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Chapter 7: Problem 31

Why is it much harder to explain the line spectra of polyelectronic atoms and ions than it is to explain the line spectra of hydrogen and hydrogenlike ions?

Short Answer

Expert verified
Explaining the line spectra of polyelectronic atoms and ions is more challenging than that of hydrogen and hydrogen-like ions primarily due to the electron-electron interactions among multiple electrons in polyelectronic systems. These interactions result in more complex energy levels and make it difficult to predict and calculate energy level transitions and the resulting line spectra, as opposed to hydrogen and hydrogen-like ions with only one electron and simpler energy levels.

Step by step solution

01

Understand the line spectra

The line spectra are the unique patterns of light emitted or absorbed by atoms and ions when their electrons undergo transitions between different energy levels. In other words, the line spectra are the fingerprints of atoms and ions that can be used to identify elements in a sample.
02

Describe the hydrogen atom

A hydrogen atom consists of one proton and one electron. The energy levels in a hydrogen atom can be described by the Schr枚dinger equation and characterized by a single quantum number, \(n\). When the electron moves from one energy level to another, it will emit or absorb a photon with the energy difference between these two levels.
03

Describe the hydrogen-like ions

Hydrogen-like ions are ions with just one electron, such as He^+, Li^2+, and so on. They are similar to hydrogen atoms in terms of their energy levels and line spectra. The interactions and energies can be calculated using the same framework as the hydrogen atom, considering the different charge of the nucleus. The Schr枚dinger equation can be applied, and the energy levels are also characterized by a single quantum number, \(n\), just as in the hydrogen atom.
04

Introduce polyelectronic atoms and ions

In polyelectronic atoms and ions, there are multiple electrons present. As such, their energy levels become much more complex and challenging to describe, as there can be several quantum numbers involved and each electron can interact not only with the nucleus but also with each other. This introduces additional terms in their Hamiltonian, making the Schr枚dinger equation more complex.
05

Explain the electron-electron interactions

In polyelectronic atoms and ions, the electrons will experience not only the electrostatic attraction to the nucleus but also the electrostatic repulsion between themselves. The electron-electron repulsion causes shifts in the energy levels and can lead to complex splitting and mixing of the energy levels. This makes it harder to identify and predict the energy level transitions and the resulting line spectra.
06

Summarize the main reason

To sum it up, explaining the line spectra of polyelectronic atoms and ions is more difficult mainly because of the electron-electron interactions that occur among multiple electrons in these systems. These interactions lead to more complex energy levels and make it much more challenging to predict and calculate the energy level transitions and the resulting line spectra, compared to hydrogen and hydrogen-like ions that have just one electron and simpler energy levels.

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Most popular questions from this chapter

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

In the second row of the periodic table, \(\mathrm{Be}, \mathrm{N}\), and \(\mathrm{Ne}\) all have endothermic (unfavorable) electron affinities, whereas the other second-row elements have exothermic (favorable) electron affinities. Rationalize why Be, \(\mathrm{N}\), and Ne have unfavorable electron affinities.

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

An unknown element is a nonmetal and has a valence electron configuration of \(n s^{2} n p^{4}\). a. How many valence electrons does this element have? b. What are some possible identities for this element? c. What is the formula of the compound this element would form with potassium? d. Would this element have a larger or smaller radius than barium? e. Would this element have a greater or smaller ionization energy than fluorine?
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