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Agenda

- Electronegativity and Polar Covalent Bonds

• Lesson: PPT

• Handouts: 1.PPTHANDOUT,

• Text: 1. P. 70-73, 102-108 -69 - Pauling’s Bonding Continuum, Ionic Character

• HW: 1. Finish all the worksheets 2. P. 73 # 1- 6, P. 107 # 1, pg 108 # 1,4-6

The basic units: ionic vs. covalent • Ionic compounds form repeating units. • Covalent compounds form distinct molecules. • Consider adding to NaCl(s) vs. H2O(s):

H O

H Cl Na

Na Cl

Cl

Cl

Na

Na H

O H

• NaCl: atoms of Cl and Na can add individually forming a compound with million of atoms.

• H2O: O and H cannot add individually, instead molecules of H2O form the basic unit.

POLAR COVALENT BONDS • Sometimes when atoms of two different elements

form a bond by sharing an electron pair there is an UNEQUAL sharing of electrons

• When this occurs the bond is called a POLAR BOND

• The unequal sharing results from the difference in electronegativities of the two atoms. The one with the greater electronegativity exerts a greater attraction for the electrons

I’m not stealing, I’m sharing unequally • We described ionic bonds as stealing electrons • In fact, all bonds share – equally or unequally. • Note how bonding electrons spend their time:

• Point: the bonding electrons are shared in each compound, but are not always shared equally.

• The greek symbol indicates “partial charge”.

H2 HCl LiCl

+ – 0 0 + –

covalent (non-polar) polar covalent ionic

H H H Cl [Li]+ [ Cl ]–

Electronegativity • Recall that electronegativity is “a number that describes the

relative ability of an atom, when bonded, to attract electrons”.

• The periodic table has electronegativity values. • We can determine the nature of a bond based on EN

(electronegativity difference). • EN = higher EN – lower EN NBr3: EN = 3.0 – 2.8 = 0.2 (for all 3 bonds). • Basically: a EN below 0.4 = covalent, 0.4 - 1.7 = polar

covalent, above 1.7 = ionic • Determine the EN and bond type for these: HCl, CrO, Br2,

H2O, CH4, KCl

Electronegativity Answers

HCl: 3.0 – 2.1 = 0.9 polar covalent

CrO: 3.5 – 1.6 = 1.9 ionic

Br2: 2.8 – 2.8 = 0 covalent

H2O: 3.5 – 2.1 = 1.4 polar covalent

CH4: 2.5 – 2.1 = 0.4 covalent

KCl: 3.0 – 0.8 = 2.2 ionic

The bonding continuum developed by Pauli, can be used as a guideline to determine if a bond is true

covalent polar covalent or ionic.

In order to truly know if a substance is ionic or covalent, experimental data is needed to verify that the properties do apply.

• Why do some solids dissolve in water but others do not?

• Why are some substances gases at room temperature, but others are liquid or solid?

• What gives metals the ability to conduct electricity, what makes non-metals brittle?

• The answers have to do with …

Intermolecular forces

Questions

Agenda

- Intramolecular and Intermolecular Forces & Molecular Polarity • Lesson: PPT • Handouts: 1.PPTHANDOUT, • Text: 1. P. 96-98,109-117 Intramolecular and

Intermolecular Hydrogen Bonding, Effect on the Properties of Ionic, True Covalent and Polar Covalent Compounds

• • HW: 1. Finish all the worksheets 2. P. 73 # 7 - 8 ; P. 113

mini investigation ;P. 115 # 4,5 ; P. 118 # 1-3 ; • P. 89-92 # 36 – 38, 40, 65 (inter and intra),69,70

Holding it together Q: Consider a glass of water. Why do molecules of water

stay together? A: there must be attractive forces.

Intramolecular forces occur between atoms

Intermolecular forces occur between molecules

• We do not consider intermolecular forces in ionic bonding because there are no molecules.

• We will see that the type of intramolecular bond determines the type of intermolecular force.

Intramolecular forces are much

stronger

Intramolecular Forces

• Forces of electrostatic attraction within a molecule

• Occur between the nuclei of the atoms and their electrons making up the molecule (i.e. covalent bonds)

• Must be broken by chemical means

• Form new substances when broken

Intermolecular forces • Forces of attraction between two molecules (i.e.

London dispersion forces, dipole – dipole interactions or hydrogen bonds)

• These forces are much weaker than Intramolecular forces or bonds and are much easier to break

• Physical changes ( changes of state) break or weaken these forces

• Do not form new substances when broken

• These forces affect the melting and boiling points of substances, the capillary action and surface tension, as well as the volatility and solubility of substances

Types of Intermolecular forces: London Dispersion Forces

• Non-polar molecules do not have dipoles like polar molecules. How, then, can non-polar compounds form solids or liquids?

• London forces result from a type of tiny dipole. • These forces exist between all molecules. • They are masked by stronger forces (e.g. dipole-dipole) so

are sometimes insignificant, but they are important in non-polar molecules.

• Because electrons are moving around in atoms there will be instants when the charge around an atom is not symmetrical.

• The resulting tiny dipoles result in attractions between atoms and/or molecules.

London dispersion forces

• These forces are based on the simultaneous attraction of the electrons of one molecule by the positive nuclei of neighbouring molecules

• The strength of the force is directly related to the number of electrons and protons in a given molecule

• The greater the number of electrons and protons the greater the force

• ”Van der Waal” force

London forces

Instantaneous dipole: Induced dipole:

Eventually electrons are situated so that tiny dipoles form

A dipole forms in one atom or molecule, inducing a dipole in the other

Dipole - Dipole Interactions • Occur between polar molecules having dipoles

• Molecules with dipoles are characterized by oppositely charged ends that are due to an unequal distribution of charge on the molecule

• The polarity of a molecule is determined by both the polarity of the Intramolecular bond and the shape of the molecule

• These forces are based on the simultaneous attraction of the electrons of one dipole by the dipoles of neighbouring molecules

• The strength of the force is related to the polarity of the given molecule

• ”Van der Waal” force

Dipole - Dipole attractions

H Cl

+ –

• Molecules are attracted to each other in a compound by these +ve and -ve forces

+ –

+ –

Hydrogen Bonds • These forces are a type of dipole – dipole interaction

• Occur between Hydrogen atoms in one molecule and highly electronegative atoms [F, O, and N] where there are usually unshared pairs of electrons present

• Q- Calculate the EN for HCl and H2O

• The high EN of NH, OH, and HF bonds cause these to be strong forces.

• Also, because of the small size of hydrogen, it’s positive charge can get very close to the negative dipole of another molecule.

HCl: EN = 3.0-2.1 = 0.9 H2O: = 3.5 – 2.1 = 1.4

Hydrogen bonding

• It is so strongly positive that it will sometimes exert a pull on a “lone pair” in a non-polar compound

• Hydrogen bond are the strongest of the Intermolecular forces and are about 1/10 the strength of a covalent bond

Ionic Forces

Ionic forces may be both inter and intra since a crystalline lattice is formed.

For convenience sake ionic substances are referred to by the smallest ratio of atoms present in the lattice.

Why oil and water don’t mix • Lets take a look at why oil and water don’t mix (oil is

non-polar, water is polar)

+ – + +

– +

+ – +

+ – + +

– +

+ – +

+ – +

+ – +

The partial charges on water attract, pushing the oil (with no partial charge) out of the

way.

Predicting boiling points using the Strength of Intermolecular forces

• Molecules that are isoelectronic have the

same strength of London dispersion forces

• More polar molecules have stronger dipole – dipole interaction and higher melting and boiling points

• The greater the number of electrons per molecule, the stronger the London forces and hence the higher the melting and boiling point

•

Electronegativity & physical properties • Electronegativity can help to explain properties of

compounds like those in the lab.

+ – + – • Lets look at HCl: partial charges keep

molecules together.

• The situation is similar in NaCl, but

the attraction is even greater (EN =

2.1 vs. 0.9 for HCl).

• Which would have a higher melting/boiling point? NaCl because of its greater EN. • For each, pick the one with the lower boiling point a) CaCl2,

CaF2 b) KCl, LiBr c) H2O, H2S. Explain.

– +

+ –

CaCl2 would have a lower melting/boiling point: CaCl2 = 3.0 – 1.0 = 2.0 CaF2 = 4.0 – 1.0 = 3.0

LiBr would have a lower melting/boiling point:

KCl = 3.0 – 0.8 = 2.2 LiBr = 2.8 – 1.0 = 1.8

H2S would have a lower melting/boiling point: H2O= 3.5 – 2.1 = 1.4 H2S = 2.5 – 2.1 = 0.4

Note: other factors such as atomic size

within molecules also affects melting and

boiling points. EN is an important factor

but not the only factor. It is most useful when comparing atoms and molecules of similar

size.

Polar Molecules

• Polar bonds may cause the whole molecule to be polar.

• Polar molecule is a molecule in which the uneven distribution of electrons results in a positive charge at one end and a negative charge at the other end

• Non-polar molecule is a molecule in which the electrons are equally distributed among the atoms, resulting in no localized charges

Molecular Dipole • A result of the bond dipoles in a molecule.

• Bond dipoles may or may not cancel out thereby producing either molecules that are nonpolar, if they cancel, or polar, if they do not cancel

Example1: • If the polar bonds are evenly (or symmetrically)

distributed, the bond dipoles cancel and do not create a molecular dipole. For example, the three bonds in a molecule of BF3 are significantly polar, but they are symmetrically arranged around the central fluorine atom. No side of the molecule has more negative or positive charge than another side, and so the molecule is nonpolar.

Example: A water molecule is polar because (1) its O-H bonds are significantly polar, and (2) its bent geometry makes the distribution of those polar bonds asymmetrical. The side of the water molecule containing the more electronegative oxygen atom is partially negative, and the side of the molecule containing the less electronegative hydrogen atoms is partially positive.

Predicting Molecular Polarity – General Steps

Go over Tutorial 1 and Table 3 on pg. 106 Pg. 228( Grade 12 Textbook)

Step 1: Draw a reasonable Lewis structure for the substance. Step 2: Identify each bond as either polar or nonpolar. (If the difference in electronegativity for the atoms in a bond is greater than 0.4, we consider the bond polar. If the difference in electronegativity is less than 0.4, the bond is essentially nonpolar.) • If there are no polar bonds, the molecule is nonpolar. • If the molecule has polar bonds, move on to Step #3.

Predicting Molecular Polarity – General Steps

Step 3: If there is only one central atom, examine the electron groups around it.

• If there are no lone pairs on the central atom, and if all the bonds to the central atom are the same, the molecule is nonpolar. (This shortcut is described more fully in the Example that follows.)

• If the central atom has at least one polar bond and if the groups bonded to the central atom are not all identical, the molecule is probably polar. Move on to Step #4.

Step 4: Draw a geometric sketch of the molecule.

Predicting Molecular Polarity – General Steps

Step 5: Determine the symmetry of the molecule using the following steps.

• Describe the polar bonds with arrows pointing toward the more electronegative element. Use the length of the arrow to show the relative polarities of the different bonds. (A greater difference in electronegativity suggests a more polar bond, which is described with a longer arrow.)

• Decide whether the arrangement of arrows is symmetrical or asymmetrical

Predicting Molecular Polarity – General Steps

• If the arrangement is symmetrical and the arrows are of equal length, the molecule is nonpolar.

• If the arrows are of different lengths, and if they do not balance each other, the molecule is polar.

• If the arrangement is asymmetrical, the molecule is polar.

Predicting Molecular Polarity- Exercises

Decide whether the molecules represented by the following formulas are polar or nonpolar. (You may need to draw Lewis structures and geometric sketches to do so.)

a. CO2 b. OF2 c. CCl4 d. CH2Cl2 e. HCN

Predicting Molecular Polarity- Solutions

a. The Lewis structure for CO2 is • The electronegativities of carbon and oxygen are 2.55

and 3.44. The 0.99 difference in electronegativity indicates that the C-O bonds are polar, but the symmetrical arrangement of these bonds makes the molecule NONPOLAR.

• If we put arrows into the geometric sketch for CO2, we see that they exactly balance each other, in both direction and magnitude. This shows the symmetry of the bonds.

Predicting Molecular Polarity- Solutions

b. The Lewis structure for OF2 is

The electronegativities of oxygen and fluorine, 3.44 and 3.98, respectively, produce a 0.54 difference that leads us to predict that the O-F bonds are polar. The molecular geometry of OF2 is bent. Such an asymmetrical distribution of polar bonds would produce a POLAR molecule.

Predicting Molecular Polarity- Solutions

c. The molecular geometry of CCl4 is tetrahedral. Even though the C-Cl bonds are polar, their symmetrical arrangement makes the molecule

NONPOLAR.

Predicting Molecular Polarity- Solutions d. The Lewis structure for CH2Cl2 is • The electronegativities of hydrogen, carbon, and chlorine

are 2.20, 2.55, and 3.16. The 0.35 difference in electronegativity for the H-C bonds tells us that they are essentially nonpolar. The 0.61 difference in electronegativity for the C-Cl bonds shows that they are polar. The following geometric sketches show that the polar bonds are asymmetrically arranged, so the molecule is POLAR. (Notice that the Lewis structure above incorrectly suggests that the bonds are symmetrically arranged. Keep in mind that Lewis structures often give a false impression of the geometry of the molecules they represent.)

Predicting Molecular Polarity- Solutions

e. The Lewis structure and geometric sketch for HCN are the same: • The electronegativities of hydrogen, carbon, and

nitrogen are 2.20, 2.55, and 3.04. The 0.35 difference in electronegativity for the H-C bond shows that it is essentially nonpolar. The 0.49 difference in electronegativity for the C-N bond tells us that it is polar. Molecules with one polar bond are always POLAR.

Agenda

• Day 27 - Molecular Polarity- Computer Lab

• Lesson: PPT

• Handouts: 1.PPTHANDOUT,

• Text: 1. P. 96-98,109-117 Intramolecular and Intermolecular Hydrogen Bonding, Effect on the Properties of Ionic, True Covalent and Polar Covalent Compounds

• HW: 1. Finish all the worksheets 2. P. 73 # 7 - 8 ; P. 113 mini investigation ;P. 115 # 4,5 ; P. 118 # 1-3 ;

• P. 89-92 # 36 – 38, 40, 65 (inter and intra),69,70

Testing concepts 1. Which attractions are stronger: intermolecular or

intramolecular? 2. How many times stronger is a covalent bond compared

to a dipole-dipole attraction? 3. What evidence is there that nonpolar molecules attract

each other? 4. Which chemical in table 3 on pg. 113 has the weakest

intermolecular forces? Which has the strongest? How can you tell?

5. State the difference between London Dispersion forces Dipole-Dipole attractions.

6. A) Which would have a lower boiling point: O2 or F2? Explain. B) Which would have a lower boiling point: NO or O2? Explain.

7. Which would you expect to have the higher melting point (or boiling point): C8H18 or C4H10? Explain.

8. What two factors causes hydrogen bonds to be so much stronger than typical dipole-dipole bonds?

9. So far we have discussed 4 kinds of intermolecular forces: ionic, dipole-dipole, hydrogen bonding, and London forces. What kind(s) of intermolecular forces are present in the following substances:

a) NH3, b) SF6, c) PCl3, d) LiCl, e) HBr, f) CO2

(hint: consider EN and molecular shape/polarity) Challenge: Ethanol (CH3CH2OH) and dimethyl ether

(CH3OCH3) have the same formula (C2H6O). Ethanol boils at 78 C, whereas dimethyl ether boils at -24 C. Explain why the boiling point of the ether is so much lower than the boiling point of ethanol.

H – bonding and boiling point

• See Mini- Investigation on pg. 113 – • Q – Why does BP as period , why are some BP

high at period 2?

-200

-150

-100

-50

0

50

100B

oilin

g p

oin

t

Period

Predicted and actual boiling points

Group 4

Group 5

Group 6

Group 72 3 4 5

H – bonding and boiling point- Answers

Boiling points increase down a group (as period increases) for two reasons: 1) EN tends to decrease and 2) size increases. A larger size means greater London forces.

Boiling points are very high for H2O, HF, and NH3 because these are hydrogen bonds (high EN), creating large intermolecular forces

Testing concepts- Answers 1. Intramolecular are stronger. 2. A covalent bond is 100x stronger. 3. The molecules gather together as liquids or solids at low

temperatures. 4. Based on boiling points, CH4 (-162) has the weakest

forces, H2O has the strongest (100). 5. London forces

– Are present in all compounds – Can occur between atoms or molecules – Are due to electron movement not to EN – Are transient in nature (dipole-dipole are more

permanent). – London forces are weaker

6. A) F2 would be lower because it is smaller. Larger atoms/molecules can have their electron clouds more easily deformed and thus have stronger London attractions and higher melting/boiling points.

B) O2 because it has only London forces. NO has a small EN, giving it small dipoles.

7. C8H18 would have the higher melting/boiling point. This is a result of the many more sites available for London forces to form.

8. 1) a large EN, 2) the small sizes of atoms.

Testing concepts- Answers

9. a) NH3: Hydrogen bonding (H + N), London.

b) SF6: London only (it is symmetrical).

c) PCl3: EN=2.9-2.1. Dipole-dipole, London.

d) LiCl: EN=2.9-1.0. Ionic, (London).

e) HBr: EN=2.8-2.1. Dipole-dipole, London.

f) CO2: London only (it is symmetrical)

Challenge: In ethanol, H and O are bonded (the large EN

results in H-bonding). In dimethyl ether the O is bonded

to C (a smaller EN results in a dipole-dipole attraction

rather than hydrogen bonding).

Testing concepts- Answers

Agenda

- Model Building • Lesson: • Handouts: 1.Model Building Activity • Text: 1. P. 96-98,109-117 Intramolecular and

Intermolecular Hydrogen Bonding, Effect on the Properties of Ionic, True Covalent and Polar Covalent Compounds

• HW: 1. Finish all the worksheets 2. P.88 –93 # 1-27, 31-35, 39, 50-60,65-67, 72-74

• P. 128-131 # 6-10, 38-40,42-46,54-56; pg 138-143- as many question as possible.

•