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Chapter 7: Problem 55

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.

Short Answer

Expert verified
The ionization energy of neon is positive and has a larger magnitude than the electron affinity of fluorine, which is negative. This is because neon has a very stable electron configuration, making it more difficult to remove an electron, while fluorine has a high electron affinity as it is one electron short of obtaining a stable electron configuration.

Step by step solution

01

Write the electron configuration for neon and fluorine

First, we need to write the electron configurations for neon and fluorine. Check the periodic table and find out the atomic numbers of neon and fluorine. The atomic number of neon (Ne) is 10, so its electron configuration is: \(1s^2 2s^2 2p^6\) The atomic number of fluorine (F) is 9, so its electron configuration is: \(1s^2 2s^2 2p^5\)
02

Write the equations for Ionization Energy and Electron Affinity processes

To find the first ionization energy of neon, we are removing one electron from the neutral atom. The equation for this process is: Ne(\(1s^2 2s^2 2p^6\)) 鈫� Ne\(^+\)(\(1s^2 2s^2 2p^5\)) + e\(^-\) For the electron affinity of fluorine, we are adding one electron to the neutral atom. The equation for this process is: F(\(1s^2 2s^2 2p^5\)) + e\(^-\) 鈫� F\(^-\)(\(1s^2 2s^2 2p^6\))
03

Determine the sign for Ionization Energy and Electron Affinity

When energy is required to remove an electron, the Ionization Energy is positive. When adding an electron the energy is released; hence, the Electron Affinity is negative. Therefore, the ionization energy of neon is positive, while the electron affinity of fluorine is negative.
04

Compare the magnitudes of Ionization Energy and Electron Affinity

We can expect that the magnitude of the ionization energy of neon and the electron affinity of fluorine won't be equal. This is because Neon is a noble gas with a completely filled electron shell (2p subshell is full), which makes it very stable, and it requires more energy to remove one electron from this stable configuration. Conversely, fluorine is one electron short of obtaining a stable electron configuration (2p subshell needs one more electron to become full). So, fluorine has a high electron affinity due to its high effective nuclear charge, which attracts and stabilizes the added electron. However, it will not be the same as the ionization energy of neon. So, the ionization energy of neon has a larger magnitude than the electron affinity of fluorine. In conclusion, the ionization energy of neon is positive and has a larger magnitude than the electron affinity of fluorine, which is negative. The ionization energy is larger because neon's electron configuration is very stable, making it more challenging to remove an electron.

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Most popular questions from this chapter

The electron affinity of lithium is a negative value, whereas the electron affinity of beryllium is a positive value. Use electron configurations to account for this observation.

We can draw an analogy between the attraction of an electron to a nucleus and seeing a lightbulb-in essence, the more nuclear charge the electron "sees," the greater the attraction. (a) Within this analogy, discuss how the shielding by core electrons is analogous to putting a frosted-glass lampshade between the lightbulb and your eyes, as shown in the illustration. (b) Explain how we could mimic moving to the right in a row of the periodic table by changing the wattage of the lightbulb. (c) How would you change the wattage of the bulb and/or the frosted glass to mimic the effect of moving down a column of the periodic table? [Section 7.2]

Consider the gas-phase transfer of an electron from a sodium atom to a chlorine atom: $$ \mathrm{Na}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(g) $$ (a) Write this reaction as the sum of two reactions, one that relates to an ionization energy and one that relates to an electron affinity. (b) Use the result from part (a), data in this chapter, and Hess's law to calculate the enthalpy of the above reaction. Is the reaction exothermic or endothermic? (c) The reaction between sodium metal and chlorine gas is highly exothermic and produces \(\mathrm{NaCl}(\mathrm{s})\), whose structure was discussed in Section 2.7. Comment on this observation relative to the calculated enthalpy for the aforementioned gas-phase reaction.

Why did Mendeleev leave blanks in his early version of the periodic table? How did he predict the properties of the elements that belonged in those blanks?

Make a simple sketch of the shape of the main part of the periodic table, as shown (a) Ignoring \(\mathrm{H}\) and \(\mathrm{He}\), write a single straight arrow from the element with the smallest bonding atomic radius to the element with the largest. (b) Ignoring \(\mathrm{H}\) and \(\mathrm{He}\), write a single straight arrow from the element with the smallest first ionization energy to the element with the largest. (c) What significant observation can you make from the arrows you drew in parts (a) and (b)? [Sections \(7.3\) and 7.4]
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