Slide 1 / 47 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website: www.njctl.org Slide 2 / 47 AP Chemistry Unit 4: Presentation B Chemical Bonding: Lewis Structures, Hybridization, and Bond Order www.njctl.org Slide 3 / 47 Chemical Bonding The nature of the bonding in water leads to a bent shape causing it to be able to dissolve many of the nutrients and chemicals life depends on.
Slide 4 / 47 Molecular Bonding The nature of the bonding within a molecule influences many properties of the substance Some of the properties influenced by the bonding Melting Point and Boiling Point Viscosity (resistance to flow) Solubility Vapor Pressure Molar Absorptivity (light absorbed per M) Slide 5 / 47 Lewis Structures Covalent bonds are formed by atoms sharing electrons between nuclei so as to have a full valence shell. e-e- e-e- e-e- e- e- e-e- O O C e- e- e-e- e-e- Both O and C require 8 electrons for a full valence shell (s 2 p 6 ) Shared pairs of electrons can be represented by lines and un- bonded electrons can be represented as dots. O C O Slide 6 / 47 Lewis Structures A proper lewis structure uses only the valence electrons available from the atoms in the molecule AND distributes the electrons so each atom has a full valence shell. Guidelines for writing lewis structures Guideline One: Determine the ordering of atoms in the molecule
Slide 7 / 47 Lewis Structures Guideline One: Determine the ordering of atoms in the molecule Typically, the least electronegative atom is the central atom. Cl C Cl 4 Cl C Cl S O 2 O S O Cl But not always... often it's the less abundant atom H 2 O H O H N H 3 H N H H In hydrocarbons, the carbon atoms will form a congo line or chain... CH 3 CH 2 CH 2 OH C C C O H Slide 8 / 47 Lewis Structures Guidelines for writing lewis structures Guideline Two: Determine the number of valence electrons in the molecule NH 3 = 8 CCl 3 H = 26 If the molecules is an ion, one must either subtract or add electrons to the valence electron count. NO 3- = 23 +1 = 24 NH 4+ = 9-1 = 8 Slide 9 / 47 Lewis Structures Guidelines for writing lewis structures Guideline Three: Form a single bond (2 shared electrons) between all elements and then distribute electrons such that all atoms have a full valence shell, saving the central atom for last. Example: H 2 O (8 ve) Notice H needs only 2 electrons for a full valence shell. H - O - H Notice that C does not have a Example: CO 2 (16 ve) full valence shell and therefore adjustments will need to be O - C - O made to this structure
Slide 10 / 47 Lewis Structures: Octet Rule The "Octet Rule" refers to the fact that a full valence shell for most elements is a full outer s and p orbital or 8 electrons. Some elements do not follow this as shown below. H = 2 Be = 4 B = 6 In addition, elements in period 3 or below can have expanded octets or more than 8 valence electrons. Slide 11 / 47 Lewis Structures Guidelines for writing lewis structures Guideline Four: If an atom is short of an octet, additional electrons must be shared between the nuclei forming "Pi" bonds. O - C - O O C O Pi bonds Note: Pi bonds are formed from valence electrons in "p" orbitals. Slide 12 / 47 Lewis Structures Guidelines for writing lewis structures Guideline Five: If all atoms have a full valence shell but valence electrons remain, they are to be added to the central atom in pairs. F 34 valence electrons S F F F Extra pair of un-bonded electrons is added to central atom. As we will see shortly, these extra electrons influence the properties of the molecule significantly
Slide 13 / 47 1 Which of the following molecules would have 10 valence electrons in the lewis structure? A NH 4 + B CN- C H 2 O D NO 2 - E N 2 O Slide 14 / 47 2 How many valence electrons can be used in the lewis structure for NO + ? A 6 B 8 C 10 D 12 E None of these Slide 15 / 47 3 Which of the following molecules has a central atom with an expanded octet? A SO 2 B SCl 2 C PF 3 D XeF 2 E CO 32-
Slide 16 / 47 4 Which of the following molecules would require Pi bonds in the lewis structure? I. NO 3- A I only II. CO 32- B II only III. HCN C III only D I and II only E I, II, and III Slide 17 / 47 5 How many unbounded pairs of electrons are on the central atom in ClO 3 -? A 1 B 2 C 3 D 4 E None of these Slide 18 / 47 6 Which of the following molecules would have a lewis structure most similar to CO 2 ? A SO 2 B CS 2 C NO 2- D CO 32- E H 2 O
Slide 19 / 47 7 Below is a skeleton for the lewis structure for alphaketoglutarate, a kreb's cycle intermediate. After finishing the lewis structure, how many Pi bond are needed to complete the structure? A 0 B 1 H H O O O C - C - C - C - C C 2 O O H H D 3 E 4 Slide 20 / 47 8 Which of the following would contain the largest number of Pi bonds? A CH 4 B CO 32- C C 2 H 2 D SF 6 E C 3 H 6 Slide 21 / 47 Resonance Structures When "Pi" bonds can be formed in more than one location, the electrons are thought to be shared across all of the possible locations. This is shown by writing resonance structures. O One pi bond is needed but could N be formed from electrons shared O O by any of three oxygens. Resonance structures O O O N N N O O O O O O
Slide 22 / 47 Resonance Structures The bonds involved in resonance are equivalent in strength and in length. In essence, the pi bond electrons are shared across all of the bonds in which we find resonance. O O O N N N O O O O O O EQUALS O Pi bond electrons shared across all three bonds. N O O Slide 23 / 47 9 Which of the following molecules demonstrate resonance structures? I. NO 2- A I only II. CH 3 COO- (both O attached to C) B II only III. CH 3 CH 2 OH C III only D I and II only E I, II, and III Slide 24 / 47 10 How many resonance structures would be needed to represent SO 3 ? A 0 B 1 C 2 D 3
Slide 25 / 47 11 All bonds that demonstrate resonance are equal in length but not in strength. True False Slide 26 / 47 12 The C-O bonds in the carbonate ion (CO 32- ) would consist of … A 3 single bonds B 2 single bonds of longer length and 1 double bond of shorter length C 3 double bonds D 3 bonds equal in length but shorter than a single bond E 3 bonds equal in length but longer than a single bond Slide 27 / 47 13 Which of the following require no resonance structures to represent? A NO + B SO 2 C CH 3 COOH (both O attached to C) D NO 3 - E All require resonance structures
Slide 28 / 47 Bond Order The bond order refers to the number of bonds between two atoms in a molecule. It is calculated by adding up the bonds attached to the atom divided by the number of atoms attached to that atom. O Bond order of N-O bonds = 4/3 = 1.33 N O O Bond order of N-N bond = 3/1 = 3 N N Note: The higher the bond order, the stronger and shorter the bond. Slide 29 / 47 14 Which of the following contains bonds of the lowest order? A N 2 B SO 2 C SO 3 D CF 4 E All have the same bond order Slide 30 / 47 15 Which of the following would have a bond order of 1.5? A CO 32- B NO 2- C CO 2 D CS 2 E NH 3
Slide 31 / 47 16 Which of the following is true regarding bond order? A The higher the bond order the longer and weaker the bond B The higher the bond order the longer and stronger the bond C The higher the bond order the shorter and stronger the bond D The higher the bond order the shorter and weaker the bond Slide 32 / 47 17 Which of the following carbon molecules would have the shortest C-O bond lengths? A CO 2 B CO 32- C CH 3 OH D CO E All have the same bond lengths Slide 33 / 47 Hybridization In order to explain observations in molecular bonding, it has been proposed that atoms will hybridize s and p orbitals to create new orbitals of equal energy which are then involved in bonding. Carbon is known in nature to form compounds in which it must form 4 bonds. However, it's electron configuration suggests it could only share 2 electrons resulting in just 2 bonds. [Ne] __ ___ ___ ___ 2s 2p If the s and p orbitals were hybridized, four degenerate orbitals would be formed each with an electron that could be shared. [Ne] __ __ __ __ (sp 3 hybrid orbitals)
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