Resonance example:
- Carbonate ion - CO_{3}^{2-}
- 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:
- SF_{4}
- 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.
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
VSEPR predicts geometry based on these assumptions in a few simple, sequential, steps:
- 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). [model]
These three electron pair geometries can lead to five molecular geometries:
- Linear (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 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, CH_{2}O)
- 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: 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, CH_{4}) [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, NH_{3}) [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, H_{2}O) [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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© R A Paselk
Last modified 12 March 2013