Module 3
Predicting Geometry
John Pollard
University of Arizona
The molecular structure of a compound is dependent on the position of
the atoms, the atom connectivity and molecular geometry. All three
factors are dependent on valency of the atoms involved and on the
affinity that the atoms have for electrons which is related to their
ionization potentials. To illustrate, let’s look at 3 cases and illustrate
some simple rules that can be used to build the Lewis structures.
Case 1- H2O
Rule 1: Choose the central atom; never H as it only forms 1
bond. The central atom tends to be the atom with the lowest
1st ionization energy.
In the case of water, O is the central atom. It has a higher
ionization potential than H but H is never going to be the
central atom.
H
O
Periodic Behavior
1st IE increases
H
Rule 2: Count up the total number of valence electrons. In the case of H2O, there are 6
from O and 1 from each H giving a total of 8. Due to spin pairing, we can treat the
electrons as pairs. 8 electrons is 4 pairs.
Rule 3: Connect all the peripheral atoms to the central atom with single bonds (1 pair
each)
H
O
H
This uses 2 pairs and leaves 2 left to be place.
Rule 4: Use the remaining pairs to satisfy the octet rule in each atom. Start with the
outside atoms and work in (H is excluded of course as it will always have 1 bond). Any
left over pairs of electrons should be place as lone pairs on the central atom. Multiple
bonds are formed from lone pairs to complete the unfilled octets of atoms.
H
O
H
This is now the complete Lewis structure for H2O. The octet for O is satisfied by placing
2 lone pairs of electrons on it.
Case 2- CH2O
Rule 1: Choose the central atom; never H as it only forms 1 bond. The central atom
tends to be the atom with the lowest 1st ionization energy.
In CH2O, C has the lowest ionization potential (it is further to the left and down on the
periodic table) so it is the central atom. The other atoms are then place around it.
O
C
H
H
Rule 2: Count up the total number of valence electrons. 4 for C, 6 for O and 1 for each
H makes 12 total or 6 pairs.
Rule 3: Connect all the peripheral atoms to the central atom with single bonds (1 pair
each)
O
C
H
H
This uses 3 pairs and leaves 3 left to use.
Rule 4: Use the remaining pairs to satisfy the octet rule in each atom. Start with the
outside atoms and work in (H is excluded of course as it will always have 1 bond). Any
left over pairs of electrons should be place as lone pairs on the central atom. Multiple
bonds are formed from lone pairs to complete the unfilled octets of atoms.
O
C
H
H
As is always the case, one bond to H completes its bonding. According to the rule, the
remaining 3 pairs are placed on O to complete its octet. The only problem is that C still
does not have a complete octet. In the case where a central atom cannot complete its
octet and there are no more electron pairs to distribute, lone pairs from the outside atoms
should be moved in to form multiple bonds. In this case, moving 1 lone pair to a bonding
pair creates a C=O double bond while completing the octets of both.
O
O
C
C
H
H
H
H
correct structure
Case 3- O3
Rule 1: Choose the central atom; never H as it only forms 1 bond. The central atom
tends to be the atom with the lowest 1st ionization energy.
In this case, all the atoms are the same so placing one O in the center with the other 2
around it is a good start.
O
O
O
Rule 2: Count up the total number of valence electrons. 6 x 3 = 18 total or 9 pairs
Rule 3: Connect all the peripheral atoms to the central atom with single bonds (1 pair
each)
O
O
O
There are now 7 pairs remaining
Rule 4: Use the remaining pairs to satisfy the octet rule in each atom. Start with the
outside atoms and work in (H is excluded of course as it will always have 1 bond). Any
left over pairs of electrons should be place as lone pairs on the central atom. Multiple
bonds are formed from lone pairs to complete the unfilled octets of atoms.
O
O
O
O
O
O
O
O
O
resonance hybrid
There are actually two options for the final structure. One places the double bond on the
“left” and the other on the “right. These structures are energetically equivalent and as a
result, a hybrid of the two structures best represents the distribution of charge in O3. In
the resonance hybrid, 3 electron pairs are delocalized over 2 bonds. This delocalization is
a stabilizing factor for the molecule as electron-electron repulsions are minimized
resulting in a lowering of the overall potential energy of the system. The resonance
hybrid structure of O3 is verified by experimental data. Both oxygen-oxygen bonds in O3
are found to be equivalent in length which indicates that they have an equal charge
distribution and bonding modality.
Resonance hybrids are seen in other important compounds. The following 2 cases
explore other situations where hybrids are seen as the best description of the charge
distribution in molecules.
Case 1- Benzene is another example of a compound with energetically equivalent
resonance structures which result in a resonance hybrid description for the charge
distribution. As with O3, electrons are delocalized over the entire structure which tends
to lead to a reduction in potential energy and enhanced stability.
Benzene C6H6
Resonance
Hybrid
Case 2- NO3¯ (the nitrate ion)
Let’s build this structure first.
Rule 1: Choose the central atom; never H as it only forms 1 bond. The central atom
tends to be the atom with the lowest 1st ionization energy.
Nitrogen has the lowest ionization potential and is the central atom.
O
N
O
O
Rule 2: Count up the total number of valence electrons
Each O contributes 6 (18 total), N contributes 5, and one more is added for the negative
charge on the molecule. This gives 6+6+6+5+1 = 24 total or 12 pairs.
*Whenever a molecule is negatively charged, the same value of valence electrons are
added to the total (-1 adds 1, -2 adds 2….). If a molecule has a positive charge, valence
electrons are removed from the total count (+1 charge removes 1, +2 removes 2…).
Rule 3: Connect all the peripheral atoms to the central atom with single bonds (1 pair
each)
O
N
O
O
Rule 4: Use the remaining pairs to satisfy the octet rule in each atom. Start with the
outside atoms and work in (H is excluded of course as it will always have 1 bond). Any
left over pairs of electrons should be place as lone pairs on the central atom. Multiple
bonds are formed from lone pairs to complete the unfilled octets of atoms.
O
N
O
O
There are 3 energetically equivalent resonance hybrids are possible.
O
O
O
O
N
N
N
O
O
O
O
O
As with O3 and benzene, NO3¯ is stabilized by delocalizing charge over multiple bonds.
Electron Repulsion and Molecular Geometry
Charge delocalization also plays a dominant role in the 3D geometry that
the atoms assume in molecules. The lowest energy molecule geometry
results from the high electron density regions around atoms spacing as far
apart as possible. Referred to as Valence Shell Electron Pair Repulsion
or VSEPR, is the theory that will guide us in predicting the lowest energy
geometry for molecules.
To demonstrate, let’s look at some cases.
Case 1- CO2
The Lewis structure for CO2 is
O C O
In determining the structure, we look at the number of regions of electron density around
the central atom. This is equal to the number of atoms bonded plus the number of lone
pairs. In this case, carbon has 2 regions as it is bonded to 2 atoms. Therefore, the oxygen
atoms will space themselves as far as possible which places them 180º from each other
resulting in a linear geometry. In general, any central atom surrounded by 2 regions of
electron density will exhibit a linear geometry with the coordinated atoms being 180º
from each other.
Case 2- CH2O
As previously determined, the Lewis structure for CH2O is
O
C
H
H
In this case, the central carbon has 3 regions of electron density around it (bonded to 3
atoms). The lowest energy geometry spaces the surrounding atoms approximately 120º
from each other resulting in a trigonal planar geometry. In the case of CH2O, the H-CH angle is 118º due to the extra electron density from the C=O double bond and oxygen
lone pairs electrons (but this does not change the molecular geometry label of trigonal
planar).
Case 3- O3
As previously shown, the Lewis structure of O3 is:
O
O
O
The central oxygen has 3 regions of electron density around it (2 bonded atoms and 1
lone pair). Therefore, the 3 regions space themselves into a trigonal planar electron pair
geometry but since the molecular geometry only takes into account atom positions, the
molecular geometry is considered bent. The electron pair geometry and molecular
geometry are only the same for a molecule if there are no lone pairs on the central atom.
In general, lone pairs have a stronger repulsion effect than bonded pairs so the actual OO-O bond angle is 116.8º, slightly smaller than 120º.
Case 4- CH2F4
The Lewis structure for CH2F2 is:
H
H C F
F
A question that is often asked about a structure like this is whether or not it matters if the
H atoms are placed “across” or “next to” each other in the Lewis structure.
Determination of the molecular geometry will show us that it really doesn’t matter.
The central carbon has 4 regions electron density around it. The lowest energy
configuration here puts the atoms in a geometry called tetrahedral, where all the angles
are approximately 109.5º. This is why the placement of the outside atoms in the 2D
Lewis structure is irrelevant; they are all approximately at an equal distance and “next to”
each other. To represent the 3D structure in 2D (as you are seeing now) chemists often
use perspective cues that imply atoms coming “out of the plane” or “into the plane”.
Wedged lines imply coming out of the plane of the paper or computer screen and dashed
lines imply going back into the plane of the paper or computer screen.
H
H
C
F
F
Case 5- NH3
The Lewis structure for NH3 is:
H N H
H
The central atom N has 4 regions of electron density around it which indicates that the
electron pair geometry is tetrahedral. The molecular geometry is trigonal pyramid and
due to the increased repulsive effects of the lone pair on the nitrogen, the H-N-H bond
angles are around 107º. The 3D Lewis structure below shows one of the tetrahedral
positions being occupied by the lone pair of the nitrogen.
N H
H
H
Case 6- H2O
The Lewis structure for water is:
H
O
H
The oxygen has 4 regions of electron density around it giving an electron pair geometry
of tetrahedral. With 2 of the regions being lone pairs, the molecular geometry is bent (as
on O3). The H-O-H angle is pinched by the presence of the 2 lone pairs an is 105º.
Molecules without “central” atoms.
VSEPR can be used to determine the local geometry around any atom. In the case of
molecules like hydrocarbons or many biologically relevant molecules, there is no one
central atom. Despite this, VSEPR can be used to determine the local geometry around
any atom in these molecules. To illustrate this, let’s look at some of these molecules.
Case 1- C2H4 (ethylene)
Ethylene is a widely used compound in industry and is important hormone in biology.
The Lewis structure of ethylene is:
H
approx 120
o
H
C C
H
H
Each carbon has 3 regions of electron density respectively and are therefore trigonal
planar. All the atoms in ethylene are in the same plane making the molecule relatively
flat.
Case 2- Atenolol (C14H22N2O3)
Atenolol is the #4 most widely used pharmaceutical drug on the market. It is a betablocker used to lower blood pressure, reduce chest pain and the risk of recurrent heart
attacks. It would be impossible to draw the Lewis structure of Atenolol from scratch
without having very specific structural information (so don’t worry about having to draw
molecule this complicated). Let’s look at the structure and apply VSEPR to determine
the local geometries around various atoms. Here is the Lewis structure.
H
H
H
C
H
H
O
H
H
N
C
C
H
O
C
H
C
H
H
H
C
H
H
C
C
C
C
C
H
O
H
H
C
H
C
H
N
H
H
H
It is common with large molecules such as this to simplify the structure by drawing what
is called a line structure. In the line structure, all the hydrogens attached to carbons re
left out and only the letters for non carbon atoms are indicated. Carbons are implied by
the bends of intersecting lines. If there is a C≡C triple bond (which is linear) the carbons
are implied at the end of each ≡. The line structure for Atenolol looks like this.
H
O
H
N
O
O
H
N
H
Just as with the previously discussed molecules, determining the geometry around an
atom begins by adding the number of bonded atoms with the lone pairs. This gives you
the regions of electron density and directly correlates to the electron pair geometry. If
there are lone pairs, then the molecular geometry will be different than the electron pair
geometry. Here are the regions of electron density and associated molecular geometries
for some select atoms on Atenolol.
4 regions of e
bent
H
H
O
N
O
O
H
4 regions of e
tetrahedral
4 regions of e
tetrahedral
N
3 regions of e
trigonal planar
3 regions of e
trigonal planar H
4 regions of e
trigonal pyramid
Here are summaries of geometries associated with VSEPR.
# eregion
s
2
Example
e- pair
geometry
Molecular
geometry
Linear
(180o)
Linear
Trigonal
Planar
(~120o)
3
118o
Trigonal planar
Trigonal
Planar
(< 120o)
3
Bent or Angular
# eregion
s
4
Example
Cl
F
F
e- pair
geometry
Molecular
geometry
Tetrahedral
(109o)
Cl
4
Tetrahedral
Tetrahedral
(< 109o)
lone pair of electrons
in tetrahedral position
N
H
4
107.8o
H
H
Trigonal Pyramid
Tetrahedral
(< 109o)
O
H
H
104.5o
Bent or Angular
Structures, structures, structures.
Draw the Lewis structures and determine the electron pair and molecular geometries for
the following molecules. Identify any structures that have resonance hybrids and draw
them.
a) COCl2 b) CO32- c) NF3 d) N2O e) CH2Cl2 f) HNO3 g) CH2O2 h) SCl2
i) N2O4 j) CH3OH k) PH3 l) H3O+ m) CS2 n) C2H2
Nitrogen Power
In addition to NH3, nitrogen forms three other compounds with hydrogen
that are useful as rocket propellants: N2H4, N2H2 and N4H4 (which does
not form a ring with the N’s). Draw the Lewis structures of each and use
these to comment on the bond strengths and lengths of the nitrogennitrogen bonds in each (in general, triple bonds are stronger than double
which are stronger than single). Also, assign an electron pair and
molecular geometry to each N in all three structures.
Beginning of Life
Some scientists postulate that many of the organic molecules
important for life on early earth arrived on meteorites. In 1969, the
Murchison meteorite landed in Australia in 1969 and was found to
contain 92 different amino acids including 21 found in life forms
found on earth. A skeleton structure (only the single bonds are shown)
of one of these alien amino acids is shown below. Redraw the
complete Lewis structure. Also, draw a 3D image of this molecule
using wedges and dashed lines (this is a complex one so to practice, start by drawing the
3D structure of CH3OH). Assign molecular geometries to all the C and N atoms.
H
H3N
C
C
CH2 O
CH3
O
Trippin’
The Lewis structures for mescaline, a hallucinogenic substance, and dopamine, a
neurotransmitter, are shown below. Determine the molecular geometry around the atoms
indicated by arrows. Also, propose a reason for mescaline’s ability to disrupt nerve
impulses.
NH2
NH2
O
O
OH
O
OH
mescaline
dopamine
Rhubarb Pie
Oxalic acid (H2C2O4) is found in toxic concentrations in rhubarb leaves. The acid forms
two ions, HC2O4¯ and C2O42-. Draw the Lewis structures for all 3 and identify any
resonance hybrids.
Chloral
Chloral, Cl3C-CH=O, reacdts with water to form the sedative chloral hydrate, Cl3CCH(OH)2. Draw the Lewis structures for these substances and describe the change in
molecular shape, if any, that occurs around each of the carbon atoms during the reaction.
Your are Lord of the Rings
Each of the following formulas are for compounds where the carbons form a ring. Draw
each Lewis structure and determine if any of them have resonance hybrids.
a) C3H4 b) C3H6 c) C4H6 d) C4H4 e) C6H6