1. Chemical Bonding I: Basic Concepts Chapter 9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Gilbert Lewis - Atoms combine in order to achieve a more stable electron configuration => Maximum stability is achieved when an atom is isoelectric with a noble gas When atoms interact to form a chemical bond, only the outer regions are in contact => Valence electrons Lewis dot symbols - symbol of element and one dot for each valence electron

3. 9.1 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding. 1A 1ns 1 2A 2ns 2 3A 3ns 2 np 1 4A 4ns 2 np 2 5A 5ns 2 np 3 6A 6ns 2 np 4 7A 7ns 2 np 5 Group# of valence e - e - configuration

4. Focus on main group elements (representative elements) (Groups 1A => 8A)

5. Li + F [Li + ] [ F ] - The Ionic Bond 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 => [He] => [Ne] Li [Li + ] + e - e - + F [ F ] - F - Li + + [Li + ] [ F ] - An ionic bond is the electrostatic force that holds together an ionic compound => large crystals, not individual molecules

6. Ca+ O [Ca 2+ ] [ O ] 2-2- Ionic bonds form when a metal bonds to a non-metal. 1s 2 2s 2 2p 4 [Ar] [Ne] 2 Ca (s) + O 2 (g) 2 CaO (s) Split into oxygen atoms [Ar]4s 2 Transfer 2 electrons from Ca to O

7. Li[Li + ] Sometimes more than 2 atoms are involved 1s 2 2s 1 1s 2 2s 2 2p 4 4 Li (s) + O 2 (g) 2 Li 2 O (s) 2 + O2 [ O ] 2-2- [Ne] [He] 3 Mg (s) + 2 N 2 (g) Mg 3 N 2 (s) Mg + 2 N [Mg 2+ ] [ N ] 3 - 1s 2 2s 2 2p 3 [Ne] [Ne]3s 2 3 32 6 electrons transferred We can predict which compounds will be ionic based on ionization energies and electron affinities. BUT these are measured in the gas phase and ionic compounds are generally solids => very different environment.

8. Lattice energy (E) increases as Q increases and/or as r decreases. cmpd lattice energy MgF 2 MgO LiF LiCl 2957 3938 1036 853 Q= +2,-1 Q= +2,-2 r F < r Cl Electrostatic (Lattice) Energy E = k Q+Q-Q+Q- r Q + is the charge on the cation Q - is the charge on the anion r is the distance between the ions Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Coulomb’s Law More charge Shorter distance Multiple charge Small ions

9. Born-Haber Cycle for Determining Lattice Energy  H overall =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 oooooo Think of this as a variation of a conservation of energy problem. Form ions Form atoms in gas phase Form crystalline lattice Hess’ Law

10. Lattice Energy is always positive!! In a series, as lattice energy increases, so does the melting point.

11. A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Why should two atoms share electrons? FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond Covalent compounds contain only covalent bonds. Share 2 electrons Valence electrons that are NOT involved in bonding (non-bonding)

12. Interactions between molecules in covalent compounds are weaker than bonds in the molecule and in ionic compounds.

13. 8e - H H O ++ O HH O HHor 2e - Lewis structure of water Double bond – two atoms share two pairs of electrons single covalent bonds O C O or O C O 8e - double bonds Triple bond – two atoms share three pairs of electrons N N 8e - N N triple bond or Share 2 e - Share 4 e - Share 6 e -

14. Bond Type Bond Length (pm) C-CC-C 154 CCCC 133 CCCC 120 C-NC-N 143 CNCN 138 CNCN 116 Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond

15. H F F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron rich region electron poor region e - riche - poor ++ --

16. Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest Electronegativity - relative, F is highest X (g) + e - X - (g) Gas phase

17. Electronegativity is a relative measure of the attraction an atom has for the shared electrons in a bond => numerical scale => the higher the number, the stronger the attraction 0.7 1.0 4.0 Willing to give up electrons Wants to fill its orbitals => not measured directly, but calculated using bond energies, atom size, ……..

18. 9.5

20. Non-metal atom electronegativities Metal atom electronegativities > Easier to empty valence shell Easier to fill valence shell Fr [Fr + ] + e - e - + F [ F ] - F is the highest electronegativity because atom is very small => valence shell is close to the nucleus => needs 1 electron to fill orbital Fr is the lowest electronegativity because atom is very large => valence shell does not have much interaction with the nucleus

21. Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - Increasing difference in electronegativity Classification of bonds by difference in electronegativity DifferenceBond Type 0Covalent  2 Ionic 0 < and <2 Polar Covalent Example: FrF F = 4.0 Fr = 0.7 Δ = 3.3 Example: CS 2 S=C=S C = 2.5 S = 2.5 Δ = 0.0

22. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN and NH bonds in H 2 NNH 2. Cs – 0.7Cl – 3.03.0 – 0.7 = 2.3Ionic H – 2.1S – 2.52.5 – 2.1 = 0.4Polar Covalent N – 3.0 3.0 – 3.0 = 0Covalent S HH N – 3.0H – 2.13.0 – 2.1 = 0.9Polar Covalent N = N HH HH Bonds involving 2 atoms of the same element MUST be covalent.

23. 1.Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. 2.Count total number of valence e -. Add 1 for each negative charge. Subtract 1 for each positive charge. 3.Complete an octet for all atoms except hydrogen 4.If structure contains too many electrons, form double and triple bonds on central atom as needed. Writing Lewis Structures Element most willing to share. Start with atoms bonded to the central atom. Add lone pairs to central atom.

24. Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center FNF F Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 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 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 6 electrons20 electrons F F F N

25. Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 – C is less electronegative than O, put C in center OCO O Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -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 C and O atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons Step 5 - Too many electrons, form double bond and re-check # of e - 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24 Double Bond - - - 2 Charge

26. Two possible skeletal structures of formaldehyde (CH 2 O) HCOH H CO H 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. formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. Shared between 2 atoms Extra bond => + formal charge Remove bond => - formal charge

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

28. C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 H CO H formal charge on C = 4 -0 -0 -½ x 8 = 0 formal charge on O = 6 -4 -4 -½ x 4 = 0 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - 00 9.7

29. Formal Charge and Lewis Structures 9.7 1.For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. 2.Lewis structures with large formal charges are less plausible than those with small formal charges. 3.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. Which is the most likely Lewis structure for CH 2 O? HCOH +1 H CO H 00

30. What are the formal charges on the atoms of the carbonate ion (CO 3 2- ) ? OCO O Double Bond - - - 2 Charge - - O3O3 O1O1 O2O2 O 1 6 – 4 -1/2 (4) = 0 O 2 6 – 6 -1/2 (2) = -1 O 3 6 – 6 -1/2 (2) = -1 C 4 – 0 -1/2 (8) = 0 Total of formal charges = 0 -1 -1 0 = -2 = ion charge

31. 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. OOO + - OOO + - OCO O -- OCO O - - OCO O - - What are the resonance structures of the carbonate (CO 3 2 -) ion? Use double headed arrow In reality, all bonds are determined to be equivalent. = Equilibrium

32. C C C CC C H H H H H H C C C CC C H H H H H H Resonance structures – neither structure is correct Animal analogy – Rhinoceros looks like a cross between two mythical animals (griffin & unicorn) => it does not oscillate between the two forms Benzene – C 6 H 6 Kekulé structures In benzene, all C – C bonds lengths are 140 pm Normal C – C bond length is 154 pm and C = C bond length is 133 pm Ring structure

33. Exceptions to the Octet Rule The Incomplete Octet HHBe Be – 2e - 2H – 2x1e - 4e - BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 9.9 Be, B & Al are the most common examples Cannot complete an octet with only 4 electrons

34. Exceptions to the Octet Rule Odd-Electron Molecules N – 5e - O – 6e - 11e - NO N O The Expanded Octet (central atom with principal quantum number n > 2) SF 6 S – 6e - 6F – 42e - 48e - S F F F F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 9.9 Third row and below

35. Expanded Octet Rule Exceptions  Larger atoms can have involvement with unfilled d orbitals  The larger the central atom, the more atoms can surround it (especially if surrounding atoms are small (ex: F)

36. What is we have a choice between an expanded octet with fewer formal charges and one following the octet rule with more formal charges? For 3 rd row and beyond, the preference is for the structure that obeys the octet rule. OSO O -- OSO O - SO 4 2- Satisfies octet rule Most likely structure - - O - OR - - O Expanded octet

37. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) +  H 0 = 436.4 kJ Cl 2 (g) Cl (g) +  H 0 = 242.7 kJ HCl (g) H (g) +Cl (g)  H 0 = 431.9 kJ O 2 (g) O (g) +  H 0 = 498.7 kJ OO N 2 (g) N (g) +  H 0 = 941.4 kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond Cl H Cl H Single bonds Double bond Triple bond

38. Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g)  H 0 = 502 kJ OH (g) H (g) +O (g)  H 0 = 427 kJ Average OH bond energy = 502 + 427 2 = 464 kJ Compare strength of single and multiple bonds.

39. Bond Energies (BE) and Enthalpy changes in reactions  H 0 = total energy input – total energy released =  BE(reactants) –  BE(products) Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. Note: We are not simply taking products - reactants.

40. Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH1436.4 FF 1156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF2568.21136.4  H 0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ HHFFHF+HF+

41. H 2 (g) + Cl 2 (g) 2HCl (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) HH1 436.4 Cl 1 242.7 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) HCl2 431.9  H 0 = 436.4 + 242.7 – 2 x 431.9 = = -184.7 kJ

42. 2H 2 (g) + O 2 (g) 2H 2 O (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) HH2 436.4 1 498.7 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) HO4 464.5  H 0 = (2 x 436.4) + 498.7 – (4 x 464.5) = = -486.5 kJ OO