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Ch. 8 Bonding and Molecular Structure

Ch. 9 Orbital Hybridization

The content of Ch. 8 – 9 will be taught through both this PPT and

LM 8-9.1

Text references: 8.1, 8.2, 8.4-8.9, 9.1-9.2

The rest of this PPT…• Focuses on substances with covalent bonds

– Molecules – Polyatomic ions

• For any molecule or PAion, you will learn to…– draw the Lewis Dot Structure (LDS) – identify the 3D molecular geometry– determine the polarity– Identify hybridization and sigma/pi bonding– (These are all components of LM 8-9.1)

Chapter 8

LM 8-9.1 Part I

Lewis Dot Structures

How to draw LDS for molecules and ions:1. Write formula for compound.2. Write symbol for least electronegative atom in center

and write symbols for all other atoms around it.3. Add valence e- from all atoms to find out total # of e- you

must assign to the compound.4. Draw one line (single bond) from central atom to all

outer atoms. (1 line = 2 e-) Subtract 2e- for each line you’ve drawn from the total # valence e-.

5. Give all OUTER atoms their complete “octet.” Subtract this value from total.

6. If there are ANY remaining e-, assign them in pairs around the central atom.

7. If there are no more e- but the central atoms still needs more to complete the octet, remove a lone pair from an outer atom and turn this into a double bond. Repeat if you need to form a triple bond. Ex. #1 Ex. #2

Example #1

• Draw LDS for nitrogen trifluoride:

Formula = NF3

Total valence e- = 5 + 3(7) = 26e-

Remaining e- = 26 – 6 = 20e-

••••• •

••• •••

••••

• •

Remaining e- = 20 – 18 = 2e-

••

Click this link and go to Lewis Dot Structures for more info.

Example #2

• Draw LDS for carbon dioxide:

Formula = CO2

O = C = O

Since C did not have enough e- to complete the octet, each oxygen must donate a lone pair of e- to make a double bond.

Electron Deficient Atoms

• Elements from Groups I, II, & III will not hold 8 e- when bonded

• Group I has 2 e- in bonding

• Group II has 4 e- in bonding

• Group III has 6 e- in bonding

1 valence e-

2 valence e-

3 valence e-

Expanded Octet

• Some atoms (usually group V and higher) can POTENTIALLY take on more than 8 e- in bonding– Group IV – VIII must always have at least

8 e-, however some will take more

• Ex. AsF5

• As has 5 fluorine atomsbonded to it and thus issharing 10 total e- with the outer atoms

F As F

Click on this link and go to Expanded Octet for more info.

Resonance Structures• When there are equivalent LDS for a molecule

or ion we say that the structure experiences resonance.

• Applies to a symmetric molecule that contains at least one double bond

• The double bond could be in multiple places and still have the same structure.

Click this link and go to Resonance for more info.

Why Resonance Structures?

• Electrons in molecules w/double bonds are often delocalized between two or more atoms.

• Electrons in a single Lewis structure are assigned to specific atoms-a single Lewis structure is insufficient to show electron delocalization.

• Composite of resonance forms more accurately depicts electron distribution. Molecule or ion doesn’t actually switch between resonance structures. The structure behaves as a blend or composite of all possible structures

• Proof of resonance?– Bond lengths in resonance structures shorter than single

bonds/longer than double bonds– Bonds are stronger than single bonds/weaker than double

Resonance Structures

• Draw all resonance structures for ozone (O3)

Getting a 3D perspective of molecules…For a preview of what’s to come, click on

this link and click on VSEPR Model.

Molecular Geometry

Chapter 8

LM 8-9.1 Part II

Applying VSEPR to Balloons

• Valence Shell Electron Pair Repulsion = model for predicting molecular geometry (3D arrangement of BP and LP)

• Watch what your instructor can do w/balloons…

VSEPR• Valence Shell Electron Pair Repulsion model for

predicting molecular geometry• Model describes the 3D arrangement of atoms in

molecule based on # of lone pairs (LP) and bond pairs (BP) of e- on a central atom.

• A molecule adopts the geometry that minimizes the repulsive force among the e- pairs on central atom

• By minimizing repulsion around central atom you get the most stable geometry of a molecule (minimize repulsion to maximize stability)

For the next slide, draw LDS for:

• BF3, O3

• CH4, NH3, H2O

• PCl5, SF4, ClF3, XeF2

• SF6, BrF5, XeF4What does each grouping have in common? What changes as you go from one grouping to the next? How

might this effect the placement of e- or “geometry” according to VSEPR theory?

Applying VSEPR Theory• Lowest energy arrangement when there are:

– 2BP, 0LP = – 3BP, 0LP = – 4BP, 0LP = – 5BP, 0LP = – 6BP, 0LP =

• Bond angle = angle formed when 3 atoms bond together

Linear 180o

Trigonal Planar 120o

Tetrahedral 109.5o

Trigonal bipyramidal 120o, 90o

Octahedral 90o

Linear Shapes

Examples;

Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

Trigonal Planer Geometry

Examples;

Tetrahedral Geometry

Examples;

Trigonal Bipyramidal Geometry

Examples;

Octahedral Geometry

Examples;

SiF63-

Applying VSEPR Theory (cont’d)• Lowest energy arrangement when there are lone

pairs??? Leads to molecular geometry• LP’s push harder on surrounding BP’s

– LP can spread out more in nonbonding orbital than BP can (BP more localized/constrained in space)

– LP’s make bond angles slightly smaller than expected– The bond angle tends to decrease as # of LP

increases• Steric # (total # of e- pairs) determines e- pair

arrangement to minimize repulsion; #BP vs. LP determines name of molecular geometry (as opposed to electronic geometry); LP influence bond angles

• See how this works…try link above

Steric # = 3 (3 Total Pair of E-)

• 3BP, 0LP

• 2BP, 1LP

Examples;NO2

Trigonal Planar

Steric # = 4 (4 Total Pair of E-) • 4BP, 0LP

• 3BP, 1LP

• 2BP, 2LP

Examples;NH3

Examples;H2OClO2

Tetrahedral

Trigonal pyramidal

<109.5o

<<109.5o

Steric # = 5 (5 Total Pair of E-)• 5BP, 0LP

• 4BP, 1LP

Examples;

Trigonal bipyramidal

Seesaw

Steric # = 5 (contd)

• 3BP, 2LP

• 2BP, 3LP

Examples;ClF3

SeO32-

Examples;XeF2

T-Shaped

Linear

Steric # = 6 (6 Total Pair of E-) • 6BP, 0LP

• 5BP, 1LP 4BP, 2LPExamples;

Examples;XeF4

Octahedral

Square Pyramidal

Square Planar

<90o 90o

What if there’s >1 central atom?

• Number each atom in the chain (this one has 4 central atoms)

• Count the number of BP and LP on each central atom

• Identify the geometry of each central atom (there’s no geometry for the whole structure)

1 2 3 4

• C1: 4BP, 0LP tetrahedral• C2, 3, 4: 3BP, 0LP trigonal planar (each one)

Chapter 8Lab 8-9.1 Part III

Molecular Polarity

To be polar or not to be polar, that is the question!

Bond Polarity vs. Molecular Polarity• Bond polarity – refers to

distribution of e- cloud around 2 atoms in a bond– Polar if unequal distribution

(dipole-one end of bond slightly pos and the other slightly neg) (has a dipole moment)

– Nonpolar if equal distribution (has no dipole moment)

• Determined by difference in EN values of atoms

More info on polarity at this link and select Partial Charges and Bond Dipoles

Chemical Bond Formation• Chemical bond – results when a chem

rxn occurs between 2 or more atoms and the valence e- reorganize so that a net attractive force occurs between them– Ionic Bonds

• Transfer of valence e-• Electrostatic attractn of opp. charged ions• Usu. btwn M + NM, M + PAion or PAion + NM

– Covalent Bonds• Sharing of valence e-• Electrostatic attractn of e- of one atom to nucleus

of another• Usu. btwn NM + NM

Covalent Bond Formation

Electronegativity (EN)• Definition: a value that represents the

relative ability of an atom to attract e- towards itself in a bond– EN is NOT an energy– Scale ranges from 0 - 4

• What this means: larger EN = atom has greater attraction for e- in a bond

Electronegativity (EN)• Trend:

– Down Group = decrease– L to R in Period = increase

• Explain Trend:– Down Group = inc shielding e- = decrease pull

on own valence e- = decrease pull on any e- in bond

– L to R in Period = inc Z = increase pull on own valence e- = increase pull on any e- in bond

Electronegativity (EN)• Deviations:

Group 8: Most Grp 8 elements tend to have EN = 0 b/c they have full outer en levels. This means they tend not to form bonds like other elements. Some Grp 8 elements do have EN values, though, b/c they are large enough to form bonds.

Is bond ionic or polar?• Difference in EN values between two atoms will

determine if bond is Ionic or Covalent (found in Ch. 8.7)

• On a continuum of EN:

• Can also use distance on PT as indicator of bond type: the closer two atoms are, the more covalent the bond; the farther apart they are, the more ionic the bond

0.4 1.7 40

Nonpolarcovalent

Polarcovalent

100% covalent 0% covalent

0% ionic 100% ionic

Is bond ionic or polar?

• Molecular polarity – related to distribution of e- cloud around entire molecule – Polar molecule – experiences unequal

distribution of e- cloud (aka: dipole – one end of molecule slightly + and other end slightly -) (has a dipole moment)

– Nonpolar molecule – experiences equal distribution of e- cloud (no dipole moment)

• Determined by symmetry of molecule (geometry and arrangement of LP)

• LP tend to cause asymmetry in structure which results in a shift in the e- cloud

Bond Polarity vs. Molecular Polarity

What is a dipole moment?• A substance possesses a dipole moment if its centers of

positive and negative charge

do not coincide and cancel each other out.

= e x d

(expressed in Debye units)

(you won’t need to use this)

• The larger the dipole moment, the more polar is the bond or

molecule (due to a greater asymmetry in the distribution of

electrons around the bond or molecule)

• The greater the EN, the larger the dipole moment for a bond.

——++

polarpolar

Why is water polar?• H—O bonds (polar bonds) are not

“balanced” around the structure (the pull of each bond does not cancel the other)

• Lone pairs cause asymmetry (imbalance)

• E- cloud of molecule is shiftedsuch that there is greaterneg chargenear O and greater pos charge by H

General Rule for Molecular Polarity• If all outer atoms are same and structure is

symmetrical (dipole moments of bonds “cancel” each other out b/c same magnitude but opposite direction) molecule is NP (no overall dipole moment)– If 0 LP then NP (if all outer atoms same)

• If structure is asymmetrically arranged (has LP or different outer atoms) molecule is Polar (has dipole moment)– If has LP* then polar– *Exception: Square planar (4 BP, 2LP) and Linear (2 BP,

3 LP) are NP even though they have LP– LP’s cancel each other out

Attractive forces in & btwn molecules

• Intramolecular forces = attractive forces within the molecule (bonds)

• Intermolecular forces (IMF) = attractive forces between molecules (sticky factors)

• Properties of covalent compounds are attributed to their IMF– Ex. Polar cmpds tend to have higher MP and

BP due to stronger IMF

HybridizationThe “real story” behind bonding…

For a preview of what’s to come, click on this link and select

Hybridization.

Chapter 9Lab 8-9.1 Part IV

Draw the LDS of CH4:

What is the e- configuration for C? for H?

Carbon = 2s2 2p2 H = 1s1

s px py pz s s

Is there really room for each H to share e- with Carbon?

Carbon = 2s2 2p2 H = 1s1

s px py pz s s

Is there really room for Hydrogen to share 4 e- with Carbon?

No! Carbon must make room for 2 more e- so it promotes an e- (moves it to a p orbital)

s px py pz

Here’s what it would look like so far:

Carbon = 2s2 2p2H = 1s1

And then if Hydrogen bonds with Carbon:

Remember that each p orbital is on an X, Y or Z axis at 90o angles from one another.

Also remember that the s orbital is smaller in radius than the p orbitals.

Evidence that something “different” is happening…

• Each C-H bond in methane has the same length (109 pm)

• Each H-C-H bond angle is the same (109.5o)• How is this possible if :

– the s and p orbitals are different sizes from one another (this would lead to bonds having different lengths)?

– The s and p orbitals are at 90o not at 109.5o angles

Then answer is….Hybridization

• All of the s, p atomic orbitals mix to create new hybrid orbitals

• # of new hyrbrid orbitals = # of atomic orbitals that were mixedEX. If s and p atomic orbitals mix get 2 new hybrid orbitals, each called sp hybrid

• Each hybrid orbital is exactly like the other and the geometry is based on these (not atomic orbitals)

More on Hybridization…• Identify the hybridization if the following atomic

orbitals are mixed:– s + p + p– s + p + p + p + d

• What geometry could you expect from the above hybridization?

sp2 hybridization

sp3d hybridization

3 hybrids (3BP or 2BP, 1LP)

5 hybrids (5BP or 4BP, 1LP, or 3BP, 2 LP or 2BP, 3LP)

Hybrids can contain LP or BP. # BP & LP determines geometry. Even if you know hybridization, must know what each orbital contains to determine geometry.

http://www.mikeblaber.org/oldwine/chm1045/notes/Geometry/Hybrid/Geom05.htm

Sulfur difluoride1. Draw the LDS.

2. Identify the geometry.

3. Is the molecule polar?

4. What is the hybridization of the central atom?

•• ••

Yes, it is polar

2BP+2LP = 4 hybrids = sp3 hybridization

Xenon tetrafluoride1. Draw the LDS.

2. Identify the geometry.

3. Is the molecule polar?

4. What is the hybridization of the central atom?

Squar Planar

No, it is not polar

4BP+2LP = 6 hybrids = sp3d2 hybridization

F F•

••

Sigma and Pi Bonding• Draw LDS and Label all sigma and pi bonds

in each structure:

• Ethane (C2H6)

• Ethene (Ethylene) (C2H4)

• Ethyne (Acetylene) (C2H2)

Go to this link for more info about this section.

Sigma and Pi Bonding• Ethane (C2H6)

Sigma and Pi Bonding• Ethene (Ethylene) (C2H4)

Sigma and Pi Bonding• Ethyne (Acetylene) (C2H2)

Sigma and Pi Bonding• Sigma Bonding () – head to head overlap

of hybrid or unhybridized orbitals

• Pi Bonding () – side to side overlap of unhybridized p bonds (contains 2 e- total)

*All single bonds consist of bonds.*One of the double bonds is a bond.

*The second and third bond of a double/triple bond are bonds.*Can only occur when central atom is sp or sp2 hybridized

Benzene (C6H6)

b/c of resonance we write…

Covalent Bond Characteristics

• Bond Order: The number of bonds between two atoms

• C—C• C==C • C C

• Bond Energy: Energy a bond must absorb to break (sign?)

• Bond Length: Avg distance between the centers of two nuclei (Next slide…)

• How are they all related?

What’s the BO in a resonance structure? BO = 1

BO = 2

BO = 3

Determining Bond Length

Factors Affecting Bond Length

Larger atoms = longer bonds

How are they all related?___BL = ___BE

___BO = ___BE

___BO = ___BL

indirect

direct

indirect

Using Bond Energies to find Hrxn

• Draw the dot structures of the reactants and the products

• Determine the energy needed to BREAK all the bonds in the reactants (Endothermic, positive value)

• Determine the energy change to MAKE all the bonds in the products (Exothermic, negative value)

• Add them together to get an approximate value for the H of the reaction

Using Bond Energies to find Hrxn

• Determine the H for the combustion of methane gas. (use BE values on next slide)

• What is the Hrxn when you use Hf values?

CH4 + 2O2 CO2 + 2H2O

BEreactants = 4(C-H) + 2(O=O) = 4(413kJ) + 2(498 kJ)

BEproducts = 2(C=O) + 4(H-O) = 2(-745 kJ) + 4(-463 kJ)

BEreactants = 2648 kJ

Notice these are pos energies b/c bonds are

breaking!!!

BEproducts = -3342 kJ

H Rxn = BEproducts + BEreactants = -694 kJ

Notice these are neg energies b/c bonds are

being made!!!

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total e

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placement of e

bond pairs bp of e

lone pair of e

bonded valence e group

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