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Chapter 7: Problem 50

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
If the electron affinity for an element is a negative number, it means that the anion of the element is more stable than the neutral atom. This is because a negative electron affinity indicates an exothermic process where energy is released upon electron addition, leading to a lower energy state and increased stability for the anion compared to the neutral atom. This stability arises from the more stable electron configuration achieved by the anion.

Step by step solution

01

Understand Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to a neutral atom in the gaseous state to form an anion. It is a measure of the atom's ability to attract and accept an additional electron. A highly negative electron affinity means that the energy released when an electron is added to the atom is large, which indicates that the atom has a strong attraction for electrons.
02

Recognize the Significance of a Negative Electron Affinity Value

A negative electron affinity means that energy is released when an electron is added to the element, and the formation of the anion is considered to be an exothermic process. An exothermic process indicates that the system is losing energy, making it more stable after the process has occurred.
03

Compare the Stability of the Anion and Neutral Atom

When energy is released after an electron is added to the atom, the new anion formed will have a lower energy state and increased stability compared to the neutral atom. This is because atoms strive to achieve lower energy states, usually by filling their outer electron shells. When an atom gains an electron and forms an anion, the added electron can help the atom achieve a more stable electron configuration, thereby result in a more stable anion.
04

Conclusion

So, if the electron affinity for an element is a negative number, it means that the anion of the element is more stable than the neutral atom. This is because the addition of an electron to the atom results in the release of energy, leading to a more stable, lower-energy anionic state.

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Most popular questions from this chapter

(a) As described in Section 7.7 , the alkali metals react with hydrogen to form hydrides and react with halogens to form halides. Compare the roles of hydrogen and halogens in these reactions. Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. (b) What is the oxidation number and electron configuration of calcium in each product?

An element \(\mathrm{X}\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} . \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(1000^{\circ} \mathrm{C} ) .\) Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\) . (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(\mathrm{X}\) is a metal, nonmetal, or metalloid? (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element \(\mathrm{X}\).

Arrange each of the following sets of atoms and ions, in order of increasing size: \((\mathbf{a}) \mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se} ;(\mathbf{b}) \mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+};\) \((\mathbf{c}) \mathrm{Ca}, \mathrm{Ti}^{4+}, \mathrm{Sc}^{3+} ;(\mathbf{d}) \mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}.\)

Note from the following table that there is a significant increase in atomic radius upon moving from Y to La, whereas the radii of \(Z r\) to Hf are the same. Suggest an explanation for this effect.

Write the electron configurations for the following ions, and determine which have noble-gas configurations: \((\mathbf{a})\mathrm{Ru}^{3+}\) \((\mathbf{b}) \mathrm{As}^{3-},(\mathbf{c}) \mathrm{Y}^{3+},(\mathbf{d}) \mathrm{Pd}^{2+},(\mathbf{e}) \mathrm{Pb}^{2+},(\mathbf{f}) \mathrm{Au}^{3+}.\)
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