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Chemistry II

by: Abtin Notetaker

Chemistry II CHEM 1041

Abtin Notetaker
GPA 3.979

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This is a brief study guide/notes that focused on the main points of chapter 10. I would like to apologize in advance because I was not able to write the questions for the "For Practice" problems d...
General Chemistry Lecture (CHEM 1041)
Dr. Janice Denton
Class Notes
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This 8 page Class Notes was uploaded by Abtin Notetaker on Sunday January 24, 2016. The Class Notes belongs to CHEM 1041 at University of Cincinnati Blue Ash College taught by Dr. Janice Denton in Spring 2016. Since its upload, it has received 26 views. For similar materials see General Chemistry Lecture (CHEM 1041) in Chemistry at University of Cincinnati Blue Ash College.


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Date Created: 01/24/16
Chapter 10 Summarization VESPER theory- the theory is based on the idea that lone pairs, single bonds, multiple bonds and even single electrons repel one another through columbic forces. The repulsion between electron groups on interior atoms of a molecule determine the geometry of the molecule. There are many different types of geometry depending on the number of groups od electron groups. Linear (Two Electron Groups): Through the repulsion of the electron groups, it creates a 180 bond angle. Trigonal Planar (Three Electron Groups): The three electron groups assume a 120 bond angle which maximizes their separation. Tetrahedral (Four Electron Groups): The tetrahedron is a three- dimensional shape that maximizes their separation in 3-D when the bond angle is 109.5. Trigonal Bipyramidal (Five Electron Groups): In this structure, five electron groups are bonding with the central atom. The angles in the trigonal bipyramidal structure are not all the same. The angles between the equatorial positions (trigonal planar plane) are 120 while the angle between the the axial positions (two bonds on either side of the trigonal plane) are 90. Octahedral (Six Electron Group): This is when there are six electron groups around the central atom and the bond angles in this geometry are all 90. Practice 10.1 See page 384 for the question. Solution: The molecular geometry of CCl4 is Tetrahedral because there are four electron groups. There are many cases where the electron geometry and the molecular geometry are the same, but in some cases that is not true. 1. Consider a the molecule H2O. When you draw the lewi structure, you would guess that the molecule is Linear because there are two sigma bonds. Despite this, we actually have a electron geometry of tetrahedral because there will be two lone pairs which would act as two separate groups. Unlike a trigonal pyramidal, the molecule does not have three sigma bonds, but two. So the molecule is *Bent* with a bond angle of <109.5 which is less than the ideal bond angle of a normal tetrahedron. 2. When we have a molecule that would be considered a tetrahedral through their electron geometry (a lone pair and 3 sigma bonds), its molecular geometry would be trigonal pyramidal with a bond angle <109.5. *The bond angle is less than the ideal bond angle of the shape of the molecule through their electron geometry, which in this case is a tetrahedral* The list goes on, and the thing to keep in mind is that the electron geometry of a molecule is the same as the molecular geometry. The bond angle of the molecule (if it has lone pairs of electrons) would be less than the ideal bond angle of the shape of the electron geometry (if tetrahedral <109.5 etc) See Page 389 as a table of guidance when choosing your electron and molecular geometry. 10.2 For Practice (pg 390) Solution: Bent For Practice 10.3: (pg 390) Solution: The molecule (I3)- is Linear and that is because the 3 lone pair electrons cancel each other repulsion therefore causing the molecule to be Linear. “Representing Molecular Geomatries on Paper” When representing molecular geometries on paper, a straight line represents a bond in plane of paper, a Hatched wedge or squiggly line represents a bond going into the page, while a solid wedge is a bond coming out of the page. Refer to page 391 for examples of all different molecular geometries on paper. “Predicting the shapes of Larger Molecules” All large molecules have interior atoms (or atoms that bond with various atoms). You treat all those atoms as you would when trying to predict the shapes of one molecule. Refer to page 391 for example of the molecule of glycine. For Practice 10.4: See Page 392 for the question. Solution: The interior atom of Carbon on the very left is tetrahedral. The Carbon in the middle is trigonal planar, while the Oxygen on the right is bent. Molecular Shape and Polarity: 1. When determining polarity of a molecule, we focus on two things. First, we try to determine the bonds between the central atom and the other atoms are polar (using electronegativity). If the difference between the two are between .5-2, then the bonds are polar. If they are less than . 5 or greater than 2, then the molecule can not be polar. 2. Despite this, just because a bond is polar, does not mean that the molecule is polar. After we have concluded that the bonds are polar, we look measure the net dipole moment. The dipole moment between two atoms can be describe as the pull from the central out by the outer atoms. Now the net dipole moment is the total force that the dipole is exerting on the molecule. If the structure was Linear, (CO2 for example), the net dipole moment would be 0 because both Oxygen are pulling the Carbon with the exact same force, so the Carbon would not move, therefore having no net dipole moment. See page 394 for a brief explanation of the net dipole of the different shapes. Also keep in mind that even if a molecule is Linear but the central atom is bonding with two different elements, there would be a net dipole because the dipole moments of those two elements are different. Practice 10.5 (See page 394 for problem) Solution: CF4 is nonpolar because the net dipole of cancels Valence Bond Theory: Hybridization of Atomic Orbitals: Hybridization- a procedure that allows us to combine the standard atomic orbitals to form new atomic orbitals called hybrid orbitals that correspond more closely to the actual distribution of electrons in chemically bonded atoms. When you Hybridize a molecule, you don’t worry about it’s molecular geometry but rather it’s electron geometry. Through it’s electron geometry, you can figure out its Hybridization scheme (See page 407). After that, you draw out your central atoms and their sigma bonds (single bonds) with your terminal atoms. Remembering to label sigma bonds and their Hybridization scheme. Afterwards, you draw out your pi bonds (double and triple bonds). (See page 409 For more steps on drawing out your Pi bonds). I want to apologize for not being able to bring examples from the textbook or Professor Denton’s notes. There would have been issues with copyright. Despite this, there are lots of pages that I listed that would help re enforce my point and hopefully you guys like this summary.


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