The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 \(\mathrm{pm}\) and 424.7 \(\mathrm{kJ} / \mathrm{mol}\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen (the O-H bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1\(s\) character? c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain. d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. e. Estimate the bond order for OH. f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.

Step by step solution

01

Identify the atomic orbitals involved

The exercise mentions that the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen.

02

Draw the sigma bonding MO

The sigma bonding molecular orbital (σ) is formed by the in-phase, or head-on, overlap of the \(p_{z}\) orbital of oxygen and the 1\(s\) orbital of hydrogen. Draw the σ orbital as the two overlapping orbitals, indicating the regions where there is a positive or negative phase (usually with different shading or colors).

03

Draw the sigma antibonding MO

The sigma antibonding molecular orbital (σ*) is formed by the out-of-phase, or head-on, overlap of the \(p_{z}\) orbital of oxygen and the 1\(s\) orbital of hydrogen. Draw the σ* orbital as the two orbitals overlapping, but with a node between them, indicating where there is a positive or negative phase (again with different shading or colors). #b. Which of the two MOs has the greater hydrogen 1\(s\) character?#

04

Compare the characters of the MOs

In general, the bonding molecular orbitals have more character of the atom with the lower energy atomic orbitals, while the antibonding molecular orbitals have more character of the atom with the higher energy atomic orbitals. Since the hydrogen 1\(s\) orbital is lower in energy than the oxygen \(p_{z}\) orbital, the bonding MO (σ) will have more hydrogen 1\(s\) character. #c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain.#

05

Consider the symmetry of orbitals

Orbitals can form molecular orbitals (bonding or antibonding) only if they have suitable symmetry. In the case of the 2\(p_{x}\) orbital from oxygen and the 1\(s\) orbital of hydrogen, they do not have the appropriate symmetry to overlap or interact effectively. Therefore, the 2\(p_{x}\) orbital of oxygen cannot form MOs with the 1\(s\) orbital of hydrogen. #d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels.#

06

Build the MO energy-level diagram

Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, we just need to consider the 2\(p_{z}\) orbital interacting with the 1\(s\) orbital of hydrogen. First, draw the energy-level diagram with the atomic orbitals of oxygen and hydrogen on different sides of the diagram. Next, draw the bonding MO (σ) and antibonding MO (σ*) between the atomic orbitals. After that, place the correct number of electrons in the energy levels (one from hydrogen and 6 from oxygen, in which 2 electrons are filled in the \(p_{z}\) orbital). Put one electron in the bonding MO (σ) and one electron in the antibonding MO (σ*). #e. Estimate the bond order for OH.#

07

Calculate the bond order

Bond order is calculated using the formula: Bond order = (No. of electrons in bonding MOs - No. of electrons in antibonding MOs) / 2 For OH, we have 1 electron in the bonding MO (σ) and 1 electron in the antibonding MO (σ*). Therefore, Bond order = (1 - 1) / 2 = 0 Thus, the bond order for OH is 0. This result is unexpected, considering that OH exists in reality. It suggests that other orbitals or interactions may be significant in the actual OH molecule. #f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.#

08

Compare the bond order of OH and \(\mathrm{OH}^{+}\)

Since our calculated bond order for OH is 0, which is unexpected, it is not easy to predict the bond order for \(\mathrm{OH}^{+}\) based on our simplified MO diagram. In reality, other orbitals or interactions may be significant to the actual molecule, changing its bond order. A more detailed analysis and calculation would be necessary to make a reliable prediction.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!