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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 electron affinity of Bromine is negative because adding an electron to its 4p subshell (electron configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦�) increases its stability and releases energy. However, Krypton has a stable electron configuration with a filled 4p subshell (electron configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦�), and adding an electron would require it to enter a new energy level, which disrupts its stability. This process requires an input of energy, resulting in a positive electron affinity. The difference in electron affinity between Bromine and Krypton is due to the difference in their electron configurations and the stability of their electron arrangements.

Step by step solution

01

Determine the electron configurations of Bromine and Krypton

To explain the difference in electron affinity, we need to first find the electron configurations of Bromine (Br) and Krypton (Kr). For Bromine (Br, atomic number 35): 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦� For Krypton (Kr, atomic number 36): 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦�
02

Analyze the stability of the electron configurations

Now that we have the electron configurations, let's analyze the stability. Bromine: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦� Krypton: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦� Observe that Krypton has a filled 4p subshell, making its electron configuration more stable. In the case of Bromine, there is one electron less in the 4p subshell, and gaining one more electron would result in a filled 4p subshell, adding stability.
03

Explain the difference in electron affinity

Electron affinity refers to the energy change when an electron is added to a neutral atom. Generally, atoms with a higher electron affinity will more readily accept an electron. Negative electron affinity means that energy is released when an electron is added, while a positive value means energy needs to be input for an electron to be added. In the case of Bromine, adding an electron to complete its 4p subshell increases its stability, which in turn releases energy. Thus, the electron affinity of Bromine is negative. Krypton, on the other hand, has a stable electron configuration with the filled 4p subshell. Adding another electron would require it to enter a new energy level, which would disrupt the stability of the previously filled 4p subshell. This process requires an input of energy, resulting in a positive electron affinity. In summary, the difference in electron affinity between Bromine and Krypton is due to the difference in their electron configurations and the stability of their electron arrangements.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Electron Configurations
Electron configurations lay the foundational groundwork for understanding many aspects of chemical behavior. The configuration of an atom determines how it will interact with other atoms and what kind of chemical reactions it could participate in. For instance, the electron configuration for a Bromine atom is 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦�, which means it is one electron short of having a full 4p subshell. On the other hand, Krypton's configuration is 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4辫鈦�, which indicates its outermost subshell is completely filled, endowing it with greater stability.

When an atom like Bromine, which is on the brink of achieving a stable octet, gains one more electron, it reaches a more stable state. This directly influences its electron affinity, as the atom releases energy upon gaining that electron, signifying a negative electron affinity. Meanwhile, Krypton's reluctance to accept an additional electron due to its already complete outer shell reflects in a less favorable, or positive, electron affinity.
Stability of Electron Arrangements
The stability of an atom's electron arrangement is pivotal in its chemistry. Atoms strive for the most stable electron configuration, often resembling the nearest noble gas. The arrangement of electrons across the various subshells and orbitals follows the Aufbau principle, where electrons fill lower energy orbitals before moving to higher ones. This explains why Bromine with its electron configuration ending in 4辫鈦� is more inclined to accept an electron, completing the 4辫鈦� configuration akin to Krypton. Thus, it possesses a negative electron affinity.

In contrast, a noble gas like Krypton, with a full outer shell, has no low-energy states available for an extra electron. Adding an electron to Krypton forces it into a new, less stable energy level, increasing its potential energy and demanding an energy input鈥攈ence, its positive electron affinity. These principles are central to understanding how electron configurations determine stability and reactivity of elements.
Energy Change in Atomic Processes
At the heart of electron affinity is the concept of energy change during atomic processes. Electron affinity measures the amount of energy released or absorbed when an electron is added to a neutral atom. For Bromine, this process is exothermic鈥攅nergy is discharged because the electron addition leads to a more stable electronic structure.

Krypton's case is different; here, adding an electron is endothermic, requiring an energy contribution. The energy change can be associated with the electron entering a higher principal energy level, as Krypton's existing orbitals are already fully occupied. Understanding these energy changes is fundamental in explaining why elements behave differently in chemical reactions and why elements in the same group can have varying electron affinities despite having similar electronic structures.

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Most popular questions from this chapter

(a) What is an isoelectronic series? (b) Which neutral atom is isoelectronic with each of the following ions: \(\mathrm{Al}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Br}^{-}, \mathrm{Sn}^{2+}\)

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

Little is known about the properties of astatine, \(\mathrm{At}\), because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. (a) Do you expect the element to be a gas, liquid, or solid at room temperature? Explain. (b) What is the chemical formula of the compound it forms with Na?

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-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, 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) If the core electrons were totally effective at shielding the valence electrons and the valence electrons provided no shielding for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in \(\mathrm{P}\) ? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the 3 s electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a \(\mathrm{P}\) atom, which orbital will it come from? Explain.
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