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Chemistry

Week 13

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Element quiz

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Acid naming answers

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Steps for drawing lewis structures

  1. Atom farthest away from fluorine is center (hydrogen is never in center).
  2. Total number of valence electrons.
  3. Divide by 2 to find number of pairs.
  4. Place bonds between atoms.
  5. Place remaining pairs around atoms.
  6. Is octet rule satisfied? May need to adjust for double and triple bonds.

* If it is a polyatomic ion, make sure to consider the charge when totaling the electrons.

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Section 3, Question 39

39. Draw the Lewis structure for ethylene, C2H4.

Answer:

  1. Furthest away from fluorine in middle, hydrogen always on outside.
  2. Total valence: carbon: 4(x2), hydrogen: 1(x4). Total: 12
  3. 12/2= 6 pairs
  4. Start with single bonds between them all. 2 leftover.
  5. Carbon has not satisfied octet, so add another bond.

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Section 3, Question 41

41. Draw the Lewis structure for the NH4+.

Answer:

  1. Furthest away from fluorine in the middle, hydrogen always on outside.
  2. Total valence: nitrogen: 5, hydrogen: 1(x4). Total: 9, but because of positive charge, take away one electron, so total=8.
  3. 8/2= 4 pairs
  4. Single bonds, bracket
  5. Octet satisfied.

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Resonance structures

When more than one valid Lewis structure can be written.

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Section 3, Question 43

43. Draw the Lewis resonance structure for NO2-.

Answer:

  1. Furthest away from fluorine in middle.
  2. Total valence: nitrogen: 5, oxygen: 6 (x2), total: 17 electrons. Due to negative charge, add one electron. Total: 18 electrons.
  3. 18/2 = 9 pairs
  4. Single bonds first, fill in leftover pairs as lone.
  5. Nitrogen needs to satisfy octet. Double bond needed.

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Section 3, Question 47

47. Draw the expanded octet Lewis structure for ClF3

Answer:

  1. Furthest from fluorine in middle.
  2. Total valence: Cl: 7, F: 7 (x3), Total: 28
  3. 28/2=14
  4. Single bonds, then fill in lone pairs
  5. Chlorine has 10 rather than 8

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Section 3, Question 37

37.Draw the Lewis structure for BH3.

  1. Furthest away from fluorine in the middle, hydrogen always outside.
  2. Total valence electrons Boron- 3, Hydrogen- 1 (x3), Total: 6
  3. 6/2 = 3 pairs
  4. No more available.
  5. Boron is an exception to the rule of octet.

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VSEPR Model: Valence Shell Electron Pair Repulsion

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Section 4, QUESTION 56

56. Determine the molecular shape, bond angle, and hybrid orbitals for BF3.

Answer: Find total pairs on central atom and shared pairs.

3, 3 therefore…

trigonal planar, 120 degrees, sp2

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Section 4, Question 57

57. Determine the molecular shape, bon angle, and hybrid orbitals for OCl2.

Answer: Find total pairs on central atom and shared pairs.

4, 2 therefore…

Bent, 104.5 degrees, sp3

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Section 4, Question 64

64. Compare the size of an orbital that has a shared electron pair with one that has a lone pair.

Answer: A lone pair occupies more space than a shared

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Modeling!

Total Pairs:

Shared Pairs:

Lone Pairs:

Shape:

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Intermolecular forces or van der waals forces

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How strong?

Strongest:

Medium:

Weakest:

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Section 5, Question 75

75. Identify each molecules as polar or nonpolar: SCl2, CS2, and CF4.

Answer:

SCl2 polar

CS2 nonpolar

CF4 nonpolar

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Station 4: gecko

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gecko

Why can’t it walk on teflon?

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Practice balancing a chemical equation

H2 (g) + Cl2 (g) → HCl (g)

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homework

Tuesday: Read Chapter 9, Section 1 Questions 1, 4, 7, 11, 12

Wednesday: Read Chapter 9, Section 2 Questions 14, 18, 21, 25, 29, 30

Thursday: Read Chapter 9, Section 3, Questions 35, 40, 45, 46, 50-52

Friday: Study elements 46-60, be able to tell me the symbol if I give you the element.

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