1. CHEMICAL BONDING LEWIS THEORY OF BONDING

3. Results from the transfer of electrons from a metal to a non- metal. A chemical bond between oppositely charged ions Held together by electrostatic attraction

4. IONIC BONDING

5. Formed when an orbital from 2 different atoms overlap Electrons must have opposite spins

6. COVALENT BONDING

7. CHEMICAL BONDS Bond Type Single Double Triple # of e’s 2 4 6 Notation — =  Bond order 1 2 3 Bond strength Increases from Single to Triple Bond lengthDecreases from Single to Triple

8. TYPES OF BONDING CONDITIONS BETWEEN ELEMENTS Low Electronegativity and low Ionization energy (Metals) High electronegativity and High Ionization energy (Non-metals) Low Electronegativity and low Ionization energy (Metals) Metallic bondingIonic bonding (transferring of electrons between atoms) High electronegativity and High Ionization energy (Non-metals) Ionic bondingCovalent bonding (sharing of electrons between atoms)

9. Electronegativity

10. LEWIS DIAGRAMS OR STRUCTURES A convention developed to “show” the relationship between atoms when they form bonds. Why is it necessary? predict where the electrons are in a molecule needed to predict the shape of a molecule

11. LEWIS DIAGRAMS FOR IONIC COMPOUNDS  Identify the number of valence shell electrons and determine the charge on the ion using the “stable octet rule”.  Write the elemental symbol, place dots to represent the electrons in the valence shell, enclose in square brackets and write the ionic charge as a superscript. [Na] + or [ Cl ] -

12. STRUCTURAL DIAGRAMS FOR COVALENT COMPOUNDS Draw the Lewis Diagram for nitrogen trifluoride (NF 3 ). Step 1. Count the valence electrons N = 5 F = 7 5 + 3( 7) = 26 valence electrons

13. STRUCTURAL FORMULA FOR COVALENT COMPOUNDS Step 2. Write a skeletal structure. Use the least electronegative atom in the centre Electronegativity: N = 3.0 & F = 4.0 FF F N = a pair of e - (a single bond)

14. Step 3. Complete the octets for each terminal atom (except H) FF F N    

15. Step 4. Assign any additional electrons as lone pairs on the central atom  FF F N   

16. Example 2. COCl 2 (24 electrons) CCl O    

17. Step 5. Make multiple bonds where necessary to complete the octets.  CCl O      CCl O    

18. Example 3. Chlorate ion, ClO 3 - ((1 x 7) + (3 x 6) + 1) = 26 ClOO O  :   : 

19. In some covalent compounds, the bonds between atoms occur because one atom has donated both electrons to the covalent bond. This is called a coordinate covalent bond. N : H H H H+H+ + N H H H H + Nitrogen supplies the two lone pair electrons to this N-H bond. The H + ion has no electrons.

20. To determine the number of coordinate covalent bonds – subtract the bonding capacity (lone valence electrons) from the number of bonds the atom has. N H H H H + Nitrogen Bonds  4 Bonding capacity  3 Coordinate bonds  4-3=1

21. In some compounds the SCH3U guidelines may not “work”. On occasion, both elements have the same electronegativity or there may be two or more possible Lewis Structures. e.g. CS 2 (both electronegativities = 2.5) is it S=C=S or C=S=S ? Exceptions to the Octet Rule

22. In such situations, one determines the Formal Charge. The option with the lowest formal charge has the most stable and viable structure. The Formal Charge for an atom is the number of valence electrons in the free neutral atom minus the number of valence electrons assigned to the atom in the Lewis structure.

23. Formal Charge = (# valence electrons)-(# of bonds)-(# of unshared e - )

24. CSS Valence electrons466 Electrons assigned646 Formal Charge-220 C=S=S  

25. SCS Valence electrons646 Electrons assigned646 Formal Charge000 S=C=S  

26. In some structures the Lewis structure does not represent the true structure of the compound. Bond order is the number of shared pairs of electrons between two atoms. (i.e. – the number of bonds between two atoms) As the bond order increases... The length of the bond decreases. The energy associated with breaking the bond increases.

27. The number of shared electrons in a bond affects its length and energy. Bond TypeBond Order Bond Length (pm) (10 -12 m) Bond Energy (kJ/mol) C-O1143351 C=O2121745 C-C1154348 C=C2134615 C 3120812 C-N1143276 C=N2138615 C N3116891

28. CHO 2 - is a polyatomic ion with the following Lewis structure.  The C-O bond lengths are experimentally determined to be between C-O and C=O. The bond order is neither 1 or 2, but considered to be somewhere in between (i.e.- 1.5).  The Lewis structure does not support the experimental data. C H O O

29. The actual structure is a resonance hybrid of the two resonance structures. C H O O C H O O The resonance structure does not “flip-flop” back and forth between the two. It is a hybrid form of the two.

30. The bond order for NO 3 - is, (1+1+2)/3=1.33. The resonance structure is... N O O O N O O O N O O O

31. CC C CC C H H H HH H CC C CC C H H H HH H Benzene has a bond order of 1.5. (1+2+1+2+1+2)/6=1.5

32. The work on quantum theory in conjunction with the success of Lewis structures resulted in the inevitable connections between the two areas of study. Linus Pauling, a friend of Gilbert Lewis, connected the two with the valence bond theory.

33. PRACTICE COMPLETE THE LEWIS DIAGRAM AND CHEMICAL BONDING WORKSHEETS

34. e-pairsNotationName of VSEPR shapeExamples 2AX 2 LinearHgCl 2, ZnI 2, CS 2, CO 2 3AX 3 Trigonal planarBF 3, GaI 3 AX 2 ENon-linear (Bent)SO 2, SnCl 2 4AX 4 TetrahedralCCl 4, CH 4, BF 4 - AX 3 E(Trigonal) PyramidalNH 3, OH 3 - AX 2 E 2 Non-Linear (Bent)H 2 O, SeCl 2 5AX 5 Trigonal bipyramidalPCl 5, PF 5 AX 4 EDistorted tetrahedral (see-sawed) TeCl 4, SF 4 AX 3 E 2 T-ShapedClF 3, BrF 3 AX 2 E 3 LinearI 3 -, ICl 2 - 6AX 6 OctahedralSF 6, PF 6 - AX 5 ESquare PyramidalIF 5, BrF 5 AX 4 E 2 Square PlanarICl 4 -, BrF 4 -