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How do we use Lewis dot structures to illustrate bonding in CO2? We we know that the oxygen atoms will repel each other so carbon will be in the center with an oxygen at each end. How many valence electrons do each of those atoms have? Well we know oxygen has six valence electrons because it is in periodic table group 6A. Carbon is in group 4A of the periodic table so it has 4 valence electrons.
This Lewis dot structure demonstrates that oxygen has 2 unpaired electrons and has 2 unpaired electrons available for covalent bonds. Carbon has 4 unpaired electrons available for 4 covalent bonds.
The illustration shows how to use a Lewis dot structure to represent the bonding in the CO2 molecule.
Lewis Dot Structure for Methanol
The chemical symbol for methanol is CH30H or CH40. Hydrogen can only form one bond so it can not be the central atom. Hydrogen is always on the outside of a molecule with which it is sharing electrons. We know therefore that Carbon and Oxygen will be in the center with 3 hydrogen atoms surrounding the carbon atom and a single hydrogen bonding the oxygen atom.
Here we demonstrate the valence electrons with dots. Carbon has 4, hydrogen 1 and oxygen 6. It is easy to see that when the electrons are shared each atom will have its outer shell completed. With sharing each hydrogen will have 2 electrons, carbon 8 electrons and oxygen 8 electrons in the outer shell. The molecule is balanced and happy 
.
When the dots are connected between the shared electrons, the Lewis dot structure is completed. This is the Lewis dot structure for methanol.
Transcription by James C. Gray MD FACOG
Video by Janet Gray Coonce, MS
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How does one draw the Lewis dot structure for the nitrate anion? The formula of the nitrate anion is NO3–
The first step is to identify the central atom, in this case nitrogen, and write the symbol for nitrogen, N. The 3 oxygen atoms, symbol O, are distributed around the central atom, N. Now look at the periodic table to determine the number of valence electrons. Nitrogen is in Group VA and has 5 valence electrons. Oxygen is in Group 6 and has 6 valence electrons. Oxygen is more electronegative since it has more valence electrons and is to the right of nitrogen in the periodic table. In this illustration the valence electrons are distributed around each of the atoms. The octet rule predicts that the electrons will want to form 4 pairs around each atom. Notice that oxygen has 2 electrons available to form a pair and nitrogen has 3 lone electrons.
Since this is an anion and has a negative charge, an extra electron must be added to one of the more electronegative oxygen atoms. This extra electron is indicated at the arrow in the drawing above. The minus sign indicates that this electron accounts for the negative charge of the anion. This electron was donated by the associated cation which has a positive charge.
Now it’s time to “connect the dots.”
Our working diagram is on the left. In order to satisfy the octet rule on each atom, nitrogen formed a double bond with one of the oxygens, it donated an electron to another oxygen and the 3rd oxygen gained an electron from its associated cation. The octet rule is fulfilled for each atom and the overall net charge on the molecule is negative 1.
There are 3 possible Lewis structures for the nitrate anion. These are referred to as resonant structures and reflect the dynamic relationship of the electron sharing in the overall molecule.
Transcription by James C. Gray, MD FACOG
Video by Janet Gray Coonce, MS
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C2H4 is the symbol for the simple organic compound, ethylene. It is the simplest alkene (hydrocarbon with carbon-carbon double bonds). This illustration demonstrates how to draw the Lewis dot structure for C2H4.
The first step is to write the symbols for carbon (C) adjacent to each other in the center and the hydrogens (H) will have to surround the carbons. The hydrogen atoms will be distributed evenly on the outside of the molecule so we draw 2 symbols for H outside each carbon. The next step is to draw the valence electrons. Hydrogen has 1 carbon has 4.
In this diagram, the dots were connected to show that the valence electrons are being shared in covalent bonds. Each of the lines represent 2 electrons being shared. Remember that hydrogen needs 2 valence electrons (a duet) to complete it’s outer shell. Carbon needs 8 valence electrons (an octet) to complete its outer shell. As we look over this diagram, we can see that by sharing in covalent bonds, every atom has a full outer shell of valence electrons (2 around each hydrogen and 8 around each carbon).
C2H2 is the symbol for the simple organic compound, acetylene. It is the simplest alkyne (hydrocarbon with carbon-carbon triple bonds). This illustration demonstrates how to draw the Lewis dot structure for C2H2.
To begin, a C for each central carbon was drawn in the center. The hydrogens must go on the outside and they must be as far apart from each other as possible, so we draw one on each side of the molecule in line with the carbon atoms. Next we put in a dot for each electron and connect the dots.
CH2O is the symbol for the simple organic compound, formaldehyde. It is the simplest aldehyde (hydrocarbon with a carbon atom connected to an oxygen atom by a double bond and to a hydrogen atom by a single bond). This illustration demonstrates how to draw the Lewis dot structure for CH2O.
On the left the attempt to draw the Lewis structure failed when we attempted to join one of the hydrogens to oxygen. However, on the right, we know it is the correct Lewis structure because all atoms have their complete set of valence electrons. There are 8 surrounding oxygen and carbon (a complete octet) and hydrogen has its full set of 2 (duet).
Solutions to problems:
Lewis CHCl3
CHCl3 = Chloroform, also known as trichloromethane and methyl trichloride
Lewis HCN
Lewis Hydrogen Cyanide
Lewis H2O2
Lewis Hydrogen Peroxide
Lewis H2CO2 or HCO2H
Lewis Formic Acid
Transcription by James C. Gray MD FACOG
Video by Janet Gray Coonce, MS
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Next to draw the Lewis dot structure for ethanol, the alcohol which is found in fermented beverages. Fermentation is the organic chemical reaction whereby yeast metabolize sugars which are found in fruits to ethanol and carbon dioxide. The molecular formula for ethanol is C2H6O. There would be 2 ways of writing the Lewis structural formula C2H6O. When compounds have the same molecular formula but different structural formulas they are referred to as isomers. We want to draw the Lewis dot structure for ethanol.
Another way of writing the molecular formula for ethanol helps to illustrate exactly which isomer we are referring to and how we should draw the Lewis dot structure. Writing the formula as CH3CH2OH indicates we are specific that we are referring to the isomer of C2H6O which is ethanol.
By writing the formula in this way we know that there is a CH3 group, a CH2 group and an OH group. Writing the formula in this way (CH3–CH2 –OH) tells us which of the structural formulas we are referring to and how to draw the Lewis dot structure.
When we draw the Lewis dot structure, we first write the central atoms. Remember hydrogens always go on the outside. We start with the CH3 group, then CH2, then OH. Next step is to place dots for the valence electrons. Hydrogen gets 1, carbon gets 4 and oxygen gets 6. The lone electron is available for bonding.
The Lewis structure is complete when we connect the dots to show the sharing of electron pairs between different atoms.
Here is the Lewis dot structure for the other isomer of C2H6O, dimethyl ether. Another way of writing this molecular formula is CH3-O-CH3.
Here is the Lewis dot structure for dimethyl ether. Although it has the same molecular formula as ethanol, it is a different compound with different physical properties because of the different way the atoms are arranged. Dimethyl ether is a gas, not a liquid. Dimethyl ether is used as a refrigerant and is used to freeze warts much the same way as liquid nitrogen. It is not something anyone would want to attempt to drink. It is an isomer of C2H6O which is the same molecular formula as ethanol.
Transcription by James C. Gray MD FACOG
by Janet Gray Coonce MS
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Now let’s draw the Lewis dot structure for PCl5, phosphorous pentachloride.
The first step is to draw phosphorous (P)Pin the center. Phosphorus is in group VA in the periodic table and has 5 valence electrons. Chlorine (Cl) is a halogen. The halogens are in periodic table group VIIB and have 7 valence electrons. The 5 chlorine (Cl) atoms are on the right in the periodic table and are therefore more electronegative. They will be distributed evenly around the central phosphorus (P) atom. The next step is to distribute the valence electrons around each of the atoms. I recommend doing this on the Lewis dot drawing by drawing dots to represent the valence electrons of each atom.
In this drawing, I have drawn 7 valence electrons around each chlorine atom and 5 valence electrons around the phosphorus. The lone electron of each chlorine atom aligns with the lone electrons on the phosphorus atom.
In this step a line is drawn to connect the dots representing the sharing of the lone electrons in a bond between the P and Cl atoms.
This is the same drawing but it was cleaned up to make a neater presentation for the Lewis dot structure for PCl5. Now if we count electrons around each atom, we see that each of the chlorine atoms has a complete octet of 8 valence electrons. This means that the chlorine atoms are stable and are therefore “happy.” 
But look at the phosphorus atom there are 2 x 5 = 10 valence electrons. This is referred to as an expanded octet. Atoms in period 3 and below in the periodic table may expand their octet because they have d orbitals. This is in contrast elements with atomic numbers of 1 – 10. That includes the elements in period 1 (H and He) and elements in period 2 (Li, Be, B, C, N, O, F and Ne). The first 10 elements of the periodic table do not have d orbitals and cannot expand their valence electrons outside of their duet (H and He) or octet for those in period 2.
Transcribed by James C. Gray, MD FACOG
Videos by Janet Gray Coonce, MS
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In this tutorial I want to help you visualize the electronic and molecular geometries predicted by the valence-shell electron-pair repulsion, VSEPR, models. I will demonstrate how to represent 3 dimensional models of electronic and molecular geometry in 2 dimensional drawings.
There are 3 regions of electron density in the acetylene (C2H2) molecule. Each carbon is flanked on the outside by a single bond to hydrogen. The two central carbon atoms are sharing a triple bond. So each of the carbon atoms have 2 electron dense areas trying to get as far apart as possible. When 2 electron dense areas connected to a central atom get as far apart as possible, they will align themselves across a straight line. Acetylene therefore has linear molecular geometry. The bond angle of a straight line is 180 degrees.
Carbon dioxide, C02 is another example of a central atom with two areas of electron density repelling each other. They oxygen atoms are getting as far apart from each other as possible and form a straight line. It is linear electronic and linear molecular geometry.
If there are 3 identical atoms connected to a central atom, each will repel each other and get as far apart as possible. They will assume a trigonal planar electronic and trigonal planar molecular geometry. All 3 of the atoms bound to the central atom will be in the same plane and will be equal distant. In the case of 3 the bond angle will be 120 degrees (360/3).
Here is a central atom with 3 electron dense areas but only a lone pair of electrons is in one of the 3 areas. The lone electrons will be more electronegative than the atoms sharing electrons in a bond. The bonded atoms will be repelled by the force of the lone pair (blue arrows) and the bond angle will be less than 120 degrees. The molecular shape will be bent. The electronic geometry is still trigonal planar.
When there are 4 atoms bonding to a central atom and 4 regions of electron density, the electronic and molecular shape is tetrahedral. The atoms will arrange themselves equally around the central atom in 3 dimensions. In the 2 dimensional representation, the solid wedge represents an atom projecting up from the plane of the paper. The dotted wedge represents an atom projecting below the plane of the paper as illustrated.
If one of the 4 areas of electron density was a lone pair, the molecular shape would be a trigonal pyramid (3 atoms attached to central atom). The electronic shape would still be tetrahedral (4 areas of electronic density around the central atom).
If 2 of the 4 areas of electron density is represented by a lone pair of electrons and 2 electrons shared in a bond with another atom, then the less electronegative atoms in the bond will be repelled by the more electronegative regions occupied by the lone electrons. The electronic shape would still be tetrahedral but the molecular shape would be bent.
If a central atom shares 5 electrons by forming covalent bonds with 5 other atoms the electronic and molecular geometry is a trigonal bipyramid.
If one of the 5 areas of electron density is occupied by a lone pair of electrons, the electronic shape would still be trigonal bipyramid but the molecular shape would be a “see-saw.”
If 2 areas are occupied by lone pairs the molecular geometry will be T-shaped.
If there are 3 pairs of unshared electrons and 2 bonds, the molecular geometry will be linear. The electronic geometry is still trigonal bipyramid. This is the shape when 5 regions surrounding a central atom try to repel each other in order to get as far apart as possible.
With 6 regions all repelling each other to get as far apart as possible, the molecular and electronic geometry is octahedral.
If one of the 6 regions is a lone pair, the molecular geometry would be a square pyramid. The electronic geometry would still be an octahedral.
If 2 of the regions are occupied by a lone pair of electrons then the molecular geometry would be square planar. There are still 6 regions of electron density so the electronic geometry is still octahedral.
Review: Regions of electron density (RHED) surrounding a central atom are of 4 types:
1. single bond = a RHED consisting of 1 pair of shared electrons
2. double bond = a RHED consisting of 2 pairs of shared electrons
3. triple bond = a RHED consisting of 3 pairs of shared electrons
4. a RHED consisting of lone pair of unshared electrons.
Here is the Lewis structure for carbon dioxide, CO2. How many electron regions around the central carbon atom? Answer: There is a double bond on each side. A double bond is a single region. Therefore we know 1 + 1 = 2. So the answer is there are 2 regions. What about the molecular shape? If the electron regions associated with the more electronegative oxygen atoms repel each other and try to get as far apart as possible, what molecular shape will the CO2 molecule have? Linear. A straight line. A bond angle of 180 degrees. What is the geometry of the electron regions? They are also in a straight line. Linear.
Summary: CO2 has 2 electron regions (RHED) around the central carbon atom with no lone pairs of unshared electrons.
Electronic geometry: Linear
Molecular geometry: Linear
Here is the Lewis structure for ammonia gas, NH3. How many regions of high electron density (RHED) are there around the central nitrogen atom? Count them. There are 3 single bonds and a single lone pair of electrons. 3 + 1 = 4. If all of these electronic regions get as far apart as possible what will be the 3 dimensional shape? There are 4 points and 4 points in 3 dimensions is a tetrahedral. Therefore the electronic shape is tetrahedral. However, there are only 3 atoms around the central nitrogen. What is the MOLECULAR geometry? The 3 hydrogen atoms form a triangle in one plane and the nitrogen is above it. It is a trigonal pyramid.
Summary: Ammonia gas, NH3 has 4 electron regions (RHED) around the central nitrogen atom with 1 pair of unshared electrons.
Electronic geometry: Tetrahedral
Molecular geometry: Trigonal pyramid
Here is the Lewis structure for water, H20.
There are 4 RHED, 2 lone pairs and 2 single bonds. The 4 electronic regions will get as far apart as possible in a tetrahedral electronic geometry.
There are only 2 atoms attached to the central oxygen atom and they will be repelled by the lone pairs. Although there are only 2 atoms bonded to oxygen and they are trying to get as far apart from each other as possible, the three atoms are not in a straight line. The hydrogen atoms are also being repelled by the lone pairs of electrons. The molecular geometry is not a straight line. The molecular geometry is bent.
Summary: Water, H20, has 4 electron regions around the central oxygen atom with 2 pair of unshared electrons and 2 single bonds.
Electronic geometry: Tetrahedral
Molecular geometry: Bent
Summary of molecular geometries for each of the electronic geometries:
A. Trigonal Planar Electronic geometry = 3 RHED
1. 3 bonds + 0 lone pair = trigonal planar geometry
2. 2 bonds + 1 lone pair = bent molecular geometry
B. Tetrahedral Electronic Geometry = 4 RHED
1. 4 bonds + 0 lone pair = tetrahedral molecular geometry
2. 3 bonds + 1 lone pair = trigonal pyramid molecular geometry
3. 2 bonds + 2 lone pair = bent molecular geometry
C. Trigonal Bipyramid Electronic Geometry = 5 RHED
1. 5 bonds + 0 lone pair = octahedral molecular geometry
2. 4 bonds + 1 lone pair = see-saw or distorted tetrahedral molecular geometry
3. 3 bonds + 2 lone pair = T-shaped molecular geometry
4. 2 bonds + 1 lone pair = linear molecular geometry
D. Octahedral Electronic Geometry = 6 RHED
1. 6 bonds + 0 lone pairs = Octahedral molecular geometry
2. 5 bonds + 1 lone pair = Square Pyramid molecular geometry
3. 4 bonds. + 2 lone pair = Square Planar molecular geometry
Transcription by James C. Gray MD FACOG
by Janet Gray Coonce, MS
If you plan to view the video on your cell phone, consider your data plan and whether you should wait until you have a WiFi connection to avoid cellular charges.
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
by Janet Gray Coonce, MS
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Transcription begins at 9 minutes 45 seconds into the video.
Now lets use the VSEPR theory to predict the electronic and molecular geometry of carbon dioxide.
The first thing we have to do is to draw a Lewis dot structure. The first step is to determine which of the atoms is the least electronegative and put that atom in the center. Looking at the periodic table, the electronegativity increases from left to right. Carbon is in group IV and oxygen is in group VI, therefore, carbon is less electronegative than oxygen.
So to draw the Lewis dot structure, carbon goes in the center with oxygen on either side. The next step is to determine the number of valence electrons of each atom. Remember that valence electrons are the electrons available for bonding. We can determine the number of electrons by looking at the periodic table and the group number of the atoms. Carbon is in group IV and all members of group IV have 4 valence electrons. In the Lewis dot structure the 4 electrons are indicated by 4 dots surrounding the carbon atom as illustrated. Oxygen is a member of group VI and all members of group VI have six valence electrons—6 electrons in the outermost shell. 6 dots are placed around each oxygen atom in the Lewis dot structure as illustrated. There are a total of 6 + 4 + 6 = 16 valence electrons in the carbon dioxide molecule. Remember that the valence electrons are illustrated by dots in the Lewis dot structure and we want each of the atoms to have a complete octet after bonding.
Completing the octet:
In order for oxygen to have a complete octet, it needs 2 more electrons. This is determined by looking a the periodic table. It is in group VI and has 6 electrons in its valence shell. Each oxygen atom needs 8 – 6 = 2 electrons to complete its octet. This is convenient because carbon has 4 valence electrons and it needs 8 – 4 = 4 electrons to complete its octet.
So if we draw a line between the lone electrons, then by sharing the electrons in a bond, each atom will “feel” like they have a complete octet of valence electrons. There are 2 pairs of shared electrons connecting each oxygen atom to carbon. Each pair of shared electrons is a bond so carbon has 2 double bonds, each connecting the carbon to an oxygen atom.
The Lewis dot structure is re-drawn showing how to illustrate the 2 double bonds.
Each of the double bonds represent a single region of electron density. The electronic geometry is represented by the purple balloons. The molecular geometry is represented by the atomic model. The black ball represents carbon. The green balls represent oxygen. The VSEPR theory predicts that the regions of electronegativity will repel each other and assume a geometric shape which will place them as far apart as possible.
The VSEPR theory therefore predicts the geometry to be linear and the bond angle is therefore equal to 180 degrees.
Now notice that carbon dioxide and sulfur dioxide have very different molecular geometries. Carbon dioxide has 2 regions of electron density. Those regions will repel each other and get as far apart as possible. This results in a linear electronic and a linear molecular geometry.
Sulfur dioxide has a very different molecular and electronic geometry compared to carbon dioxide. Sulfur dioxide has a non-bonded electron pair on the central sulfur atom. This lone pair of electrons is a region of electron density. There are 2 other regions of electron density, namely the bonds between sulfur and oxygen. Three resonant structures are possible for sulfur dioxide as illustrated in the Lewis dot structure. There are 3 regions of electron density in each of the resonant structures. The lone pair of electrons are more electronegative than the bonds to oxygen. This increased electronegative force results in a bent molecular and electronic geometry for sulfur dioxide.
This affects the dipole moment of the molecule. A dipole moment refers to a difference in charge between different poles of the molecule. Sulfur dioxide has a dipole moment where the end of the molecule closest to oxygen is more electronegative that the end of the molecule closest to sulfur. This is because oxygen is more electronegative than sulfur. This can be predicted by noting the position of sulfur and oxygen on the periodic table. The valence electrons of oxygen are closer to it’s nucleus than the valence electrons of sulfur are from it’s nucleus. Oxygen will therefore tend to pull the electrons away from sulfur and result in a polar difference as shown in the Lewis dot structure below as a bond dipole symbol ( |—>).
Each of the oxygen molecules have a partial negative charge relative to the central sulfur atom as indicated by the |—> symbol. Together that gives the lower end of the molecule a net partial negative charge relative to the top of the molecule as indicated by the downward directed |—> symbol. This predicts that the molecule would have a dipole moment and that sulfur dioxide would be polar.
In the case of carbon dioxide there are polar covalent bonds but they are equal and in opposite directions. There is a partial negative force on each end of the molecule but they are equal and symmetric. There is no difference in charge between ends of the molecule. Carbon dioxide therefore has no dipole moment.
In this “tug-of-war”model demonstration of carbon dioxide, an equal force is applied in opposite directions. There is no movement. There is no dipole moment.
In this “tug-of-war” demonstration using a model of sulfur dioxide, there is an equal force applied toward each of the oxygen molecules. There is a net movement in the direction of the large arrow. There is a dipole moment because there is a difference in charge between the top and the bottom of the molecule. The molecule is more positive above and more negative below. This difference in forces is illustrated by the movement in this demonstration. The molecular geometry is not linear, it is bent.
So for a molecule consisting of 2 elements, A and B combined in a compound AB2, and there are no lone pairs of electrons, there will be 2 regions of electron density. The more electronegative atoms (B) will repel each other equally and get as far apart from each other as possible. The bond angle will be 180 degrees. The electronic geometry will be linear. The molecular geometry will be linear. The net dipole will be zero in this case since the atoms on each end are the same. The molecule would not be polar.
If the atoms were all different, then the bond dipoles would not be equal and symmetric. The molecule would be polar. Using the “tug-of-war” demonstration, there would be movement due to the difference in forces.
A molecule with 3 electron densities getting as far apart as possible is illustrated here by the theoretical compound AB3. An example would be AlCl3. There are no lone pairs of electrons (unshared electrons) so the bond angles will be 120 degrees and equal. The electronic geometry is trigonal planar. The molecular geometry is trigonal planar. Since the 3 bonds are between the same atoms (A-B), the dipole forces would all be equal, and there would be no net dipole.
If one of the atoms were different then the forces would not be equal. In this example, H2CO, the oxygen end is more electronegative than the hydrogen end. There is a net dipole and the molecule is polar.
Learning how to predict the polarity of a molecule is a very important concept in chemistry. This allows us to make predictions about solubility, reactivity, boiling point, melting point, and other properties.
Part III of VSEPR Theory begins at video time 21:18
Transcription by James C. Gray MD FACOG
Chemistry Video by Janet Gray Coonce MS
If you plan to view the video on your cell phone, consider your data plan and whether you should wait until you have a WiFi connection to avoid cellular charges.
Review the notes after viewing the video:
If a central atom has 2 sigma ‘s’ bonds and no lone pairs then the molecular shape is linear with a predicted 180◦ bond angle, in other words the orbitals are arranged in a straight line with the central atom. This central atom is said to be sp hybridized. This illustration shows 2 sp hybridized orbitals getting as far apart as possible. The geometry is linear.
What do I mean by sp?
An s orbital is in the shape of a sphere and a p orbital is in the shape of a dumbbell. These shapes define the probability of finding an electron within that space. In an s orbital the electron can be found in a sphere surrounding the nucleus. In a p orbital the electron can be found on opposite sides of the central nucleus. In this illustration, the probability of finding an electron in the py orbital is defined by a dumbbell shape with the bulbs of the dumbbell above and below the central atom.
There are 3 orientations of the p orbital. One on the x, y, and z axis. In this illustration they are designated px, py, and pz.
Here colored play dough is used to create a 3 dimensional representation of these orbitals. The s orbital is represented by the blue sphere and the p orbitals are represented by the yellow dumbbell shapes.
Here the blue s and yellow px orbitals were mixed to form 2 green hybridized sp orbitals. These two sp hybridized orbitals will orient themselves so that they will be as far apart as possible. They will have a linear arrangement, a 180◦ bond angle.
Now the play dough orbitals are assembled. The empty p orbitals, represented by yellow, are perpendicular to each other in the same plane. The sp hybridized orbitals, represented by the green are centered and perpendicular to the plane of the yellow p orbitals.
Example: Ethyne (acetylene) C2H2
Each carbon in the molecule acetylene can be represented by the electronic orbital configuration of this model. See the video and blog "Sigma and Pi Bonds" to see how these orbitals form a triple bond.
A sigma bond occurs when the electron overlap occurs between the nuclei of the atoms.
A single bond is a sigma bond
A double bond is a sigma bond and a pi bond.
A triple bond is a sigma bond and 2 pi bonds.
Trigonal Planar Molecular Geometry, Example Aluminum Bromide
When a central atom is involved with 3 sigma bonds, the orbitals will align themselves as far apart as possible around the central nucleus as illustrated here. The bond angles are 120◦. There is one s orbital and 2 p orbitals. The orbitals are sp2 hybridized, the geometry is trigonal planar. It is all on the same plane. An example would be AlBr3.
Bent Molecular Geometry, Trigonal Planar Electron Geometry.
If the atom has 2 sigma bonds and a lone pair of electrons, it is still sp2 hybridized. There are 3 electron regions in the same plane with predicted bond angle of slightly less than 120◦. This “bent” sp2 hybridization configuration is predicted due to the higher electronegativity of the lone pair of electrons compared to the sigma bonds. The electronegativity of the lone pair of electrons will repel and therefore bend the angle between the sigma bonds to less than 120 degrees. This is the effect of valence shell electron pair repulsion (VSEPR). Even with a lone electron pair and 2 sigma bonds, there are 3 electron regions, it is still referred to as sp2 hybridization. An example of a molecule with 2 sigma bonds and a lone pair of electrons is S02.
Tetrahedral Molecular Geometry
With 4 sigma bonds and no lone pairs there are 4 electron regions and the molecular shape is tetrahedral. The predicted bond angle is 109.5◦. This central atom is sp3 hybridized. Methane (CH4) is an example of a molecule with sp3 hybridization with 4 sigma bonds.
Trigonal Pyramid Molecular Geometry
This atom has 3 sigma bonds and a lone pair. There are 4 areas of electron density. It is sp3 hybridized and the predicted bond angle is less than 109.5◦. It has a trigonal pyramid geometry. An example is NH3, ammonia gas.
If there are 2 lone pair of electrons and 2 sigma bonds there are still 4 areas of electron density. It is sp3 hybridized. A “bent” molecular shape is predicted due to the electronegative repulsion by the 2 lone pair of electrons as predicted by the VSEPR theory. Therefore the predicted bond angle is less than 109.5◦. An example is H20.
Trigonal Bipyramid Molecular Geometry
PCl5 gas is an example of a molecule with 5 dsp3 orbitals with trigonal bipyramid geometry. It is not geometrically possible to arrange 5 bonds equal distant around a central atom. In this model I use two colors to represent the chloride atoms. On the equatorial plane the chloride atoms are dark green. There are 3 bonds on the equatorial plan and the bond angles are equal to 120◦. On the axial plane the chloride atoms are colored red. The bond angle is 90◦ between the atoms on the axial plane and those on the equatorial plane.
See-Saw Molecular Geometry
SF4 gas is an example of a molecule with 5 dsp3 orbitals but one of the orbitals contain a lone pair of electrons. The bond angle is still 90◦ between the atoms on the axial plane (red) and those on the equatorial plane (dark green). The angle between the sigma bonds on the equatorial plane (dark green atoms) are bent and therefore are less than 120◦.
T-Shaped Molecular Geometry
ClF3 is a T-shaped dsp3 hybridized molecule. It has 3 sigma bonds and 2 pair of lone electrons. The lone electrons are in dsp3 hybridized orbitals on the equatorial plane. The bond angle is still 90◦ between the atoms on the axial plane (red) and those on the equatorial plane (dark green). All 4 atoms in chlorine trifluoride are halogens from group VIIA in the periodic table. All of them need only one electron to complete the octet rule.
Rocket science factoid: Although ClF3 is “happy” that the valence shell of each atom is filled, all of them would be much happier if they could each get an electron from something other than another member of the halogen group. A halogen would much prefer a metal to share its electrons. Chlorine trifluoride is highly reactive and was considered to be used in rocket fuel as an oxidant (removes electrons from another substance). It was pressurized and put in metal containers lined with metal fluoride. However, it was too reactive. If conditions (such as heat) caused the protectant metal fluoride coating to come off, the atoms in the chlorine trifluoride molecule will steal electrons from (react with) the metal container in which the pressurized liquid ClF3 was stored. In other words it could potentially burn and blow up the container!
Tetrahedral Electrical Geometry but Linear Molecular Geometry
If a dsp3 central atom has 2 sigma bonds and 3 pairs of lone electrons, it will have linear geometry. The bond angle is 180◦ which is a straight line. XeF2 is an example of a molecule with this configuration. It is a strong fluorinating agent.
Octahedral Molecular Geometry
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◦.
Octahedral Electrical Geometry with Square Pyramid Molecular Geometry
If one of the sigma bonds are replaced with a lone pair of electrons, the molecule would have a square pyramid geometry. BrF5 is an example. Like ClF3 it is also an interhalogen compound and very reactive.
Square Planar Molecular Geometry
If 2 of the sigma bonds are replaced with a lone pair of electrons, the molecule is still d2sp3 hybridized but with a square planar geometry. XeF4 is an example.
Tetrahedral and Square Planar Geometry Compared
Contrast the sp3 hybridized tetrahedral model on the left with the d2sp3 hybridized square planar model on the right. The geometry is completely different.
Now lets use the valence bond theory to make predictions about these molecules. The central carbon has 4 electron regions surrounding it so we know it is tetrahedral and sp3 hybridized. We expect a bond angle of 109.5◦. There are no empty p orbitals because all 3 p orbitals were used to hybridize into the tetrahedral molecular shape.
The carbons in this Lewis dot structure have 3 bonds 120◦ apart and are sp2 hybridized. There are 3 sigma bonds and one pi bond. The pi bond forms from the p orbital above and below the plane of these atoms.
Each carbon in this molecule have 2 sp orbitals in a linear geometry with 2 overlapping p orbitals connecting the carbons to each other. So each carbon is sp linear and the overall molecule is linear.
Summary:
Hybridized orbitals geometry Example
sp linear (180◦) C2H2 = acetylene
sp2 trigonal planar AlBr3 = aluminum tribromide
sp3 tetrahedral planar (109.5◦) CH4 = methane
dsp3 trigonal bipyramid (90◦, 120◦) PCl5
= phosphorus pentachloride
d2sp3 octahedral (90◦) SF6 = Sulfur hexafluoride
Transcription by James C. Gray MD FACOG
by Janet Gray Coonce, MS
If you plan to view the video on your cell phone, consider your data plan and whether you should wait until you have a WiFi connection to avoid cellular charges.
Review the notes after watching the video:
The two main principles of the VSEPR theory (Valence Shell Electron Pair Repulsion Theory) are:
1. Regions of high electron density (RHED) repel one another
2. Unshared RHED (“lone pairs”) take up a little more room than shared RHED.
Regions of high electron density (RHED) repel one another
There are two types of electron density regions:
1. Shared regions (purple balloons)
2. Unshared regions (yellow balloons)
Bond with 2 Shared Electrons
If there are 2 shared electrons between two atomic nuclei, then there is one region of high electron density (RHED).
Bond with 4 shared electrons
If there are 4 shared electrons between two atomic nuclei, then there is still only one REGION of high electron density (RHED). If there are 6 electrons between 2 nuclei, there is still just one RHED. If there is a resonance bond between 2 nuclei, there is still just one RHED.
Atom with 2 paired Electrons Occupying 1 RHED
UNSHARED pair or LONE PAIR of electrons are 2 non-bonded electrons on a central atom. In this illustration, the red ball represents the central atom, the yellow balloon represents the RHED occupied by the UNSHARED pair of electrons. The two dots represent the two valence electrons which are not participating in a bond, they just occupy a RHED and therefore take up space around the central atom. As a matter of fact the RHED occupied by UNSHARED electrons take up MORE space than RHED occupied by SHARED electrons.
Note of clarification: It is understood in the above illustration that the red ball representing the nucleus has other RHED other than the one illustrated. If the atom only had 2 electrons and only 1 RHED, the atom would be helium and both of the electrons would be in an s orbital (negative force field) surrounding the atom. In other words the nucleus would be in the center of the balloon and protected from forming any bonds. Helium is a noble gas. It is inert and does non reactive under normal conditions.
3 Dimensional Representation of Single Bond Geometry
If there are 2 shared electrons between two atomic nuclei, then there is one region of high electron density (RHED) and it represents a single bond between the atoms. A single bond is represented by a single line. It is a SINGLE RHED. That RHED is represented by the purple balloon. The single bond is represented by the single line between the atoms.
3 Dimensional Representation of Double Bond Geometry
If there are 4 shared electron between two atomic nuclei, then there is still only one REGION of high electron density (RHED). These electrons are shared in pairs and each represents a bond. Therefore 4 shared electrons represents 2 bonds, a double bond. Double bonds are represented by 2 lines drawn between the atoms. These two bonds are physically located within a SINGLE RHED. That RHED is represented by the purple balloon.
3 Dimensional Representation of Triple Bond Geometry
If there are 6 electrons between 2 nuclei, there is still just one RHED. The electrons are shared in pairs. There are 3 pairs of electrons being shared. This is a TRIPLE BOND and is represented by 3 parallel lines between the atoms. These three bonds and the 6 electrons are physically located within a SINGLE RHED. That RHED is represented by the purple balloon.
Resonance Bond in Benzene
A resonance bond is when there is a non-integer bond order between the pair of atoms. The bond order refers to the number of chemical bonds between a pair of atoms. When a pair of electrons is shared with other pairs of atoms (example bond between carbon atoms in the benzene ring) then the number of bonds per atomic pair will not be a whole number. This would be an example of a molecule with resonance bonding. A resonance bond is represented by a solid line and a dotted line between the atoms. The non-integer pairs of electrons are physically located within a SINGLE RHED. That RHED is represented by the purple balloon.
Linear Molecular Geometry (2 RHED)
If a central atom is bonded to 2 other atoms, the shared electrons between each of the atoms are going to repel one another. Electrons are negatively charged, like charged particles repel, therefore they are going to arrange themselves to get as far away from each other as possible. Therefore the bond angle between the bonds is 180◦. In other words, all three atoms and the 2 RHED between them will be arranged in a straight line. The molecular geometry is linear. The electronic geometry is linear.
Trigonal Planar Geometry (3 RHED)
If there are 3 RHED around a central atom repelling each other, we would predict the bond angle between the central atom and each of these regions to be 120◦ (360 degrees in a circle divided by 3). The electronic geometry and the molecular geometry is trigonal planar.
Trigonal Electronic Geometry with Bent Molecular Geometry (3 RHED)
If instead of 3 shared regions, there were 2 shared and 1 unshared pair of electrons, the repulsion forces would not be equal. The RHED of the unshared pair is more electro-negative than the 2 shared RHED. The electronic geometry is still trigonal planar because there 3 RHED. However, the molecular geometry is “bent” meaning that the bond angle is compressed and therefore less than 120◦. Sulfur dioxide (SO2) is an example of a molecule with this symmetry.
Tetrahedral Molecular and Electronic Geometry (4 RHED)
If there are 4 RHED with single bonds, then the electronic and molecular geometry is tetrahedral, not planar. The bond angles of 4 equal bonds surrounding a central atom are each 109.5◦. Methane (CH4) is an example of a molecule with tetrahedral molecular geometry.
Tetrahedral Electronic and Trigonal Pyramid Molecular Geometry (4 RHED)
Nitrogen is an example of an atom with a lone pair of electrons and 3 unpaired electrons each of which can be shared in a covalent bond. In ammonia gas, NH3, the nitrogen is covalently bonded with 3 hydrogen atoms. The electronic geometry is still tetrahedral because there are 4 separate RHED which are repelling each other. However there are only 4 atoms and the molecular geometry between the central nitrogen and the 3 hydrogen atoms is a trigonal pyriamid. Since the RHED occupied by the lone pair of electrons is more electronegative, the bond angles are predicted by the VSEPR theory to be bent to less than 109.5◦.
Trigonal Bipyramid Electronic and Molecular Geometry (5 RHED)
An example of this geometry is phosphorus pentachloride (PCl5). In this example, phosphorus has 5 RHED, each bonded to chlorine. It is not possible to arrange 5 regions around the central phosphorus atom with identical bond angles. This is a trigonal bipyrimid electronic and molecular structure. The bond angles are 120◦ between the 3 bonds on the equatorial plane. The bond angles are 90◦ between the bonds on the equatorial plane and those on the 2 bonds on the axial plane. The bond angle between the central phosphorus atom and the 2 hydrogen atoms on the axial plane is 180◦ , a straight line.
Model Representation of Trigonal Bipyramid Molecular Geometry (5 RHED)
It may help to visualize the planes using a model. The green balls represent the atoms on the equitorial plane. The blue balls the atoms on the axial plane. Notice that the blue balls are in a straight line with the central atom between them. The green balls are 120 degrees apart. The green balls are at a 90 degree angle to the axial plane.
2 Dimensional Diagram and 3 Dimensional Compared
In the 2 dimensional representation in the background, the solid wedge represents the green ball coming toward you. The dotted wedge represents the green ball going away from you. The angle from one blue ball to the other around the central atom is 180 degrees. The 90 degree angle between the blue and the green balls is labeled on the 2 dimensional model. The 120 degree angle between the green balls is also labeled on the 2 dimensional illustration.
Octahedral Geometry (6 RHED) Illustrated in 2 and 3 Dimensions
An example of a molecule with 6 RHED is Sulfur hexafluoride (SF6). Visually compare and understand the second 2 dimensional drawing of the 3 dimensional models. The balloons in the first image represent the electronic geometry of the 6 RHED. The 3rd image is of a molecular model which demonstrates the relationship of the atoms, the molecular geometry. In this case both the electronic and the molecular geometry is octahedral. The second image demonstrates how to illustrate this geometry in 2 dimensions.
The VSEPR (valence shell electron pair repulsion) theory predicts both the molecular and the electronic geometry based on the chemical formula.
Electronic Geometry: the shape of the arrangement of ALL RHED surrounding the central atom
Molecular Geometry: The shape of the arrangement of atomic nuclei in the molecule
If you know the formula of a molecule and can draw the Lewis dot structure of that molecule, then you can predict both the electronic and the molecular geometry of that molecule.
Electronic Geometry (Trigonal Planar) of Sulfur dioxide (SO2)
Molecular Geometry (Trigonal Planar) of Sulfur dioxide (SO2) is Bent (< 120 degrees)
Illustrating how the lone unpaired electrons repel the less negative electrons being shared in the 2 bonds and bend the bond angles to less than 120 degrees.
Part II will begin at 9 minutes 48 seconds into the YouTube video.
Transcribed by James C. Gray MD FACOG
Janet Gray Coonce 