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Chemistry Unit 6

Chemistry Unit 6. Chemical Bonding. Warm-up. Why do atoms bond and what part do ions play in this?. Objectives. TSWBAT: Identify types of bonds and their characteristics. Agenda. Go over tests Go over warm-up Notes on Bonding Practice Wrap-up. Three Main Types of Bonds.

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Chemistry Unit 6

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  1. Chemistry Unit 6 Chemical Bonding

  2. Warm-up • Why do atoms bond and what part do ions play in this?

  3. Objectives • TSWBAT: Identify types of bonds and their characteristics.

  4. Agenda • Go over tests • Go over warm-up • Notes on Bonding • Practice • Wrap-up

  5. Three Main Types of Bonds • Ionic Bond – Atoms transfer electrons to fill their valence shells, oppositely charged ions are formed, opposites attract. • Occurs between a metal and a nonmetal • Covalent Bond – Atoms share electrons to fill their valence shells. • Occurs between nonmetals • Metallic Bonds – Atoms share a “sea of electrons.” • Occurs between atoms of a metal

  6. Why Do Atoms Bond? • To become more stable • like the noble gases. • Octet Rule – atoms tend to gain, lose or share electrons in order to acquire a full shell of valence electrons. (usually 8)

  7. Ionic Bonding • Ion – a charged particle • A neutral atom becomes an ion when it loses or gains an electron. • If an atom loses an electron, it becomes a (+) ion called a cation. • If an atom gains an electron, it becomes a (-) ion called an anion.

  8. Ionic Bonding • Example Na Cl To become more stable, sodium must loose one electron To become more stable, chlorine must gain one electron

  9. Cl Na Ionic Bonding • Example Sodium loses an electron and becomes an Na+1 ion. Chlorine gains an electron and becomes a Cl-1 ion. Opposites attract, and an ionic compound is formed… NaCl

  10. Try Another Example Br Al Aluminum will become more stable if it gets rid of three electrons. Bromine will become more stable if it receives one electron. Are both atoms more stable as a result of this transfer? No, Al must donate two more… where?

  11. Br Br Br Al Aluminum & Bromine Now, each atom has a full valence shell… all are more stable.

  12. Br Br Br Al Aluminum and Bromine Aluminum donated 3 e-, so it becomes Al+3 Each bromine accepted 1 e-, so they each become Br-1 The compound that forms is AlBr3

  13. Let’s Wrap it Up • Ionic bonds are held together by electrostatic forces. • The result of an ionic bond is called an ionic compound. • Ionic bonds form between a metal and a nonmetal atom due to large differences in electronegativity. (1.7 or greater) • The nonmetal’s EN is so much greater than the metal’s, that it removes the electrons, forming oppositely charged ions!

  14. For Example: Na and O EN of Na = 0.9 EN of O = 3.5

  15. Why does Sodium and Oxygen form an ionic bond? 3.5 EN of O - 0.9 EN of Na 2.4 Difference in EN • Difference in electronegativity is 2.4(>1.7) • An ionic bond will form. • Oxygen has a greater electronegativity, and is able to yank electrons away from sodium.

  16. Warm-up 3/26/2013 Do the following warm-up by yourself. This will be handed in and graded as a warm-up quiz. Barium and Bromine Magnesium and Chlorine Potassium and Sulfur Sodium and Oxygen Lithium and Fluorine Aluminum and Nitrogen • For the pairs listed to the right, show: • Symbol with charges • Lewis Dot Structures • Name

  17. Covalent Bonding O O Each atom of Oxygen needs two more electrons to become more stable, so they will share two pairs of electrons. A diatomic molecule of oxygen is formed. O2

  18. H H O Try another example Oxygen needs two electrons to become more stable. Each atom of hydrogen needs one more electron to become more stable. All atoms become more stable (have full valence shells). A molecule of water is made. H2O

  19. Let’s Wrap it Up… Again! • Covalent bonds are held together by a mutual need for the shared electrons (electronegativity) • Covalent bonding occurs when a sharing of electrons results in an overlap of valence orbitals. Each electron is attracted to the positive charge of the opposite nucleus. • The result of a covalent bond is called a molecule.

  20. Polar and Nonpolar Covalent Bonds • A covalent bond occurs between nonmetals because there is only a slight difference in their electronegativites. (less than 1.7 diff in EN) The atoms share the electrons fairly equally. • If one nonmetal has a greater EN than the other, it can “hog” the electrons, forming a POLAR covalent bond. (>0  <1.7 diff. in EN) • If the nonmetals have equal EN, they will share equally and form NONPOLAR covalent bond. (0 diff. in EN)

  21. For Example: N and O EN of O = 3.5 EN of N = 3.0

  22. Why does Nitrogen and Oxygen form a Covalent Bond? 3.5 EN of Oxygen • 3.0 EN of Nitrogen 0.5 = difference in EN Difference in EN is less than 1.7, therefore a covalent bond will form. Difference in EN is greater than 0, therefore the covalent bond will be polar. (Unequal sharing of e-)

  23. One Final Example If Chlorine bonds with Chlorine (a diatomic molecule), the difference in EN would be “0”, thus a nonpolar covalent bond will form. (Equal sharing of e-)

  24. Writing Ionic Formulas Calcium Chloride • Locate the metal on the periodic table and write the element symbol with its oxidation number. Ca +2

  25. Writing Ionic Formulas • Locate the nonmetal on the periodic table and write the element’s symbol with its oxidation number. Cl-1

  26. Ca+2 Cl-1 • Find the common factor between the two oxidation numbers. • In this case, 2. • Decide how many of each ion is needed to make the charge equal to the common factor. • In this case, 1 calcium ion (+2) and 2 chlorine ions (-1 and –1 = -2). Compounds are neutral. • Use this number of ions as the subscript for the element, and write the formula. • In this case, Ca Cl2.

  27. Writing Ionic Formulas Part 2 Aluminum Oxide • Locate the metal on the periodic table and write the element symbol with its oxidation number. Al +3

  28. Writing Ionic Formulas Part 2 • Locate the nonmetal on the periodic table and write the element’s symbol with its oxidation number. O-2

  29. Al+3 O-2 • Find the common factor between the two oxidation numbers. • In this case, 6. • Decide how many of each ion is needed to make the charge equal to the common factor. • In this case, 2 aluminum ions (+3 and +3 = +6) and 3 oxygen ions (-2 and -2 and -2 = -6). Compounds are neutral. • Use this number of ions as the subscript for the element, and write the formula. • In this case, Al2O3.

  30. Try these examples on your own. • Sodium and Oxygen • Lithium and Sulfur • Aluminum and Chlorine • Potassium and Nitrogen • Magnesium and Fluorine

  31. Naming Ionic Compounds • Write the name of the metal. • Write the name of the nonmetal with the ending changed to –ide. Example: Nitrogen = nitride Sulfur = sulfide Oxygen = oxide Chlorine = chloride Phosphorus = phosphide Iodine = iodide Fluorine = fluoride Bromine = bromide

  32. Naming Ionic Compounds Al2S3 • Write the name of the metal. Aluminum • Write the name of the nonmetal, changing the ending to –ide. Sulfide • Name the compound. Aluminum Sulfide

  33. Naming Ionic Compounds BaCl2 • Write the name of the metal. Barium • Write the name of the nonmetal, changing the ending to –ide. Chloride • Name the compound. Barium Chloride

  34. Covalent Molecules Use prefixes to designate the number of atoms of each element used in the molecule.

  35. Writing Covalent Formulas Dinitrogen Pentoxide • Write the symbol of each element. N O • Add the subscript as indicated by the prefixes. N2O5

  36. Writing Covalent Formulas Carbon Dioxide • Write the name of each element. C O • Add the subscripts as indicated by the prefixes. CO2

  37. Try these examples on your own. • Sulfur dioxide • Sulfur monoxide • Carbon tetrachloride • Dihydrogen dioxide • Nitrogen triiodide

  38. Naming Covalent Molecules NH3 • Write the name of the first nonmetal using its subscript as a prefix. Nitrogen (No prefix written for the first element IF it is a one.) • Write the name of the second nonmetal using its subscript as a prefix and change the ending to -ide. Trihydride • Name of the molecule: Nitrogen trihydride

  39. Naming Covalent Molecules Pt.2 P2O5 • Write the name of the first nonmetal using the subscript as a prefix. Diphosphorus • Write the name of the second nonmetal using the subscript as a prefix and change the ending to -ide. Pentoxide • Name the molecule: Diphosphorus Pentoxide

  40. Try a few examples on your own. • CO • CO2 • SF2 • PI3 • H2O

  41. Polar and Nonpolar Covalent Bonds • A covalent bond occurs between nonmetals because there is only a slight difference in their electronegativites. (less than 1.7 diff in EN) The atoms share the electrons fairly equally. • If one nonmetal has a greater EN than the other, it can “hog” the electrons, forming a POLAR covalent bond. (>0  <1.7 diff. in EN) • If the nonmetals have equal EN, they will share equally and form NONPOLAR covalent bond. (0 diff. in EN)

  42. Transition MetalsWtg. Formulas / Nmg. Compounds • Most transition metals can form ions with more than one charge. • Examples: Copper atoms can become Cu +1 and Cu +2 ions Iron atoms can become Fe +2 and Fe +3 ions

  43. Writing Formulas w/Transition Metals Iron (III) Oxide • Write the symbol for the transition metal. Ex. Fe • Take the number in parentheses and write it as the oxidation number. Ex. Fe +3

  44. Writing Formulas w/Transition Metals Iron (III) Oxide • Write the symbol for the nonmetal. Ex. O • Look up its oxidation number on the periodic table, and add it to the symbol. Ex. O -2

  45. Writing Formulas w/Transition Metals Fe +3 O –2 • Find the common factor between the two oxidation numbers. In this case = 6 • Decide how many of each ion is needed to make the charge equal to the common factor. In this case 2 Fe and 3 O ions. • Use this number of ions as the subscript for the element, and write the formula. Fe2O3

  46. Copper (I) Sulfide • Write the symbol for the transition metal. Ex. Cu • Take the number in parentheses, and write it as the oxidation number. Ex. Cu +1

  47. Copper (I) Sulfide • Write the symbol for the nonmetal. Ex. S • Look up its oxidation number on the periodic table, and add it to the symbol. Ex. S -2

  48. Copper (I) Sulfide Cu +1 S –2 • Find the common factor between the two oxidation numbers. In this case = 2 • Decide how many of each ion is needed to make the charge equal to the common factor. In this case 2 Cu and 1 S ion. • Use this number of ions as the subscript for the element, and write the formula. Cu2S

  49. Naming Compounds w/Transition Metals FeO • Look up the nonmetal on the periodic table. Find its oxidation number. Oxygen O-2 • Look up the metal on your ion chart. Find the possible oxidation numbers. Fe +2 or Fe +3

  50. Fe +2 or Fe +3 O-2 • Decide which ion will form in the proper ratio with the known charge on the oxygen ion. FeO • Iron bonds in a 1 to 1 ratio with oxygen, therefore, the iron ion must have a +2 charge. • Name the compound, indicating the oxidation number of the metal in parenthesis. Iron (II) Oxide

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