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Chapter 7: Problem 53

While the electron affinity of bromine is a negative quantity, it is positive for Kr. Use the electron configurations of the two elements to explain the difference.

Short Answer

Expert verified
The difference in electron affinity between bromine and krypton can be explained by their electron configurations. Bromine, with an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\), has one empty spot in its 4p subshell. Adding an electron makes it more stable, releasing energy and resulting in a negative electron affinity. In contrast, krypton's electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\), with a fully filled 4p subshell. Adding an electron requires it to go into a higher energy, less stable 5s subshell, thus requiring energy input and causing a positive electron affinity.

Step by step solution

01

Determine the electron configurations of bromine and krypton

To analyze the difference in electron affinity between bromine and krypton, we first need to know their electron configurations. Bromine (Br) has an atomic number of 35, while krypton (Kr) has an atomic number of 36. Their electron configurations are as follows: Bromine (Br): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\) Krypton (Kr): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\)
02

Analyze the electron configurations

Looking at the electron configurations, we can see that bromine has 5 electrons in its 4p subshell, while krypton has a completely filled 4p subshell with 6 electrons. Bromine has one empty spot in its 4p subshell, which means adding an electron to bromine will complete its 4p subshell, making it more stable. This process releases energy, resulting in a negative electron affinity. On the other hand, krypton's 4p subshell is already full. Adding an additional electron would require it to go into the next higher energy subshell (5s), which is less stable. This process requires energy, causing the electron affinity of krypton to be positive.
03

Relate electron configurations to electron affinity

In summary, the difference in electron affinity between bromine and krypton can be explained by their electron configurations: - Bromine readily accepts an additional electron to complete its 4p subshell, releasing energy in the process. This results in a negative electron affinity. - Krypton's 4p subshell is already full, so adding an electron requires it to go into a higher energy subshell (5s), making the atom less stable. This process requires energy input, resulting in a positive electron affinity.

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Most popular questions from this chapter

When magnesium metal is burned in air (Figure 3.5), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a \(6.3-\mathrm{g} \mathrm{Mg}\) ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(\mathrm{~g})\) and the reaction goes to \(\mathrm{com}\) pletion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(\mathrm{~g})\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

(a) Because an exact outer boundary cannot be measured or even calculated for an atom, how are atomic radii determined? (b) What is the difference between a bonding radius and a nonbonding radius? (c) For a given element, which one is larger?

(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 \(\left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) where \(\mathrm{Z}\) 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 (c)?

In the chemical process called electron transfer, an electron is transferred from one atom or molecule to another (We will talk about electron transfer extensively in Chapter 20.) A simple electron transfer reaction is $$ \mathrm{A}(g)+\mathrm{A}(g) \longrightarrow \mathrm{A}^{+}(g)+\mathrm{A}^{-}(g) $$ In terms of the ionization energy and electron affinity of atom A, what is the energy change for this reaction? For a representative nonmetal such as chlorine, is this process exothermic? For a representative metal such as sodium, is this process exothermic? [Sections \(7.4\) and \(7.51\)

Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in Kr? Which will be closer to the nucleus? Explain.
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