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Chapter 9: Problem 86

Use the MO model to explain the bonding in \(\mathrm{BeH}_{2}\). When constructing the MO energy-level diagram, assume that the Be's \(1 s\) electrons are not involved in bond formation.

Short Answer

Expert verified
Using the MO model, the bonding in \(\mathrm{BeH}_{2}\) involves the overlap of beryllium's \(2s\) orbital with two hydrogen atoms' \(1s\) orbitals, forming a bonding molecular orbital (\(σ\)). All four electrons occupy the bonding orbital, while the antibonding orbital (\(σ^{*}\)) remains empty, suggesting a stable bonding interaction and overall stability of the \(\mathrm{BeH}_{2}\) molecule.

Step by step solution

01

Determine the electron configuration of beryllium and hydrogen atoms

Beryllium has four electrons, with the electron configuration \(1s^{2} 2s^{2}\). Since the problem states that the \(1s\) electrons are not involved in bond formation, we will only consider the \(2s\) electrons of beryllium. Hydrogen has one electron and its electron configuration is \(1s^{1}\). Since there are two hydrogen atoms in \(\mathrm{BeH}_{2}\), there is a total of 2 \(1s\) electrons from the hydrogen atoms.
02

Construct the MO energy-level diagram

To construct the MO energy-level diagram for \(\mathrm{BeH}_{2}\), we need to combine beryllium's \(2s\) orbitals with hydrogen's \(1s\) orbitals. When we combine these atomic orbitals, we create molecular orbitals, in this case a bonding orbital (\(σ\)) and an antibonding orbital (\(σ^{*}\)). The energy of the bonding orbital is lower than the atomic orbitals, and the energy of the antibonding orbital is higher than the atomic orbitals.
03

Populate the MO energy-level diagram with electrons

Now that we have constructed the MO energy-level diagram, we need to fill it with electrons according to the Aufbau principle, which states that electrons fill the lowest energy orbitals available. There are a total of 4 electrons in \(\mathrm{BeH}_{2}\) (2 electrons from Be's \(2s\) orbital and 2 electrons from two H atoms' \(1s\) orbitals). We will start by adding the electrons to the lowest energy orbital (\(σ\)). Since each orbital can hold up to 2 electrons, the bonding orbital will contain all 4 electrons. The antibonding orbital (\(σ^{*}\)) will remain empty.
04

Analyze the bonding in \(\mathrm{BeH}_{2}\) using the MO model

Based on the MO energy-level diagram and the placement of electrons in the molecular orbitals, we can conclude that \(\mathrm{BeH}_{2}\) has one filled bonding orbital (\(σ\)). This suggests that the bond formation in \(\mathrm{BeH}_{2}\) involves the overlap of beryllium's \(2s\) orbital with hydrogen's \(1s\) orbitals, resulting in a stable bonding interaction. No electrons are present in the antibonding orbital (\(σ^{*}\)), which also supports the stability of \(\mathrm{BeH}_{2}\).
05

Summary

In this exercise, we have used the MO model to explain the bonding in \(\mathrm{BeH}_{2}\). We determined the electron configurations of beryllium and hydrogen, combined their atomic orbitals to form molecular orbitals, populated the MO energy-level diagram with electrons, and analyzed the bonding interactions in \(\mathrm{BeH}_{2}\). The MO model helps us understand the bonding and stability of this molecule, showing that the bond formation involves the overlap of the beryllium's \(2s\) orbital with the hydrogen's \(1s\) orbitals.

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Most popular questions from this chapter

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-}\), and NO using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and NO than the hybrid orbital model?

The diatomic molecule OH exists in the gas phase. 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 molecular orbitals result from the overlap of a lower-energy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of \(\mathrm{OH}\). Explain.

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{aligned} \mathrm{NCl}_{3}(g) & \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta H &=375 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) & \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & \Delta H &=158 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

Consider three molecules: \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). Molecule A has a hybridization of \(s p^{3}\). Molecule B has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule C consists of two \(\sigma\) bonds and two \(\pi\) bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.
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