Module 2 Practice Set: Energetics, Newman Projections, and Molecular Orbital Theory

Current Status
Not Enrolled
Price
20 USD
Get Started

Molecular Orbital Theory Diagram Sample Questions

Molecular Orbital Theory Diagram:

Provide a molecular orbital diagram for the orbitals that result from the mixing of the relevant atomic and/or atomic hybrid orbitals. In each case, identify which orbital resulted from the constructive and destructive overlap of the atomic orbital wavefunctions.

hydrogen atom + hydrogen atom -> hydrogen molecule (H2)

Answer Key:

The addition of two hydrogen radicals involves two total electrons, one coming from each hydrogen. The addition of these two radicals creates a sigma bond between the two hydrogen atoms.

Each hydrogen group contributes one electron to the newly formed bond between them. The pairing of these electrons to form the bond is energetically favorable, because each electron enters the bonding orbital which is more stable than the orbitals the individual electrons started in. We know that the electrons enter the bonding orbital because electrons will always choose to fill the lowest energy orbital first.

The diagram below illustrates each electron going into the bonding (σ) orbital to form the bond. The orbitals in the middle of the diagram represent the individual orbitals the radical on each hydrogen atom started in.

The bonding orbital has constructive interference, and orbital overlap. Orbital overlap is required for the formation of a bond, so all bonding orbitals will have that characteristic. Both of these are stabilizing characteristics, which is why the bonding orbital is lower in energy than the starting orbitals.

The anti-bonding orbital has destructive interference, and no orbital overlap. Both of these are destabilizing characteristics, which is why the anti-bonding orbital is higher in energy than the starting orbitals.

Note that there is no need to label the individual orbitals the radicals came from with 1s, 2s, 2p, etc. in radical addition.

 

Bond Order:

Determine the bond order of F2.

Answer Key:

The first thing we need to do is fill out an M.O. diagram for F2. Fluorine has 9 total electrons, and 7 valence electrons in its valence shell which is the second shell. Therefore, we will need the 1s, 2s, and 2p orbitals for this diagram. The atomic number of fluorine is greater than 8, therefore the σ 2p orbital will be lower energy than the π 2p orbitals and the σ* 2p orbital will be higher in energy than the π* 2p orbitals.

The M.O. diagram is divided into two sides coming together to form the σ and π bonding and anti-bonding orbitals. Therefore, each side of the diagram represents the electrons of one fluorine atom.

Each orbital can hold a maximum of 2 electrons, so there will be 2 electrons in the 1s orbital of each fluorine, 2 electrons in each 2s orbital, and 5 electrons in the 2p orbitals. The bonding orbitals are filled first, then the anti-bonding orbitals are filled if any electrons are left over. The σ and σ* orbitals are all filled for the 1s and 2s orbitals because there are 4 total electrons in those subshells, 2 from each fluorine atom. In the 2p subshell, there are 5 electrons from each fluorine atom. These electrons will fill the lowest energy bonding orbitals first, so 2 electrons go into the σ 2p orbital, and 4 electrons will fill the π 2p orbitals. After all of the bonding orbitals are filled, 2 electrons will fill each π* 2p orbital because those are the lowest energy anti-bonding orbitals.

Once the M.O. diagram is filled out, the bond order of F2 can be calculated. The equation to find bond order can is shown below.

The total number of bonding electrons includes both σ and π. The total number of anti-bonding electrons includes σ* and π*. Therefore, we can rewrite the equation to include those details to find the bond order.

A bond order of 1 means that a single bond will form between the two fluorine atoms.

 

Molecular Orbital Theory Orbital Depictions Sample Question

For each of the following molecules, draw the sigma, sigma*, pi, and pi* orbitals for the following bonds/molecules.

Answer Key:

Sigma (σ): The sigma orbital is the bonding orbital of the sigma bond between the two carbon atoms. In a sigma bonding orbital, one small lobe and one large lobe are pointed towards each other. The two lobes need to be in-phase, which means they have to both be shaded or unshaded. Being in-phase allows the two lobes to overlap to form a sigma orbital which allows the formation of a sigma bond.

Sigma* (σ*): The sigma* orbital is the anti-bonding orbital of the sigma bond between the two carbon atoms. In a sigma anti-bonding orbital, two large lobes or two small lobes are pointed at each other. The two lobes are also out of phase, which means one is shaded and one is unshaded. Being out of phase prevents the two lobes from overlapping to form a sigma orbital which prevents the formation of a sigma bond.

Pi (π): The pi orbital is the bonding orbital of the pi bond between the two carbon atoms. The pi bond is made of two pi electrons, one electron coming from each atom. The pi electrons are in p-orbitals, and the individual p-orbitals are drawn on each atom with one lobe above the plane, and one lobe below the plane. In the pi bonding orbital, the lobes are in-phase which means they are both shaded and unshaded in the same plane. This allows the orbitals to overlap and form pi orbitals above and below the plane which allows the formation of a pi bond.

Pi* (π*): The pi* orbital is the anti-bonding orbital of the pi bond between the two carbon atoms. In the pi anti-bonding orbital, the lobes of the p-orbitals are are out of phase which means they are shaded and unshaded in opposite planes. This prevents the lobes from overlapping and prevents the formation of pi orbitals above and below the plane which prevents the formation of the pi bond.

 

3D Orbital Depictions Sample Questions

Draw each molecule in an appropriate 3D manner, showing a depiction of the sigma (σ) and pi (π) bonding molecular orbitals present in each of the following molecules.

Answer Key:

The sigma bond between the carbon atoms is represented by a Csp2-Csp2 sigma bonding orbital. It results from the in-phase orbital overlap of an sp2 hybridized orbital from each carbon.

The sigma bonds between the carbon and hydrogen atoms are represented by Csp2-H sigma bonding orbitals. They result from the in-phase orbital overlap of an sp2 hybridized orbital from the carbons, and an s-orbital from the hydrogens.

Note that the sigma orbitals between the carbon atoms are larger than sigma orbitals between the carbon and hydrogen atoms. This results from the fact that the combination/overlap of two hybridized orbitals is larger than the overlap of one hybridized orbital and one s-orbital.

 

Newman Projections Sample Questions

Draw a Newman projection for the following molecule down the C2-C3 bond, numbered from left to right.

Answer Key:

The first thing we need to do is define the axis of the Newman Projection. The C2-C3 axis is shown in the image below:The next thing we need to do is determine whether the molecule is in a staggered or eclipsed conformation. To do this, we check if the groups that are in the same plane are on the same side or opposite sides of the defined axis. In this example, we will pick the two groups in the plane. We can see that they are on opposite sides of the C2-C3 axis, therefore this molecule is in a staggered conformation.

The molecule shown above will then be translated into a Newman Projection. C2 is the front carbon, and is represented by the middle of the Y shape. C3 is the back carbon, and is represented by the circle around the Y shape. The front carbon is in a Y shape because the group in the plane on the front carbon is pointed down, so it needs to be pointed down in the Newman Projection as well.

The C2-C3 axis is defined from left to right. Because the axis is from the left, the dashed groups will be on the left of the Newman Projection. This means the wedged groups will be on the right. The groups in the plane of the molecule will always be in the straight up and straight down positions of the Newman Projection.

 

Full Rotation of Newman Projections Sample Questions

Instructions:

  • Newman Projection: For each of the following molecules, draw the Newman Projection for the specified C-C bond.
  • 360-Degree Rotation:
    • After creating the initial Newman Projection, illustrate the complete 360-degree rotation of the Newman Projection.
    • Label each conformation (e.g., staggered, eclipsed) appropriately.
  • Potential Energy Surface:
    • Draw the potential energy diagram corresponding to the full 360-degree rotation of the Newman Projection.
    • Ensure that the energy changes are clearly marked for each conformation.

Draw a Newman projection for the following molecule down the C2-C3 bond, numbered from left to right.

Answer Key:

The first thing we need to do is define the axis of the Newman Projection. The C2-C3 axis is shown in the image below:The next thing we need to do is determine whether the molecule is in a staggered or eclipsed conformation. To do this, we check if the groups that are in the same plane are on the same side or opposite sides of the defined axis. In this example, we will pick the two groups in the plane. We can see that they are on opposite sides of the C2-C3 axis, therefore this molecule is in a staggered conformation.

The molecule shown above will then be translated into a Newman Projection. C2 is the front carbon, and is represented by the middle of the Y shape. C3 is the back carbon, and is represented by the circle around the Y shape. The front carbon is in a Y shape because the group in the plane on the front carbon is pointed down, so it needs to be pointed down in the Newman Projection as well.

The C2-C3 axis is defined from left to right. Because the axis is from the left, the dashed groups will be on the left of the Newman Projection. This means the wedged groups will be on the right. The groups in the plane of the molecule will always be in the straight up and straight down positions of the Newman Projection.

The rotation of a Newman Projection happens 60 degrees at a time. Every 60 degrees it rotates, it switches the type of conformation it is in. The first Newman Projection is in a staggered conformation, so after the first 60 degree rotation it will become eclipsed. This will happen for the full 360 degrees.

When rotation Newman Projections, choose either the front carbon or back carbon to rotate and do not move the other carbon. In this example, the back carbon is rotated while the front carbon is kept still. The rotation of the back carbon can happen clockwise or counterclockwise. In this example the rotation happens in the clockwise direction.

The names of the Newman Projections come from the relative distance between the groups with the largest electron clouds on the front and back carbons of the axis. The group with the largest electron cloud on the front carbon is the methyl group, and the group with the largest electron cloud on the back carbon is the methyl group. In the initial Newman Projection, those two groups are staggered and are as far apart as possible which minimizes the steric repulsion. Therefore it is the staggered-anti conformation. In the fourth Newman Projection, the two groups are eclipsing each other which maximizes the steric repulsion. This makes that projection eclipsed-syn. The remaining staggered conformations have gauche interactions between the two methyl groups, therefore they are labeled staggered-gauche. The remaining eclipsed conformations are simply eclipsed, because the two methyl groups are not eclipsing each other.

The potential energy surface (PES) for the full rotation of the Newman Projection shown above is a representation of the energy of each conformation. The staggered-anti conformation has the lowest energy because the two methyl groups are a full 180 degrees away from each other which minimizes steric repulsion. The staggered-gauche conformations have higher energy than the staggered-anti conformation because the two methyl groups are adjacent to each other, so there is elevated steric repulsion compared to when they are 180 degrees away from each other.

The eclipsed-syn conformation has the highest energy because the two methyl groups are directly eclipsing each other which maximizes steric repulsion. The eclipsed conformations have lower energy than the eclipsed-syn conformation because the two methyl groups are not eclipsing each other, so there is less steric repulsion compared to when they are eclipsing each other.

The staggered conformations exist in the “dips” or “valleys” of the PES because they are true conformations and can exist for extended periods of time. The eclipsed conformations are transition states, so they exist on the “peaks” in the PES and are marked with the transition state symbol.

*The staggered-anti conformations can start at an energy of zero or they can start further up on the y-axis.

 

 

Course Description:

Key Features:

  • Molecular Orbital Theory Diagrams:
    • Practice identifying leaving groups in various molecules.
    • 10 questions
  • Molecular Orbital Theory Orbital Depictions
    • Practice drawing orbital depictions for various bonding and anti-bonding orbitals.
    • 20 questions
  • 3D Orbital Depictions
    • Practice drawing 3D orbital depictions for various molecules.
    • 20 questions
  • Newman Projections
    • Practice drawing Newman Projections in both eclipsed and staggered conformations for various molecules.
    • 20 questions
  • Newman Projections – Full Rotations
    • Practice drawing Newman Projections in both eclipsed and staggered conformations for various molecules.
    • Practice drawing a full 360 degree rotation of various Newman Projections and naming each conformation based on its energy.
    • Practice drawing a potential energy surface for each 360 degree rotation of various Newman Projections
    • 10 questions

Detailed Answer Keys:

  • Step-by-step worked solutions for every problem.
  • Comprehensive explanations that clarify key concepts and problem-solving strategies.

Enhanced Learning Experience:

  • Structured practice that mirrors advanced topics in molecular orbital theory and conformational analysis.

Recommended Prerequisite:

  • Completion of Module 2: Energetics, Newman Projections, and Molecular Orbital Theory instructional videos.
  • Completion of Module 1: Formal Charge, Partial Charge, Electron Density, Hybridization, and Resonance Structures.

Who Should Engage:

  • Learners seeking to deepen their understanding of molecular orbital theory and Newman Projections.
  • Ideal for students preparing for advanced organic chemistry exams.

Benefits:

  • Extensive practice with a variety of problems to ensure thorough understanding.
  • Detailed answer keys with step-by-step explanations to reinforce learning.
  • Improved ability to predict molecular behavior and conformations.

Ratings and Reviews

0.0
Avg. Rating
0 Ratings
5
0
4
0
3
0
2
0
1
0
What's your experience? We'd love to know!
No Reviews Found!
Show more reviews
What's your experience? We'd love to know!
Scroll to Top