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Chapter 9: Problem 43

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. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

Short Answer

Expert verified
In conclusion, the stable diatomic species predicted by the molecular orbital model are H₂⁺, H₂, H₂⁻, He₂²⁺, and He₂⁺.

Step by step solution

01

Molecular orbitals for H₂ species

For Hydrogen, we have the 1s orbitals which combine to form the σ(1s) bonding orbital and the σ*(1s) antibonding orbital. Let's calculate the bond order for each H₂ species. a. For H₂⁺ species: - We have 1 electron in σ(1s) bonding orbital and 0 electrons in σ*(1s) antibonding orbital.
02

Calculate bond order for H₂⁺

Bond Order = \(\frac{(1-0)}{2}\) = \(\frac{1}{2}\) Since the bond order is greater than zero, H₂⁺ is a stable diatomic species. a. For H₂: - We have 2 electrons in σ(1s) bonding orbital and 0 electrons in σ*(1s) antibonding orbital.
03

Calculate bond order for H₂

Bond Order = \(\frac{(2-0)}{2}\) = 1 Since the bond order is greater than zero, H₂ is a stable diatomic species. a. For H₂⁻: - We have 2 electrons in σ(1s) bonding orbital and 1 electron in σ*(1s) antibonding orbital.
04

Calculate bond order for H₂⁻

Bond Order = \(\frac{(2-1)}{2}\) = \(\frac{1}{2}\) Since the bond order is greater than zero, H₂⁻ is a stable diatomic species. a. For H₂²⁻: - We have 2 electrons in σ(1s) bonding orbital and 2 electrons in σ*(1s) antibonding orbital.
05

Calculate bond order for H₂²⁻

Bond Order = \(\frac{(2-2)}{2}\) = 0 Since the bond order is equal to zero, H₂²⁻ is not a stable diatomic species.
06

Molecular orbitals for He₂ species

For Helium, we have the 1s orbitals which combine to form the σ(1s) bonding orbital and the σ*(1s) antibonding orbital, similar to the hydrogen species. Let's calculate the bond order for each He₂ species. b. For He₂²⁺: - We have 2 electrons in σ(1s) bonding orbital and 0 electrons in σ*(1s) antibonding orbital.
07

Calculate bond order for He₂²⁺

Bond Order = \(\frac{(2-0)}{2}\) = 1 Since the bond order is greater than zero, He₂²⁺ is a stable diatomic species. b. For He₂⁺: - We have 2 electrons in σ(1s) bonding orbital and 1 electron in σ*(1s) antibonding orbital.
08

Calculate bond order for He₂⁺

Bond Order = \(\frac{(2-1)}{2}\) = \(\frac{1}{2}\) Since the bond order is greater than zero, He₂⁺ is a stable diatomic species. b. For He₂: - We have 2 electrons in σ(1s) bonding orbital and 2 electrons in σ*(1s) antibonding orbital.
09

Calculate bond order for He₂

Bond Order = \(\frac{(2-2)}{2}\) = 0 Since the bond order is equal to zero, He₂ is not a stable diatomic species. In conclusion, the stable diatomic species are H₂⁺, H₂, H₂⁻, He₂²⁺, and He₂⁺.

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Most popular questions from this chapter

The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide (CO \(_{2}\) ), and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right.\) ). The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}{ }^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) EX- plain your answer.

The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.

Describe the bonding in the \(\mathrm{O}_{3}\) molecule and the \(\mathrm{NO}_{2}^{-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in these two species?

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?
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