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Chapter 6: Problem 67

(a) For an \(\mathrm{He}^{+}\) ion, do the \(2 s\) and \(2 p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?

Short Answer

Expert verified
In \( \mathrm{He}^{+} \), \( 2s \) and \( 2p \) orbitals have the same energy. In \( \mathrm{He} \), \( 2s \) is lower in energy than \( 2p \).

Step by step solution

01

Understand the Ion Configuration

The \( \mathrm{He}^{+} \) ion is a helium atom that has lost one electron. This means it has one electron left. The electronic configuration is \( 1s^1 \). In this scenario, the \( 2s \) and \( 2p \) orbitals are unoccupied and their energy levels are determined primarily by the nuclear charge and electron shielding, which are minimal for a single-electron species.
02

Assess Energy Levels in \( \mathrm{He}^{+} \)

In the case of \( \mathrm{He}^{+} \), which is a hydrogen-like ion, the energy levels depend only on the principal quantum number \( n \). For \( n=2 \), both \( 2s \) and \( 2p \) orbitals have the same energy in a hydrogen-like ion due to the absence of electron-electron repulsion and other perturbations.
03

Consider Adding an Electron to Form He

If we add an electron to transform \( \mathrm{He}^{+} \) into \( \mathrm{He} \), the electronic configuration becomes \( 1s^2 \). Here, the situation differs because electron-electron interactions and other effects can split the energy levels of the \( 2s \) and \( 2p \) orbitals. This splitting generally causes the \( 2s \) orbital to have lower energy than the \( 2p \) orbital due to penetration and shielding effects.
04

Analyze Change in Energy Levels Upon Addition

In the \( \mathrm{He} \) atom, the \( 1s^2 \) configuration leads to electronic repulsion that affects the energy levels of any higher energy orbitals. As a result, in neutral \( \mathrm{He} \), the \( 2s \) orbital is often found at a lower energy compared to the \( 2p \) orbital.

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Key Concepts

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

2s and 2p Orbitals
The concept of 2s and 2p orbitals is fundamental to understanding atomic structure. These orbitals belong to the second energy level of an atom and are derived from the principal quantum number, n=2.
The 2s orbital is spherical in shape, and the 2p orbitals have a dumbbell shape, oriented in three different axes (x, y, z). Despite differences in shape, these orbitals can have the same energy in certain cases, such as in hydrogen-like ions.
  • Shape and Orientation: The 2s orbital is non-directional and spherical, easing electron cloud penetration closer to the nucleus.
  • Energy Considerations: In a system like the e}^{+}$ ion, due to minimal electron-electron interactions, these orbitals have identical energy levels.
Understanding orbitals helps us grasp how electrons occupy space around an atom and predict chemical properties.
Electron Configuration
Electron configuration refers to the distribution of electrons in an atom or ion's orbitals. It is written as a series of numbers and letters, such as 1s虏, denoting how electrons occupy different atomic orbitals.
Each set of an electron configuration provides vital details about the overall energy and stability of an atom or ion.
  • He^+ Configuration: This ion has a configuration of 1s鹿, containing one electron and exposing the roles of unoccupied 2s and 2p orbitals.
  • He Configuration: As helium gains a second electron to form He, the electron configuration becomes 1s虏. This causes new inter-electronic interactions which can modify energy levels.
This configuration determines the chemical behavior and reactivity of elements, directly impacting electron interactions inside atoms.
Hydrogen-like Ions
Hydrogen-like ions are species with a single electron orbiting a nucleus (like a hydrogen atom). They have simplified electronic interactions, making them ideal for studying quantum mechanics and energy levels.
In a hydrogen-like ion, energy levels are solely determined by the principal quantum number, n.
  • Principal Quantum Number: Energy levels depend only on n, which leads to degeneracy within the same n level, meaning 2s and 2p have equal energy.
  • Minimal Interactions: With just one electron, there are no electron-electron repulsions. Therefore, these ions do not experience level splitting seen in multi-electron systems.
Analyzing these ions allows for a clearer understanding of fundamental atomic theory.
Electron-Electron Interactions
Electron-electron interactions are forces that occur between electrons within an atom. These interactions significantly influence energy levels and chemical properties.
In atoms with more than one electron, like He, the electron-electron repulsion changes the simple structure presented in hydrogen-like ions.
  • Orbital Energy Shift: Adding electrons increases repulsion, raising energy levels. This leads to differences in energy between 2s and 2p orbitals.
  • Shielding: Electrons in inner orbitals (1s虏) can shield electrons in 2s and 2p, affecting their energy differently and resulting in a split in energy levels.
Understanding these interactions is critical to predicting the atom's behavior and reactivity.

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Most popular questions from this chapter

When the spectrum of light from the Sun is examined in high resolution in an experiment similar to that illustrated in Figure 6.9 , dark lines are evident. These are called Fraunhofer lines, after the scientist who studied them extensively in the early nineteenth century. Altogether, about 25,000 lines have been identified in the solar spectrum between \(1000 \mathrm{nm}\) and \(295 \mathrm{nm}\). The Fraunhofer lines are attributed to absorption of certain wavelengths of the Sun's "white" light by gaseous elements in the Sun's atmosphere. (a) Describe the process that causes absorption of specific wavelengths of light from the solar spectrum. (b) To determine which Fraunhofer lines belong to a given element, say, neon, what experiments could a scientist conduct here on Earth?

Molybdenum metal must absorb radiation with an energy higher than \(7.22 \times 10^{-19} \mathrm{~J}\left({ }^{\text {" }}\right.\) energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

Using the periodic table as a guide, write the condensed electron configuration and determine the number of unpaired electrons for the ground state of \((\mathbf{a}) \mathrm{Cl},(\mathbf{b}) \mathrm{Al},(\mathbf{c}) \mathrm{Zr},(\mathbf{d}) \mathrm{As},\) (e) Sb, (f) W.

The series of emission lines of the hydrogen atom for which \(n_{f}=4\) is called the Brackett series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Brackett series are observed. (b) Calculate the wavelengths of the first three lines in the Brackett series-those for which \(n_{i}=5,6,\) and 7.

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71)\) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a 55.4 -g sample of \(\mathrm{ZrO}_{2},\) calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).
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