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1.9 Hybridized Atomic Orbitals

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1 1.9 Hybridized Atomic Orbitals
Given the electron configuration for C and H, imagine how their atomic orbitals might overlap Would such orbital overlap yield methane? Copyright 2012 John Wiley & Sons, Inc.

2 1.9 Hybridized Atomic Orbitals
To make methane, the C atom must have 4 atomic orbitals available for overlapping If an electron is excited from the 2s to the 2p, will that make it suitable for making methane? If four H atoms were to come in and overlap with the 2s and 2p orbitals, what geometry would the resulting methane have? Copyright 2012 John Wiley & Sons, Inc.

3 1.9 Hybridized Atomic Orbitals
The carbon must undergo hybridization to form 4 equal atomic orbitals The atomic orbitals must be equal in energy to form four equal-energy symmetrical C-H bonds Copyright 2012 John Wiley & Sons, Inc.

4 1.9 Hybridized Atomic Orbitals
Should the shape of an sp3 orbital look more like an s or more like p orbital? Copyright 2012 John Wiley & Sons, Inc.

5 1.9 Hybridized Atomic Orbitals
To make CH4, the 1s atomic orbitals of the H atoms will overlap with the four sp3 hybrid atomic orbitals of C Copyright 2012 John Wiley & Sons, Inc.

6 1.9 Hybridized Atomic Orbitals
Draw a picture that shows the necessary atomic orbitals and their overlap to form ethane (C2H6). Draw a picture that shows the necessary atomic orbitals and their overlap to form water. Practice with conceptual checkpoint 1.19 Copyright 2012 John Wiley & Sons, Inc.

7 1.9 Hybridized Atomic Orbitals
Consider ethene (ethylene). Each carbon in ethene must bond to three other atoms, so only three hybridized atomic orbitals are needed Copyright 2012 John Wiley & Sons, Inc.

8 1.9 Hybridized Atomic Orbitals
An sp2 hybridized carbon will have three equal-energy sp2 orbitals and one unhybridized p orbital Copyright 2012 John Wiley & Sons, Inc.

9 1.9 Hybridized Atomic Orbitals
The sp2 atomic orbitals overlap to form sigma (σ) bonds Sigma bonds provide maximum HEAD-ON overlap Copyright 2012 John Wiley & Sons, Inc.

10 1.9 Hybridized Atomic Orbitals
The unhybridized p orbitals in ethene form pi (π) bonds, SIDE-BY-SIDE overlap Practice with conceptual checkpoint 1.20 Copyright 2012 John Wiley & Sons, Inc.

11 1.9 Hybridized Atomic Orbitals
The unhybridized p orbitals in ethene form pi (π) bonds, SIDE-BY-SIDE overlap of p-orbitals giving both CONSTRUCTIVE and DESTRUCTIVE interference MO theory provides a similar picture. Remember, red and blue regions are all part of the same orbital Copyright 2012 John Wiley & Sons, Inc.

12 1.9 Hybridized Atomic Orbitals
Why is sp2 hybridization not appropriate for methane (CH4)? Copyright 2012 John Wiley & Sons, Inc.

13 1.9 Hybridized Atomic Orbitals
Consider ethyne (acetylene). Each carbon in ethyne must bond to two other atoms, so only two hybridized atomic orbitals are needed Copyright 2012 John Wiley & Sons, Inc.

14 1.9 Hybridized Atomic Orbitals
The sp atomic orbitals overlap HEAD-ON to form sigma (σ) bonds while the unhybridized p orbitals overlap SIDE-BY-SIDE to form pi bonds Practice with SkillBuilder 1.7 Copyright 2012 John Wiley & Sons, Inc.

15 1.9 Hybridized Atomic Orbitals
Which should be stronger, a pi bond or a sigma bond? WHY? Which should be longer, an sp3 – sp3 sigma bond overlap or an sp – sp sigma bond overlap? Copyright 2012 John Wiley & Sons, Inc.

16 1.9 Hybridized Atomic Orbitals
Explain the different strengths and lengths below. Practice with conceptual checkpoint 1.24 Copyright 2012 John Wiley & Sons, Inc.

17 1.10 Molecular Geometry Valence shell electron pair repulsion (VSEPR theory) Valence electrons (bonded and lone pairs) repel each other To determine molecular geometry… Determine the Steric number Copyright 2012 John Wiley & Sons, Inc.

18 1.10 Molecular Geometry Valence shell electron pair repulsion (VSEPR theory) Valence electrons (bonded and lone pairs) repel each other To determine molecular geometry… Predict the hybridization of the central atom If the Steric number is 4, then it is sp3 If the Steric number is 3, then it is sp2 If the Steric number is 2, then it is sp Copyright 2012 John Wiley & Sons, Inc.

19 1.10 sp3 Geometry For any sp3 hybridized atom, the 4 valence electron pairs will form a tetrahedral electron group geometry Methane has 4 equal bonds, so the bond angles are equal How does the lone pair of ammonia affect its geometry? The bond angels in oxygen are even smaller, why? Copyright 2012 John Wiley & Sons, Inc.

20 1.10 sp3 Geometry The molecular geometry is different from the electron group geometry. HOW? Copyright 2012 John Wiley & Sons, Inc.

21 1.10 sp2 Geometry Calculate the Steric number for BF3
Electron pairs that are located in sp2 hybridized orbitals will form a trigonal planar electron group geometry What will be the molecular geometry? Copyright 2012 John Wiley & Sons, Inc.

22 1.10 sp2 Geometry How many electrons are in Boron’s unhybridized p orbital? Does this geometry follow VSEPR theory? Copyright 2012 John Wiley & Sons, Inc.

23 1.10 sp2 Geometry Analyze the steric number, hybridization, electron group geometry and molecular geometry for this imine? Copyright 2012 John Wiley & Sons, Inc.

24 1.10 sp Geometry Analyze the Steric number, the hybridization, the electron group geometry, and the molecular geometry for the following molecules BeH2 CO2 Copyright 2012 John Wiley & Sons, Inc.

25 1.10 Geometry Summary Practice with SkillBuilder 1.8
Copyright 2012 John Wiley & Sons, Inc.

26 1.11 Molecular Polarity Electronegativity Differences cause induction
Induction (shifting of electrons WITHIN their orbitals) results in a dipole moment. Dipole moment = (the amount of partial charge) x (the distance the δ+ and δ- are separated) Dipole moments are reported in units of debye (D) 1 debye = esu ∙ cm An esu is a unit of charge. 1 e- has a charge of 4.80 x esu cm are included in the unit, because the distance between the centers of + and – charges affects the dipole Copyright 2012 John Wiley & Sons, Inc.

27 1.11 Molecular Polarity Consider the dipole for CH3Cl
Dipole moment (μ) = charge (e) x distance (d) Plug in the charge and distance μ = (1.056 x esu) x (1.772 x 10-8 cm) Note that the amount of charge separation is less than what it would be if it were a full charge separation (4.80 x esu) μ = 1.87 x esu ∙ cm Convert to debye μ = 1.87 D Copyright 2012 John Wiley & Sons, Inc.

28 1.11 Molecular Polarity What would the dipole moment be if CH3Cl were 100% ionic? μ = charge (e) x distance (d) Plug in the charge and distance μ = (4.80 x esu) x (1.772 x 10-8 cm) The full charge of an electron is plugged in μ = 8.51 x esu ∙ cm = 8.51 D What % of the C-Cl bond is ionic? Is the C-Cl bond mostly ionic or mostly covalent? Copyright 2012 John Wiley & Sons, Inc.

29 1.11 Molecular Polarity Check out the polarity of come other common bonds Copyright 2012 John Wiley & Sons, Inc.

30 1.11 Molecular Polarity Why is the C=O double bond so much more polar than the C-O single bond? Copyright 2012 John Wiley & Sons, Inc.

31 1.11 Molecular Polarity For molecules with multiple polar bonds, the dipole moment is the vector sum of all of the individual bond dipoles Copyright 2012 John Wiley & Sons, Inc.

32 1.11 Molecular Polarity It is important to determine a molecule’s geometry FIRST before analyzing its polarity If you have not drawn the molecule with the proper geometry, it may cause you to assess the polarity wrong as well Would the dipole for water be different if it were linear rather than angular? Copyright 2012 John Wiley & Sons, Inc.

33 1.11 Molecular Polarity Electrostatic potential maps are often used to give a visual depiction of polarity Copyright 2012 John Wiley & Sons, Inc.

34 1.11 Molecular Polarity Practice with SkillBuilder 1.9
Copyright 2012 John Wiley & Sons, Inc.

35 1.11 Molecular Polarity Explain why the dipole moment for pentane = 0 D Copyright 2012 John Wiley & Sons, Inc.

36 Study Guide for sections 1.9-1.11
DAY 2, Terms to know: Sections hybridized atomic orbitals, steric number, degenerate orbitals, pi bond, sigma bond, electron group geometry, molecular geometry, induction, dipole moment DAY 2, Specific outcomes and skills that may be tested on exam 1: Sections Be able to predict what type of hybridization atomic orbitals will undergo for an atom depending on the bonding and steric number of the atom. Be able to explain how the ratio of s and p orbitals mixed into the hybrid atomic orbitals affects the shape and energy of the resulting hydridized orbitals. Be able to draw a picture showing the overlapping of the correct atomic orbitals for small molecules with 8 atoms or less including molecules with single, double, and triple bonds. Be able to predict where molecules will have pi bonds and sigma bonds between atoms. Be able to predict the steric number, hybridization, electron group geometry, molecular geometry, and approximate bond angle for atoms within molecules. Be able to determine the direction of dipole moment for a molecule and rate its approximate magnitude. Be able to describe how molecular geometry affects polarity or dipole. Klein, Organic Chemistry 2e

37 Practice Problems for sections 1.9-1.11
Complete these problems outside of class until you are confident you have learned the SKILLS in this section outlined on the study guide and we will review some of them next class period Klein, Organic Chemistry 2e

38 Prep for day 3 Must Watch videos: Other helpful videos:
(dissolving, Tyler DeWitt) (solubility) (drawing organic molecules) Other helpful videos: (solvation, Brightstorm) (drawing organic molecules) Read sections , Klein, Organic Chemistry 2e

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