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Chapter 9: Problem 94

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
The MO model predicts a stable BeH₂ molecule with Be forming bonds with both hydrogen atoms. The atomic orbitals involved in bonding are Be's 2s and 2p orbitals, and two H's 1s orbitals. The molecular orbitals are formed by combining Be's 2s orbital with the H's 1s orbitals, resulting in H₂σ₂s bonding MO and H₂σ₂s* antibonding MO. The MO energy-level diagram shows one filled bonding MO and no filled antibonding MO, along with filled non-bonding 2p orbitals. This confirms the stable bonding between Be and both H atoms.

Step by step solution

01

Identify the atomic orbitals involved in bonding

Since beryllium's 1s electrons are not involved in bond formation, then we are only dealing with beryllium's 2s and 2p orbitals. Hydrogen has only the 1s orbital. Therefore, the atomic orbitals involved in bonding are Be's 2s and 2p orbitals and two H's 1s orbitals.
02

Construct molecular orbitals from the atomic orbitals

To construct the molecular orbitals (MOs), we need to consider the linear combinations of atomic orbitals (LCAOs). In the case of BeH₂, the MOs will be formed by combining Be's 2s and 2p orbitals with the H's 1s orbitals. Since H's 1s orbitals are spherically symmetric, we can only combine them with Be's 2s orbital, as the symmetry of orbitals must be conserved. The molecular orbitals formed from these combinations are as follows: 1. The Be 2s orbital combines constructively with the two H 1s orbitals, forming the H₂σ₂s bonding MO. 2. The Be 2s orbital combines destructively with the two H 1s orbitals, forming the H₂σ₂s* antibonding MO. The MO combinations can be represented in the MO energy-level diagram.
03

Fill in electrons into the MO energy-level diagram

Starting with the lowest energy MOs and following the Pauli Exclusion Principle and Hund's rule, we can fill in electrons into the energy-level diagram. Beryllium has 4 electrons (excluding the 1s electrons) and 2 electrons each for the hydrogen atoms, leading to a total of 8 electrons in the BeH₂ molecule. Step 1: Fill the H₂σ₂s bonding MO with 2 electrons. Step 2: Fill the Be 2s orbital with 2 remaining electrons from Be. Step 3: Fill the Be 2p orbital with the remaining 4 electrons. (2 in the 2p_x orbital and 2 in the 2p_y or 2p_z orbital) Step 4: Do not fill the H₂σ₂s* antibonding MO as all electrons are already placed.
04

Analyze the MO diagram and nature of bonding

From the MO energy-level diagram and electron filling, we can infer the following: 1. BeH₂ molecule has one filled bonding MO (H₂σ₂s) while the antibonding MO (H₂σ₂s*) is not filled. 2. Both Be 2p orbitals are filled, and they constitute non-bonding orbitals. Since there are more electrons in bonding orbitals than in antibonding orbitals, the BeH₂ molecule is predicted to be stable according to the MO model. The presence of a filled bonding MO indicates that there is bonding between Be and both H atoms, while the non-bonding 2p orbitals do not participate in forming bonds. Thus, the MO model predicts a stable BeH₂ molecule with Be forming bonds with both hydrogen atoms.

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Most popular questions from this chapter

The transport of \(\mathrm{O}_{2}\) in the blood is carried out by hemoglobin. Carbon monoxide can interfere with oxygen transport because hemoglobin has a stronger affinity for \(\mathrm{CO}\) than for \(\mathrm{O}_{2}\). If \(\mathrm{CO}\) is present, normal uptake of \(\mathrm{O}_{2}\) is prevented, depriving the body of needed oxygen. Using the molecular orbital model, write the electron configurations for \(\mathrm{CO}\) and for \(\mathrm{O}_{2}\). From your configurations, give two property differences between \(\mathrm{CO}\) and \(\mathrm{O}_{2}\)

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\(\mathrm{FClO}_{2}\) and \(\mathrm{F}_{3} \mathrm{ClO}\) can both gain a fluoride ion to form stable anions. \(\mathrm{F}_{3} \mathrm{ClO}\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\) will both lose a fluoride ion to form stable cations. Draw the Lewis structures and describe the hybrid orbitals used by chlorine in these ions.

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{array}{cc} \mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta H=375 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & \Delta H=158 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ 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.)

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{H}_{2}+, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\) b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)
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