h#p://vimeo.com/1041071 Agenda -‐ Electronega<vity and Polar Covalent Bonds • Lesson: PPT • Handouts: 1.PPTHANDOUT, • Text: 1. P. 70-‐73, 102-‐108 -‐69 -‐ Pauling’s Bonding Con<nuum, 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 repea<ng units. • Covalent compounds form dis<nct molecules. • Consider adding to NaCl(s) vs. H2O(s): Na Cl Na Cl Cl Na Cl Na H O H O H 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 Some<mes when atoms o f 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 electronega<vi<es of the two atoms. The one with the greater electronega<vity exerts a greater a#rac<on 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 <me: H2 δ0 H H δ0 covalent (non-polar) HCl δ+ H Cl δ– polar covalent LiCl [Li]+[ + – ionic • Point: the bonding electrons are shared in each compound, but are not always shared equally. • The greek symbol δ indicates “par<al charge”. Cl ]– Electronega<vity • Recall that electronega<vity is “a number that describes the rela<ve ability of an atom, when bonded, to a#ract electrons”. • The periodic table has electronega<vity values. • We can determine the nature of a bond based on ΔEN (electronega<vity diﬀerence). • ΔEN = higher EN – lower EN NBr3: ΔEN = 3.0 – 2.8 = 0.2 (for all 3 bonds). • Basically: a ΔEN below 0.4 = covalent, 0.4 -‐ 1.7 = polar covalent, above 1.7 = ionic • Determine the ΔEN and bond type for these: HCl, CrO, Br2, H2O, CH4, KCl Electronega<vity 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 con<nuum 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 proper<es do apply. Predic<ng Molecular Polarity – General Steps Go over Tutorial 1 and Table 3 on pg. 106 Step 1: Draw a reasonable Lewis structure for the substance. Step 2: Iden<fy each bond as either polar or nonpolar. (If the diﬀerence in electronega<vity for the atoms in a bond is greater than 0.4, we consider the bond polar. If the diﬀerence in electronega<vity is less than 0.4, the bond is essen<ally nonpolar.) • If there are no polar bonds, the molecule is nonpolar. • If the molecule has polar bonds, move on to Step #3. Polar Molecules • Polar bonds may cause the whole molecule to be polar. • Polar molecule is a molecule in which the uneven distribu<on of electrons results in a posi<ve charge at one end and a nega<ve charge at the other end • Non-‐polar molecule is a molecule in which the electrons are equally distributed among the atoms, resul<ng in no localized charges Predic<ng Molecular Polarity – General Steps Step 3: Draw a geometric sketch of the molecule. Use the VSEPR THEORY to guide you. Step 4: Determine the symmetry of the molecule using the following steps. • Describe the polar bonds with arrows poin<ng toward the more electronega<ve element. Use the length of the arrow to show the rela<ve polari<es of the diﬀerent bonds-‐ (the bond dipoles ). (A greater diﬀerence in electronega<vity suggests a more polar bond, which is described with a longer arrow.) • Decide whether the arrangement of arrows is symmetrical or asymmetrical Predic<ng 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. VSEPR THEORY Valence Shell Electron Pair Repulsion Theory CONCLUSION 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. • If the central atom has at least one polar bond and if the groups bonded to the central atom are not all iden<cal, the molecule is probably polar. USE the VSEPR THEORY AND MOLECULAR SYMMETRY to determine the molecular polarity. Molecular Dipole in a molecule. A result of the bond dipoles • • 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 nega<ve or posi<ve charge than another side, and so the molecule is nonpolar. Predic<ng 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 Predic<ng Molecular Polarity-‐ Solu<ons a. The Lewis structure for CO2 is • The electronega<vi<es of carbon and oxygen are 2.55 and 3.44. The 0.99 diﬀerence in electronega<vity 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 direc<on and magnitude. This shows the symmetry of the bonds. Predic<ng Molecular Polarity-‐ Solu<ons b. The Lewis structure for OF2 is The electronega<vi<es of oxygen and ﬂuorine, 3.44 and 3.98, respec<vely, 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 distribu<on of polar bonds would produce a POLAR molecule. Predic<ng Molecular Polarity-‐ Solu<ons c. The molecular geometry of CCl4 is tetrahedral. Even though the C-‐Cl bonds are polar, their symmetrical arrangement makes the molecule NONPOLAR. Predic<ng Molecular Polarity-‐ Solu<ons d. The Lewis structure for CH2Cl2 is • The electronega<vi<es of hydrogen, carbon, and chlorine are 2.20, 2.55, and 3.16. The 0.35 diﬀerence in electronega<vity for the H-‐C bonds tells us that they are essen<ally nonpolar. The 0.61 diﬀerence in electronega<vity 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. (No<ce that the Lewis structure above incorrectly suggests that the bonds are symmetrically arranged. Keep in mind that Lewis structures omen give a false impression of the geometry of the molecules they represent.) Predic<ng Molecular Polarity-‐ Solu<ons e. The Lewis structure and geometric sketch for HCN are the same: • The electronega<vi<es of hydrogen, carbon, and nitrogen are 2.20, 2.55, and 3.04. The 0.35 diﬀerence in electronega<vity for the H-‐C bond shows that it is essen<ally nonpolar. The 0.49 diﬀerence in electronega<vity 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 Proper<es of Ionic, True Covalent and Polar Covalent Compounds • HW: 1. Finish all the worksheets 2. P. 73 # 7 -‐ 8 ; P. 113 mini inves<ga<on ;P. 115 # 4,5 ; P. 118 # 1-‐3 ; • P. 89-‐92 # 36 – 38, 40, 65 (inter and intra),69,70 • • • • Agenda -‐ Model Building • Lesson: • Handouts: 1.Model Building Ac<vity • Text: 1. P. 96-‐98,109-‐117 Intramolecular and Intermolecular Hydrogen Bonding, Eﬀect on the Proper<es 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 ques<on as possible. • 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 Proper<es of Ionic, True Covalent and Polar Covalent Compounds • • HW: 1. Finish all the worksheets 2. P. 73 # 7 -‐ 8 ; P. 113 mini inves<ga<on ;P. 115 # 4,5 ; P. 118 # 1-‐3 ; • P. 89-‐92 # 36 – 38, 40, 65 (inter and intra),69,70 Questions • 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 bri#le? • The answers have to do with … Intermolecular forces Holding it together Q: Consider a glass of water. Why do molecules of water stay together? A: there must be a#rac<ve forces. Intramolecular forces are much stronger Intramolecular forces Intermolecular forces occur occur between atoms 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 • Forces of electrosta<c a#rac<on 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 two molecules (i.e. Forces of a#rac<on between London dispersion forces, dipole – dipole interac<ons 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 mel<ng and boiling points of substances, the capillary ac<on and surface tension, as well as the vola<lity 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 <ny dipole. • These forces exist between all molecules. • They are masked by stronger forces (e.g. dipole-‐dipole) so are some<mes 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 resul<ng <ny dipoles result in a#rac<ons between atoms and/or molecules. London dispersion forces • These forces are based on the simultaneous of one molecule by a#rac<on of the electrons the posi<ve 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 A dipole forms in one atom or molecule, inducing a <ny dipoles form dipole in the other Dipole -‐ Dipole Interac<ons • Occur between polar m olecules having dipoles • Molecules with dipoles are characterized by oppositely charged ends that are due to an unequal distribu<on 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 a#rac<on 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 δ+ δ– H Cl • Molecules are a#racted to each other in a compound by these +ve and -‐ve forces δ+ δ– δ+ δ + δ – δ– Hydrogen Bonds These forces are a type o f dipole – dipole interac<on • • Occur between Hydrogen atoms in one molecule and highly electronega<ve atoms [F, O, and N] where there are usually unshared pairs of electrons present • Q- Calculate the ΔEN for HCl and H2O HCl: ΔEN = 3.0-‐2.1 = 0.9 H2O: = 3.5 – 2.1 = 1.4 • The high ΔEN of NH, OH, and HF bonds cause these to be strong forces. • Also, because of the small size of hydrogen, it’s posi<ve charge can get very close to the nega<ve dipole of another molecule. Hydrogen bonding • It is so strongly posi<ve that it will some<mes 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 lavce is formed. For convenience sake ionic substances are referred to by the smallest ra<o of atoms present in the lavce. 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) δ+ – δ+ δ δ+ – δ+ δ δ+ – δ+ δ The par<al charges on water a#ract, pushing the oil (with no par<al charge) out of the way. δ+ δ– δ+ – δ+ δ δ+ – δ+ δ δ+ δ+ – δ+ δ δ+ – δ+ δ Predic>ng 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 interac<on and higher mel<ng and boiling points • The greater the number of electrons per molecule, the stronger the London forces and hence the higher the mel<ng and boiling point • Electronega<vity & physical proper<es • Electronega<vity can help to explain proper<es of compounds like those in the lab. • Lets look at HCl: par<al charges keep molecules together. • The situation is similar in NaCl, but the attraction is even greater (ΔEN = 2.1 vs. 0.9 for HCl). δ+ δ– δ+ δ– δ+ δ– – + + – • Which would have a higher mel<ng/boiling point? NaCl because of its greater ΔEN. • 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 mel<ng/boiling point: CaCl2 = 3.0 – 1.0 = 2.0 CaF2 = 4.0 – 1.0 = 3.0 LiBr would have a lower mel<ng/boiling point: KCl = 3.0 – 0.8 = 2.2 LiBr = 2.8 – 1.0 = 1.8 H2S would have a lower mel<ng/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 mel<ng and boiling points. ΔEN is an important factor but not the only factor. It is most useful when comparing atoms and molecules of similar size. Tes<ng concepts 1. Which a#rac<ons are stronger: intermolecular or intramolecular? 2. How many <mes stronger is a covalent bond compared to a dipole-‐dipole a#rac<on? 3. What evidence is there that nonpolar molecules a#ract 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 a#rac<ons. 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 mel<ng 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 ΔEN 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 Predicted and actual boiling points 100 Group 4 Boiling point 50 0 Group 5 -50 Group 6 -100 -150 2 3 4 5 Group 7 -200 Period • See Mini- Investigation on pg. 113 – • Q – Why does BP↑ as period ↑, why are some BP high at period 2? H – bonding and boiling point- Answers Boiling points increase down a group (as period increases) for two reasons: 1) ΔEN tends to increase 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 ΔEN), crea<ng 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 ΔEN – Are transient in nature (dipole-‐dipole are more permanent). – London forces are weaker Testing concepts- Answers 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 a#rac<ons and higher mel<ng/boiling points. B) O2 because it has only London forces. NO has a small ΔEN, giving it small dipoles. 7. C8H18 would have the higher mel<ng/boiling point. This is a result of the many more sites available for London forces to form. 8. 1) a large ΔEN, 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: ΔEN=2.9-‐2.1. Dipole-‐dipole, London. d) LiCl: ΔEN=2.9-‐1.0. Ionic, (London). e) HBr: ΔEN=2.8-‐2.1. Dipole-‐dipole, London. f) CO2: London only (it is symmetrical) Challenge: In ethanol, H and O are bonded (the large ΔEN results in H-‐bonding). In dimethyl ether the O is bonded to C (a smaller ΔEN results in a dipole-‐dipole a#rac<on rather than hydrogen bonding).
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