Chem 109 - General Chemistry - Spring 2011

Lecture Notes 28: 4 April

Return Exam 2

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

Draw a correct Lewis Structure.
Determine the Steric Number = the number of bonded atoms + the number of lone pairs.
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). [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

Exceptions to the "Octet Rule." There are two categories of exceptions to the octet rule among representative elements:

Small atoms in Periods I & II which cannot accomodate or normally don't accomodate a full set of p-electrons. Thus hydrogen and helium have no ground state p-orbitals and so can only carry a maximum of 2 electrons (a duet). The first three atoms of Period II form a transition between hydrogen (duet) and carbon (octet). Thus lithium, because of its very small size, often forms covalent rather than ionic compounds, with a single bond (e.g. LiH). Similarly beryllium can form compounds with a quartet of electrons around it (e.g. BeH₂) and boron can accomodate a sextet (e.g. BH₃).
Atoms in the p-block of Periods 3 and higher can have "expanded valence shells" with 10 or 12 electrons in the outermost shell by using some of their "empty" d-orbitals to hold the extra electrons.

To help determine if the octet rule is followed recall Clark's Method (abbreviated) for determining bonding in covalent Lewis Structures:

Add up all of the valence electrons in the structure (remember to add one electron for each negative charge, or subtract one for each positive charge)

If e^- = 6y + 2 where y = # atoms other than H, then octet rule is followed with single bonds only.
If e^- < 6y + 2 then probably have multiple bonding with the number of multiple bonds = /2 (remember a triple bond is 2 multiple bonds!). However, note the exceptions below with small atoms (H, Li, Be, and B).
If e^- > 6y + 2 then have an expanded valence shell. Note that if = 2, then pentavalent (10 electrons in the valence shell) , and if = 4, then hexavalent (12 electrons in the valence shell).

If you can draw more than one structure, then chose the most symmetrical.

If two or more structures are equally symmetrical, then you probably have resonance and should show all structures connected by double arrows.

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:
- 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:
- steric number = 3, so trigonal planar electronic geometry, & three bonded atoms so
- trigonal planar molecular geometry

Expanded valence shells - Quick overview

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. Two additional electronic geometries are possible:

Trigonal bipyramidal with angles of 90° & 120° results when the valence shell is expanded to accomodate 10 electrons in five pairs (steric number = 5).
Octahedral with angles of 90° results when the valence shell is expanded to accomodate 12 electrons in six pairs (steric number = 6).

These two electron pair geometries can lead to six new molecular geometries in addition to another way to make a linear molecule: