VSEPR: valence shell electron pair repulsion

VSEPR theory enables us to take our 2-dimensional Lewis structures into 3-dimensional shapes. The entire theory relies on the fact that electron pairs (lone pairs, single bonds, double bonds, or triple bonds) tend to repel each other such that atoms and lone pairs are maximally separated from one another.

The first step in working with the theory is to draw the Lewis structure for the compound. Let's take the example of water: H₂O.
We take special note of the central atom: the oxygen atom. For the central atom, we need to count how many "electron pairs" we have. An electron pair can be a lone pair, a single bond, a double bond, or a triple bond. The total number of electron pairs determines the electron geometry.

The electron geometry for water is tetrahedral.
In contrast to the electron geometry, the molecular shape only takes into account how the atoms are arranged around the central atom. When the central atom has no lone pairs, the electron geometry is the same as the molecular shape. However, if the central atom has lone pairs, the molecular shape is different than the electron geometry.

For water, the central atom (oxygen) has 2 lone pairs, which affect the way the hydrogen atoms are arranged.

The molecular shape for water is bent (or V-shaped).
You can use the images below to help you figure out the electron geometries and molecular shapes.

Molecular shape and molecular polarity

For both molecules, (1) draw the Lewis structure, (2) determine the electron pair geometry, and (3) determine the molecular shape.
  1. BF₃
  2. NF₃
hint
Start by drawing the Lewis structures for both molecules.

Then determine the electron geometry and the molecular shape of the molecules. Remember that electron geometry takes into account all the atoms and lone pairs around the central atom, whereas molecular shape only takes into account the atoms around the central atom.
Lewis Structures
Electron Pair Geometry
The electron pair geometry is determined by counting the total number of bonds and lone pairs around the central atom. This number is called the steric number (SN).

For BF₃, the central boron has 3 bonds around it, so the electron pair geometry is trigonal planar (SN = 3).

For NF₃, the central nitrogen has 3 bonds around it and 1 lone pair, so the electron pair geometry is tetrahedral (SN = 4).
molecular shapeS
The molecular shape is determined by the arrangement of the atoms around the central atom. Lone pairs can affect the shape, but we are interested only in the arrangement of the atoms.

For BF₃, since there no lone pairs around the central boron, the electron pair geometry is the same as the molecular shape. Therefore, BF₃ is trigonal planar in shape.

For NF₃, since there a lone pair on the central nitrogen, the electron pair geometry is different from the molecular shape. NF₃ is trigonal pyramid in shape because of the presence of the lone pair.
Between BF₃ and NF₃, which molecule is polar?
hint
Once you have drawn the Lewis structure and determined the molecular shape, draw the bond dipoles based on the electronegativities of the atoms in each bond.
Solution
Once we have drawn the Lewis structure and determine the molecular shape, we need to draw in bond dipoles.

Bond dipoles are arrows (vectors) that we draw between two atoms (or every bond). The positive end of the arrow originates at the less electronegative atom and the negative end of the arrow points toward the more electronegative atom. Remember that electronegativity increases from left-to-right and decreases from top-to-bottom on the periodic table.
For BF₃, the bond dipoles (in blue) point from the boron to the fluoride atoms. For the NF₃, the bond dipoles (in blue) point from the nitrogen to the fluoride ions.

If the bond dipoles cancel out (pointing in equal and opposite directions) the molecule is nonpolar overall. This is the case for trigonal planar BF₃.

If the bond dipoles do not cancel out, the molecule is polar.  This is the case for trigonal pyramidal NF₃. The molecular dipole (the sum of the bond dipoles) is drawn in red.
Determine if IBr₂⁻ is polar or nonpolar.
solution
First, draw the Lewis structure.

The electron pair geometry for IBr₂⁻ is trigonal bipyramid because the central iodine has 2 atoms and 3 lone pairs around it (SN = 5, dsp³).

The molecular shape for IBr₂⁻ is linear because of the 3 lone pairs on the iodine.

After this, we can draw the bond dipoles (shown in blue) for each of the I-Br bonds. These bond dipoles cancel out completely because the bond dipoles are pointing in opposite direction. So the overall molecule is nonpolar.
How to get the molecular shape?
Examples: CCl₄, SCl₄, and XeCl₄
How to determine molecular polarity?
Examples: CCl₄, SCl₄, and XeCl₄

More practice problems

Draw the Lewis structure for each compound and then determine the electron geometry, molecular shape, and hybridization of the central atom.
  1. SF₂
  2. NNO
  3. SeF₆
  4. ClF₃
  5. IF₅
  6. PCl₅
solution 1
SN = 4
Electron geometry = tetrahedral
Molecular shape = bent or V-shaped
Hybridization = sp³
solution 2
**Middle structure is best representation of NNO**
SN = 2
Electron geometry = linear
Molecular shape = linear
Hybridization = sp
solution 3
SN = 6
Electron geometry = octahedral
Molecular shape = octahedral
Hybridization = d²sp³
solution 4
SN = 5
Electron geometry = trigonal bipyramid
Molecular shape = T-shaped
Hybridization = dsp³
solution 5
SN = 6
Electron geometry = octahedral
Molecular shape = square pyramid
Hybridization = d²sp³
solution 6
SN = 5
Electron geometry = trigonal bipyramid
Molecular shape = trigonal bipyramid
Hybridization = dsp³
From the compounds above, determine which are polar and which are nonpolar.
polar
SF₂, NNO, ClF₃, and IF₅

The bond dipoles are drawn in orange.

The (net) molecular dipoles are draw in blue.
nonpolar
SeF₆ and PCl₅

The bond dipoles are drawn in orange.

Since the bond dipoles all cancel out there are no (net) molecular dipoles drawn.