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The hydrogen molecular ion, \(\mathrm{H}_{2}^{+},\) can be detected spectroscopically. Write the electron configuration of the ion in molecular orbital terms. What is the bond order of the ion? Is the hydrogen-hydrogen bond stronger or weaker in \(\mathrm{H}_{2}^{+}\) than in \(\mathrm{H}_{2} ?\)

Short Answer

Expert verified
Electron configuration: \( \sigma_{1s}^{1} \). Bond order: \( \frac{1}{2} \). The bond in \( \mathrm{H}_{2}^{+} \) is weaker than \( \mathrm{H}_{2} \).

Step by step solution

01

Electron Configuration for \\ \( \mathrm{H}_{2}^{+} \)

The ion \( \mathrm{H}_{2}^{+} \) has one proton from each hydrogen and a single electron, as one electron is removed. The electron will occupy the lowest available molecular orbital which is the bonding sigma orbital (\( \sigma_{1s} \)). Therefore, the electron configuration is \( \sigma_{1s}^{1} \).
02

Calculate Bond Order

The bond order is calculated using the formula: \( \text{Bond Order} = \frac{1}{2} (\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals}) \). For \( \mathrm{H}_{2}^{+} \), there is 1 electron in the bonding orbital \( \sigma_{1s} \) and 0 in antibonding, so Bond Order = \( \frac{1}{2}(1 - 0) = \frac{1}{2} \).
03

Compare Bond Strength with \( \mathrm{H}_{2} \)

The bond order of \( \mathrm{H}_{2} \) is 1 (since \( \mathrm{H}_{2} \) has 2 electrons in the \( \sigma_{1s} \) bonding orbital and 0 in antibonding, Bond Order = \( \frac{1}{2}(2 - 0) = 1 \)). Since \( \mathrm{H}_{2}^{+} \) has a lower bond order of \( \frac{1}{2} \), the bond in \( \mathrm{H}_{2}^{+} \) is weaker than the bond in \( \mathrm{H}_{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Electron configuration describes the distribution of electrons in atomic or molecular orbitals. For \( \mathrm{H}_{2}^{+} \), the ion consists of two protons and one electron—the molecule loses one electron from the diatomic hydrogen molecule \( \mathrm{H}_{2} \). This lone electron occupies the lowest energy level, which is the bonding sigma orbital. Thus, the electron configuration for \( \mathrm{H}_{2}^{+} \) is represented as \( \sigma_{1s}^{1} \). This signifies that one electron is in the \( \sigma_{1s} \) bonding molecular orbital. Understanding this distribution is crucial as it influences properties like bonding and spectral characteristics.
Bond Order
Bond order is an essential concept in molecular orbital theory as it gives insight into the stability of the bond between atoms. It relates directly to the strength and length of a bond. The formula for bond order is:
  • \( \text{Bond Order} = \frac{1}{2} (N_{b} - N_{a}) \),
  • where \( N_{b} \) is the number of electrons in bonding orbitals,
  • and \( N_{a} \) is the number of electrons in antibonding orbitals.
In the case of \( \mathrm{H}_{2}^{+} \), you have 1 electron in a bonding orbital and 0 in antibonding orbitals, resulting in a bond order of \( \frac{1}{2} \). This lower bond order indicates a weaker bond compared to a bond order of 1 as observed in \( \mathrm{H}_{2} \).
Spectroscopy
Spectroscopy is a technique used to detect and analyze molecules through their interaction with light. This technique can detect the presence of the \( \mathrm{H}_{2}^{+} \) ion, as molecules absorb specific wavelengths of light which leads to electronic transitions. By studying these transitions, scientists can determine the energy levels of molecular orbitals and observe the configuration of electrons. Spectroscopic methods thus serve as powerful tools in determining molecular structure and detecting ions like \( \mathrm{H}_{2}^{+} \).
Hydrogen Ion
The hydrogen ion \( \mathrm{H}_{2}^{+} \) is interesting because it holds a singular electron between two protons. This makes it a one-electron system, often used as a simple model for studying molecular bonding and quantum mechanical effects. Unlike the neutral hydrogen molecule \( \mathrm{H}_{2} \), which has a stable electron pair, \( \mathrm{H}_{2}^{+} \) lacks this stability and thus results in a weaker bond. The uniqueness and simplicity of \( \mathrm{H}_{2}^{+} \) make it crucial in exploring theoretical principles.
Bond Strength
Bond strength is an indication of how strongly atoms are bonded together in a molecule. This is directly related to bond order; a higher bond order generally means stronger bond strength. In \( \mathrm{H}_{2}^{+} \), the bond order is \( \frac{1}{2} \), whereas, in \( \mathrm{H}_{2} \), it is 1. This comparison shows that the single electron bond in \( \mathrm{H}_{2}^{+} \) is inherently weaker than the two-electron bond in \( \mathrm{H}_{2} \). Understanding bond strength is important to infer the reactivity and stability of molecules.

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Most popular questions from this chapter

Consider the three fluorides \(\mathrm{BF}_{4}, \mathrm{SiF}_{4},\) and \(\mathrm{SF}_{4}\) (a) Identify a molecule that is isoelectronic with \(\mathrm{BF}_{4}\) (b) Are \(\operatorname{sir}_{4}\) and \(\mathrm{SF}_{4}\) isoelectronic? (c) What is the hybridization of the central atom in \(\mathrm{BF}_{4}^{-}\) and \(\mathrm{SiF}_{4} ?\)

Which of the homonuclear, diatomic molecules of the second-period elements (from \(\mathrm{Li}_{2}\) to \(\mathrm{Ne}_{2}\) ) are paramagnetic? Which have a bond order of \(1 ?\) Which have a bond order of \(2 ?\) Which diatomic molecule has the highest bond order?

What is the connection between bond order, bond length, and bond energy? Use ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) as examples.

Melamine is an important industrial chemical, used to make fertilizers and plastics. (a) The carbon-nitrogen bond lengths in the ring are all the same length (about \(140 \mathrm{pm}\) ). Explain. (b) Melamine is made by the decomposition of urea, \(\left(\mathrm{H}_{2} \mathrm{N}\right)_{2} \mathrm{CO}\) \(6\left(\mathrm{H}_{2} \mathrm{N}\right)_{2} \mathrm{CO}(\mathrm{s}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{N}_{6}(\mathrm{s})+6 \mathrm{NH}_{3}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{g})\) Calculate the enthalpy change for this reaction. Is it endo- or exothermic? \(\left[\Delta_{f} H^{\circ} \text { for melamine(s) }=\right.\) \(-66.1 \mathrm{kJ} / \mathrm{mol}\) and for urea(s) \(=-333.1 \mathrm{kJ} / \mathrm{moll}\)

Bromine forms a number of oxides of varying stability. (a) One oxide has \(90.90 \%\) Br and \(9.10 \%\) O. Assuming its empirical and molecular formulas are the same, draw a Lewis structure of the molecule and specify the hybridization of the central atom (O). (b) Another oxide is unstable BrO. Assuming the molecular orbital diagram in Figure 9.16 applies to BrO, write its electron configuration (where Br uses \(4 s\) and \(4 p\) orbitals). What is the highest occupied molecular orbital (HOMO) for the molecule?

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