl

Cl

••••Cl

••••

••

••••

•• •• ••

••Cl

••••

••Cl••

Phosphorous pentachloride

[ICl4]-1P – 5e-

5Cl – 35e-

40e-PCl5

Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens.

Draw Lewis structures for PCl3 PCl5 and the carbonate anion. Determine the AXE designation for each.

Draw Lewis structues and determine the AXE designation for each.

P

Cl

Cl

••

••Cl

••••

••

••••

•• •• ••

••Cl

••••

••Cl••

5 electron groups5 bonding 0 non-bonding

O C O

O ][ 2- 3 electron groups3 bonding0 non-bonding

••

•• ••

••

P

Cl

ClCl

••••

••••

••

••

4 electron groups3 bonding1 non-bonding

AX3E1

AX5E0

AX3E0

]BrO3[–

Valence e- = 7 + 3(6) + 1 = 26

O Br O

O

MUST look to see if its an ion and add the necessary electron!

H C N

HCNValence e- = 1 + 4 + 5 = 10

Carbon is central atom, watch for hydrogen--1 bond

–

Write the Lewis structure of the carbonate ion (BrO3-)

and hydrogen cyanide, give AXE designation.


AX3E1

Electron Domains

2 electron groups2 bonding0 non-bonding AX2

Write resonance structures for the carbonate ion, CO3

-.Write the Lewis structure of the carbonate ion (CO3

2-).Step 1 – C is less electronegative than O, put C in centerStep 2 – Count valence electrons (C and O)

Valence electrons = 4 + 6 + 6 + 6 + 2 = 24 valence electronsStep 3 - Arrange the atoms draw bonds between C and O atoms and complete octet on C and O atoms.


AX3

Electron Domains “resonance structures”All equally good and plausible

Which structure is correct?

A concept called “resonance” is used when more than one plausible Lewis structure can be drawn.

Measured bond lengths show they are equal!

O O O O O O••••••

••

••••

••••

••••

••

••2 equally good Lewisstructures

Which structure is correct?

Example: Ozone, O3

O O O O O O••••••

••

••••

••••

••••

••

••

Both are!

O O O••••••••

••a resonance hybrid structure

Write resonance structures for the nitrate ion, NO3-.

Write resonance structures for the nitrate ion, NO3-.

PLAN:

Count valence e- of atoms = 5 + (3X6) + 1 = 24 e-Most electronegative atom in centerSurround and get an octet around N


AX3

Electron Domains

O C O O C O

Both are two plausible structures for CO2

When more than one Lewis structure is plausible, we apply the concept of FORMAL CHARGE to figure out the best Lewis structure!

VS

Which is the best one?

USE FORMAL CHARGE

1. The best structure is one that minimizes total formal charge. Net charge of ion or molecule must equal total formal charge.

2.! Also, the best Lewis structure places negative charge on the most electronegative atom.

Assigned Atoms = all from lone pair e! + ! ( bonded e! )

# Valence e- 6 4 6 6 4 6 # of Assinged e- 6 4 6 5 4 7 Formal Charge 0 0 0 +1 0 !1

O C O O C O

This structure wins!

To use the concept of formal charge, we determine the formal charge for each atom.

Atom Formal charge = # valence e- - Assigned e- to Atom

Example: Write 3 plausible Lewis structures for the thiocyanate ion [SCN]–


S C N[ ] –S C N[ ] –

S C N[ ] –

1 2 3

3-plausible Lewis structures which one is best?

# of Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e-S C N

it’s a -1 ion1. Count the Valence e-

2. Draw the Lewis Structures With Least Electronegative Atom as central atom.

N ] –-2S C N[ ] –

FCS = 6 - 4 -2 = 0FCC = 4 - 0 - 4 = 0FCN = 5 - 6 - 2 = -1

S C[FCS = 6 - 2 -3 = 1FCC = 4 - 0 - 4 = 0FCN = 5 - 6 - 1 = -2

S C N[ ] –

FCS = 6 - 6 -1 = -1FCC = 4 - 0 - 4 = 0FCN = 5 - 2 - 3 = 0

0 0 -1 -1 0 0 0+1

1. Formal charge must sum to charge of ion or molecule.2. N is more electronegative than C or S, it should have a the most negative charge in the “best structure”.3. The most plausible structure has the least amount of formal charge.

Structure on the left is “best” structure!S C N[ ] –0 0 -1


Write resonance structures for the nitrate ion, NCO-

and determine the most plausible Lewis structure.EXAMPLE: NCO- has 3 possible resonance forms -

A B C

formal charges

-2 0 +1 -1 0 0 0 0 -1

Forms B and C have negative formal charges on N and O; this makes them more preferred than form A.Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

Write resonance structures for the nitrate ion, NCO-

and determine the most plausible Lewis structure.

Chemists use Valence Shell Electron Pair Repulsion Theory to predict the shapes of molecules using these five electron group geometries.

1. Draw Lewis Structure from chemical formula.

2. Count all electron domains to get AXE code.

4. Match the number of bonding and non-bonding domains to the proper VSEPRT geometry.

3. Group domains into bonding and non-bonding pairs of electrons.

VSEPRT explains the geometry of molecules but NOT how covalent bonds are formed with that geometry.

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

VSEPRT Valence BondTheoryVSEPRTLewis Structure

The electron geometry is the geometry of all electron groups, whereas the “molecular geometry” describes the geometry of only the atoms bonded to the central atom.

AX3E1 = Tetrahedral electron geometery with 109.5˚ bond angles.

Molecular Geometry is trigonal pyramidal bond angles <109.5˚

Molecular Geometry

Electron GroupGeometry

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

The goal is to understand geometry (via VSEPRT) and to relate it to a picture of covalent bonding in molecules.

VB Theory

OctahedralLinearTrigonalPyramidal Tetrahedral Trigonal

Bipyramidal

sp3d2sp sp2 sp3 sp3d

The 3-D geometry of a molecule is one of five basic arrangements of electron groups (domains).

Trigonal Planar

Trigonal Bipyramidal Octahedral

Linear Tetrahedral

The total number of electron groups (domains) defines one of the five basic geometries.

2 EG

3 EG 4 EG

5 EG 6 EG

The electron geometry is the geometry of all electron domains, whereas the “molecular geometry” describes the geometry of only the atoms bonded to the central atom.

AX3E1 = Tetrahedral electron geometery with 109.5˚ bond angles.

Molecular Geometry is trigonal pyramidal bond angles <109.5˚

How Predict Geometry Using VSEPRT1.Draw a plausible Lewis structure for the molecule.

2.Determine the total number of electron domains and identify them as bonding or lone pairs.

3.Use the total number of electron domains to establish the electron geometry from one of the five possible geometric shapes.

4.Establish the AXnEm designation to establish the molecular geometry (or do both electron and molecular geometry together simultaneously)

5. Remember bond angles in molecules are altered by lone pairs of electrons (repulsion forces reduce angles).

6. Molecules with more than one central atom can be handled individually.

2 Electron Groups = Linear Electron Geometry and 1-Possible Molecular Geometry

Other Examples:CS2, HCN, BeF2

Bond Angle

AX2E0 = AX2

Cl ClBe

S C N

O C O

A = Central AtomX = # of BondedDomains

E = # Non-Bonded Domains

Examples:SO2, O3, PbCl2, SnBr2 A

Examples:SO3, BF3, NO3

-, CO32-

AX3A

3-Electron Domain

3 Electron Groups = Trigonal Planar Electron Geometry and 2-Possible Molecular Geometries

AX2E1

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

AX4

AX3E1 AX2E2

4 Electron Groups = Tetrahedral Electron Geometry and 3-Possible Molecular Geometries

Bond Angle

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

AX5 AX4E1

AX2E3

AX3E2

AxialPosition

EquatorialPosition

5 Electron Groups = Trigonal Bipyramial Electron Geometry and 4-Possible Molecular Geometries

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-AX4E2

AX6

AX5E1

6 Electron Groups = Octahedral Electron Geometry and 3-Possible Molecular Geometries

Predicting Molecular ShapesDraw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.

<109.50

Predicting Molecular ShapesDraw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.

2. Count the electron domains and find electron geometry and molecular from core 5 electron domain shapes (using AXE designation and sub-shapes)

5. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair.

3. There are 4 electron domains so the electron geometry is tetrahedral

4. The designation is AX3E1 so the molecular geometry is trigonal pyramidal.

1. Count the valence electrons and draw Lewis structure for PF3: VE = 5 + 3(7) = 26 e-

Predicting Molecular Shapes with Two, Three, or Four Electron Groups

(b) For COCl2, C has the lowest EN and will be the center atom.There are 24 valence e-, 3 atoms attached to the center atom.

124.50

1110

Type AX3

5. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

2. Count the electron domains and establish electron geometry from 5 shapes3. There are 3 electron domains so the electron geometry is trigonal planar4. The molecular geometry designation is AX3E0 so the molecular geometry is also trigonal planar (no lone pairs).

1. Draw the Lewis structure

Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.

Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.

(a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal.

(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

More Than One Central Atom

• In acetic acid, CH3COOH, there are three central atoms.• We assign the geometry about each central atom

separately.

What is the geometryaround these atoms?

Take one atom at a time and apply the rules of electron domains.

ethaneCH3CH3

ethanolCH3CH2OH

More Than One Central Atom

Determine the shape around each of the central atoms in acetone, (CH3)2C=O.

Find the shape of one atom at a time after writing the Lewis structure.

tetrahedral tetrahedral

trigonal planar

>1200

<1200

Predicting the Molecular Shape With Multiple Central Atoms Electronegativity is an element’s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule. The units are dimensionless (all relative measurements to Li).

RankFONClBr

Differences in elements electronegativity between bonding atoms result in the formation of polar-covalent bonds and net dipole moments in molecules.

Net Dipole MomentNo Net Dipole Moment

Polar BondPolar Bond

Polar Bond Polar Bond

Think of the dipole moment as a molecule with separated charges + and -.

Draw a Lewis structure, show the AXE designation, determine electron and molecular geometry and whether polar or non-polar of:

CCl3H

CCl4

CH4

Draw a Lewis structure, show the AXE designation, determine electron and molecular geometry and whether polar or non-polar of:

CCl4AX4Tetrahedral EGTetrahedra MG

CH3ClPolar bondPolar MoleculeHas Dipole Moment

CCl4Polar bondsNot Polar MoleculeNo Dipole Moment

CH3Cl

For a poly-atomic molecule we must consider the vector sum of polar bonds in the molecule to see if there is a net dipole moment.

No NetDipoleMoment

DipoleMoment

DipoleMoment

DipoleMoment

No NetDipoleMoment

From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable:

(a) Ammonia, NH3

(b) Boron trifluoride, BF3

(c) Carbonyl sulfide, COS (atom sequence SCO)

10-

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Predicting the Polarity of Molecules

(a) Ammonia, NH3 (b) Boron trifluoride, BF3

(c) Carbonyl sulfide, COS (atom sequence SCO)

PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable:

PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity.

SOLUTION: (a) NH3

ENN = 3.0

ENH = 2.1

bond dipoles molecular dipole

The dipoles reinforce each other, so the overall molecule is definitely polar.

10-


Sample Problem 10.9 Predicting the Polarity of Molecules

(b) BF3 has 24 valence e! and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar.

F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar.

1200

(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar ("EN), so the molecule is polar overall.

11-


Chapter 11

Theories of Covalent Bonding

11-


Theories of Covalent Bonding

11.1 Valence Bond (VB) Theory and Orbital Hybridization

11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds

11.3 Molecular Orbital (MO)Theory and Electron Delocalization

This only!

Valence Bond Theory explains covalent bonding by the spatial overlap of atomic orbitals on bonding atoms and the sharing of electron pairs.

Electrons that must have opposite spins.

1s1 + 1s1

Bonding in F2

1s1 + 2p5

Bonding in H2

Bonding in HF

2p5 + 2p5

It’s natural to think of using “pure atomic orbitals” to describe bonding in some molecules. It works well for some....but fails with carbon.

Bonding in F2

1s1 + 1s22s22p5

Bonding in HF

2p5 + 2p5

1s22s22p5 + 1s22s22p5

1s1 + 2p5

sp3

hybridized orbitals

hybridization fixes the problem!

Bonding in carbon presents a problem as combining atomics orbitals fails. Valance Bond Theory solves this by allowing the blending or mixing of pure atomic orbitals in a process called hybridization.

Pure atomic orbitals (valence orbitals)

only two bond are possible with use of atomic orbitals only....we don’t observe CH2

By hybridizing 4 bonds are possible.

Hybridization combines or mixes different numbers of pure atomic orbitals that match one of the VSEPRT geometries. For example 1 pure s orbital + 1 p-orbital combine to give and 2 “sp hybrids” that when superimposed form a linear geometry for bonding.

s-orbital + p-orbital --> 2 sp hybrid orbitals -->

s-orbital + Two p-orbital --> 3 sp2 hybrids = Trig Planar

2-superimposed sphybrid orbitals

s-orbital + Three p-orbitals -> Four sp3 hybrids = Tetrahedral

sp3 hybrid orbitals

The process of combining pure atomic orbitals to form “hybrid orbitals” on central bonding atoms in a molecule is called hybridization.

1. The number of hybrid orbitals obtained equals the number of atomic orbitals mixed.

2. The name of and shape of a “hybrid orbital” varies with the types of atomic orbitals mixed. (s + p vs s + two p)3. Each hybrid orbital has a specific geometry that matches one of five VSEPRT shapes (show below).

sp3d2

Octahedral

sp sp2 sp3 sp3d

Linear TrigonalPlanar

Tetrahedral Trigonal Bipyramidal

Some generalized rules and comments on VBT and the formation of hybridized orbitals.

sp3d2

Octahedral

sp sp2 sp3 sp3d

Linear Trigonalplanar

Tetrahedral Trigonal Bipyramidal

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

Connect the dots and it becomes easy to see and understand.

Valence Bond Theoryexplains how bonds are made

ElectronGeometry

Molecular Geometry AXnEm Hybridization

Linear Linear AX2 sp

Trigonal planar

Trigonal planar V-shaped bent

AX3

AX2E1sp2

Tetrahedral

Tetrahedral Trigonal pyramidal

V-shaped bent

AX4

AX3E1 AX2E2

sp3

Trigonal bipyramidal

Trigonal bipyramidalSeesaw

T-shaped Linear

AX5

AX4E1

AX3E2

AX2E3

sp3d

OctahedralOctahedral

Square pyramidal Square planar

AX6

AX5E1

AX4E2

sp3d2

Determine the VSEPRT geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule?

Determine the electron domain, molecular geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule?

tetrahedral, 180, sp3

sp3

sp2

sp

sp2

bent, <109.5, sp3

trig planar 120˚, sp2

linear 180˚, sp

Atomic OrbitalsMixed

# Hybrid OrbitalsFormed

HybridShape

Linear AX2

Trig Planar AX3

Tetrahedral AX4

Trig Bypyr AX5

Octahedral AX6

s + p s + 2 p s + 3 p s + 3 p + d s + 3 p + 2d

Two sp Three sp2 Four sp3 Five sp3d Six sp3d2

Orbitals Leftover for Pi bonds

Two p one p none Four d Three d

Linking VSEPRT To Valence Bond Theory Hybrids

2s

--The number of hybrid orbitals formed is equal to the number of “pure orbitals” combined!

--When superimposed the “sp-hybrid” give us bonding orbitals for a linear molecules.

An sp hybrid is formed from the combination of a one pure 1s orbital and a one 2p orbital from a central bonding atom producing two new orbitals called sp orbitals.

s-orbital p-orbitalTwo sp hybrid orbitals

sp hybrid orbitals superimposed

Hybridization

s + p Hybridization = 2 sp

Example sp hybrid: Show the bonding scheme and hybridized orbitals used in BeCl2

2 unhybridized unoccupied p-orbitals

After hybridization we have on the central atom, 2 pure p-orbitals and two sp hybrids.

2 “left-over” p-orbitals

hybridization

Isolated Be Atom

Hybridized Be Atom

Show the bonding scheme and hybridized orbitals in BeCl2

two sp hybrids on Be

two lone p-orbitals

sp2 = Triginal planar geometry, 120˚ bond angle

3-atomic orbitals, s and two p’s combine to form 3-sp2 hybrid orbitals

An sp2 hybrid is formed from the combination of a one pure 1s orbital and a two 2p orbitals from a central bonding atom producing two new orbitals called sp2 orbitals.

Superimposed Hybrid orbitals form a triginal planar geometry

Example 2: sp2 hybridizaton scheme BF3.

Boron Orbital Box Diagram

Boron Hybrid Box Diagram

Bonding of pure p-orbital in F with sp2 hybridized orbitals in BF3

Tetrahedral geometry = sp3 hybrid orbitals

sp3 = Tetrahedral geometry = 109.5˚ bond angle

Note the number of hybrids formed is the number of atomic orbitals combined!

combine to generatefour sp3 orbitals

which are representedcollectively as: sp3

Example: sp3 orbital hybridization: CH4.

the four sp3 hybrid orbitals form a tetrahedral shape

sp3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp3 orbitals on each carbon atom. End to end overlap with a 1s orbital from H gives four sigma bond in CH4.

CH4

This is the ground stateconfiguration of valence atomic orbitals

Example 3: sp3 hybrid orbitals in H2O.

What is the electronic geometry?What is the molecular geometry?What orbitals contribute to bonding?

Note the lone pairs occupy 2-of the sp3 orbitals

sp3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp3 orbitals. The e- are distributed throughout the hybrids ready for bonding. End to end overlap with a 1s orbital from H gives four sigma bond in CH4.

sp3 is tetrahedral shape. In water we have AX2E2

What is the electron geometry, the molecular geometry at each carbon atom? Use that information to determine the hybridization around each carbon atom in nicotinic acid? How many sigma and pi bonds are in nicotinic acid?

Example 2: sp3 hybridization in NH3.

Tetrahedral Electron Geometry AX3E1Trigonal Pyramidal Molecular Geometry

sp3d hybridization in PCl5.

Isolated P atom

Trigonal Bipyramidal Electron Geometry AX5E0Trigonal BiPyramidal Molecular Geometry

The sp3d2 hybrid orbitals in SF6Octahedral Electron Geometry AX6E0Octahedral Molecular Geometry

ElectronGeometry

Molecular Geometry AXnEm Hybridizaton

Linear Linear AX2 spTrigonal planar

Trigonal planar V-shaped bent

AX3

AX2E1sp2

Tetrahedral

Tetrahedral Trigonal pyramidal

V-shaped bent

AX4

AX3E1 AX2E2

sp3

Trigonal bipyramidal

Trigonal bipyramidalSeesaw

T-shaped Linear

AX5

AX4E1

AX3E2

AX2E3

sp3d

OctahedralOctahedral

Square pyramidal Square planar

AX6

AX5E1

AX4E2

sp3d2

Describe the types of bonds and orbitals in acetone, (CH3)2CO and in CO2 and in HCN?

Molecular formula

Lewis structure

VSEPRTGeometry

Hybrid orbitals

Step 1 Step 2 Step 3

Describe the types of bonds and orbitals in acetone, (CH3)2CO.PLAN:

Draw the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of geometries predicted from VSEPRT. Draw the orbitals and show overlap.

SOLUTION:

sp3 hybridized

sp3 hybridized

sp2 hybridized

# bonds$ bond

sp hybrid:Ethylyne: HC!CH:Linear

sp hybrid orbitals

Lone p orbitals that are not hybridized

Sigma bonds (" bonds) and Pi bonds (# bonds)are two different types of covalent chemical bonds that form as a result of end to end spatial overlap of atomic orbitals or hybridized orbitals (" bonds) or side to side overlap on bonding atoms (# bonds)

Lone p orbitals that were not hybridized on each carbon atom are able to form Pi bonds in a “side to side” overlap. A pair of electrons is shared in this region of space.

sp2 hybrid orbitals on each carbon atom use end to end overlap to form a sigma bond.

# bonds overlap side to side

sp hybrid:Ethylyne: HC!CH:Linear

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