| Chem 109 |
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Summer 2002 |
| Lecture Notes:: 8 July |
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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
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 is based on three assumptions (there are more advanced versions, but unnecessary for us):
VSEPR predicts geometry based on these assumptions in a few simple, sequential, steps:
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). [model]
These three electron pair geometries can lead to five molecular geometries:
- Linear (carbon dioxide, carbon monoxide)
- CO;
- valence electrons = 4 + 6 = 10
- 6y + 2 = 14, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)
- LS = :C:::O:
- Considering C as the central atom, have one bonded atom and one lone-pair, therefore
- steric number = 2, so linear electronic geometry, and two atoms so
- linear molecular geometry
- 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= :O::C::O:
- 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 (carbocation, CH3+; formaldehyde, CH2O)
- carbocation, CH3+
- valence electrons = 4 + 3x1 - 1 = 6
- three bonds possible, since only three pairs, single bonds because only have H's bound to C.
- LS: from symmetry C will be central atom, therefore=
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- 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
- 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: from symmetry C will be central atom, therefore=
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- 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
- 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=
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- 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
- 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=
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- 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
- 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=
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- 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
Exceptions to the "Octet Rule." There are two categories of exceptions to the octet rule among representative elements:
To help determine if the octet rule is followed recall Clark's Method (abbreviated) for determining bonding in covalent Lewis Structures:
Small atoms:
- Linear with angles of 180° ( BeCl2) Note that this is an example where the "hi-lo" rules will lead us astray. The "1.7 rule" for electronegativities on the other hand predict covalent bonding. Also, the extreme small size of Be also leads to a prediction of greater covalent bond charecter than expected for other alkaline earth elements.
- valence electrons = 2 + 2x7 = 16
- from symmetry Be will be central atom, therefore only two pairs on Be with single bonds to Cl
- LS:
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- steric number = 2, so linear electronic geometry, so
- linear molecular geometry
- Trigonal planar with angles of 120° ( BCl3)
- valence electrons = 3 + 3x7 = 24
- from symmetry B will be central atom, therefore only three pairs on B with single bonds to Cl
- LS:
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- steric number = 3, so trigonal planar electronic geometry, & three bonded atoms so
- trigonal planar molecular geometry
Representative atoms with empty d-shells can also have what are sometimes referred to as expanded valence shells. In these cases the d-orbitals also participate in bonding enabling more bonds to be formed. Thus two additional electroninc geometries are possible:
- Trigonal bipyramidal with angles of 90° & 120° results when the valence shell is expanded to accomodate 10 electrons in five pairs.
- Octahedral with angles of 90° results when the valence shell is expanded to accomodate 12 electrons in six pairs.
These two electron pair geometries can lead to six new molecular geometries in addition to another way to make a linear molecule:
- Trigonal bipyramidal with angles of 90° & 120° (PCl5)
- Seesaw with angles of 90° & 120° (SF4)
- T-shaped with angles of 90° (ClF3)
- Linear with angles of 180° (I3-)
- Octahedral with angles of 90° (AsF6-)
- Tetragonal pyramidal with angles of 90° (ICl5)
- Square planar with angles of 90° (XeF4)
Solutions for the examples above are available on the supplement.
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© R A Paselk
Last modified 8 July 2002