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Chapter 7: Problem 49

Write an equation for the first electron affinity of helium. Would you predict a positive or a negative energy value for this process? Is it possible to directly measure the first electron affinity of helium?

Short Answer

Expert verified
The first electron affinity of helium can be represented by the equation: He(g) + e⁻ ⟶ He⁻(g) Since helium is a noble gas with a stable electron configuration (1s^2), its energy value for the first electron affinity would be positive, indicating the process is not favorable. It is not possible to directly measure the first electron affinity of helium due to its unwillingness to gain electrons and the instability of the resulting helium anion.

Step by step solution

01

Understanding Electron Affinity

Electron affinity is the energy change associated with the addition of an electron to a gaseous atom. A more negative electron affinity value indicates a more stable state after capturing the electron. Most elements have a negative electron affinity; however, some elements, like noble gases, have a positive electron affinity, meaning they resist capturing an electron. The first electron affinity is the energy change associated with gaining one electron to form a negative ion.
02

Electron Configuration of Helium

Helium is a noble gas with an electron configuration of 1s^2. It has a completely filled s-orbital, which gives it maximum stability. Thus, helium has a very low tendency to gain or lose electrons.
03

Writing an Equation for the First Electron Affinity of Helium

The equation for the first electron affinity of helium is: He(g) + e⁻ ⟶ He⁻(g) This equation represents the addition of an electron to a gaseous helium atom, forming a helium anion (He⁻).
04

Predicting the Energy Value for the First Electron Affinity of Helium

As helium is a noble gas with a completely filled energy level, it has no affinity for extra electrons. In the case of helium, the electron-electron repulsion in a helium anion is stronger than the attraction between the new electron and the helium nucleus. Therefore, we would predict a positive energy value for the first electron affinity of helium, indicating that the process is not favorable.
05

Measuring the First Electron Affinity of Helium

It is not possible to directly measure the first electron affinity of helium because helium, being a noble gas, has a stable electron configuration and does not readily gain electrons. Any attempt to add an electron would lead to an unstable helium anion, which quickly loses the electron due to the strong electron-electron repulsion. The experimental difficulties in forming and stabilizing a helium anion make it impossible to directly measure the first electron affinity of helium.

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Most popular questions from this chapter

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. exo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm} .\) (a) What is the energy of a photon of this light, in joules? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(1.72 \times 10^{-18} \mathrm{~J}\). What is the first ionization energy of \(\mathrm{Hg}\), in $\mathrm{kJ} / \mathrm{mol} ?$ (d) Using Figure 7.10 , determine which of the halogen elements has a first ionization energy closest to that of mercury.

Compare the elements bromine and chlorine with respect to the following properties: (a) electron configuration, (b) most common ionic charge, \((\mathbf{c})\) first ionization energy, (d) reactivity toward water, \((\mathbf{e})\) electron affinity, \((\mathbf{f})\) atomic radius. Account for the differences between the two elements.

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Lithium is added to water. (b) Calcium is added to water. (c) Potassium reacts with chlorine gas. (d) Rubidium reacts with oxygen.

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+,\) respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Consider the following equation: $$ \mathrm{Al}^{3+}(g)+e^{-} \longrightarrow \mathrm{Al}^{2+}(g) $$ Which of the following statements are true? (i) The energy change for this process is the second electron affinity of Al atom since \(\mathrm{Al}^{2+}(g)\) is formed. (ii) The energy change for this process is the negative of the third ionization energy of the Al atom. (iii) The energy change for this process is the electron affinity of the \(\mathrm{Al}^{2+}\) ion.
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