Chem 107 

Fall 2009 
Lecture Notes: 22 October 


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Molecular Geometry The importance of molecular shape: recognition at the molecular level in organisms. Shape and electron density are extraordinarily important to the interaction of biomolecules  Examples
 Sarin (nerve gas)
 "drugs"
 estrogen mimics ("feminization" of various animal populations  birth control complication)
Lewis Structures enable us to predict bonding patterns for compounds of the representative elements, but how can we predict their shapes? We will add another tool, VSEPR Theory, to our chemical toolbox  a simple way to predict the geometry of bonds around a central atom (for larger molecules predict one center at a time).
VSEPR Theory
VSEPR (Valence Shell Electron Pair Repulsion) Theory is based on three assumptions (there are more advanced versions, but unnecessary for us):
 Electron pairs will orient around a central point to minimize repulsion.
 Lonepairs of electrons will have greater repulsion than bonded pairs of electrons (note that the atoms are ignored in terms of repulsion).
 Repulsion is strong at 90° and weaker at 120° (weakest at 180°).
VSEPR predicts geometry based on these assumptions in a few simple, sequential, steps:
 Draw a correct Lewis Structure.
 Determine the Steric Number = the number of bonded atoms + the number of lone pairs = "electron clouds" in valance shell of central atom.
 Maximize the angles between electron pairs, placing the lone (unbonded) pairs at the extremes.
For central atoms with eight outer electrons (octets) there are three possible electron pair geometries:
 Linear with angles of 180° ( a single pair and a triple bond, or two double bonds).
 Trigonal planar with angles of 120° (one double bond and two single pairs).
 Tetrahedral with angles of 109.5° (four single pairs).
These three electron pair geometries can lead to five molecular geometries.
Five molecular geometries
Linear
 Carbon monoxide, CO
 valence electrons = 4 + 6 = 10
 6y + 2 = 14, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multibonds (2 double or 1 triple)
 LS = :C:::O:
 Considering C as the central atom, have one bonded atom and one lonepair, therefore
 steric number = 2, so linear electronic geometry, and two atoms so
 linear molecular geometry
 Carbon dioxide, CO_{2}
 valence electrons = 4 + 2x6 = 16
 6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multibonds (2 double or 1 triple)
 LS: from symmetry C will be central atom, therefore=
 Considering C as the central atom, have 2 bonded atoms and no lonepairs, therefore
 steric number = 2, so linear electronic geometry, and
 linear molecular geometry
Trigonal planar
 Formaldehyde, CH_{2}O
 valence electrons = 4 + 2x1 + 6 = 12
 6y + 2 = 6 x 2 + 2 = 14; so molecule has 2 fewer electrons than required for all single bonds, 1 double bond
 LS: from symmetry C will be central atom, therefore=
 Considering C as the central atom, have 3 bonded atoms and no lonepairs, therefore
 steric number = 3, so trigonal planar electronic geometry, and 3 atoms so
 trigonal planar molecular geometry and a model showing single vs. double bonds:
Tetrahedral
 Methane, CH_{4}
 valence electrons = 4 + 4x1= 8
 four bonds possible, since only 4 pairs, single bonds because only have H's bound to C.
 LS: from symmetry C will be central atom, therefore=
 Considering C as the central atom, have 4 bonded atoms and no lonepairs, therefore
 steric number = 4, so tetrahedral electronic geometry, and 4 atoms so
 tetrahedral molecular geometry = another view =
Trigonal pyramidal
 Ammonia, NH_{3}
 valence electrons = 5 + 3x1= 8
 only 4 pairs, single bonds because only have H's bound to N, 3 bonds, since only 3 H's
 LS: from symmetry N will be central atom, therefore=
 Considering N as the central atom, have 3 bonded atoms and one lonepair, therefore
 steric number = 4, so tetrahedral electronic geometry, but only 3 atoms so
 trigonal pyramidal molecular geometry = view from beneath N=
Bent
 Water, H_{2}O
 valence electrons = 6 + 2x1= 8
 only 4 pairs, single bonds because only have H's bound to O, 2 bonds, since only 2 H's
 LS: from symmetry O will be central atom, therefore=
 Considering O as the central atom, have 2 bonded atoms and 2 lonepairs, therefore
 steric number = 4, so tetrahedral electronic geometry,
 but only 2 atoms, so bent molecular geometry =
Polarity:So now we can predict bonding and shape in representative group molecules (and thus most biomolecules). How can we predict electron density and thus charge distribution? Need two bits of information:
 Shape (based on VSEPR Theory)
 Electron distribution within a bond (based on electronegativity).
 So what are the considerations for determining charge distribution in a molecule?
 A bond between two atoms of different electronegativities is polar.
 The more electronegative atom has a partial negative charge (δ) while the less electronegative atom has a partial positive charge (δ+)
 The two atoms together comprise a dipole.
 The dipole is symbolized by an arrow with the positive sign (tail) at the electropositive atom (lower EN), and the arrow pointing toward the electronegative atom (higher EN).
Examples:
Molecule Geometry Structure Electronegativities Bond Dipoles Molecular Dipole Model Carbon monoxide linear C=O
EN_{C}= 2.5, EN_{O}= 3.5 Carbon dioxide linear O=C=O
EN_{C}= 2.5, EN_{O}= 3.5 None: two dipoles are of equal magnitude, but opposite in direction and cancel. Water bent
EN_{H}= 2.1, EN_{O}= 3.5 Ammonia trigonal pyramidal
EN_{H}= 2.1, EN_{N}= 3.0 Ammonium ion tetrahedral
EN_{H}= 2.1, EN_{N}= 3.0 None: four dipoles are symmetrically arranged to cancel each other out and give a spherically charged but nonpolar ion.

© R A Paselk
Last modified 21 October 2009