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(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 π molecular orbital is zero due to the electron density distribution being above and below the internuclear axis. (b) Similarities between π2p 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 π2p MO from 2px atomic orbitals lying in the xz plane and π2p MO from 2py atomic orbitals lying in the yz plane. (c) π2p MOs are lower in energy than π2p* MOs because π2p MOs have constructive overlap of p atomic orbitals, leading to increased electron density and stronger bonding, while π2p* 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 π molecular orbital

To understand the probability of finding an electron on the internuclear axis if it occupies a π molecular orbital, we need to understand the electron density distribution in these orbitals. The π 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 π molecular orbital is zero.
02

b) Similarities and differences between π2p MOs made from 2px and 2py atomic orbitals

To find the similarities and differences between the π2p MOs made from 2px and 2py atomic orbitals, we'll analyze the shape and orientation of these orbitals. 1. Similarities: - Both π2p 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. - π2p MOs originating from 2px and 2py atomic orbitals have the same energy levels. 2. Differences: - π2p MO originating from 2px atomic orbitals lies in the xz plane, whereas π2p MO originating from 2py atomic orbitals lies in the yz plane. - The electron density distribution in π2p MO made from 2px atomic orbitals is in the x-direction, while the distribution in π2p MO made from 2py atomic orbitals is in the y-direction.
03

c) Why π2p MOs are lower in energy than π2p* 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 π2p MOs, the notation "π2p" refers to the bonding π molecular orbitals, while "π2p*" refers to the antibonding π molecular orbitals. The reason why π2p MOs are lower in energy than π2p* MOs lies in the nature of the overlap between the atomic orbitals. In π2p 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 π2p* 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

For both atoms and molecules, ionization energies (Section 7.4) are related to the energies of orbitals: The lower the energy of the orbital, the greater the ionization energy. The first ionization energy of a molecule is therefore a measure of the energy of the highest occupied molecular orbital (HOMO). See the "Chemistry Put to Work" box on Orbitals and Energy. The first ionization energies of several diatomic molecules are given in electron-volts in the following table: \begin{tabular}{ll} \hline Molecule & \(I_{1}(\mathrm{eV})\) \\ \hline \(\mathrm{H}_{2}\) & \(15.4\) \\ \(\mathrm{~N}_{2}\) & \(15.6\) \\ \(\mathrm{O}_{2}\) & \(12.1\) \\ \(\mathrm{~F}_{2}\) & \(15.7\) \\ \hline \end{tabular} (a) Convert these ionization energies to \(\mathrm{kJ} / \mathrm{mol} .(\mathrm{b})\) On the same plot, graph \(I_{1}\) for the \(\mathrm{H}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) atoms (Figure 7.11) and \(I_{1}\) for the molecules listed. (c) Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? (d) Use molecular orbital energy-level diagrams to explain the trends in the ionization energies of the molecules.

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Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: (a) 3, (b) 4 , (c) 5, (d) 6

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