# Lewis Structures for Covalent Molecules-Octet Variations, cont.

Resonance example:

• Carbonate ion - CO32-
• valence electrons = 4 + 3 (6) + 2 = 24
• 6y + 2 = 26, but e- = 24, therefore expect one multiple bond.
• LS =
• However, other equally symmetrical structures are possible, so:

Expanded Valence Shell Example:

• SF4
• valence electrons = 6 + 4(7) = 34
• 6y + 2 = 32, but e- = 34, therefore expect expanded valence shell with one extra electron pair.
• LS =

Additional exercises on Lewis Structures are available in the Lewis Structure Module.

For a modern view of bonding illustrated with QuickTime movies based on quantum calculations you may enjoy the Supplement.

# 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 (Valence Shell Electron Pair Repulsion) Theory

Based on three assumptions (there are more advanced versions, but unnecessary for us):
• Electron pairs will orient around a central point to minimize repulsion.
• Lone-pairs 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:

1. Draw a correct Lewis Structure.
2. Determine the Steric Number = the number of bonded atoms + the number of lone pairs.
3. 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:

1. Linear with angles of 180° ( a single pair and a triple bond, or two double bonds).
2. Trigonal planar with angles of 120° (one double bond and two single pairs).

3. Tetrahedral with angles of 109.5° (four single pairs). [model]

#### These three electron pair geometries can lead to five molecular geometries:

• Linear (carbon dioxide)
• CO2
• valence electrons = 4 + 2x6 = 16
• 6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (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 lone-pairs, therefore
• steric number = 2, so linear electronic geometry, and
• linear molecular geometry
• Trigonal planar (formaldehyde, CH2O)
• formaldehyde, CH2O
• 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: O can only have two bonds, so C will be central atom, therefore =
• Considering C as the central atom, have 3 bonded atoms and no lone-pairs, 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, CH4) [model]
• 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 lone-pairs, therefore
• steric number = 4, so tetrahedral electronic geometry, and 4 atoms so
• tetrahedral molecular geometry, and rotated for a different view:
• Trigonal pyramidal (ammonia, NH3) [model]
• 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 lone-pair, therefore
• steric number = 4, so tetrahedral electronic geometry, but only 3 atoms so
• trigonal pyramidal molecular geometry and rotated to view molecule from below:
• Bent (water, H2O) [model]
• 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 lone-pairs, therefore
• steric number = 4, so tetrahedral electronic geometry, but only 2 atoms so
• bent molecular geometry

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