1. Chemical Bonding I: Basic Concepts Chapter 9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acknowledgement Thanks to the McGraw-Hill companies, Inc. for allowing usage in the classroom.

2. 9.1 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding (or chemical reaction). 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

3. 9.1 Lewis Dot Symbols for the Representative Elements & Noble Gases

4. Rules for Drawing Lewis dot Symbol 1. Count the total number of valence electrons in the atom or ion. For a neutral atom, the group number is the valence electron number; for an ion, subtract each charge. Example: N atom has 5 valence electrons; while N 3- has 5 – (-3) = 5+3 = 8 valence electrons. Mg atom has 2 valence electrons; while Mg 2+ has 2 – (+2) = 2-2 = 0 valence electrons. 2. The number of valence electrons for Noble gases is 8 except for He, which is 2. 3. Assume there is an imaginary and invisible square outside the atom or ion, fill in electrons according to Hund’s rule by spreading electrons out first prior to pair them together. 4. For ions, put atom symbol inside the [ ] and then put charges outside as superscript. The Lewis dot symbols for ions shown in the next slide are not preferred. Example: ¨ [Mg] 2+ [:N:] 3- ¨ There are 4 pairs of paired electrons in the Lewis dot symbol for N 3-. There are 8 paired electrons in the Lewis dot symbol for N 3-. http://www.pogil.org/uploads/media_items/lewis-dot-structures-of-atoms-and-ions.original.pdf http://www.saskschools.ca/curr_content/chem20/trends/atomradm.html

5. What are the correct Lewis dot symbols of (A)F,F -, (B)S,S 2-, (C) O,O 2-, (D) N,N 3- ?

6. 9.2 The Ionic Bond Li+ F Li + F - 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 - Chemical Bonds Two general types: Ionic bonds---- between ions of opposite charges Covalent bonds---- sharing of electrons

7. The Octet Rule Atoms tend to gain, lose, or share electrons until they are surrounded by 8 valence electrons. An octet = 8 electrons = 4 pairs of electrons = noble gas arrangement = very stable Exception: H = 2 electrons only Example: Na [Ne]3s 1 loses 1 e- ----> Na + [Ne] Cl[Ne]3s 2 3p 5 gains 1 e- --> Cl - [Ar]

8. 9.3 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. E lattice LiF(s) Li + (g) + F - (g) Lattice Energy E lattice increases as charge (Q + Q - ) on ions increases E lattice decreases as inter-ionic distances increase (charge is much more important than distance)

9. Which of the following is a correct order of lattice energy? (a)NaCl < AlCl 3 < MgCl 2 (b) LiF < LiCl < LiBr (c) Na 2 O < MgO < Al 2 O 3 (d) Ga 2 O 3 < CaO < K 2 O (e) LiCl < NaCl < KCl Hint: Lattice Energy E lattice increases as charge (Q + Q - ) on ions increases E lattice decreases as inter-ionic distances increase (Charge effect is much more important than distance effect)

10. A covalent bond is a chemical bond in which two or more electrons are shared by two nonmetal atoms. Why should two atoms share electrons? To satisfy the octet rule. FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond 9.4

11. 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 9.4

12. Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond 9.4 Bond strength (or bond enthalpy): See Table 9.4 triple > double> single

13. H F F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two nonmetal atoms. Thus, polarity can not be used in the ionic bonds. electron rich region electron poor region e - riche - poor ++ -- 9.5 Bond polarity Nonpolar bond: two identical atoms form a bond & equally share the bond’s electron pair. This is the definition used in CHEM 1411. Electron density is pulled more tightly around the F end of the molecule.

14. Rules for Bond Polarity 1. Bond polarity can be calculated by taking the electron negativity difference (E. D.) between two atoms. The difference is either zero or positive. It’s always done by taking the bigger value – smaller value. 2. The greater the difference, the greater the bond polarity. 3. General rules: (a) E.D. < 0.5  nonpolar bond (b) If 0.5 = E.D. < 1.9 (or 2.0)  polar bond (c) 1.9 (or 2.0) = E.D. or greater  ionic bond Note: Other chemists use different numbers. Recall metal and nonmetal atoms form an ionic bond. See slide titled as Classification of bonds by difference in electronegativity 4. In CHEM 1411, if the E.D. is zero, then the bond is nonploar; if it’s nonzero and there is no metal involved, then it’s a polar bond.

15. Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable Electronegativity – defined, relative, F is the most electronegative X (g) + e - X - (g) 9.5 the term used to describe the relative attraction of an atom for the electrons in a bond Atom of the bond with the greater electronegativity will carry the partial negative charge

16. Example: Which of the following is a polar covalent bond? (a)The HN bond in NH 3. (b) The SiSi bond in Cl 3 SiSiCl 3. (c) The CaF bond in CaF 2. (d) The OO bond in O 2. (e) The NN bond in H 2 NNH 2. Hint: The reason that (C) is not a polar covalent bond is that it is actually an ionic bond.

17. 9.5 The Electronegativities of Common Elements

18. Some Important Rules for Electronegativity: Rememorize the periodic trend. The most electronegative element: F (4.0) Next most electronegative element: O (3.5) Next two most electronegative: N (3.0) and Cl (3.0) Least electronegative element: Cs (0.7) Also H (2.1) C (2.5) Br (2.8) Na (0.9)

19. Non polar 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 0Non polar Covalent  2 2 Ionic 0 < and <2Polar Covalent 9.5 H-HNaClHCl

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 = 0Non polar Covalent 9.5 Practice problem of chap9:Q11

21. 1.Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. Arrange atoms & connect with single bonds – Single bond = 2 electrons 2. Count total number of valence e -. Add 1 for each negative charge. Subtract 1 for each positive charge. 3. Complete an octet for atoms bonded to the central atom except hydrogen (H=2). Then add leftover electrons to central atom. Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding. 4. If the central atom has fewer than 8 electrons, try adding double and triple bonds on central atom as needed. Writing Lewis Structures Can be done two ways: (1) Lewis dot symbol method used in section 9.4 and (2) the method introduced here, section 9.6, the formula or math method 9.6

22. 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 9.6

23. Write the Lewis structure of the carbonate ion (CO 3 2- ). Practice the nitrate ion carbonate ion (NO 3 - ). 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 9.6 Step 5 - Too few 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

24. 9.7 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. FC = [ Val. e - ] -[(nonbonded e - ) + 1/2(bonded e - )] Remember: Formal charges are NOT real charges; however, the sum of all formal charges is the true charge on the ion (positive or negatively charged) or molecule/compound (zero charge or electrically neutral). ** Recall oxidation numbers are NOT real charges; however, the sum of all oxidation numbers is the true charge on the ion (positive or negatively charged) or molecule/compound (zero charge or electrically neutral). See Chapter 4.

25. Rules for Formal Charges 1. Formal charges are used for book-keeping only and they are NOT the real charges; however, the sum of all formal charges is the true charge on the ion (positive or negatively charged) or molecule/compound (zero charge or electrically neutral). 2. Formal charges can assist in determining which Lewis structure is the best choice. This requires reasoning from electron negativity view point. That is, more electronegative atom gets negative formal charge; while less electronegative atom gets the positive formal charge. 3. The best Lewis structure is the one without charge separation; if charge separation is unavoidable, then the value should be as small as possible. 4. If the Lewis structure is written wrongly, then the formal charges will be wrongly accordingly and vise versa.

26. Practice problem of chap9:Q12

27. Resonance Structures Why necessary? Lewis structures are not adequate More than one Lewis Structure is needed to represent the real molecule 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 + - Example: experimentally, ozone has two identical bonds whereas the Lewis Structure requires one single (longer) and one double bond (shorter). Sulfur dioxide has resonance structures, similar to ozone.

28. Resonance Structures: connect different resonance structures with actual molecule is not depicted by resonance structures actual molecule does not go between resonance structures Molecules that can be depicted by resonance structures are more stable than expected 9.8 Resonance in Benzene: C 6 H 6

29. OCO O -- OCO O - - OCO O - - 9.8 What are the resonance structures of the carbonate (CO 3 2 -) ion? Which of the following has resonance structure? (a) NH 4 + (b) CO 2 (c) AlI 3 (d) C 6 H 6 (e) H 2 O

30. Exceptions to the Octet Rule The Incomplete Octet (too few electrons on central atom) 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

31. Exceptions to the Octet Rule Odd-Electron Molecules Result: unpaired electron N – 5e - O – 6e - 11e - NO N O The Expanded Octet (too many e- on central atom; 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

32. Which of the following is correct concerning the lone pairs on the underlined atoms in compounds? (a)ICl, 3(b) H 2 S, 6(c) CH 4, 4 (d) CaH 2, 2(e) SCl 2, 4 Which of the following obeys the octet rule? (a)BF 3 (b) SF 4 (c) NO (d) PF 5 (e) NO 3 - FBF F

33. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy (always positive). 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 9.10 Bond energy

34. 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. Do NOT confuse with the calculation method made by enthalpy of formation, which is done by using  products –  reactants. 9.10 Endothermic reaction Exothermic reaction

35. 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 9.10