Chapter 4 Compounds and Their Bonds

Transcription

1 Chapter 4 Compounds and Their Bonds 4.1 Octet Rule and Ions Octet Rule An octet is 8 valence electrons. is associated with the stability of the noble gases. He is stable with 2 valence electrons (duet). valence electrons He 2 2 Ne 2, 8 8 Ar 2, 8, 8 8 Kr 2, 8, 14, Ionic and Covalent Bonds Metals Form Positive Ions Atoms that are not noble gases form octets to become more stable. by losing, gaining, or sharing valence electrons. by forming ionic bonds or covalent bonds. Metals form positive ions by a loss of their valence electrons. with the electron configuration of the nearest noble gas. that have fewer electrons than protons. Group 1A metals ion 1+ Group 2A metals ion 2+ Group 3A metals ion Formation of a Sodium Ion, Na + Charge of Sodium Ion, Na + With the loss of its valence electron, the sodium ion has a 1+ charge. Na atom Na + ion 11p + 11p + 11e - 10e , 8 5 6

2 Formation of Mg 2+ Charge of Magnesium Ion, Mg 2+ Magnesium achieves an octet by losing its two valence electrons. With the loss of two valence electrons, magnesium forms a positive ion with a 2+ charge. Mg atom Mg 2+ ion 12p + 12p + 12e- 10e A. The number of valence electrons in aluminum is 1) 1e -. 2) 2e -. 3) 3e -. B. The change in electrons for octet requires a 1) loss of 3e -. 2) gain of 3e -. 3) a gain of 5e -. C. The ionic charge of aluminum is 1) 3-. 2) 5-. 3) 3+. D. The symbol for the aluminum ion is 1) Al 3+. 2) Al 3-. 3) Al +. A. The number of valence electrons in aluminum is 3) 3 e -. B. The change in electrons for octet requires a 1) loss of 3e -. C. The ionic charge of aluminum is 3) 3+. D. The symbol for the aluminum ion is 1) Al Formation of Negative Ions Formation of a Chloride, Cl - In ionic compounds, nonmetals in Groups 5A (15), 6A (16) and 7A (17) Chlorine achieves an octet by adding an electron to its valence electrons. achieve an octet arrangement by gaining electrons. form negatively charged ions with 3-, 2-, or 1- charges

3 Charge of a Chloride Ion, Cl - Ionic Charge from Group Numbers By gaining 1 electron, the chloride ion has a 1- charge. Chlorine atom, Cl Chloride ion, Cl 17p + 17p + 17e - 18e The charge of a positive ion is equal to its Group number. Group 1A(1) = 1+ Group 2A(2) = 2+ Group 3A(3) = 3+ The charge of a negative ion is obtained by subtracting 8 or 18 from its Group number. Group 6A(16) = 6-8 = 2- or 16-18= Some Ionic Charges A. The number of valence electrons in sulfur is 1) 4e -. 2) 6e -. 3) 8e -. B. The change in electrons for octet requires a 1) loss of 2e -. 2) gain of 2e -. 3) a gain of 4e -. C. The ionic charge of sulfur is 1) 2+. 2) 2-. 3) A. The number of valence electrons in sulfur is 2) 6e -. B. The change in electrons for octet requires a 2) gain of 2e -. Chapter 4 Compounds and Their Bonds 4.2 Ionic Compounds C. The ionic charge of sulfur is 2)

4 Ionic Compounds Salt Is An Ionic Compound Ionic compounds consist of positive and negative ions. have attractions called ionic bonds between positively and negatively charged ions. have high melting and boiling points. are solid at room temperature. Sodium chloride or table salt is an example of an ionic compound Ionic Formulas Charge Balance for NaCl, Salt An ionic formula consists of positively and negatively charged ions. is neutral. has charge balance. total positive charge = total negative charge The symbol of the metal is written first, followed by the symbol of the nonmetal. In NaCl, a Na atom loses its valence electron. a Cl atom gains an electron. the symbol of the metal is written first, followed by the symbol of the nonmetal Charge Balance in MgCl2 Charge Balance in Na 2 S In MgCl 2, a Mg atom loses 2 valence electrons. two Cl atoms each gain 1 electron. subscripts indicate the number of ions needed to give charge balance. In Na 2 S, two Na atoms lose 1 valence electron each. one S atom gains 2 electrons. subscripts show the number of ions needed to give charge balance

5 Writing Ionic Formulas from Charges Charge balance is used to write the formula for sodium nitride, a compound containing Na + and N 3. Na + 3 Na + + N 3 = Na 3 N Na + 3(+1) + 1(3-) = 0 Formula from Ionic Charges Write the ionic formula of the compound with Ba 2+ and Cl. Write the symbols of the ions. Ba 2+ Cl Balance the charges. Ba 2+ Cl two Cl - needed Cl Write the ionic formula using a subscript 2 for two chloride ions. BaCl Select the correct formula for each of the following ionic compounds. A. Na + and O 2-1) NaO 2) Na 2 O 3) NaO 2 B. Al 3+ and Cl - 1) AlCl 3 2) AlCl 3) Al 3 Cl C. Mg 2+ and N 3-1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 A. Na + and O 2-2) Na 2 O check: 2Na + + O 2- = 2(1+) + 1(2-) = 0 B. Al 3+ and Cl - 1) AlCl 3 check: Al 3+ + Cl - = (3+) + 3(1-) = 0 C. Mg 2+ and N 3-3) Mg 3 N 2 check: 2Mg 2+ +2N 3- = 2(3+) + 2(3-) = Chapter 4 Compounds and Their Bonds 4.3 Naming and Writing Ionic Formulas Naming Ionic Compounds with Two Elements To name a compound that contains two elements, identify the cation and anion. name the cation first, followed by the name of the anion with an ide ending

6 Charges of Representative Elements Names of Some Common Ions Complete the names of the following ions: Ba 2+ Al 3+ K + N 3 O 2 F P 3 S 2 Cl Ba 2+ Al 3+ K + barium aluminum potassium N 3 O 2 F nitride oxide fluoride P 3 S 2 Cl phosphide sulfide chloride Examples of Ionic Compounds with Two Elements Formula Ions Name cation anion NaCl Na + Cl - sodium chloride K 2 S K + S 2- potassium sulfide MgO Mg 2+ O 2- magnesium oxide CaI 2 Ca 2+ I - calcium iodide Al 2 O 3 Al 3+ O 2- aluminum oxide Write the names of the following compounds: 1) CaO 2) KBr 3) Al 2 O 3 4) MgCl

7 Write the names of the following compounds: 1) CaO calcium oxide 2) KBr potassium bromide 3) Al 2 O 3 aluminum oxide 4) MgCl 2 magnesium chloride Write the formulas and names for compounds of the following ions: Br S 2 N 3 Na + Al Transition Metals Form Positive Ions Na + Br S 2 N 3 NaBr sodium bromide Na 2 S sodium sulfide Na 3 N sodium nitride Most transition metals and Group 4 (14) metals form 2 or more positive ions. However, Zn 2+,Ag +, and Cd 2+ form only one ion. Al 3+ AlBr 3 aluminum bromide Al 2 S 3 aluminum sulfide AlN aluminum nitride 39 Metals That Form More Than One Cation Naming Ionic Compounds with Variable Charge Metals The name of metals with two or more positive ions (cations) use a Roman numeral to identify ionic charge

8 Naming Variable Charge Metals Naming FeCl 2 Transition metals with two different ions use a Roman numeral after the name of the metal to indicate ionic charge. To name FeCl 2 : 1. Determine the charge of the cation using the charge of the anion (Cl - ). Fe ion + 2 Cl - = Fe ion + 2- = 0 Fe ion = Name the cation by the element name and add a Roman numeral in parentheses to show its charge. Fe 2+ = iron(ii) 3. Write the anion with an ide ending. FeCl 2 = iron(ii) chloride Naming Cr 2 O 3 To name Cr 2 O 3 : 1. Determine the charge of cation from the anion (O 2- ). 2 Cr ions + 3 O 2- = 0 2 Cr ions + 3 (2-) = 0 2 Cr ions - 6 = 0 2 Cr ions = 6+ Cr ion = 3+ = Cr Name the cation by the element name and add a Roman numeral in parentheses to show its charge. Cr 3+ = chromium(iii) 3. Write the anion with an ide ending. chromium(iii) oxide = Cr 2 O 3 Select the correct name for each. A. Fe 2 S 3 1) iron sulfide 2) iron(ii) sulfide 3) iron(iii) sulfide B. CuO 1) copper oxide 2) copper(i) oxide 3) copper(ii) oxide Guide to Writing Formulas from the Name 48

9 Writing Formulas Writing Formulas Write a formula for potassium sulfide. 1. Identify the cation and anion. potassium = K + sulfide = S 2 2. Balance the charges. K + S 2 K + 2(1+) + 2(1-) = K + and 1 S 2 = K 2 S Write a formula for iron(iii) chloride. 1. Identify the cation and anion. iron (III) = Fe 3+ (III = charge of 3+) chloride = Cl 2. Balance the charges. Fe 3+ Cl Cl = (3+) + 3(1-) = 0 Cl 3. 1 Fe 3+ and 3 Cl = FeCl What is the correct formula for each of the following? A. Copper(I) nitride 1) CuN 2) CuN 3 3) Cu 3 N B. Lead(IV) oxide 1) PbO 2 2) PbO 3) Pb 2 O 4 The correct formula is A. Copper(I) nitride 3) Cu 3 N Need 3 Cu + and N 3- B. Lead(IV) oxide 1) PbO 2 Need Pb 4+ and 2 O Chapter 4 Compounds and Their Bonds 4.4 Polyatomic Ions Polyatomic Ions A polyatomic ion is a group of atoms. has an overall ionic charge. Some examples of polyatomic ions are NH + 4 ammonium OH hydroxide NO 3 nitrate NO 2 nitrite CO 2 3 carbonate PO 3 4 phosphate HCO 3 hydrogen carbonate (bicarbonate) 53 54

10 Some Compounds with Polyatomic Ions Some Names of Polyatomic Ions The names of common polyatomic anions end in ate. NO 3 nitrate PO 3 4 phosphate with one oxygen less end in ite. NO 2 nitrite PO 3 3 phosphite with hydrogen attached use the prefix hydrogen (or bi). HCO 3 hydrogen carbonate (bicarbonate) HSO 3 hydrogen sulfite (bisulfite) Names and Formulas of Common Polyatomic Ions Naming Compounds with Polyatomic Ions The positive ion is named first, followed by the name of the polyatomic ion. NaNO 3 sodium nitrate K 2 SO 4 potassium sulfate Fe(HCO 3 ) 3 iron(iii) bicarbonate or iron(iii) hydrogen carbonate (NH 4 ) 3 PO 3 ammonium phosphite Guide to Naming Compounds with Polyatomic Ions Some Compounds with Polyatomic Ions 59 60

11 Match each formula with the correct name. A. MgS 1) magnesium sulfite MgSO 3 2) magnesium sulfate MgSO 4 3) magnesium sulfide B. Ca(ClO 3 ) 2 1) calcium chlorate CaCl 2 2) calcium chlorite Ca(ClO 2 ) 2 3) calcium chloride Match each formula with the correct name. A. MgS 3) magnesium sulfide MgSO 3 1) magnesium sulfite MgSO 4 2) magnesium sulfate B. Ca(ClO 3 ) 2 1) calcium chlorate CaCl 2 3) calcium chloride Ca(ClO 2 ) 2 2) calcium chlorite Name each of the following compounds: A. Mg(NO 3 ) 2 B. Cu(ClO 3 ) 2 C. PbO 2 D. Fe 2 (SO 4 ) 3 E. Ba 3 (PO 3 ) 2 Name each of the following compounds: A. Mg(NO 3 ) 2 magnesium nitrate B. Cu(ClO 3 ) 2 copper(ii) chlorate C. PbO 2 lead(iv) oxide D. Fe 2 (SO 4 ) 3 iron(iii) sulfate E. Ba 3 (PO 3 ) 2 barium phosphite Writing Formulas with Polyatomic Ions The formula of an ionic compound containing a polyatomic ion must have a charge balance that equals zero (0). Na + and NO 3 -> NaNO 3 with two or more polyatomic ions has the polyatomic ions in parentheses. Mg 2+ and 2NO 3 -> Mg(NO 3 ) 2 subscript 2 for charge balance Select the correct formula for each. A. aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 B. copper(ii) nitrate 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) C. iron(iii) hydroxide 1) FeOH 2) Fe 3 OH 3) Fe(OH) 3 D. tin(iv) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) 65 66

12 Select the correct formula for each. A. aluminum nitrate 3) Al(NO 3 ) 3 B. copper(ii) nitrate 2) Cu(NO 3 ) 2 C. iron(iii) hydroxide 3) Fe(OH) 3 D. tin(iv) hydroxide 1) Sn(OH) 4 Write the correct formula for each. A. potassium bromate B. calcium carbonate C. sodium phosphate D. iron(iii) oxide E. iron(ii) nitrite Flowchart for Naming Ionic Compounds Write the correct formula for each. A. potassium bromate KBrO 3 B. calcium carbonate CaCO 3 C. sodium phosphate Na 3 PO 4 D. iron(iii) oxide Fe 2 O 3 E. iron(ii) nitrite Fe(NO 2 ) Name the following compounds: A. Ca 3 (PO 4 ) 2 B. FeBr 3 C. Al 2 S 3 D. Zn(NO 2 ) 2 E. NaHCO 3 Name the following compounds: A. Ca 3 (PO 4 ) 2 Ca 2+ PO 3 4 calcium phosphate B. FeBr 3 Fe 3+ Br iron(iii) bromide C. Al 2 S 3 Al 3+ S 2 aluminum sulfide D. Zn(NO 2 ) 2 Zn 2+ NO 2 zinc nitrite E. NaHCO 3 Na + HCO 3 sodium hydrogen carbonate or sodium bicarbonate 71 72

13 Write the formulas for the following: A. calcium nitrate B. iron(ii) hydroxide C. aluminum carbonate D. copper(ii) bromide E. lithium phosphate Write the formulas for the following: A. calcium nitrate Ca 2+, NO 3 Ca(NO 3 ) 2 B. iron(ii) hydroxide Fe 2+, OH Fe(OH) 2 C. aluminum carbonate Al 3+, CO 3 2 Al 2 (CO 3 ) 3 D. copper(ii) bromide Cu 2+, Br CuBr 2 E. lithium phosphate Li +, PO 4 3 Li 3 PO Chapter 4 Compounds and Their Bonds 4.5 Covalent Compounds Covalent Bonds Covalent bonds form when atoms share electrons to complete octets. between two nonmetal atoms. between nonmetal atoms from Groups 4A (14), 5A (15), 6A (16), and 7A (17) Forming Octets in Molecules In a fluorine, F 2,, molecule, each F atom shares 1 electron. attains an octet. 78

14 Carbon Forms 4 Covalent Bonds Multiple Bonds In a CH 4 (methane) molecule, 1 C atom shares electrons with 4 H atoms to attain an octet. In a nitrogen molecule, N 2, each N atom shares 3 electrons. each N attains an octet. the bond is a multiple bond called a triple bond. the name is the same as the element. each H atom shares 1 electron to become stable, like helium Naming Covalent Compounds Guide to Naming Covalent Compounds In the names of covalent compounds, prefixes are used to indicate the number of atoms (subscript) of each element. (mono is usually omitted) Naming Covalent Compounds Naming Covalent Compounds What is the name of SO 3? 1. The first nonmetal is S sulfur. 2. The second nonmetal is O, named oxide. 3. The subscript 3 of O is shown as the prefix tri. SO 3 -> sulfur trioxide The subscript 1 (for S) or mono is understood. Name P 4 S The first nonmetal, P, is phosphorus. 2. The second nonmetal, S, is sulfide. 3. The subscript 4 of P is shown as tetra. The subscript 3 of O is shown as tri. P 4 S 3 -> tetraphosphorus trisulfide 83 84

15 Select the correct name for each compound. A. SiCl 4 1) silicon chloride 2) tetrasilicon chloride 3) silicon tetrachloride B. P 2 O 5 1) phosphorus oxide 2) phosphorus pentoxide 3) diphosphorus pentoxide C. Cl 2 O 7 1) dichlorine heptoxide 2) dichlorine oxide 3) chlorine heptoxide 86 Select the correct name for each compound. Write the name of each covalent compound. A. SiCl 4 3) silicon tetrachloride B. P 2 O 5 3) diphosphorus pentoxide C. Cl 2 O 7 1) dichlorine heptoxide CO CO 2 PCl 3 CCl 4 N 2 O Guide to Writing Formulas for Covalent Compounds Write the name of each covalent compound. CO CO 2 PCl 3 CCl 4 N 2 O carbon monoxide carbon dioxide phosphorus trichloride carbon tetrachloride dinitrogen oxide 89 90

16 Example: Writing Formulas for Covalent Compounds Write the formula for carbon disulfide. STEP 1: STEP 2: Elements are C and S No prefix for carbon means 1 C Prefix di = 2 Formula: CS 2 Write the correct formula for each of the following: A. phosphorus pentachloride B. dinitrogen trioxide C. sulfur hexafluoride Write the correct formula for each of the following: A. phosphorus pentachloride 1 P penta = 5 Cl PCl 5 B. dinitrogen trioxide di = 2 N tri = 3 O N 2 O 3 Identify each compound as ionic or covalent, and give its correct name. A. SO 3 B. BaCl 2 C. (NH 4 ) 3 PO 3 D. Cu 2 CO 3 E. N 2 O 4 C. sulfur hexafluoride 1 S hexa = 6 F SF Study Tip: Ionic or Covalent A compound is ionic if the first element in the formula or the name is a metal or the polyatomic ion NH 4+. K 2 O K is a metal; compound is ionic; potassium oxide covalent if the first element in the formula or the name is a nonmetal. N 2 O N is a nonmetal; compound is covalent; dinitrogen oxide Identify each compound as ionic or covalent and give its correct name. A. SO 3 covalent sulfur trioxide B. BaCl 2 ionic barium chloride C. (NH 4 ) 3 PO 3 ionic ammonium phosphite D. Cu 2 CO 3 ionic copper(i) carbonate E. N 2 O 4 covalent dinitrogen tetroxide 95 96

17 Identify each compound as ionic or covalent and give its correct name. A. Ca 3 (PO 4 ) 2 B. FeBr 3 C. SCl 2 D. Cl 2 O E. N 2 Identify each compound as ionic or covalent and give its correct name. A. Ca 3 (PO 4 ) ionic Ca 2+ PO 3 4 calcium phosphate B. FeBr 3 ionic Fe 3+ Br iron(iii) bromide C. SCl 2 covalent 1S 2 Cl sulfur dichloride D. Cl 2 O covalent 2 Cl 1 O dichlorine oxide E. N 2 covalent element nitrogen Determine if each is ionic (I) or covalent (C ), and write the formula. A. calcium nitrate B. boron trifluoride C. aluminum carbonate D. dinitrogen tetroxide E. copper(i) phosphate Determine if each is ionic (I) or covalent (C ), and write the formula. A. calcium nitrate (I) Ca 2+, NO 3 Ca(NO 3 ) 2 B. boron trifluoride (C) 1 B, 3 F BF 3 C. aluminum carbonate (I) Al 3+, CO 3 2 Al 2 (CO 3 ) 3 D. dinitrogen tetroxide (C) 2 N, 4 O N 2 O 4 E. copper(i) phosphate (I) Cu +, PO 4 3 Cu 3 PO Chapter 4 Compounds and Their Bonds 4.6 Electronegativity and Bond Polarity Electronegativity The electronegativity value indicates 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

18 Some Electronegativity Values for Group A Elements Electronegativity`increases Nonpolar Covalent Bonds A nonpolar covalent bond Low values Electronegativity decreases ` High values occurs between nonmetals. is an equal or almost equal sharing of electrons. has almost no electronegativity difference (0.0 to 0.4). Examples: Electronegativity Atoms Difference Type of Bond N-N = 0.0 Nonpolar covalent Cl-Br = 0.2 Nonpolar covalent H-Si = 0.3 Nonpolar covalent Polar Covalent Bonds Comparing Nonpolar and Polar Covalent Bonds 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: Electronegativity Atoms Difference Type of Bond O-Cl = 0.5 Polar covalent Cl-C = 0.5 Polar covalent O-S = 1.0 Polar covalent Ionic Bonds Electronegativity and Bond Types An ionic bond occurs between metal and nonmetal ions. is a result of electron transfer. has a large electronegativity difference (1.8 or more). Examples: Electronegativity Atoms Difference Type of Bond Cl-K = 2.2 Ionic N-Na = 2.1 Ionic S-Cs = 1.8 Ionic

19 Predicting Bond Types Use the electronegativity difference to identify the type of bond [nonpolar covalent (NP), polar covalent (P), or ionic (I)] between the following: A. K-N B. N-O C. Cl-Cl D. H-Cl Use the electronegativity difference to identify the type of bond [nonpolar covalent (NP), polar covalent (P), or ionic (I)] between the following: A. K-N 2.2 ionic (I) B. N-O 0.5 polar covalent (P) C. Cl-Cl 0.0 nonpolar covalent (NP) D. H-Cl 0.9 polar covalent (P) Chapter 4 Compounds and Their Bonds 4.7 Shapes and Polarity of Molecules VSEPR Guide to Predicting Molecular Shape In the valence-shell electron-pair repulsion theory (VSEPR), the electron groups around a central atom are arranged as far apart from each other as possible. have the least amount of repulsion of the negatively charged electrons. have a geometry around the central atom that determines molecular shape

20 Four Electron Groups In a molecule of CH 4, there are 4 electron groups around C. repulsion is minimized by placing 4 electron groups at angles of 109, which is a tetrahedral arrangement. the shape with four bonded atoms is tetrahedral. Three Bonding Atoms and One Lone Pair In a molecule of NH 3, 3 electron groups bond to H atoms, and the fourth one is a lone (nonbonding) pair. repulsion is minimized with 4 electron groups in a tetrahedral arrangement. with 3 bonded atoms, the shape is pyramidal Two Bonding Atoms and Two Lone Pairs In a molecule of H 2 O, 2 electron groups are bonded to H atoms and 2 are lone pairs (4 electron groups). 4 electron groups minimize repulsion in a tetrahedral arrangement. the shape with 2 bonded atoms is bent. Shapes with 4 Electron Groups Electron Pairs Bonded Atoms Lone Pairs Molecular Shape Tetrahedral CH Pyramidal NH 3 Example Bent H 2 O Study Tip State the number of electron groups, lone pairs, and use VSEPR theory to determine the shape of the following molecules or ions. 1) tetrahedral 2) pyramidal 3) bent A. PF 3 B. H 2 S C. CCl 4 To determine shape, 1. draw the electron-dot structure. 2. count the electron pairs around the central atom. 3. count the bonded atoms to determine shape. 4 electron pairs and 4 bonded atoms = tetrahedral 4 electrons pairs and 3 bonded atoms = pyramidal 4 electron pairs and 2 bonded atoms = bent

21 Polar Molecules A. PF 3 4 electron groups, 1 lone pair, (2) pyramidal B. H 2 S 4 electron groups, 2 lone pairs, (3) bent C. CCl 4 4 electron groups, 0 lone pairs, (1) tetrahedral A polar molecule contains polar bonds. has a separation of positive and negative charge called a dipole, indicated with δ + and δ -. has dipoles that do not cancel. δ + δ - H Cl H N H dipole H dipoles do not cancel Nonpolar Molecules Determining Molecular Polarity A nonpolar molecule contains nonpolar bonds. Cl Cl H H or has a symmetrical arrangement of polar bonds. O=C=O Cl Cl C Cl Cl dipoles cancel STEP 1: STEP 2: STEP 3: Example:.. H O: H Write the electron-dot formula. Determine the shape. Determine if dipoles cancel or not. H 2 O H 2 O is polar dipoles do not cancel Identify each of the following molecules as 1) polar or 2) nonpolar. Explain. A. PBr 3 B. HBr C. Br 2 D. SiBr 4 Identify each of the following molecules as 1) polar or 2) nonpolar. Explain. A. PBr 3 1) pyramidal; dipoles do not cancel; polar B. HBr 1) linear; one polar bond (dipole); polar C. Br 2 2) linear; nonpolar bond; nonpolar D. SiBr 4 2) tetrahedral; dipoles cancel; nonpolar

22 Chapter 4 Compounds and Their Bonds 4.8 Attractive Forces in Compounds Ionic Bonds In ionic compounds, ionic bonds are strong attractive forces. hold positive and negative ions together Dipole-Dipole Attractions Dispersion Forces In covalent compounds, polar molecules exert attractive forces called dipole-dipole attractions. form strong dipole attractions called hydrogen bonds between hydrogen atoms bonded to F, O, or N, and other very electronegative atoms. Dispersion forces are weak attractions between nonpolar molecules. caused by temporary dipoles that develop when electrons are not distributed equally Attractive Forces 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-dipole attractions. Dispersion forces are weak interactions and very little energy is needed to change state

23 Melting Points of Some Substances Identify the main type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion A. NCl 3 B. H 2 O C. Br-Br D. KCl E. NH Identify the main type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion 2 A. NCl 3 3 B. H 2 O 4 C. Br-Br 1 D. KCl 3 E. NH 3 135