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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 (\(蟽\)). All four electrons occupy the bonding orbital, while the antibonding orbital (\(蟽^{*}\)) 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 (\(蟽\)) and an antibonding orbital (\(蟽^{*}\)). 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 (\(蟽\)). Since each orbital can hold up to 2 electrons, the bonding orbital will contain all 4 electrons. The antibonding orbital (\(蟽^{*}\)) 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 (\(蟽\)). 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 (\(蟽^{*}\)), 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

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C} \equiv \mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43 .\) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?

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.

Why must all six atoms in \(\mathrm{C}_{2} \mathrm{H}_{4}\) lie in the same plane?
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