VSEPR Theory Part III
by Janet Gray Coonce, MS
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Transcription notes for Part III begins at Video time 21:18.
If instead of 3 equal regions of electron density (single bonds), we have only 2 bonds and an unshared pair of electrons, we could represent that by the notation, AB2u.
The electronic geometry would still be trigonal planar, but as predicted by the VSEPR theory, the lone pair of electrons will repel or bend the less electronegative atoms sharing electrons. The molecular geometry is bent. The bond angle will therefore be close to but less than 120 degrees.
Consider a compound represented by the formula AB4. An example is methane, CH4.
Here we have 4 valence shell electron pair regions repelling each other to get as far apart from the other regions as possible. The orbitals are sp3 hybridized.. This is a tetrahedral electronic and molecular geometry. There are no lone pairs of electrons so the electronic and molecular geometry are the same.
The 3 dimensional model illustrates tetrahedral geometry. All atoms bonded to the central atom are equal and symmetric so there is no net dipole. All the bond angles are the same and equal to 109.5 degrees.
If one of the atoms were different, then there would be unequal charges surrounding the central atom resulting in a net dipole. An example would be methanol, CH30H. The oxygen atom represented by the silver ball is more electronegative than the hydrogens (green) at the other end of the central carbon atom (blue). Methanol is therefore a polar compound.
In summary, if all atoms around the central atom are the same, we would predict no net dipole. If one of them are different, we would predict a dipole with the electrons being pulled toward the more electronegative region. A dipole would also be predicted if 2 of the atoms were different.
If one of the regions in the four sp3 hybridized orbitals were occupied by a lone pair of electrons, the formula would be represented by AB3u.
There still would be 4 regions of electron density, 3 bonds and one pair of unshared electrons. Ammonia gas, NH3 is an example.
The electronic geometry is tetrahedral as represented by the balloon model. There are still 4 regions of electron density. There are 3 pairs of shared electrons (purple balloons) which participate in bonds and one unshared pair of lone electrons represented by the yellow balloon. The lone pair of electrons are more electronegative and will repel the less negative shared pairs of electrons. The electronegative forces of the lone pair will repel the electron regions involved in bonding with the atoms and result in trigonal pyramid molecular geometry.
This is the molecular model of a AB3u molecule. Note that the 4 atoms are not in the same plane. The atoms in the molecule are geometrically arranged in a trigonal pyramid configuration. We would expect there to be a dipole with the lone pair of electrons above the carbon atom to be much more electronegative than the hydrogen atoms at the other end of the molecule. We would therefore predict that ammonia gas would be a polar molecule.
If there are 4 regions of electronic density (sp3 hybridized orbitals) surrounding a central atom with 2 pairs of electrons shared in bonds and 2 lone pair of unshared electrons, we could represent this theoretical molecule by the formula AB2u2. A familiar example is H20.
The electronic geometry is tetrahedral. The unshared electrons will repel the less electronegative atoms (B) which will result in a bent molecular geometry. The bond angle (B-A-B) will therefore be less than 109.5 degrees which is the angle which would exist if the molecular geometry were not bent but tetrahedral.
Here force is applied to the green atoms to illustrate how the forces from the lone pairs of electrons repel the green atoms. This illustrates why the VSEPR theory predicts a bond angle of less than 109.5 degrees.
Using the balloons to show the tetrahedral electronic geometry of H20 notice that the 2 lone pair (yellow balloons) of unshared electrons take up more space on one end of the oxygen atom than the space taken up by the bonds sharing electrons with hydrogen (purple balloons). These will create even more electronegative force than the single pair of unshared electrons we saw in NH3. We would therefore predict an even smaller bond angle than we saw in NH3. That is the bond angle is quite a bit less than 109 degrees which would exist if all 4 regions of electronegativity surrounding the central atom were equal. So the H-O-H bond is bent even more than the H-N-H bonds we saw in NH3. We would also predict water to be polar with the more electronegative forces in the region of the unpaired electrons and the less electronegative forces bonding to the hydrogen atoms.
Think about this for a minute. What can you predict? Since we predict NH3 to be polar, what do you predict would happen if you bubbled NH3 gas through a polar liquid such as distilled water (pH = 7)? Would the resultant solution be an acid or a base?
If there are 5 equal electron regions (dsp3 hybridized) participating in bonds surrounding a central atom the formula can be represented by AB5. PCl5 gas is an example of a molecule with 5 equal bonds between a central atom of phosphorus and 5 atoms of chlorine. The electronic and molecular geometry is trigonal bipyramid. TheB-A-B bond angles in the axial plane are identical and equal to 90 degrees. The B-A-B bond angles in the equatorial plane are also identical and equal to 120 degrees.
If one of the 5 electron regions of a molecule with bipyramid electronic geometry has a non-bonded lone pair of electrons it could be represented by the formula AB4u. SFl4 gas is an example.
While the electronic geometry is still trigonal bipyramid, the molecular geometry is a “see-saw.”
Here the AB4u molecular geometry, “see-saw” is demonstrated.
If two of the 5 electron regions of a molecule with bipyramid electronic geometry has a non-bonded lone pair of electrons it could be represented by the formula AB3u2. The electronic geometry is trigonal bipyramid. The orbitals are dsp3 hybridized. The molecular geometry is “T-shaped.”
ClF3 is an example of a molecule with the AB3u2 configuration. The bond angles are 90 degrees and the atoms are in a T-shaped configuration.
If 3 of the regions in a molecule with 5 electron regions have a lone pair of electrons, the configuration is expressed by the notation AB2U3. The orbitals are dsp3 hybridized. The electronic geometry is still trigonal bipyramid.
The molecular geometry is linear, the bond angles are 180 degrees. XeF2 is an example of a molecule with this configuration. It is a strong fluorinating agent.
If there are 6 equal electronic bonding regions around a central atom, the configuration can be expressed by the notation AB6.
A central atom with 6 sigma bonds is d2sp3 hybridized and is octahedral in shape. An example is SF6. The central atom has 2d + 1s + 3p = 6 hybridized orbitals. The bond angles are all equal at 90 degrees. The electronic geometry is octahedral. The molecular geometry is octahedral.
If one of the sigma bonds is replaced by a pair of unshared electrons, the configuration is expressed by the notation AB5u. There are still 6 regions of electron density. The orbitals are d2sp3 hybridized.
The molecule would have a square pyramid geometry. BrF5 is an example. Like ClF3 it is also an interhalogen compound and very reactive.
If a second of the sigma bonds is replaced by a pair of unshared electrons, the configuration is expressed by the notation AB4u2. There are still 6 regions of electron density. The orbitals are still d2sp3 hybridized.
The molecule is still has d2sp3 hybridized electronic orbitals in an octahedral geometric configuration but the molecular geometry is square planar. All 5 atoms are in the same plane with the bonded atoms forming a square around the central atom. XeF4 is an example.
Transcription by James C. Gray MD FACOG
Janet Gray Coonce 