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Chemistry-140 Lecture 22

Chemistry-140 Lecture 22. Chapter 9: Bonding & Molecular Structure: Fundamental Concepts. Chapter Highlights bonding types Lewis symbols & octets ionic bonding & ionic lattices covalent bonding & Lewis dot structures resonance structures breaking the octet rule & formal charge

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Chemistry-140 Lecture 22

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  1. Chemistry-140 Lecture 22 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts • Chapter Highlights • bonding types • Lewis symbols & octets • ionic bonding & ionic lattices • covalent bonding & Lewis dot structures • resonance structures • breaking the octet rule & formal charge • bond order & bond energy • VSEPR molecular shape & polarity

  2. F- Li+ Chemistry-140 Lecture 22 • Ionic bond: Term given to the electrostatic (charge-based) attractive forces which hold oppositely charged ions together Ionic, Covalent & Metallic Bonding

  3. : H H Chemistry-140 Lecture 22 • Covalent bond: The sharing of electrons between two atoms that acts to hold the atoms together Ionic, Covalent & Metallic Bonding

  4. ::: ::: M M M Chemistry-140 Lecture 22 • Metallic bond: Is found in metals. Atoms of the metal are bound to several neighbors, holding the atoms together but allowing electrons to move freely. Ionic, Covalent & Metallic Bonding

  5. S Chemistry-140 Lecture 22 • Lewis symbols = electron-dot symbols • Lewis symbols place one dot for each valence electron around the symbol of the element. • For Example: • S: [Ne]3s23p4 Lewis Symbols

  6. Ne Chemistry-140 Lecture 22 • Octet rule: Atoms tend to gain, lose, or share enough electrons to become surrounded by eight valence electrons • Attain the closed shell configuration of a Group 18 inert gas • Ne: [He]2s22p6 Lewis Octets

  7. Chemistry-140 Lecture 22 • The ionic bond is formed when ions of opposite charge (anions and cations) are attracted and held to one another by electrostatic attractions • Na(g) + Cl(g) NaCl(g) • Na + Cl Na+ + Cl - Ionic Bonding

  8. Chemistry-140 Lecture 22 Energetics of Forming a Na-Cl Ionic Bond Na(g) Na+(g) + e-DE = +496 kJ/mol Cl(g) + e- Cl-(g) DE = -349 kJ/mol Therefore electron transfer costs 147 kJmol-1 !!!!!

  9. E = k Chemistry-140 Lecture 22 • But, Coulomb’s law... • where: Q1 and Q2 are the charges on the cation and anion, d is the distance between the nuclei (sum of the ionic radii) • and k is a constant = 8.99 x 109 J-m/C2 • E = (6.022 x 1023)(8.99 x 109) • E = Energetics of Forming a Na-Cl Ionic Bond -411 kJ/mol

  10. Chemistry-140 Lecture 22 Therefore: the overall DE for the reaction Na(g) + Cl(g) NaCl(g) DE = (-411 + 147) kJ/mol = Energetics of Forming a Na-Cl Ionic Bond -264 kJ/mol

  11. Chemistry-140 Lecture 22 • In order to maximize the attractions among ions, ionic solids exist in lattices, which are regularly repeating three-dimensional arrays of ions Ionic Lattices

  12. Chemistry-140 Lecture 22 • Coordination number: the number of close contacts in the lattice array (equals 6 for this Na+ ion below) Coordination Number

  13. Chemistry-140 Lecture 22 • Lattice energy: the energy required to separate the crystalline solid into the constituent gaseous ions. • NaCl(s) Na+(g) + Cl-(g) • It is a measure of the stability of the crystalline state • Note: The lattice energy for NaCl(s) is -786 kJ/mol as compared to -264 kJ/mol that we calculated!! Lattice Energies

  14. E = k Chemistry-140 Lecture 22 • In summary lattice energies: • increase as charges of the ions increase • increase as sizes of the ions decrease • Coulomb’s Law!!! • increase with increasing coordination number Lattice Energies

  15. Chemistry-140 Lecture 22 Na+(g) + Cl-(g) Born-Haber Cycle Step 4 Step 3 Step 5 Na(g) + Cl(g) Step 1 Step 2 Na(s) + 1/2 Cl2(g) NaCl(s) DHfo

  16. Chemistry-140 Lecture 23 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts • Chapter Highlights • bonding types • Lewis symbols & octets • ionic bonding & ionic lattices • covalent bonding & Lewis dot structures • resonance structures • breaking the octet rule & formal charge • bond order & bond energy • VSEPR molecular shape & polarity

  17. H H H H = Cl-Cl Cl Cl Cl Cl Chemistry-140 Lecture 23 • Covalent bond: a bond formed between two atoms by sharing of electrons. • Lewis Structures for H2 and Cl2 Covalent Bonding = H-H + +

  18. or Cl-Cl Cl Cl Chemistry-140 Lecture 23 • Different multiplicities of covalent bonds are possible. • Single bonds are covalent bonds in which one pair of electrons is shared by the two atoms Covalent Single Bonds

  19. = N or N N N + N N = _ = N N N N N N = = Chemistry-140 Lecture 23 • Different multiplicities of covalent bonds are possible. • Double bonds are covalent bonds in which two pairs of electrons are shared by the two atoms. • Triple bonds are covalent bonds in which three pairs of electrons are shared by the two atoms Multiple Covalent Bonds 1.10 A 1.24 A 1.47 A

  20. Chemistry-140 Lecture 23 • Recall: for a covalent bond the bonding electrons are equally shared between two atoms. • Recall: for an ionic bond the bonding electrons are separated between the ions (electrostatic attraction). • When sharing is not equal, the bond is called a polar bond. • Equal sharing is sometimes referred to as a nonpolar bond. Bond Polarity

  21. Chemistry-140 Lecture 23 • Electronegativity: the ability of an atom in a molecule to attract electrons to itself. The higher an element's electronegativity, the better it competes for electrons. • Electronegativity is related to ionization energy and electron affinity. The scale (Pauling scale) has no units Electronegativity

  22. Chemistry-140 Lecture 23 Electronegativity Values

  23. Chemistry-140 Lecture 23 • Electronegativity difference between two atoms of a bond is related to the polarity of the bond. The greater the electronegativity difference, the more polar the bond. • > 2.0 = ionic, • < 0.5 = nonpolar • between 0.5 and 2.0 = polar Polar vs. Nonpolar Bonds

  24. d+ d- _ H F Chemistry-140 Lecture 23 Examples: d+ represents a partialpositive charge d- represents a partialnegative charge F2 4.0 - 4.0 = 0 nonpolar HF 4.0 - 2.1 = 1.9 polar LiF 4.0 - 1.0 = 3.0 ionic Bond Polarity

  25. Chemistry-140 Lecture 23 • Sum the valence electrons from all atoms in the species. • Write the atomic symbols for the atoms involved so as to show which atoms are connected to which. Draw a single bond between each pair of bonded atoms • Complete the octets of the atoms bonded to the central atom (i.e. the peripheral atoms) Drawing Lewis Structures

  26. Chemistry-140 Lecture 23 • Place leftover electrons on the central atom, even if it results in the central atom having more than an octet • If there are not enough electrons to give the central atom an octet, form multiple bonds by pulling terminal electrons from a peripheral atom and placing them into the bond with the central atom Drawing Lewis Structures

  27. Chemistry-140 Lecture 23 Question: Draw the Lewis structure for PCl3. Drawing Lewis Structures

  28. Chemistry-140 Lecture 23 Answer: Step 1: Sum the valence electrons. P has 5 and each Cl has 7 for a total of [5 + (3 x 7)] = Drawing Lewis Structures 26 valence electrons

  29. _ _ Cl P Cl _ Cl Chemistry-140 Lecture 23 Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. NOTE: In a binary (two-element) compound, the first element listed is usually the central one with the others surrounding it Drawing Lewis Structures

  30. _ _ Cl P Cl _ Cl Chemistry-140 Lecture 23 Answer: Step 3: Complete the octets on the atoms bonded to the central atom. NOTE: This accounts for 24 of the 26 valence electrons Drawing Lewis Structures

  31. _ _ Cl P Cl _ Cl Chemistry-140 Lecture 23 Answer: Step 4: Place the remaining electrons on the central atom to complete the octet. Since this gives an octet to each atom we are finished Drawing Lewis Structures

  32. Chemistry-140 Lecture 23 Question: Draw the Lewis structure for HCN. Drawing Lewis Structures

  33. Chemistry-140 Lecture 23 Answer: Step 1: Sum the valence electrons. H has 1, C has 4 and N has 5 for a total of [1 + 4 + 5)] = Drawing Lewis Structures 10 valence electrons

  34. _ _ H C N Chemistry-140 Lecture 23 Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. NOTE: Since H can only form one covalent bond it can never be the central atom. The choices are HCN or HNC. Formula is written HCN!!! Drawing Lewis Structures This accounts for 4 valence electrons

  35. _ _ H C N Chemistry-140 Lecture 23 Answer: Step 3: Complete the octets on the atoms bonded to the central atom. Drawing Lewis Structures BUT: There are only 6 valence electrons left. If we put them on N we do not achieve an octet at C !!

  36. _ = H C N = Chemistry-140 Lecture 23 Answer: Step 4: Try using multiple bonding to share the electrons between C and N. A triple bond is required to give an octet to each atom Drawing Lewis Structures

  37. Chemistry-140 Lecture 24 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts • Chapter Highlights • bonding types • Lewis symbols & octets • ionic bonding & ionic lattices • covalent bonding & Lewis dot structures • resonance structures • breaking the octet rule & formal charge • bond order & bond energy • VSEPR molecular shape & polarity

  38. Chemistry-140 Lecture 24 • Formal charges: a way of assigning a relative charge to each atom in the molecule • When several different Lewis structures seem plausible, the one in which the formal charges are minimized is generally the preferred one. Formal Charge & Lewis Structures

  39. Chemistry-140 Lecture 24 • All bonding electrons are divided equally between the atoms that form bonds • All nonbonding electrons are assigned to the atom on which they reside Assigning Formal Charge

  40. Chemistry-140 Lecture 24 • Formal charge: the number of valence electrons for the element minus the number of electrons assigned by rules 1 and 2. • formal charge on an atom in a molecule = • {# valence electrons normally found for that atom - • [(# non-bonding electrons) + 1/2(# bonding electrons)]} Assigning Formal Charge FC = VE - (NBE + 1/2BE)

  41. Chemistry-140 Lecture 24 Question: There are three possible structures for SCN-. Use formal charge to decide the most likely structure. Applying Formal Charge to Lewis Structures

  42. Chemistry-140 Lecture 24 Answer: Step 1: Sum the valence electrons. S has 6, C has 4 and N has 5 and there is an extra electron represented by the single negative charge of the ion. Total of [6 + 4 + 5 + 1] = Drawing Lewis Structures 16 valence electrons

  43. _ _ _ _ _ _ [S C N]- [C S N]- [S N C]- Chemistry-140 Lecture 24 Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. Drawing Lewis Structures

  44. _ _ _ _ _ _ [ S C N ]- [ C S N ]- [ S N C ]- Chemistry-140 Lecture 24 Answer: Step 3: Complete the octets on the atoms bonded to the central atom. Drawing Lewis Structures BUT: Each of these leaves us with only four electrons at the central atom!

  45. [S C N]- = = [S N C]- [C S N]- = = = = Chemistry-140 Lecture 24 Answer: Step 4: Use multiple bonding to share the electrons between peripheral atoms and the central atom until octets are achieved. Drawing Lewis Structures

  46. [S C N]- = = [S N C]- [C S N]- = = = = 0 0 -1 -2 +2 -1 0 +1 -2 Chemistry-140 Lecture 24 Answer: Step 5: Calculate formal charge for each atom. Drawing Lewis Structures Note: the total formal charge on each molecule is equal to the charge on the molecule

  47. [S C N]- = = [S N C]- [C S N]- = = = = 0 0 -1 -2 +2 -1 0 +1 -2 Chemistry-140 Lecture 24 Answer: Step 6: Decide on the most probable structure. Drawing Lewis Structures The structure that results in the least amount of formal charge separation throughout the molecule

  48. _ _ = O O O = O O O Chemistry-140 Lecture 24 • There are times when more than one Lewis structure involving multiple bonds seems equally stable Resonance Structures ozone • Resonance structures: structures that differ only in the placement of electrons.

  49. _ _ = O O O = O O O Chemistry-140 Lecture 24 • Resonance forms rapidly interconvert so that the structure appears to be a blend of all the forms. Resonance Structures

  50. _ _ O O O Chemistry-140 Lecture 24 • The molecule does not oscillate rapidly between two or more different forms. There is only one form of the molecule. Ozone has two equivalent O-O bonds whose length is intermediate between single and double bonds Resonance Structures

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