1. Chemical Bonding I: Basic Concepts Chapter 9 Gilbert N. Lewis

2. 화학결합 I: 공유결합 Lewis 점기호 이온결합 이온결합화합물과 격자에너지 공유결합 전기음성도 Lewis 구조의 표기 형식전하와 Lewis 구조 공명의 개념 팔전자규칙의 예외 결합엔탈피

3. Valence electrons ( 원자가전자 )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. Lewis Dot Symbols for the Representative Elements & Noble Gases

5. Li + F Li + F - The Ionic Bond( 이온결합 ) 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 [He][Ne] Li Li + + e - e - + FF - F - Li + + Li + F -

6. Electrostatic (Lattice) Energy E = k z+z-z+z- r z + is the charge on the cation z - 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.

7. 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.

8. Born-Haber Cycle for Determining Lattice Energy  H overall =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 oooooo 1.570 A 6.327 D 84% ionic o = -Lattice Energy =  H f o

10. Chemistry In Action: Sodium Chloride Mining SaltSolar Evaporation for Salt

11. A covalent bond ( 공유결합 ) is a chemical bond in which two or more electrons are shared by two atoms. Q. Why should two atoms share electrons? FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond

12. H H O ++ O HH O HHor Lewis structure of water Double bond – two atoms share two pairs of electrons single covalent bonds O C O or O C O double bonds Triple bond – two atoms share three pairs of electrons triple bond N N N Nor

13. 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( 전기음성도 ) is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest Electronegativity(Mulliken)  (IE 1 + EA) X (g) + e - X - (g) Electronegativity - relative, F is highest

17. The Electronegativities of Common Elements

18. Variation of Electronegativity with Atomic Number

19. 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 0 Covalent  2 Ionic 0 < and <2Polar Covalent

20. 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 – 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 Cs Cl H S N

21. 1. 총 valence 전자수를 산출한다. 2.Octet 을 이루기 위한 전자수를 산출한다. 수소 종류의 경우 2, 다른 원소의 경우 8 개이다. 3. 공유 전자수 = 2 의 결과 - 1 의 결과. 4. 먼저 각 원자 사이에 하나의 결합을 배당한다. 5. 나머지 결합 전자를 이중 또는 삼중결합으로 배당. 6. 나머지 비결합 전자를 옥테트를 이루게 짝으로 배당한다. 7. 형식전하를 계산한다. 형식전하의 합이 총 전하량과 같아야 한다. Lewis Structures 그리기

22. Q. Write the Lewis structure of nitrogen trifluoride (NF 3 ). FNF F Step 1 : valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 2 : 8x4=32. Step 3 : 32- 26 = 6 bonding electrons = 3 bonding pairs Step 4 : used 6 bonding electrons. 20 electrons remain. Step 6 : allocate 20 non-bonding electrons. Step 7 : formal charges 0+0+0+0 = 0.

23. Q. Write the Lewis structure of the carbonate ion (CO 3 2- ). OCO O Step 1 : 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 2 : 8x4 = 32 Step 3 - 32-24 = 8 bonding electrons = 4 bonding pairs Step 4 : use 3 pairs Step 5 : use remaing one pair to make a double bond. Step 6 : allocate 16 non-bonding electrons Step 7 : -1 + -1 + 0 + 0 = -2

24. Formal Charge( 형식전하 ) and Lewis Structures 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.

25. A resonance structure( 공명구조 ) 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 - - Q. What are the resonance structures of the carbonate (CO 3 2 -) ion?

26. Exceptions to the Octet Rule 1. The Octet-Deficient Molecule BF 3 FBF F FBF F +1

27. Exceptions to the Octet Rule 2. Odd-Electron Molecules N – 5e - O – 6e - 11e - NO N O 3. The Expanded Octet (central atom with principal quantum number n > 2) SF 6 S F F F F F F ClP O POCl 3

28. 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

29. 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

31. 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.

32. H 2 (g) + Cl 2 (g) 2HCl (g)2H 2 (g) + O 2 (g) 2H 2 O (g)

33. Q. 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

34. 다음 분자의 Lewis 구조를 모든 공명구조와 함께 그리고, 모든 원자의 형식전하를 나타내시오. (3) BF 3 (B 중심원자 ) (2) POCl 3 (P 중심원자 ) (1) H 2 SO 4 (S 중심원자 )