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Chapter 9: Problem 62

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion, and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\) (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Short Answer

Expert verified
The H鈧傗伝 ion has an electron configuration of 蟽鈧乻虏蟽鈧乻*鹿 with a bond order of 0.5, indicating weak bonding. Upon excitation with light, the electron configuration becomes 蟽鈧乻鹿蟽鈧乻*虏, yielding a bond order of -0.5. This negative bond order means the excited-state H鈧傗伝 ion is unstable due to a greater number of electrons in antibonding MOs, resulting in repulsive forces between the hydrogen atoms.

Step by step solution

01

Sketch the molecular orbitals and energy-level diagram of H鈧傗伝 ion

To sketch the molecular orbitals and energy-level diagram of the H鈧傗伝 ion, we need to consider the constructive and destructive overlap of two hydrogen atom orbitals (1s orbitals). 1. When two 1s orbitals overlap constructively, they form a bonding molecular orbital called the 蟽鈧乻 orbital. 2. When two 1s orbitals overlap destructively, they form an antibonding molecular orbital called the 蟽鈧乻* orbital. The energy-level diagram will have the 蟽鈧乻 orbital at a lower energy level than the 蟽鈧乻* orbital.
02

Write the electron configuration of H鈧傗伝 ion in terms of its MOs

A neutral H鈧� molecule has two electrons, and since the H鈧傗伝 ion has a -1 charge, it has three electrons in total. The electron configuration follows the aufbau principle, meaning that the lower energy orbitals are filled before the higher energy orbitals. So, the electron configuration of H鈧傗伝 is: 蟽鈧乻虏蟽鈧乻*鹿
03

Calculate the bond order in H鈧傗伝

Bond order is defined as `(number of electrons in bonding MOs - number of electrons in antibonding MOs)/2`. Using the electron configuration derived in step 2, we can calculate the bond order: Bond order = (2 - 1)/2 = 0.5
04

Discuss the stability of the excited-state H鈧傗伝 ion

Suppose that the ion is excited by light, and an electron moves from a lower-energy molecular orbital (蟽鈧乻) to a higher-energy molecular orbital (蟽鈧乻*). In this case, the electron configuration would become: 蟽鈧乻鹿蟽鈧乻*虏 Calculating the bond order for this excited-state configuration: Bond order = (1 - 2)/2 = -0.5 Since the bond order is negative in the excited state, the excited-state H鈧傗伝 ion would not be stable. This is because a negative bond order indicates that there are more electrons in antibonding MOs than bonding MOs, which leads to a repulsive force between the two hydrogen atoms, destabilizing the molecule.

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Key Concepts

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

Understanding the H2- Ion
The \(\mathrm{H}_{2}^{-}\) ion is a molecular species formed by the combination of two hydrogen atoms along with an extra electron. This extra electron gives the ion a negative charge, differentiating it from the neutral \(\mathrm{H}_{2}\) molecule. \The primary requirement to form an \(\mathrm{H}_{2}^{-}\) ion is the mutual interaction and overlap of the individual atomic orbitals of each hydrogen atom. \Each hydrogen atom contributes one 1s orbital, which can overlap to form molecular orbitals. \Therefore, the resultant molecular orbitals include a bonding orbital and an antibonding orbital, which are the core focus when examining this ion. By understanding the behavior and arrangement of this extra electron, you can determine the chemical bonding, stability, and other properties of \(\mathrm{H}_{2}^{-}\) ion. \This ion is an exciting example illustrating electron behavior in molecular orbital theory.
Calculating Bond Order
The concept of bond order is fundamental in molecular chemistry, providing insights into the strength and stability of a bond. It relates directly to the number of electrons shared between atoms. \When referring to \(\mathrm{H}_{2}^{-}\), the bond order can be calculated using the formula: \[\text{Bond Order} = \frac{\text{(Number of electrons in bonding MOs) - (Number of electrons in antibonding MOs)}}{2}\] Substituting in the number of electrons, we have 2 electrons in the bonding orbital and 1 in the antibonding orbital, \which results in a bond order of 0.5: \[\frac{(2 - 1)}{2} = 0.5\]
  • A bond order of 0.5 indicates a fairly weak bond, suggesting limited stability under normal conditions.
  • This fractional bond order implies that while there is a bond between the two hydrogen atoms, it is not as strong as in a diatomic hydrogen molecule (\(\mathrm{H}_{2}\)).
Understanding Electron Configuration
Electron configuration describes the arrangement of electrons in a molecular entity, which determines both its properties and chemical behavior. \In the case of the \(\mathrm{H}_{2}^{-}\) ion, the presence of three electrons (due to the extra electron beyond the neutral \(\mathrm{H}_{2}\)) demands meticulous electron arrangement. \The electron configuration follows the Aufbau principle where lower energy orbitals are filled before moving on to higher energy levels. \For \(\mathrm{H}_{2}^{-}\), this configuration is noted as \(\sigma_{1s}^{2}\sigma_{1s}^{*1}\).
  • The \(\sigma_{1s}\) orbital hosts 2 electrons, maximizing its lower energy capacity.
  • The \(\sigma_{1s}^{*}\) orbital, being higher in energy, contains the extra single electron.
This configuration plays a crucial role in dictating the molecular properties and influences the stability and bond characteristics of the ion.
Delving into Molecular Orbital Theory
Molecular Orbital Theory (MOT) is an essential concept that helps in understanding the bonding and properties of molecules like \(\mathrm{H}_{2}^{-}\). \It offers a more comprehensive description than simple valence bond theory, and explains how atomic orbitals combine to form new orbitals - molecular orbitals. \In the case of \(\mathrm{H}_{2}^{-}\), two types of molecular orbitals are formed from the overlap of the 1s orbitals of each hydrogen:
  • Bonding Molecular Orbital (\(\sigma_{1s}\)): - This orbital is formed by constructive interference, leading to increased electron density between the hydrogen nuclei.
  • Antibonding Molecular Orbital (\(\sigma_{1s}^{*}\)): - Formed by destructive interference, associated with a node between nuclei where electron density is minimal.
The framework of MOT allows chemists to rationalize why certain molecules hold together and the energies involved, thus addressing questions of stability and reactivity in molecular systems.

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Most popular questions from this chapter

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. (b) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), have? Would they be expected to have dipole moments?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the 2 s orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the z-axis), (c) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{\prime} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The \(\mathrm{CO}\) bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the Closer Look box on phases) in the plane of the paper. Two lobes should be pointing toward \(\mathrm{M}\). (d) Now draw the \(\mathrm{d}_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

Dichlorobenzene, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\), exists in three forms (isomers). called ortho, meta, and para: Which of these would have a nonzero dipole moment? Explain.

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?
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