Chapter 7 Molecular Structure: Solids and Liquids Electron Configuration of Ionic Compounds Review

Chapter 7 Slide 2 of 46 Review Formation of cations and anions.

Chapter 7 Slide 3 of 46 Metals Form Positive Ions Metals form Octets by losing all of their valence electrons. Positive ions with the electron configuration of the nearest noble gas. Positive ions with fewer electrons than protons. Group 1A(1) metals  ion 1+ Group 2A(2) metals  ion 2+ Group 3A(13) metals  ion 3+ Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 4 of 46 Formation of Mg 2+ Magnesium achieves an octet by losing its two valence electrons. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 5 of 46 Formation of Negative Ions In ionic compounds, nonmetals Achieve an octet arrangement. Gain electrons. Form negatively charged ions with 3-, 2-, or 1- charges.

Chapter 7 Slide 6 of 46 Formation of a Chloride, Cl - Chlorine achieves an octet by adding an electron to its valence electrons. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 7 of 46 Ions of Transition Metals Transition metals Form positively charged ions. Lose ____ valence electrons to form 2+ ions. Lose ____ electrons to form ions with higher positive charges. Do not form octets like representative metals.

Chapter 7 Slide 8 of 46 Ions of Transition Metals Example: Fe forms Fe 3+ and Fe 2+ Fe 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 Loss of two 4s electrons Fe 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 6 Loss of two 4s electrons and one d electron Fe 3+ 1s22s22p63s23p64s03d5 Note: The 3d subshell is half-filled and stable.

Chapter 7 Slide 9 of 46 Electron-Dot Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Why look at Electron-Dot Formulas? Ionic Compounds: Helps to determine formulas Covalent Compounds: Help us to understand molecular structures or molecular geometries.

Chapter 7 Slide 10 of 46 Molecular Geometries or Shapes

Chapter 7 Slide 11 of 46 Multiple Covalent Bonds In a single bond ______ pair of electrons is shared. In a double bond, ______ pairs of electrons are shared. In a triple bond. ______ pairs of electrons are shared.

Chapter 7 Slide 12 of 46 Electron-Dot Formulas Electron-dot formulas show The order of bonded atoms in a covalent compound. The bonding pairs of electrons between atoms. The unshared (lone, non- bonding) valence electrons. A central atom with an octet. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 13 of 46 Number of Covalent Bonds The number of covalent bonds can be determined from the number of electrons needed to complete an octet. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 7.1

Chapter 7 Slide 14 of 46 Guide to Writing Electron-Dot Formulas STEP 1 Determine the arrangement of atoms. STEP 2 Add the valence electrons from all the atoms. STEP 3 Attach the central atom to each bonded atom using one pair of electrons. STEP 4 Add remaining electrons as lone pairs to complete octets (2 for H atoms). STEP 5 If octets are not complete, form one or more multiple bonds.

Chapter 7 Slide 15 of 46 Electron-Dot Formula of SF 2 Write the electron-dot formula for SF 2. STEP 1 Determine the atom arrangement. S is the central atom. F S F STEP 2 Total the valence electrons for 1S and 2F. 1S(6e - ) + 2F(7e - ) = 20e -

Chapter 7 Slide 16 of 46 Electron-Dot Formula SF 2 STEP 3 Attach F atoms to S with one electron pair. F : S : F Calculate the remaining electrons. 20e e - = 16e - left STEP 4 Complete the octets of all atoms by placing remaining e - as 8 lone pairs to complete octets.             : F : S : F : or : F─S─F :            

Chapter 7 Slide 17 of 46 Electron-Dot Formula ClO 3 - Write the electron-dot formula for ClO 3 −. STEP 1 Determine atom arrangement. Cl is the central atom. O − O Cl O STEP 2 Add all the valence electrons for 1Cl and 3O plus 1e - for negative charge on the ion. 1Cl(7e - ) + 3 O(6e - ) + 1e − = 26e -

Chapter 7 Slide 18 of 46 Electron-Dot Formula ClO 3 - STEP 3 Attach each O atom to Cl with one electron pair. O −   O : Cl : O Calculate the remaining electrons. 26e e - = 20e - left

Chapter 7 Slide 19 of 46 Electron-Dot Formula ClO 3 - STEP 4 Complete the octets of all atoms by placing the remaining 20 e - as 10 lone pairs to complete octets.   −   − : O : : O :         │   : O : Cl : O : or : O─Cl─O :             Note: Bonding electrons can be shown by a

Chapter 7 Slide 20 of 46 Multiple Bonds In a single bond One pair of electrons is shared. In a double bond, Two pairs of electrons are shared. In a triple bond. Three pairs of electrons are shared.

Chapter 7 Slide 21 of 46 Electron-Dot Formula of CS 2 Write the electron-dot formula for CS 2. STEP 1 Determine the atom arrangement. The C atom is the central atom. S C S STEP 2 Total the valence electrons for 1C and 2S. 1C(4e - ) + 2S(6e - ) = 16e -

Chapter 7 Slide 22 of 46 Electron-Dot Formula CS 2 STEP 3 Attach each S atom to C with electron pairs. S : C : S Calculate the remaining electrons. 16e e - = 12e - left

Chapter 7 Slide 23 of 46 Electron-Dot Formula CS 2 STEP 4 Attach 12 remaining electrons as 6 lone pairs to complete octets..... : S : C : S :.... STEP 5 To complete octets, move two lone pairs between C and S atoms to give two double bonds : S : : C : : S : or : S = C = S :

Chapter 7 Slide 24 of 46 Multiple Bonds in N 2 In nitrogen N 2, Octets are achieved by sharing three pairs of electrons, which is a triple bond. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 25 of 46 Some Electron-Dot Formulas Table 7.2 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 26 of 46 Electronegativity and Polarity Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 27 of 46 Electronegativity values Indicate the attraction of an atom for shared electrons. Increases from left to right going across a period on the periodic table. Is high for the nonmetals with fluorine as the highest. Is low for the metals. Electronegativity

Chapter 7 Slide 28 of 46 Some Electronegativity Values for Group A Elements Low values High values Electronegativity increases Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 29 of 46 A nonpolar covalent bond Occurs between nonmetals. Is an equal or almost equal sharing of electrons. Has almost no electronegativity difference (0.0 to 0.4). Examples: Atoms Electronegativity Type of Bond Difference N-N = 0.0 Nonpolar covalent Cl-Br = 0.2 Nonpolar covalent H-Si = 0.3 Nonpolar covalent Nonpolar Covalent Bonds

Chapter 7 Slide 30 of 46 A polar covalent bond Occurs between nonmetal atoms. Is an unequal sharing of electrons. Has a moderate electronegativity difference (0.5 to 1.7). Examples: Atoms ElectronegativityType of Bond Difference O-Cl = 0.5Polar covalent Cl-C = 0.5Polar covalent O-S = 1.0Polar covalent Polar Covalent Bonds

Chapter 7 Slide 31 of 46 Comparing Nonpolar and Polar Covalent Bonds Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 32 of 46 Ionic Bonds An ionic bond Occurs between metal and nonmetal ions. Is a result of complete electron transfer. Has a large electronegativity difference (1.8 or more). Examples: Atoms Electronegativity Type of Bond Difference Cl-K 3.0 – 0.8 = 2.2Ionic N-Na 3.0 – 0.9 = 2.1Ionic S-Cs2.5 – 0.7= 1.8Ionic

Chapter 7 Slide 33 of 46 Predicting Bond Types Table 7.4 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 34 of 46 Range of Bond Types Table 7.5 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 35 of 46 Use the electronegativity difference to identify the type of bond between the following as: nonpolar covalent (NP), polar covalent (P), or ionic (I). A. K-N2.2ionic (I) B. N-O0.5 polar covalent (P) C. Cl-Cl0.0nonpolar covalent (NP) D. H-Cl0.9polar covalent (P) Learning Check

Chapter 7 Slide 36 of 46 Polar Molecules A polar molecule Contains _________ bonds. Has a separation of positive and negative charge called a dipole indicated with  + and  -. Has dipoles that do not cancel.  +  - H–Cl H — N —H dipole │ H dipoles do not cancel

Chapter 7 Slide 37 of 46 Nonpolar Molecules A nonpolar molecule Contains _____________ bonds. Cl–Cl H–H Or has a symmetrical arrangement of polar bonds. O=C=O Cl │ Cl –C–Cl │ Cl dipoles cancel

Chapter 7 Slide 38 of 46 Polar Bonds and Nonpolar Molecules For example, the bond dipoles in CO 2 cancel each other because CO 2 is linear.

Chapter 7 Slide 39 of 46 Determining Molecular Polarity STEP 1 Write the electron-dot formula. STEP 2 Determine the polarity of the bonds. STEP 3 Determine if any dipoles cancel or not. Example: H 2 O.. H─O: H 2 O is polar │ H dipoles do not cancel

Chapter 7 Slide 40 of 46 Polar Bonds and Polar Molecules In water, the molecule is not linear and the bond dipoles do not cancel each other. Therefore, water is a polar molecule.

Chapter 7 Slide 41 of 46 Attractive Forces Between Particles Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 42 of 46 Ionic Bonds In ionic compounds, ionic bonds Are strong attractive forces. Hold positive and negative ions together. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 43 of 46 Dipole-Dipole Attractions In covalent compounds, polar molecules Exert attractive forces called dipole-dipole attractions. Form ________ dipole attractions called hydrogen bonds between hydrogen atoms bonded to F, O, or N and typically a lone pair or unshared pair of electrons.

Chapter 7 Slide 44 of 46 Dipole-Dipole Attractions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 45 of 46 Dispersion Forces Dispersion forces are Weak attractions between nonpolar molecules. Caused by temporary dipoles that develop when electrons are not distributed equally. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 7 Slide 46 of 46 Comparison of Bonding and Attractive Forces Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table7.8

Chapter 7 Slide 47 of 46 Melting Points and Attractive Forces Ionic compounds require large amounts of energy to break apart ionic bonds. Thus, they have high melting points. Hydrogen bonds are the strongest type of dipole-dipole attractions. They require more energy to break than other dipole attractions. Dispersion forces are weak interactions and very little energy is needed to change state.