1 / 71

Chemical Bonding II

Chemical Bonding II. Lattice Energy. Remember, IE and EA are for adding/removing an electron to/from an atom in the gaseous state .

jafari
Download Presentation

Chemical Bonding II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemical Bonding II

  2. Lattice Energy • Remember, IE and EA are for adding/removing an electron to/from an atom in the gaseous state. • Ionic compounds are usually solids. The release of energy on forming the solid, called the lattice energy is the driving force for the formation of ionic compounds. • Because of high lattice energies, ionic solids tend to be hard and have high melting points. Ionic compounds are insulators in the solid state, because electrons are localized on the ions, but conduct when molten or in solution, due to flow of ions (not electrons). • Lattice energies can be calculated using Hess’s law, via a Born-Haber Cycle.

  3. Figure 9.6 The Born-Haber cycle for lithium fluoride

  4. Calculating Lattice Energy • Step 1: Convert elements to atoms in the gas state • e.g. for Li, Li (s)  Li (g) DH1 = DHatomization • for F, 1/2 F2 (g)  F (g) DH2 = 1/2 (Bond Energy) • Step 2: Electron transfer to form (isolated) ions • Li (g)  Li+ (g) + e–DH3 = IE1 • F (g) + e– F– (g) DH4 = EA1 • Step 3: Ions come together to form solid • Li+ (g) + F– (g) LiF (s) DH5 = Lattice Energy • Overall: Li (s) + 1/2 F2 (g) LiF (s) DH = DHf = S(DH1–5) • Lattice Energy = DHf – (DH1 + DH2 + DH3 + DH4)

  5. Periodic Trends in Lattice Energy Coulomb’s Law charge A X charge B electrostatic force a distance2 charge A X charge B or, electrostatic energy a distance (since energy = force X distance) • So, lattice energy increases, as ionic radius decreases (distance between charges is smaller). • Lattice energy also increases as charge increases.

  6. Figure 9.7 Trends in lattice energy

  7. I. Bonding Theory A covalent H-H bond is the net result of attractive and repulsive electrostatic forces. When bringing together two atoms that are initially very far apart. Three types of interaction occur: (1) The nucleus-electron attractions (blue arrows) are greater than the (2) nucleus-nucleus and (3) electron-electron repulsions (red arrows), resulting in a net attractive force that holds the atoms together to form an H2 molecule.

  8. A graph of potential energy versus internuclear distance for the H2 molecule. If the hydrogen atoms are too far apart, attractions are weak and no bonding occurs. A zero of energy when two H atoms are separated by great distances. A drop in potential energy (net attraction) as the two atoms approach each other. When the atoms are optimally separated, the energy is at a minimum. A minimum in potential energy at particular internuclear distance (74pm) corresponding to the stable H2 molecule and the potential energy corresponds to the negative of the bond dissociation energy. If the atoms are too close, strong repulsions occur. A increase in potential energy as the atoms approach more closely.

  9. II. Valence-Bond Theory • bond formation by overlapping orbitals: A description of covalent bond formation in terms of atomic orbital overlap is called the valence bond theory. It gives a localized electron model of bonding: core electrons and lone-pair valence electrons retain the same orbital locations as in the separated atoms, and the charge density of the bonding electrons is concentrated in the region of orbital overlap. • 2)hybridization of atomic orbitals: How do a carbon with a s orbital and three p orbitals combined with four hydrogen (s orbitals) form four bonds and all four bonds are found to be 109.5?

  10. 1s 1s   Overlap of two half-filled orbitals leads to the formation of a covalent bond. 1s-1s overlap gives a H – H single bond

  11. 2s 2p 1s      F H The 1s-2p overlap gives a H – F single bond

  12. 2s 2p 1s      F H Non-bonding electrons

  13. 2s 2s 2p 2p         F F The 2p-2p overlap gives a F – F single bond

  14. 2s 2s 2p 2p         F F Non-bonding electrons Each F atom has three pairs of non-bonding electrons.

  15. 2s 2s 2p 2p         O O Q.23 Identify the non-bonding electrons in O2 molecules. Two 2p-2p overlaps give a O=O double bond

  16. 2s 2s 2p 2p         O O Q.23 Identify the non-bonding electrons in O2 molecules. Non-bonding electrons Each O atom has two pairs of non-bonding electrons.

  17. Overlap of anempty orbitalwith afully-filled orbitalleads to the formation of aco-ordinate covalent bondordative bond

  18. Represented by an arrow  pointing from the electron pair donor to the electron pair acceptor.

  19. F3B NH3

  20. Hybridization In 1931, Linus Pauling proposed that the wave functions for the s and p atomic orbitals can be mathematically combined to form a new set of equivalent wave functions called hybrid orbitals. The mathematical process of replacing pure atomic orbitals with reformulated atomic orbitals for bonded atoms is called hybridization. In a hybridization scheme, the number of hybrid orbitals equals to the total number of atomic orbitals that are combined. The symbols identify the numbers and kinds orbitals involved.

  21. (a) NH4+ By Lewis model, the structure is  4 single bonds are formed, one of them is a dative bond.

  22. 2s 2p 1s 1s      N 3H H+ By VB Theory, Three 2p-1s(half-filled) overlaps lead to the formation of three N – H single bonds.

  23. 2s 2p 1s 1s      N 3H H+ By VB Theory, One 2s(fully-filled)-1s(vacant) overlap leads to the formation of one N  H dative bond.

  24. (b) HCN By Lewis model, the structure is H-CN  one H-C single bond and one CN triple bond.

  25. 2s 2p 2s 2p        C* By VB Theory, C  Only 2 single bonds can be formed. Promotion of a 2s electron to a 2p orbital.

  26. C* N H 2s 2s 2p 2p 1s          • The overlap of one orbital (?) of C* with an 1s orbital of Hgives theC-H single bond. • Overlaps of three orbitals (???) of C* with three 2porbitals of N give theCN triple bond.

  27. C* N H 2s 2s 2p 2p 1s          • The 2s electrons on N are non-bonding electrons. • The energy released by forming a stronger triple bond outweighs the energy required for promoting an electron from a 2s orbital to a 2p orbital.

  28. Most stableno separation of charge. (c) SO2 By Lewis model, the three possible structures are OS=O, O=SO, O=S=O

  29. 3s 3p     S By VB Theory,  Only two single bonds can be formed.  One 3p electron has to be promoted to a 3d orbital.  Expansion of Octet.

  30. 3d 3s 3s 3p 3p          S S* octet expansion By VB Theory,

  31. 3d 2s 3s 2p 3p          S* 2O  Overlaps of two half-filled orbitals (??) of S* with two half-filled 2p orbitals of an oxygen atom give a S=O double bond. A total of two S=O bonds are formed with two O atoms

  32. 3d 2s 3s 2p 3p          S* 2O Non-bonding electrons : S* 3s2 ; O 2s2 and 2p2

  33. 3d 3s 3s 3p 3p          S S* octet expansion The energy released by forming of two stronger double bonds outweighs the energy required for promoting an electron from a 3p orbital to a 3d orbital.

  34. The Concept of Resonance According to VB theory, the two less stable structures of SO2, OS=O and O=SO do ‘exist’. Each of these structures contributes in certain extent to the real structure of SO2.

  35. If represents the wave function of the real structure of SO2 molecules, then where are the wave functions of the three possible structures and a > b = c > 0

  36. More contribution Less contribution In other words, the real structure of SO2 is the resonance hydrid of the three possible structures. O=S=O  OS=O  O=SO

  37. 2s 2s 3s 2p 2p 3p            O S O* Q.24 O=SO A S=O double bond is formed by 3p(half-filled)-2p(half-filled) overlaps between S and O.

  38. 2s 2s 3s 2p 2p 3p            O S O* Q.24 O=SO A OS dative bond is formed by 3p(fully-filled)-2p(empty) overlap between S and O*

  39. 2s 2s 3s 2p 2p 3p            O S O* Q.24 O=SO Formation of dative bond is not favourable because the two unpaired 2p electrons in O are forced to pair up to give O*

  40. F-S-F (d) SF2, SF4, SF6 Molecule SF2 SF4 SF6 Most stable Lewis Structure

  41. 3s 2s 2p 3p         F S By VB Theory, Only two S-F single bonds can be formed by 3p-2p overlaps between one S atom and two F atoms  SF2 is formed. F-S-F

  42. 3d 3s 2s 3s 3p 2p 3p              F S S* By VB Theory, To form four S-F single bonds in SF4, a 3p electron in S has to be promoted to a 3d orbital.

  43. 2s 3s 2p 3p         F S 3d 3s 3p       S** By VB Theory, To form six S-F single bonds in SF6, a 3s electron in S* has to be promoted to a 3d orbital.

  44. 3s 3p     S 3d 3s 3p       S** By VB Theory, The energy released by forming more single bonds outweighs the energy required for promoting 3s and 3p electrons to 3d orbitals.

  45. F-Xe-F Q.25 Molecule XeF2 XeF4 XeF6 Most stable Lewis Structure

  46. 5s 2s 5p 2p         Xe F 5d 5s 5p      Xe* By VB Theory, To form two Xe-F bonds in XeF2, a 5p electron in Xe has to be promoted to a 5d orbital.

  47. 5d 5d 5s 5s 5p 5p            Xe** Xe* By VB Theory, To form four Xe-F bonds in XeF4, a 5p electron in Xe* has to be promoted to a 5d orbital.

  48. 5d 5s 5p 5d 5s 5p       Xe**        Xe*** By VB Theory, To form six Xe-F bonds in XeF6, a 5p electron in Xe** has to be promoted to a 5d orbital.

  49. 5d 5s 5p 5d 5s 5p       Xe**        Xe*** By VB Theory, The energy released by forming more single bonds outweighs the energy required for promoting 5p electrons to 5d orbitals.

More Related