Bonding – General Concepts

1. Bonding – General Concepts Boom

2. bonds form when the potential energy of the system will be lowered chemical bond – mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together Chemical Bonds

3. Sodium Chloride Crystal Lattice Ionic compounds form solids at ordinary temperatures. Ionic compounds organize in a characteristic crystal lattice of alternating positive and negative ions. Solid sodium chloride has a melting point ~ 800oC WOW!

4. Electronegativity: The ability of anatom in a molecule to attract shared electrons to itself.

5. Do we get it???? Let us see! Order the following bonds according to their Polarity (lowest to highest): H-H, O-H, Cl-H, S-H and F-H. (Table on pg. 334) H-H < S-H < Cl-H < O-H < F-H 0 0.4 0.9 1.4 1.9 POLARITY INCREASES

6. Ionic Bonds • Electrons are transferred • Electronegativity differences are generally greater than 1.7 • The formation of ionic bonds is always exothermic!

7. Determination of Ionic Character Electronegativity difference is not the final determination of ionic character Compounds are ionic if they conduct electricity in their molten state

8. Coulomb’s Law “The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers”

10. Sodium Chloride Crystal Lattice ΔH……Again?

11. Estimate Hf for Sodium Chloride Na(s) + ½ Cl2(g)  NaCl(s) Na(s)  Na(g) + 109 kJ Na(g)  Na+(g) + e- + 495 kJ ½ Cl2(g)  Cl(g) + ½(239 kJ) Cl(g) + e- Cl-(g) - 349 kJ Na+(g) + Cl-(g)  NaCl(s) -786 kJ Na(s) + ½ Cl2(g)  NaCl(s) -412 kJ/mol

12. Bond Length Diagram

13. Covalent Bonding Forces • Electron – electron • repulsive forces • Electron – proton attractive forces • Proton – proton repulsive forces

14. Covalent Bonds Polar-Covalent bonds • Electrons are unequally shared • Electronegativity difference between .3 and 1.7 Nonpolar-Covalent bonds • Electrons are equally shared • Electronegativity difference of 0 to 0.3 Electrostatic Potential Diagram

15. uneven distribution of charge Polarity Dipole Moment • a molecule with a imbalance in electrons has a dipole moment • polyatomic molecules have dipole moments if their polar bonds don’t cancel out

16. Determine whether each of the following bonds will be: ionic, polar covalent, OR nonpolar covalent Practice • Cl and O • 3.5-3.0=0.5 • polar covalent • Cl and Br • 3.0-2.8 • nonpolar covalent • S and H • 2.5-2.1=0.4 • polar covalent • S and Cs • 2.5-0.7=1.8 • ionic

17. For each of the following molecules, indicate which ones have a dipole moment. HCl Cl2 Practice H: 2.1 < Cl: 3.0 Dipole moment - + Cl: 3.0 = Cl: 3.0 No dipole moment

18. CH4 C: 2.5 > H: 2.1 no dipole moment H2S H: 2.1 < S: 2.5 dipole moment Practice + 4- + + + + + 2-

19. nonmetals sharing to make covalent bonds taking electrons from metals to form anions metals giving electrons to nonmetals to form cations Sizes of Ions trends are most important cation < parent atom more protons attracting electrons in anion > parent atom more electrons without additional protons Ions: Electron Configurations and Sizes

20. “Lattice energy is more exothermic with more highly charged ions”

21. isoelectronic ions: ions with the same number of electrons Ex: S2- Cl- K+ Ca2+ electrons have greater attraction to nucleus as # of protons increases so the size decreases as nuclear charge (Z) increases Sizes of Ions

22. Table of Ion Sizes

23. Give six ions that are isoelectronic with neon. N3-, O2-, F-, Na+, Mg2+, Al3+ 7 > 8 > 9 > 11 > 12 > 13 Example • Place them in order of decreasing size. • (Largest to smallest)

24. for each of the following groups, place atoms in order of decreasing size Cu, Cu+, Cu2+ all have 29 p, lower #e- mean smaller ion Cu > Cu+ > Cu2+ Ni2+, Pd2+, Pt2+ different amounts of e- and p-energy levels Pt2+ > Pd2+ > Ni2+ O, O-, O2- same #p, more electrons means larger ion O2- > O- > O Example

25. SURPRISE! • Please close your books and notes and push them forward. Also, get out something to write with and….nothing else.

26. 1. The compound most likely to be ionic is: KF CCl4 CO2 ICl CS2

27. 2. Ranking the ions S2-, Ca2+, K+, Cl- from smallest to largest gives the order as: S2-, Cl-, K+, Ca2+ Ca2+, Cl-, K+,, S2- K+, Ca2+, Cl-, S2- Cl-, S2-, K+, Ca2+ Ca2+, K+, Cl-, S2-

28. 3. The type of bonding within a water molecule is: ionic bonding polar covalent bonding nonpolar covalent bonding metallic bonding hydrogen bonding

29. 4. What type of bond would you expect in CsI?: Ionic Covalent Hydrogen Metallic van der walls

30. THAT’S ALL FOLKS! PLEASE PASS YOUR RESPONSES RIGHT TO LEFT LAST PERSON…..BRING THEM UP.

31. The compound most likely to be ionic is: KF CCl4 CO2 ICl CS2 Answer: A Bonds formed from elements with greater differences in electronegativity are most likely ionic in nature. (remember trends!)

32. Ranking the ions S2-, Ca2+, K+, Cl- from smallest to largest gives the order as: S2-, Cl-, K+, Ca2+ Ca2+, Cl-, K+,, S2- K+, Ca2+, Cl-, S2- Cl-, S2-, K+, Ca2+ Ca2+, K+, Cl-, S2- Answer E Note that all ions in this isoelectronic series have an Ar electron configuration; therefore the nuclear charge determines the size.

33. The type of bonding within a water molecule is: ionic bonding polar covalent bonding nonpolar covalent bonding metallic bonding hydrogen bonding Answer B Consider the nature of the bonding between hydrogen and oxygen in water.

34. What type of bond would you expect in CsI?: Ionic Covalent Hydrogen Metallic van der walls Answer A You tell me why???

35. Bond Length and Energy Bonds between elements become shorter and stronger as multiplicity increases.

36. Bond Energy and Enthalpy Energy required Energy released D = Bond energy per mole of bonds Breaking bonds always requires energy Breaking = endothermic (+DH) Forming bonds always releases energy Forming = exothermic (-DH)

37. Example Calculate the change in energy for the following reaction: (use table 8.4 on pg 351 in your textbook) H2(g) + F2(g)  2HF(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∆H = (DH-H + DF-F) – 2DH-F ∆H = (432 kJ + 154 kJ) – 2 (565 kJ) ∆H = -544 kJ

38. Example Using bond energies, calculate ∆H for the reaction: CH4(g) + 2Cl2(g) + 2F2(g)  CF2Cl2(g) + 2HF(g) + 2HCl(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∑ D (bonds broken) = 4 (413 kJ) + 2 (239 kJ) + 2 (154 kJ) ∑ D (bonds formed) = 2 (485 kJ) + 2 (339 kJ) + 2 (565 kJ) + 2 (427 kJ) ∆H = 2438 kJ – 3632 kJ = -1194 kJ

39. The Octet Rule Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level. Diatomic Fluorine

40. Formation of Water by the Octet Rule

41. 2nd row elements C, N, O, F observe the octet rule. 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals. Below are three examples of compounds containing elements not obeying the Octet Rule: Comments About the Octet Rule

42. Shows how valence electrons are arranged among atoms in a molecule. Reflects central idea that stability of a compound relates to noble gas electron configuration. Lewis Structures

43. Completing a Lewis Structure -CH3Cl • Make carbon the central atom • Add up available valence electrons: • C = 4, H = (3)(1), Cl = 7 Total = 14 • Join peripheral atoms • to the central atom • with electron pairs. H .. .. .. H .. • Complete octets on • atoms other than • hydrogen with remaining • electrons C .. Cl .. .. H

44. CH3Cl methyl chloride C: 1 x 4e- = 4e- H: 3 x 1e- = 3e- Cl: 1 x 7e- = 7e- 14 e- Carbon is central atom H H C Cl H duet octet duet octet duet Example

45. CH3Cl Methyl Chloride

46. Multiple Covalent Bonds:Double bonds Ethene Two pairs of shared electrons

47. Multiple Covalent Bonds:Triple bonds Ethyne Three pairs of shared electrons

48. .. Example .. .. CH2O Formaldehyde C: 1 x 4e- = 4 e- H: 2 x 1e- = 2 e- O: 1 x 6e- = 6 e- 12 O H – C - H

49. Resonance • Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. Benzene, C6H6 • The actual structure is an average of the resonance • structures. • The bond lengths in the ring are identical, and • between those of single and double bonds.

50. Resonance Bond Length and Bond Energy • Resonance bonds are shorter and stronger than single bonds. • Resonance bonds are longer and weaker than double • bonds.