Concept explainers
(a)
Interpretation:
The MO resulting from the given orbital interaction is to be drawn.
Concept introduction:
When atomic orbitals (AOs) of the same phase interact, they result in a bonding molecular orbital (MO) that is lower in energy than the individual AOs.
Answer to Problem 3.1P
The MO resulting from the given orbital interaction can be drawn as follows:
Explanation of Solution
The orbital interaction shows two orbitals that are lightly shaded. Both have the same negative phase. Since the phases of both are the same, there will be constructive interference, resulting in a bonding MO of negative phase.
The MO resulting from the given orbital interaction can be drawn as follows:
The interaction between orbitals of the same phase results in the formation of a bonding MO.
(b)
Interpretation:
Whether the resulting MO is unique compared to the one shown on the right of Figure 3-6a is to be determined.
Concept introduction:
When atomic orbitals (AOs) of the same phase interact, the resulting molecular orbital (MO) has a lower energy than the separate AOs. The phases of the interacting orbitals may be both positive or both negative. The resulting stabilization (lowering of energy) is the same for both.
Answer to Problem 3.1P
The MO resulting from the interaction shown will not be unique compared to the one shown in Figure 3-6a.
Explanation of Solution
The interaction in this case is between AOs of negative phases (light shading). Since the phases are the same, the interaction will result in constructive interference, increasing the electron density between the two nuclei. This will lower the energy of the MO compared to the individual AOs. The extent to which the energy is lowered will be the same as in case of the interaction shown in Figure 3-6a. This is because the interacting orbitals are same except for a different phase.
Therefore, the MO resulting from the interaction shown will not be unique compared to the one in Figure 3-6a.
The interaction between AOs of same phase results in a bonding MO with the same stabilization, whether their phases are both positive or both negative.
Want to see more full solutions like this?
Chapter 3 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Solve correctly please with some explanation also. Electron pair geometry(option): Linear,triagonal planar, tetrahedral,trigonal bipyramidal, octahedral.arrow_forwardThe molecule shown here contains carbons. carbons total, and hydrogens, asymmetricarrow_forwardConstruct, fully label, and populate a MO diagram for the nitrosonium anion, NO+. Based on the diagram, state which of NO+ or NO you expect to have a stronger bond.arrow_forward
- Show work with explanation needed..don't give Ai generated solution. Give Clear Detailed Solutionarrow_forwardConsider the incomplete orbital representation of O2 , below right. a. Identify which lobes are hybrid orbitals (identify the type) and which lobes arep orbitals. b. Use dotted lines to show any bonds. c. Use up or down arrows to show electron occupation of each hybrid orbital or bond.arrow_forward4. ,A: Briefly describe why the p orbital can contain a positive charge and hybridized orbital cannot. Use pictures in your answer. B: Briefly explain why a hybridized orbital can contain a negative charge, and P-orbital cannot. Use pictures in your answer.arrow_forward
- Find the correct matches for the ion NH₂ Drag and drop options on the right-hand side and submit. For keyboard navigation... SHOW MORE Electron Domain Geometry Molecular Geometry I ||| ||| Bent Trigonal planar Tetrahedralarrow_forwardProblem attachedarrow_forwardClick to see additional instructions Shown is a blank MO diagram depicting ONLY the valence shells. Fill in the electrons for the compound PS. How many electrons are in the T*3p orbitals? How many electrons are in the 03s orbital? What is the bond order for PS?arrow_forward
- Which of the following statements are TRUE for covalent pi bonds? (Choose as many as apply. Hint: There are 4 correct answers!) Pi bonds arise from head-on overlap of two orbitals. Pi bonds arise from sideways overlap of two orbitals. The electron density in a pi bond is found above and below the axis between the two bonded nuclei. The electron density in a pi bond is found in between the two bonded nuclei. A pi bond can exist independently; not every pi bond is accompanied by an associated o bond. Pi bonds always exist in conjunction with an associated o bond; they can never exist independently. The pi bonds in a molecule determine the shape of the molecule. There can be more than one pi bond between two nuclei. Every covalent bond contains at least one pi bond.arrow_forwardFigure 3 of 3 Which of the following clusters of orbitals would form the shape shown here (Figure 3) either through the formation of a single molecule or by bonding to another atom of similar electronic geometry and results in an arrangement of orbitals that is possible within the valence shell of an atom? Check all that apply. ► View Available Hint(s) two sp orbitals and two p orbitals six sp³ d² orbitals three sp orbitals two sp orbitals and four p orbitals three p orbitals six sp³ d orbitals six sp³ orbitals one sp orbital and two p orbitalsarrow_forwardHelp me pleasearrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning