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Chemistry Unit 5

Chemistry Unit 5. Chemical Bonding. 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 set of valence electrons. (usually 8). Three Main Types of Bonds.

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Chemistry Unit 5

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  1. Chemistry Unit 5 Chemical Bonding

  2. 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 set of valence electrons. (usually 8)

  3. 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 metal atoms

  4. Properties of Metallic, Molecular and Ionic Compounds

  5. 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.

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

  7. 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

  8. 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 transefer? No, Al must donate two more… where?

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

  10. 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

  11. 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 EN that it removes the metal’s valence electron. Electrons are transferred.

  12. For Example: Na and Cl EN of Na = 0.9 EN of Cl = 3.0

  13. Why does Sodium and Oxygen form an ionic bond? 3.0 EN of O - 0.9 EN of Na 2.1 Difference in EN • Difference in electronegativity is 2.1(>1.7) • An ionic bond will form. • Chlorine has a greater electronegativity, and is able to yank electrons away from sodium.

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

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

  16. Let’s Wrap it Up… Again! • Covalent bonds are held together by a mutual need for the shared electrons (electronegativity) Their orbits overlap. Each electron is attracted to the positive charge of the opposite nucleus. • The result of a covalent bond is called a molecule. • Covalent bonds form between two nonmetals due to a small (or no) difference in EN. (less than 1.7) • Neither atom’s EN is strong enough to remove the other atom’s electrons. Electrons are shared.

  17. Polar and Nonpolar Covalent Bonds • If one nonmetal has a greater EN than the other, it can “hog” the shared electrons. This forms a POLAR covalent bond. (EN difference greater than 0, but less than 1.7) • If the nonmetals have the same EN, they will share equally and form NON POLAR covalent bond. (0 diff. in EN)

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

  19. 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-)

  20. 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-)

  21. O C O Molecular Geometry • Linear molecules: atoms are connected in a straight line. • All molecules with only 2 atoms are linear. • Many molecules with 3 atoms are also linear. • Ex. O2, HCl, CO2

  22. O Molecular Geometry • Bent: bonded atoms have a bent shape due to unshared pairs of electrons. • Unshared electron pairs exert a greater repulsion force than the electron pairs in the bonds. • Ex. H2O, NH3 H H

  23. Molecular Geometry • Tetrahedral: one atom bonded to four other atoms. • The angle between any two bonds is 109.5o. • Ex. CH4 (methane) H C H H H

  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. Try these examples on your own. • BeF • Li20 • B2S3 • Mg3N2 • CaCl2

  35. 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 • Therefore, the oxidation number for the metal will be given to you as a roman numeral in the name of the compound.

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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

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

  43. 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. (Fe+2) • Name the compound, indicating the oxidation number of the metal in parenthesis. Iron (II) Oxide

  44. Fe2O3 • 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

  45. Fe2O3 • Decide which ion will form in the proper ratio with the known charge on the oxygen ion. Fe +2 or Fe +3 • Iron bonds in a 2 to 3 ratio with oxygen. Three oxygen atoms will have a charge of -6. Therefore, two iron ions must equal +6. It must be Fe +3. • Name the compound, indicating the oxidation number of the metal in parenthesis. Iron (III) Oxide

  46. Polyatomic IonsWriting Formulas / Naming Compounds • A polyatomic ion is a covalent molecule that has an ionic charge. (As opposed to being a neutral molecule.) • Poly = many • Atomic = atoms • Ion = charged particle • A charged particle that consists of more than one atom.

  47. Polyatomic Ions Examples: Sulfide Sulfate = SO4-2 Nitride Nitrate = NO3-1 Phosphide Phosphate = PO4-3 Chloride  Chlorate = ClO3-1 • Notice the ending has changed to –ate.

  48. Polyatomic Ions Examples: Sulfide Sulfite = SO3-2 Nitride Nitrite = NO2-1 Phosphide Phosphite = PO3-3 Chloride  Chlorite = ClO2-1 • Notice the ending has changed to –ite.

  49. Polyatomic ions • Not all polyaomic ions end in -ate or -ite. • Some other examples: Ammonium NH4+1 Hydroxide OH-1 • Some Polyatomic ions contain more than two elements. Ex. Acetate = C2H3O2-1

  50. Calcium Phosphite • Write the symbol for the metal. Add the oxidation number from the periodic table. Ca+2 • Write the formula for the polyatomic ion from the ion chart. Add its oxidation number. PO3-3

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