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Chemical Bonds

Chemical Bonds. Attraction between two or more atoms Interaction between valence electrons Ionic bonds Covalent bonds. Valence Electrons. Electrons in the highest (outer) electron level Have most contact with other atoms Known as valence electrons

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Chemical Bonds

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  1. Chemical Bonds • Attraction between two or more atoms • Interaction between valence electrons • Ionic bonds • Covalent bonds LecturePLUS Timberlake

  2. Valence Electrons • Electrons in the highest (outer) electron level • Have most contact with other atoms • Known as valence electrons • Outer shelI of noble gases the outer shell contains 8 valence electrons as s2p6 (except He = 2) Example: Ne 1s22s2 2p6 Ar 1s2 2s2 2p63s2 3p6 LecturePLUS Timberlake

  3. Electron Dot Structures Symbols of atoms with dots to represent the valence-shell electrons 1A 2A 3A 4A 5A 6A 7A 8A H He:      LiBe B  C  N  O : F : Ne :            Na Mg AlSiPS: Cl  : Ar :     LecturePLUS Timberlake

  4. Learning Check B1  A. X would be the electron dot formula for 1) Na 2) K 3) Al   B.  X  would be the electron dot formula  1) B 2) N 3) P LecturePLUS Timberlake

  5. Solution B1  A. X would be the electron dot formula for 1) Na 2) K   B.  X  would be the electron dot formula  2) N 3) P LecturePLUS Timberlake

  6. Octet Rule • An octet in the outer shell makes atoms stable • Electrons are lost, gained or sharedto form an octet • Unpaired valence electrons strongly influence bonding LecturePLUS Timberlake

  7. Formation of Ions from Metals • Ionic compounds result when metals react with nonmetals • Metals loseelectrons to match the number of valence electrons of their nearest noble gas • Positive ionsform when the number of electrons are less than the number of protons Group 1A metals ion 1+ Group 2A metals ion 2+ • Group 3A metals ion 3+ LecturePLUS Timberlake

  8. Formation of Sodium Ion Sodium atom Sodium ion Na  – e Na + 1s22s22p63s1 1s22s22p6 ( = Ne) 11 p+ 11 p+ 11 e- 10 e- 01+ LecturePLUS Timberlake

  9. Formation of Magnesium Ions Magnesium atom Magnesium ion  Mg  – 2e Mg2+ 1s22s22p63s2 1s22s22p6 (Ne) 12 p+ 12 p+ 12 e- 10 e- 0 2+ LecturePLUS Timberlake

  10. Cations of Some Representative Elements Group 1A Group 2A Group 3A H+ Mg2+ Al3+ Li+ Ca2+ Na+ Sr2+ K+ Ba2+ LecturePLUS Timberlake

  11. Learning Check B 2 A. Number of valence electrons in aluminum 1) 1 e- 2) 2 e- 3) 3 e- B. Change in electrons for octet 1) lose 3e- 2) gain 3 e- 3) gain 5 e- C. Ionic charge of aluminum 1) 3- 2) 5- 3) 3+ LecturePLUS Timberlake

  12. Solution B 2 A. Number of valence electrons in aluminum 3) 3 e- B. Change in electrons for octet 1) lose 3e- C. Ionic charge of aluminum 3) 3+ LecturePLUS Timberlake

  13. Learning Check B 3 Give the ionic charge for each of the following: A. 12 p+ and 10 e- 1) 0 2) 2+ 3) 2- B. 50p+ and 46 e- 1) 2+ 2) 4+ 3) 4- C. 15 p+ and 18e- 2) 3+ 2) 3- 3) 5- LecturePLUS Timberlake

  14. Solution B 3 Give the ionic charge for each of the following: A. 12 p+ and 10 e- 2) 2+ B. 50p+ and 46 e- 2) 4+ C. 15 p+ and 18e- 2) 3- LecturePLUS Timberlake

  15. Learning Check B 4 A. Why does Ca form a Ca2+ ion? B. Why does O form O2- ion? LecturePLUS Timberlake

  16. Solution B 4 A. Why does Ca form a Ca2+ ion? Loses 2 electrons to give octet 1s2 2s2 2p6 3s2 3p6 4s2 1s2 2s2 2p6 3s2 3p6 (like Ar) B. Why does O form O2- ion? Gains 2 electrons to give octet 1s2 2s2 2p4 + 2e- 1s2 2s2 2p6 2-8 (like Ne) LecturePLUS Timberlake

  17. Transition Metals • Loses electrons from the highest occupied energy level. • May also lose one or more d electrons. • Most form 2 or more positive ions LecturePLUS Timberlake

  18. Fe2+ and Fe3+ Fe atom: 1s2 2s2 2p6 3s2 3p64s2 3d6 Loss of 4s2 electrons (outermost shell) Fe2+ ion: 1s2 2s2 2p6 3s2 3p6 4s03d6 Loss of 3d1 electron Fe3+ ion: 1s2 2s2 2p6 3s2 3p6 4s03d5 A 3d5 (half-filled) subshell is stable LecturePLUS Timberlake

  19. Some Transition Metal Ions 1+ 2+ 1+ or 2+ 2+ or 3+ Ag+ Cd2+ Cu+,Cu2+ Fe2+, Fe3+ silver cadmium copper(I) ion iron(II) ion ion ion copper (II) ion iron(III) ion Zn2+ zinc ion LecturePLUS Timberlake

  20. Ions from Nonmetal Ions • In ionic compounds, nonmetals in 5A, 6A, and 7A gain electrons from metals • Nonmetal add electrons to achieve the octet arrangement • Nonmetal ionic charge: 3-, 2-, or 1- LecturePLUS Timberlake

  21. Fluoride Ions unpaired electron octet 1 - : F  + e : F :  1s22s22p5 1s22s22p6 = Ne LecturePLUS Timberlake

  22. Some Nonmetal Ions 5A 6A 7A N3 O2 F nitride __________ fluoride P3 S2 Cl ___________ __________ _________ Br  _________ I iodide ion LecturePLUS Timberlake

  23. Some Nonmetal Ions 5A 6A 7A N3 O2 F nitride oxide fluoride P3 S2 Cl phosphidesulfidechloride Br  bromide LecturePLUS Timberlake

  24. Ionic Compounds • Attraction between + ions and - ions • Electrons go from metals to nonmetals electron transfer metal nonmetal ion+ ion-– Electrons lost = Electrons gain LecturePLUS Timberlake

  25. Formulas of Ionic Compounds Formulas of ionic compounds are determined from the charges on the ions atoms ions  – Na  +  F :  Na+ : F :  NaF  sodium fluorine sodium fluoride formula Charge balance: 1+1- = 0 LecturePLUS Timberlake

  26. Writing a Formula Write the formula for the ionic compound that will form between Ba2+ and Cl. Solution: 1. Balance charge with + and – ions 2. Write the positive ion of metal first, and the negative ion Ba2+ ClCl 3. Write the number of ions needed as subscriptsBaCl2 LecturePLUS Timberlake

  27. Learning Check B 5 A. The formula for the ionic compound of Na+ and O2- is 1) NaO 2) Na2O 3) NaO2 B. The formula of a compound of aluminum and chlorine is 1) Al3Cl 2) AlCl2 3) AlCl3 C. The formula of Fe3+ and O2- is 1) Fe3O2 2) FeO3 3) Fe2O3 LecturePLUS Timberlake

  28. Solution B 5 A. The formula for the ionic compound of Na+ and O2- is 2) Na2O B. The formula of a compound of aluminum and chlorine is 3) AlCl3 C. The formula of Fe3+ and O2- is 3) Fe2O3 LecturePLUS Timberlake

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