1. Ch 9 Molecular Bonding WARM-UP: 1. Why do atoms bond? 2. How do they bond? 3. What are the deciding factors in the type of bond? 4. How does the bond type affect structure and shape?

2. Types of Bonds inside a compound Metallic Bond: a sea of electrons move freely between the __cations___ Ionic Bond: electrons are _transferred_ b/c of the large difference in electronegativities Covalent Bond: electrons are shared ; the closer the electronegativity values are to each other the more equal the sharing Polar Covalent Bond: electrons are shared unequally Non-Polar Covalent Bond: electrons are shared EQUALLY!

3. Characteristics of each bond type Metallic crystalline solid (except Hg), high melt & boil points, good conductors of heat and electricity Ionic crystalline solids, high melt & boil points, conductors when molten or aqueous Molecular (Covalent and Polar Covalent) liquids, gases, waxy solids low mp & bp, do NOT conduct elec. forms molecules

4. Determining Bond Type Predict based on Location in Periodic Table Metallic = M+M, Ionic = M+NM, Covalent = NM+NM Difference in Electronegativities Non Polar ≤ 0.4 > Polar Covalent > 1.7 ≤ Ionic Covalent Ex: prediction electronegativities Li I M+NM = I Li=.98 I = 2.66 2.66 -.98 = 1.71 CH 4 NH 3

5. Polar – Inside? Outside? Both? Polar Bond – atom to atom inside a molecule You determine by difference in electronegativity Polar Molecule – the outside of the molecule has slightly + and – ends You determine by: 1.Has a polar bond inside O=3.44 2.Is asymmetrical in shape H=2.2 1.2 = polar covalent bond BENT shape allows for + & - poles

6. Lewis Structures Ionic – show the transferred electron(s) and resulting charges Ex: magnesium bromide MgBr 2 Br Mg Br 2+

7. Lewis Structures for Molecules Show the shared pairs w/ a dash & the lone pears w/ dots 1. least electroneg atom (furtherest left on P. Table) is the CENTRAL atom 2. H is always on the END as are terminal HALOGENS 3. Put DOTS around each atom to represent the VALENCE ELECTRONS 4. Use 1 pair of e- to connect each of the atoms any leftovers go to be LONE PAIRS 5. Make sure outer atoms have 8 ELECTRONS (except H can have 2 ) 6. if central atom still needs 4 pair (8 e-) add DOUBLE or TRIPLE bonds

8. Now practice on your worksheet You can not do a formula for sulfur and oxygen because they are both negative. So just put SO in the blank Please STOP HERE TODAY with the power point

9. Go back one slide

10. STOP and go back Left arrow twice please

11. The other arrow(left)

13. Naming Covalent molecules Use your pre-fixes

14. Lewis Structure Practice lithium fluoridecalcium chloride methane (CH 4 )ammonia (NH 3 ) ethane (C 2 H 6 )ethene (C 2 H 4 ) nitrogen trifluoridecarbon disulfide

15. Polyatomic ions Phosphate ion Coordinate covalent bond = one of the atoms donates both electrons to the shared pair perchlorate ion ammonium ion nitrite ion

16. Resonance Structures Nitrate ion Ozone

17. Molecular Shapes VSEPR - Valence shell electron pair repulsion Look at the central atom 2 bonds and 0 lone pairs = LINEAR 180 o 3 bonds and 0 lone pairs = Trigonal PLANAR 120 o 4 bonds = TETRAHEDRAL 109.5 o

18. More shapes with VSEPR 1. 2 lone pairs are the most repulsive 2. 1 lone pair and 1 shared pair (bond) next most repulsive 3. 2 shared pairs (bonds) least repulsive 3 bonds & 1 unshared (lone) pair PYRAMIDAL 107 o 2 bonds & 2 unshared (lone) pairs BENT 105 o

19. Exceptions to the Octet Rule 1. Odd # of valence electrons (end up w/ x pairs and 1 leftover e-) the one leftover unpaired e- causes the cmpd to be paramagnetic 2.Boron cmpds are sometimes deficient in their octets and will be happy with 6 instead of 8 3. expanded octets: a few cmpds exceed the 8 e- P family sometimes 10 e- S can do 12 e- Xe can do 12 e-

20. Polar – Inside? Outside? Both? Polar Bond – atom to atom inside a molecule You determine by difference in electronegativity Polar Molecule – the outside of the molecule has slightly + and – ends You determine by: 1.Has a polar bond inside O=3.44 2.Is asymmetrical in shape H=2.2 1.2 = polar covalent bond BENT shape allows for + & - poles

21. Polarity Practice CH 4 NH 3 CaO H 2 O BF 3

22. Intramolecular Bonds vs Intermolecular Forces Water molecules have a polar covalent bond between each H and O. The O is more electronegative so it keeps the e- more of the time, resulting in a slight – charge on O & a slight + for the H s. The + of one H 2 O is attracted to the – of another H 2 O

23. Bond Strength Covalent Bonds have a balance between attractive and repulsive forces Bond Length = distance from center of 1 nucleus to center of the other depends on : 1. atom size 2. how many e- pairs are shared shorter bond length = stronger bond _ _ ++ ++

24. Energy and Intramolecular Bonds Energy is given off when a bond forms ( a neg. number) Energy must be added to break a bond and is called bond dissociation energy ( a + number) Ex: 2H 2 + O 2  2H 2 O If nrg added = nrg released  no nrg change overall If add is greater  endothermic If release is greater  exothermic

25. Intermolecular Forces Attraction of one molecule for another molecule weaker than a bond 1. nonpolar – nonpolar (weak) dispersion forces 2. polar – polar (stronger) dipole - dipole 3. Hydrogen bond = H of one molecule attracted to F, O, or N of another molecule

26. Molecular Orbitals Atomic orbitals describe e- location in 1 atom. orbital could be s,p,d,f, or some hybrid of these Molecular orbitals are formed when atomic orbitals overlap as atoms combine to make molecules. Ex: +  overlapping s = σ H H H 2 ss

27. Sigma bond (σ ) formed when 2 atomic orbitals combine to make a molecular orbital that is symmetrical along the axis connecting the 2 atomic nuclei (end to end) +  overlapping s = σ overlapping p ends = σ ss

28. Pi bond ( π ) Bonding e- are most likely found in sausage shaped regions above & below the nuclei They form between p – p orbitals that overlap side to side instead of end to end Ex: diatomic nitrogen N = 1s 2 2s 2 2p 3

29. 1 “s” & 1 “p”  2 “sp” HYBRID Atomic ORBITALS New atomic orbital made by combining s and p The hybrid looks more like a “p” but has both “s” and “p” character.

30. HYBRID ORBITALS In a Carbon cmpd 1s 2s 2p

31. Tetrahedral shape shows 4 equal bonds-- thus s and p orbitals must be hybridized into 4 equal orbitals.

32. Shape determines the hybridization Pyramid, or bent