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

1. Identify the valence electrons for all representative elements. 2. Rationalize why alkali metals and alkaline earth metals usually form cations and oxygen and the halogens usually form anions using Lewis dot symbols in the discussion. 3. Use Lewis dot symbols to show the formation of both ionic and molecular compounds. 4. Define lattice energy, Coulomb ’ s law and the Born-Haber cycle. 5. Demonstrate how the Born-Haber cycle is an application of Hess ’ s law and use the Born-Haber cycle to determine lattice energy for an ionic solid. 6. Identify covalent compounds, the type of covalent bonds present, and the number of lone pairs of electrons using Lewis structures.

7. Relate types of bonds to bond length and bond strength. 8. Compare and contrast various properties expected for ionic compounds versus covalent compounds. 9. Identify ionic, polar covalent and (nonpolar) covalent bonds using the concepts of electronegativity. 10. Predict the relative changes in electronegativity with respect to position on the periodic table. 11. Use the concept of electronegativity to rationalize oxidation numbers. 12. Use Lewis dot and the octet rule to write Lewis structures of compounds and ions.

13. Apply the concept of formal charge to predict the most likely Lewis structure of a compound. 14. Explain how Lewis structures are inadequate to explain observed bond length (bond types) in some compounds and how the concept of resonance must be invoked. 15. Recall several common examples in which the octet rule fails. 16. Demonstrate, using Lewis structures, the formation of a coordinate covalent (dative) bond. 17. Use Lewis structures and bond energies to predict heats of reaction. 18. Rationalize why enthalpy change for breaking chemical bonds is positive and the formation of chemical bonds is negative.

9.1 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding. 1A1A 1ns 1 2A2A 2ns 2 3A3A 3ns 2 np 1 4A4A 4ns 2 np 2 5A5A 5ns 2 np 3 6A6A 6ns 2 np 4 7A7A 7ns 2 np 5 Group# of valence e - e - configuration

9.1

9.2 Li + F Li + F - The Ionic Bond 1s 2 2s 1 1s 2 2s 2 2p 5 1s21s2 1s 2 2s 2 2p 6 [He][Ne] Li Li + + e - e - + FF - F - Li + + Li + F -

9.3 Lattice energy (E) increases as Q increases and/or as r decreases. cmpd lattice energy MgF 2 MgO LiF LiCl 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.

9.3 Born-Haber Cycle for Determining Lattice Energy  H overall =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 oooooo

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-7e- 7e-7e- FF 8e-8e- 8e-8e- F F F F Lewis structure of F 2 lone pairs single covalent bond 9.4

8e-8e- H H O ++ O HH O HHor 2e-2e- 2e-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-8e- 8e-8e- 8e-8e- double bonds Triple bond – two atoms share three pairs of electrons N N 8e-8e- 8e-8e- N N triple bond or 9.4

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 9.4

Comparison of Ionic and Covalent Compounds 9.4

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 ++ -- 9.5

Electronegativity is the ability of an atom to attract electrons toward itself in a chemical bond. Electron Affinity - measurable, Cl is highest Electronegativity - relative, F is highest X (g) + e - X - (g) 9.5

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 9.5

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 – 0.7Cl – – 0.7 = 2.3Ionic H – 2.1S – – 2.1 = 0.4Polar Covalent N – – 3.0 = 0Covalent 9.5

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 9.6

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 octet 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 9.6

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 (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 9.6 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

9.7 Two possible skeletal structures of formaldehyde (CH 2 O) HCOH H CO H An atom’s formal charge is the difference between the 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.

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 = ½ x 6 = -1 formal charge on O = 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

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 = ½ x 8 = 0 formal charge on O = 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

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. 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

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?

Exceptions to the Octet Rule The Incomplete Octet HHBe Be – 2e - 2H – 2x1e - 4e-4e- BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total =

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) = lone pairs (18x2) = 36 Total =

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 = kJ Cl 2 (g) Cl (g) +  H 0 = kJ HCl (g) H (g) +Cl (g)  H 0 = kJ O 2 (g) O (g) +  H 0 = kJ OO N 2 (g) N (g) +  H 0 = kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10

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 = = kJ 9.10

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. 9.10

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) HH FF Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF  H 0 = – 2 x = kJ 9.10