2. Electronegativity ++ –– 00 00 HClHH

3. The basic units: ionic vs. covalent Ionic compounds form repeating units. Covalent compounds form distinct molecules. Consider adding to NaCl(s) vs. H 2 O(s): H O H Cl Na Cl Na H O H H O H NaCl: atoms of Cl and Na can add individually forming a compound with million of atoms. H 2 O: O and H cannot add individually, instead molecules of H 2 O form the basic unit.

4. 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

5. 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”. H2H2 HClLiCl ++ –– 00 00 +– covalent (non-polar) polar covalent ionic HH H Cl [Li] + [ Cl ] –

6. 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 NBr 3 :  EN = 3.0 – 2.8 = 0.2 (for all 3 bonds). Basically: a  EN below 0.5 = covalent, 0.5 - 1.7 = polar covalent, above 1.7 = ionic Determine the  EN and bond type for these: HCl, CrO, Br 2, H 2 O, CH 4, KCl

7. Electronegativity Answers HCl:3.0 – 2.1 = 0.9 polar covalent CrO:3.5 – 1.6 =1.9ionic Br 2 :2.8 – 2.8 =0covalent H 2 O:3.5 – 2.1 =1.4polar covalent CH 4 :2.5 – 2.1 =0.4covalent KCl:3.0 – 0.8 =2.2ionic

8. 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) CaCl 2, CaF 2 b) KCl, LiBr c) H 2 O, H 2 S – + + – CaCl 2 would have a lower melting/boiling point: CaCl 2 = 3.0 – 1.0 = 2.0 CaF 2 = 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 H 2 S would have a lower melting/boiling point: H 2 O= 3.5 – 2.1 = 1.4 H 2 S = 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.

9. CaCl 2 would have a lower melting/boiling point: CaCl 2 = 3.0 – 1.0 = 2.0 CaF 2 = 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 H 2 S would have a lower melting/boiling point: H 2 O= 3.5 – 2.1 = 1.4 H 2 S = 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.

10. 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.