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Electronegativity part two

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Electronegativity part two

1. 1. Adding Up the Difference<br />How to determine if a bond is ionic, polar covalent, or covalent<br />
2. 2. Calculate The Difference<br />To determine how evenly (or unevenly) two atoms will share a pair of electrons, we’ll need to compare the electronegativity values of each atom.<br />If the atoms have the same electronegativity value, they’ll share the electrons evenly.<br />The greater the difference in electronegativity, the more unevenly the electrons are shared (the inequity is always in favor of the MORE electronegative atom)<br />We’ll always be looking at the absolute value of the difference.<br />
3. 3. The Rules<br />The difference in electronegativity is zero.<br />The difference in electronegativity is greater than zero.<br />The difference in electronegativity is greater than or equal to 1.9 AND one atom is a metal and the other is a non-metal. <br />Covalent <br />Polar Covalent<br />Ionic<br />
4. 4. First Things First<br />This slideshow comes with a warning: we’ll use a cutoff value of 1.9 for deciding between ionic and polar covalent bonds. You should know, though, what cutoff your textbook/teacher/instructor/professor/exam/etc. will be using. It might be the same as mine, it might not be. If you don’t know, find out. And if you don’t find out, don’t blame me if you get it wrong on the exam!<br />
5. 5. Example One: Sodium Chloride<br />Consider the compound sodium chloride. The electronegativity values for sodium and chlorine are shown below. To calculate the difference in electronegativity, we’ll consider the ABSOLUTE VALUE of the difference between the two values.<br />Sodium 0.9 Chlorine 3.0<br />Sodium chloride meets the two requirements for ionic bonding:<br />The difference in electronegativity is greater than 1.9<br />Sodium is a metal, and chlorine is a non-metal<br /> Cl 3.0<br />-Na 0.9<br /> 2.1<br />
6. 6. Example Two: Chloroethane<br />The molecule chloroethane (C2H5Cl) has three types of bonds: Carbon-Carbon, Carbon-Hydrogen, and Carbon-Chlorine.<br />We’ll use the electronegativity values shown here to calculate the difference in electronegativity between the atoms in each bond.<br />Carbon: 2.5 Hydrogen: 2.1 Chlorine: 3.0<br />
7. 7. Example Two: Chloroethane<br /> C 2.5<br /> -C 2.5<br /> 0<br /> C 2.5<br /> -H 2.1<br /> 0.4<br /> Cl 3.0<br /> -C 2.5<br /> 0.5<br />{<br />Polar Covalent<br />Covalent<br />Note that Chloroethane contains both covalent and polar covalent bonds. This is not only “okay” – it’s normal!!!<br />
8. 8. Practice One<br />ElementEN<br />Fluorine 4.0<br />Oxygen 3.5<br />Chlorine 3.0<br />Nitrogen 3.0<br />Bromine 2.8<br />Carbon 2.5<br />Sulfur 2.5<br />Iodine 2.5<br />Hydrogen 2.1<br />Phosphorous 2.1<br />Magnesium 1.2<br />Lithium 1.0<br />Sodium 0.9<br />Label each of the following bonds as ionic, polar covalent or covalent. Use the electronegativity values on the right.<br />Na—F<br />K—Cl<br />C—O<br />N—H<br />C—S<br />
9. 9. Practice One: Answers<br />Na—F is an ionic bond (4.0-0.9 = 3.1, and there is a metal and a non-metal)<br />K—Cl is an ionic bond (3.0-0.8 = 2.2, and there is a metal and a non-metal)<br />C—O is a polar covalent bond (3.5-2.5 = 1.0)<br />N—H is a polar covalent bond (3.0-2.5 = 0.5)<br />C—S is a covalent bond (2.5-2.5 =0)<br />
10. 10. Practice Two<br />ElementEN<br />Fluorine 4.0<br />Oxygen 3.5<br />Chlorine 3.0<br />Nitrogen 3.0<br />Bromine 2.8<br />Carbon 2.5<br />Sulfur 2.5<br />Iodine 2.5<br />Hydrogen 2.1<br />Phosphorous 2.1<br />Magnesium 1.2<br />Lithium 1.0<br />Sodium 0.9<br />Consider each of the following compounds. Identify EACH BOND as being ionic, polar covalent or covalent.<br />MgO<br />CO2<br />CH2O<br />NaBr<br />
11. 11. Practice Two: Answers<br />The bond between Mg and O is ionic (3.5-1.2 = 2.3)<br />The bond between C and O is polar covalent (3.5-2.5 = 1.0)<br />The bond between C and H is polar covalent (2.5-2.1 = 0.4) and the bond between C and O is polar covalent (3.5-2.5 = 1.0)<br />The bond between Na and Br is ionic (2.8-0.9 = 1.9) <br />