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Chapter 7: Problem 52

If the electron affinity for an element is a negative number, does it mean that the anion of the element is more stable than the neutral atom? Explain.

Short Answer

Expert verified
A negative electron affinity for an element generally indicates that its anion is more stable than its neutral atom, as energy is released when an electron is added to the atom. This suggests the atom has a tendency to attract electrons and form anions, often resulting in a stable electron configuration. However, other factors such as electrostatic repulsion between electrons also influence anion stability, so electron affinity alone does not solely determine anion stability.

Step by step solution

01

Understanding Electron Affinity

Electron affinity is defined as the amount of energy released or absorbed when an electron is added to a neutral atom to form a negative ion (anion). A more negative electron affinity indicates that an atom has a greater tendency to gain an electron and form an anion. In general, atoms with high electron affinity have high electronegativity and are more likely to form anions because they can easily attract electrons from other atoms.
02

Stability of Anions and Neutral Atoms

The stability of an anion depends on the balance between the energy released by the electron affinity and the energy needed to overcome the electrostatic repulsion between the outer electrons. When an electron is added to an atom, there is an attraction between the positively charged nucleus and the negatively charged electron. If the energy released through electron affinity is greater than the energy required to overcome the repulsion between the electrons, the anion will be stable.
03

Negative Electron Affinity and Anion Stability

If the electron affinity for an element is negative, it means that energy is released when an electron is added to the neutral atom to form an anion. In general, the more negative the electron affinity, the more stable the anion is. This is because a negative electron affinity indicates that the atom has a tendency to attract electrons and form anions, which usually results in a more stable electron configuration for the atom. However, it is important to note that electron affinity alone does not solely determine the stability of an anion. Other factors, such as the electrostatic repulsion between electrons, also play a crucial role in determining the overall stability of an anion. In summary, a negative electron affinity for an element suggests that its anion would generally be more stable than its neutral atom. However, the stability of an anion is determined by variables like the energy required to overcome electron repulsion and the overall electron configuration.

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Most popular questions from this chapter

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

(a) Write the electron configuration for \(\mathrm{Li}\), and estimate the effective nuclear charge experienced by the valence electron. (b) The energy of an electron in a one-electron atom or ion equals \(\left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{Z^{2}}{n^{2}}\right)\) where \(Z\) is the nuclear charge and \(n\) is the principal quantum number of the electron. Estimate the first ionization energy of Li. (c) Compare the result of your calculation with the value reported in Table 7.4 and explain the difference. (d) What value of the effective nuclear charge gives the proper value for the ionization energy? Does this agree with your explanation in \((\mathrm{c}) ?\)

Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities will have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger? Explain your answer.

Explain the following variations in atomic or ionic radii: (a) \(\Gamma>\mathrm{I}>\mathrm{I}^{+},(\mathrm{b}) \mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

Zinc in its \(2+\) oxidation state is an essential metal ion for life. \(\mathrm{Zn}^{2+}\) is found bound to many proteins that are involved in biological processes, but unfortunately \(\mathrm{Zn}^{2+}\) is hard to detect by common chemical methods. Therefore, scientists who are interested in studying \(\mathrm{Zn}^{2+}\) -containing proteins will frequently substitute \(\mathrm{Cd}^{2+}\) for \(\mathrm{Zn}^{2+}\), since \(\mathrm{Cd}^{2+}\) is easier to detect. (a) On the basis of the properties of the elements and ions discussed in this chapter and their positions in the periodic table, describe the pros and cons of using \(\mathrm{Cd}^{2+}\) as a \(\mathrm{Zn}^{2+}\) substitute. (b) Proteins that speed up (catalyze) chemical reactions are called enzymes. Many enzymes are required for proper metabolic reactions in the body. One problem with using \(\mathrm{Cd}^{2+}\) to replace \(\mathrm{Zn}^{2+}\) in enzymes is that \(\mathrm{Cd}^{2+}\) substitution can decrease or even eliminate enzymatic activity. Can you suggest a different metal ion that might replace \(\mathrm{Zn}^{2+}\) in enzymes instead of \(\mathrm{Cd}^{2+} ?\) Justify your answer.
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