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Chapter 9: Problem 64

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) Why are the \(\pi_{2 p}\) MOs lower in energy than the \(\pi_{2 p}^{*}\) MOs?

Short Answer

Expert verified
(a) The probability of finding an electron on the internuclear axis in a 蟺 molecular orbital is zero due to the electron density distribution being above and below the internuclear axis. (b) Similarities between 蟺2p MOs made from 2px and 2py atomic orbitals include the formation by lateral overlapping of p atomic orbitals, nodal planes along the internuclear axis, and the same energy levels. Differences involve the orientation and electron density distribution, with 蟺2p MO from 2px atomic orbitals lying in the xz plane and 蟺2p MO from 2py atomic orbitals lying in the yz plane. (c) 蟺2p MOs are lower in energy than 蟺2p* MOs because 蟺2p MOs have constructive overlap of p atomic orbitals, leading to increased electron density and stronger bonding, while 蟺2p* MOs have destructive overlap, decreasing electron density and reducing bonding interaction, resulting in higher energy levels.

Step by step solution

01

a) Probability of finding an electron on the internuclear axis in 蟺 molecular orbital

To understand the probability of finding an electron on the internuclear axis if it occupies a 蟺 molecular orbital, we need to understand the electron density distribution in these orbitals. The 蟺 molecular orbitals in a diatomic molecule are formed from lateral overlapping of p atomic orbitals. This overlapping creates a distribution of electron density above and below the internuclear axis but not directly on the axis. Therefore, the probability of finding an electron on the internuclear axis in a 蟺 molecular orbital is zero.
02

b) Similarities and differences between 蟺2p MOs made from 2px and 2py atomic orbitals

To find the similarities and differences between the 蟺2p MOs made from 2px and 2py atomic orbitals, we'll analyze the shape and orientation of these orbitals. 1. Similarities: - Both 蟺2p MOs are formed by the lateral overlapping of the p atomic orbitals, resulting in a plane of symmetry perpendicular to the internuclear axis. - Both have nodal planes along the internuclear axis. - 蟺2p MOs originating from 2px and 2py atomic orbitals have the same energy levels. 2. Differences: - 蟺2p MO originating from 2px atomic orbitals lies in the xz plane, whereas 蟺2p MO originating from 2py atomic orbitals lies in the yz plane. - The electron density distribution in 蟺2p MO made from 2px atomic orbitals is in the x-direction, while the distribution in 蟺2p MO made from 2py atomic orbitals is in the y-direction.
03

c) Why 蟺2p MOs are lower in energy than 蟺2p* MOs

In molecular orbital theory, an asterisk (*) denotes an antibonding molecular orbital, which implies that electrons occupying the orbital contribute to a decrease in the overall stability of the molecule. In the case of 蟺2p MOs, the notation "蟺2p" refers to the bonding 蟺 molecular orbitals, while "蟺2p*" refers to the antibonding 蟺 molecular orbitals. The reason why 蟺2p MOs are lower in energy than 蟺2p* MOs lies in the nature of the overlap between the atomic orbitals. In 蟺2p MOs, the constructive overlap of p atomic orbitals results in increased electron density between the nuclei, leading to a stronger bond and lower energy. On the other hand, in 蟺2p* MOs, the destructive overlap of p atomic orbitals creates a nodal plane between the nuclei, decreasing electron density and reducing the bonding interaction, which leads to higher energy levels.

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Most popular questions from this chapter

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(S F_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

The molecules \(\mathrm{SiF}_{4}, \mathrm{SF}_{4}\), and \(\mathrm{XeF}_{4}\) have molecular formulas of the type \(\mathrm{AF}_{4}\), but the molecules have different molecular geometries. Predict the shape of each molecule, and explain why the shapes differ.

Explain the following (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-} \cdot\) (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger \(\mathrm{O}\) - \(\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure 9.27. In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.
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