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Chemical Bonding I: The Covalent Bond

Chemical Bonding I: The Covalent Bond. Chapter 9. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs – 0.7. Cl – 3.0. 3.0 – 0.7 = 2.3. Ionic. H – 2.1. S – 2.5. 2.5 – 2.1 = 0.4. Nonpolar covalent.

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Chemical Bonding I: The Covalent Bond

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  1. Chemical Bonding I:The Covalent Bond Chapter 9

  2. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2. Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Nonpolar covalent N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent 9.5

  3. Lewis structures H H H H O O A Lewis structure is a representation of covalent bonding in which shared electrons pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms.

  4. Writing Lewis Structures • Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. • Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge. • Use valence e- determined in (2) to complete octets of atoms bonded to central atom (except hydrogen) • If central atom has less than an octet, form double and triple bonds on central atom as needed. 9.6

  5. Write the Lewis structure of nitrogen trifluoride (NF3). F N F F Step 1 – N is less electronegative than F, put N in center Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Does central atom have an octet? yes 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6

  6. Write the Lewis structure of the carbonate ion (CO32-). O C O O Step 1 – C is less electronegative than O, put C in center • Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) • -2 charge – 2e- 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on O atoms. Step 4 - Does central atom have an octet? no Step 5 - Form double bond to give central atom an octet 9.6

  7. Write the Lewis structure of the carbonate ion (CO32-). O O C C O O O O Step 1 – C is less electronegative than O, put C in center • Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) • -2 charge – 2e- 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on O atoms. Step 4 - Does central atom have an octet? no Step 5 - Form double bond(s) to give central atom an octet 9.6

  8. H H C O H C O H # bonds formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 Two possible skeletal structures of formaldehyde (CH2O) An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. 9.7

  9. H C O H C – 4 e- O – 6 e- 2H – 2x1 e- 12 e- # bonds formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 single bonds (2x2) = 4 2 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 -1 +1 formal charge on C = 4 - 2- ½ x 6 = -1 formal charge on O = 6 - 2- ½ x 6 = +1 9.7

  10. C – 4 e- H O – 6 e- C O H 2H – 2x1 e- 12 e- formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 single bonds (2x2) = 4 2 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 0 0 formal charge on C = 4 - 0- ½ x 8 = 0 formal charge on O = 6 - 4- ½ x 4 = 0 9.7

  11. Which is the most likely Lewis structure for CH2O? H C O H H C O H -1 +1 0 0 Formal Charge and Lewis Structures • For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. • Lewis structures with large formal charges are less plausible than those with small formal charges. • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. 9.7

  12. Back to CO2- O C O O Figure out which atoms have the charge.

  13. Write Lewis structures for the following: • Draw skeletal structure of compound showing what atoms are bonded to each other. In general, least electronegative element goes in the center. • Hydrogen and fluorine usually occupy the periphery (terminal positions in the Lewis structure). Sometimes the higher symmetry structure is the correct one. HCN BF3 CHFO SnO2 These correct Lewis structures should help you answer 9.23.

  14. - - + + O O O O O O A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. Expect one O-O bond to be longer than the other if ozone really does contain one double bond and one single bond. O-O : 148 pm O=O : 121 pm Experimental evidence shows that bond oxygen-to-oxygen bonds are equal in length (128 pm). 9.8

  15. What are the resonance structures of the carbonate (CO32-) ion? O O O C C C O O O - - - - O O O - - 9.8

  16. Be – 2e- 2H – 2x1e- H Be H 4e- B – 3e- 3 single bonds (3x2) = 6 3F – 3x7e- F B F 24e- Total = 24 9 lone pairs (9x2) = 18 F Exceptions to the Octet Rule The Incomplete Octet BeH2 BF3 9.9

  17. N – 5e- S – 6e- N O 6F – 42e- O – 6e- 48e- 11e- F 6 single bonds (6x2) = 12 F F Total = 48 S 18 lone pairs (18x2) = 36 F F F Exceptions to the Octet Rule Odd-Electron Molecules Odd-electron molecules are sometimes called radicals. Many radicals are highly reactive. NO The Expanded Octet (central atom with principal quantum number n > 2) SF6 9.9

  18. Write Lewis structures for the following: HCN BF3 CHFO SnO2

  19. Exceptions to the Octet Rule The Incomplete Octet BeH2 BX3 AlX3 Odd-Electron Molecules NO NO2 The Expanded Octet (central atom with principal quantum number n > 2) SF6 PF6 9.9

  20. Stability of molecules based on their covalent bond energies…

  21. DH0 = 436.4 kJ/mol H2 (g) H (g) + H (g) Cl2 (g) Cl (g) + Cl (g) DH0 = 242.7 kJ/mol DH0 = 431.9 kJ/mol HCl (g) H (g) + Cl (g) DH0 = 498.7 kJ/mol O2 (g) O (g) + O (g) O DH0 = 941.4 kJ/mol N2 (g) N (g) + N (g) O Bond Energies Single bond < Double bond < Triple bond N N The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. Bond Energy 9.10

  22. DH0 = 502 kJ/mol H2O (g) H (g) + OH (g) DH0 = 427 kJ/mol OH (g) H (g) + O (g) = 464 kJ/mol Average OH bond energy = 502 + 427 2 Average bond energy in polyatomic molecules 9.10

  23. Bond energies in thermochemistry • Energy is always required to break a bond. • Energy is always released when a bond is formed. • Whether a reaction is exothermic depends on the relative strengths of the bonds in the product(s) compared to those in the starting material(s).

  24. Bond Energies (BE) and Enthalpy changes in reactions Imagine a reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. DH0 = total energy input – total energy released = SBE(reactants) – SBE(products) endothermic exothermic 9.10

  25. 2H2 (g) + O2 (g) 2H2O (g) H2 (g) + Cl2 (g) 2HCl (g) 9.10

  26. Use bond energies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ/mol) 1 436.4 436.4 1 156.9 156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ/mol) H H F H F F 2 568.2 1136.4 DH0 = SBE(reactants) – SBE(products) DH0 = 436.4 + 156.9 – 1136.4 = -543.1 kJ 9.10

  27. Ex. 9.11 Estimate the enthalpy change for the combustion of hydrogen gas. -469 kJ/mol

  28. 9.50 (a) For the reaction 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g) Predict the enthalpy of reaction from the average bond energies in Table 9.2. (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 2) of the reactant and product molecules, and compare the result with your answer for (a). (a) -2759 kJ/mol (b) -2855 kJ/mol slightly different b/c (a) used average bond energies

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