Lect 10. Lewis Structures and Molecular Geometry

# Lect 10. Lewis Structures and Molecular Geometry - Lewis...

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Lewis Structures Molecular Geometry Molecular Polarity

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Lewis electron-dot formulas or Lewis structures . A bonding pair is shared between two atoms. A lone pair (a non-bonded electron pair) is not shared. Each atom has 8 electrons in its valence shell : : H Cl ::
Drawing “Lewis Structures” to represent covalent bonding in molecules The electrons provided by all the atoms in the molecule are available for bonding The total number of electrons in the valence shells of the atoms is determined and then distributed to the molecule according to a set of rules.

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Lewis Structures 1 The “skeleton” structure of the molecule is first drawn. 2. The number of electrons in the valence shells of the atoms in a molecule (or ion) is calculated. 3. The electrons are then distributed around each atom such that each will have a share of 8 (2 only for each H).
1 Draw a “skeleton” structure The least electronegative will be the central atom The most electronegative atoms will be at the terminal positions H and F are always terminal

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1. Calculate the number of electrons in the valence shell of the molecule or ion The number of valence electrons for each atom is the group number (i.e. N = 5, O = 6, H = 1). Adjust for charge of ions (for positive charge, remove electrons; For negative charge, add electrons)
1. Distribute the electrons such that each atom has a share of 8 Give to the terminal atoms first (the most electronegative atoms) Give 8 to the central atom If the central atom has less than 8, place double (or triple) bonds between the terminal and central atoms until they share 8 If, when each atom attains a share of 8, there are excess electrons remaining, put them on the central atom H or F never from double bonds

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N : : H H H N H H H NH 3 N in Group V has 5 valence electrons 3H has 3(1) = 3 Total valence electrons = 8
CH 2 O C in Group IV has 4 valence electrons 2H has 2(1) = 2 O in Group VI has 6 valence electrons Total valence electrons = 12 1. Place electrons on terminal O first to give it share of 8. Since C only has share of 6, place one

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## This note was uploaded on 02/03/2011 for the course CHEM 211 taught by Professor Papanastasiou during the Spring '07 term at George Mason.

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Lect 10. Lewis Structures and Molecular Geometry - Lewis...

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