Electronegativity part two
- 1. Adding Up the DifferenceHow to determine if a bond is ionic, polar covalent, or covalent
- 2. Calculate The DifferenceTo determine how evenly (or unevenly) two atoms will share a pair of electrons, we’ll need to compare the electronegativity values of each atom.If the atoms have the same electronegativity value, they’ll share the electrons evenly.The greater the difference in electronegativity, the more unevenly the electrons are shared (the inequity is always in favor of the MORE electronegative atom)We’ll always be looking at the absolute value of the difference.
- 3. The RulesThe difference in electronegativity is zero.The difference in electronegativity is greater than zero.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. Covalent Polar CovalentIonic
- 4. First Things FirstThis 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!
- 5. Example One: Sodium ChlorideConsider 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.Sodium 0.9 Chlorine 3.0Sodium chloride meets the two requirements for ionic bonding:The difference in electronegativity is greater than 1.9Sodium is a metal, and chlorine is a non-metal Cl 3.0-Na 0.9 2.1
- 6. Example Two: ChloroethaneThe molecule chloroethane (C2H5Cl) has three types of bonds: Carbon-Carbon, Carbon-Hydrogen, and Carbon-Chlorine.We’ll use the electronegativity values shown here to calculate the difference in electronegativity between the atoms in each bond.Carbon: 2.5 Hydrogen: 2.1 Chlorine: 3.0
- 7. Example Two: Chloroethane C 2.5 -C 2.5 0 C 2.5 -H 2.1 0.4 Cl 3.0 -C 2.5 0.5{Polar CovalentCovalentNote that Chloroethane contains both covalent and polar covalent bonds. This is not only “okay” – it’s normal!!!
- 8. Practice OneElementENFluorine 4.0Oxygen 3.5Chlorine 3.0Nitrogen 3.0Bromine 2.8Carbon 2.5Sulfur 2.5Iodine 2.5Hydrogen 2.1Phosphorous 2.1Magnesium 1.2Lithium 1.0Sodium 0.9Label each of the following bonds as ionic, polar covalent or covalent. Use the electronegativity values on the right.Na—FK—ClC—ON—HC—S
- 9. Practice One: AnswersNa—F is an ionic bond (4.0-0.9 = 3.1, and there is a metal and a non-metal)K—Cl is an ionic bond (3.0-0.8 = 2.2, and there is a metal and a non-metal)C—O is a polar covalent bond (3.5-2.5 = 1.0)N—H is a polar covalent bond (3.0-2.5 = 0.5)C—S is a covalent bond (2.5-2.5 =0)
- 10. Practice TwoElementENFluorine 4.0Oxygen 3.5Chlorine 3.0Nitrogen 3.0Bromine 2.8Carbon 2.5Sulfur 2.5Iodine 2.5Hydrogen 2.1Phosphorous 2.1Magnesium 1.2Lithium 1.0Sodium 0.9Consider each of the following compounds. Identify EACH BOND as being ionic, polar covalent or covalent.MgOCO2CH2ONaBr
- 11. Practice Two: AnswersThe bond between Mg and O is ionic (3.5-1.2 = 2.3)The bond between C and O is polar covalent (3.5-2.5 = 1.0)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)The bond between Na and Br is ionic (2.8-0.9 = 1.9)